# Multiple Positive Solutions to Nonlinear Boundary Value Problems of a System for Fractional Differential Equations

Authors: Chengbo Zhai; Mengru Hao
Journal: The Scientific World Journal
 (2014)

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## Abstract

By using Krasnoselskii's fixed point theorem, we study the existence of at least one or two positive solutions to a system of fractional boundary value problems given by-D0+ν1y1(t)=λ1a1(t)f(y1(t),y2(t)),-D0+ν2y2(t)=λ2a2(t)g(y1(t),y2(t)), where D0+ν is the standard Riemann-Liouville fractional derivative, ν1,ν2∈(n-1,n] for n>3 and n∈N, subject to the boundary conditions y1(i)(0)=0=y2(i)(0), for 0≤i≤n-2, and [D0+αy1(t)]t=1=0=[D0+αy2(t)]t=1, for 1≤α≤n-2, or y1(i)(0)=0=y2(i)(0), for 0≤i≤n-2, and [D0+αy1(t)]t=1=ϕ1(y1),  [D0+αy2(t)]t=1=ϕ2(y2), for 1≤α≤n-2, ϕ1,ϕ2∈C([0,1],R). Our results are new and complement previously known results. As an application, we also give an example to demonstrate our result.


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## Body

## 1. Introduction

The purpose of this paper is to consider the existence of multiple positive solutions for the following system of nonlinear fractional differential equations:(1)-D0+ν1y1(t)=λ1a1(t)f(y1(t),y2(t)),-D0+ν2y2(t)=λ2a2(t)g(y1(t),y2(t)),
where t∈(0,1), ν1,ν2∈(n-1,n] for n>3 and n∈N, and λ1,λ2>0, subject to a couple of boundary conditions. In particular, we first consider (1) subject to
(2)y1(i)(0)=0=y2(i)(0),0≤i≤n-2,[D0+αy1(t)]t=1=0=[D0+αy2(t)]t=1,1≤α≤n-2,
where f,g∈C([0,∞)×[0,∞),[0,∞)), and a1,a2∈C([0,1],[0,∞)). We then consider the case in which the boundary conditions are changed to
(3)y1(i)(0)=0=y2(i)(0),0≤i≤n-2,[D0+αy1(t)]t=1=ϕ1(y1),[D0+αy2(t)]t=1=ϕ2(y2),kkkkkkkk1≤α≤n-2,
where ϕ1,ϕ2∈C([0,1],R).Fractional differential equations arise in many fields, such as physics, mechanics, chemistry, economics, and engineering and biological sciences; see [1–11] for example. In recent years, the study of positive solutions for fractional differential equation boundary value problems has attracted considerable attention, and fruits from research into it emerge continuously. For a small sample of such work, we refer the reader to [12–20] and the references therein. The situation of at least one positive solution has been studied in many excellent monograph; see [12–19, 21] and other references therein. In [22], by means of Schauder fixed point theorem, Su investigated the existence of one positive solution to the following boundary value problem for a coupled system of nonlinear fractional differential equations:
(4)D0+αy1(t)=f(t,y2(t),D0+μy2(t)),0<t<1,-D0+βy2(t)=g(t,y1(t),D0+νy1(t)),0<t<1,y1(0)=y1(1)=y2(0)=y2(1)=0,
where 1<α,β<2,μ,ν>0,α-ν≥1,β-μ≥1.In [21], Goodrich established the existence of one positive solution to problems (1)-(2) and (1), (3) by using Krasnoselskii's fixed point theorem. Different from the above works mentioned, in this paper we will present the existence of at least two positive solutions to problems (1)-(2) and (1), (3) by using the similar method presented in [21]. Moreover, under different conditions, we also present the existence of at least one positive solution to problems (1)-(2) and (1), (3) with λ1=λ2=1.


## 2. Preliminaries

For the convenience of the reader, we present here some definitions, lemmas, and basic results that will be used in the proofs of our theorems.Definition 1 (see [23]).
Letν>0 with ν∈R. Suppose that y:[a,+∞)→R. Then the νth Riemann-Liouville fractional integral is defined to be
(5)Da+-νy(t):=1Γ(ν)∫aty(s)(t-s)ν-1ds,
whenever the right-hand side is defined. Similarly, with ν>0 and ν∈R, we define the νth Riemann-Liouville fractional derivative to be
(6)Da+νy(t):=1Γ(n-ν)dndtn∫aty(s)(t-s)ν+1-nds,
where n∈N is the unique positive integer satisfying n-1≤ν<n and t>a.Lemma 2 (see [24]).
Letg∈Cn([0,1]) be given. Then the unique solution to problem -D0+νy(t)=g(t) together with the boundary conditions y(i)(0)=0=[D0+αy(t)]t=1, where 1≤α≤n-2 and 0≤i≤n-2, is
(7)y(t)=∫01G(t,s)g(s)ds,‍
where
(8)G(t,s)={tν-1(1-s)ν-α-1-(t-s)ν-1Γ(ν),0≤s≤t≤1tν-1(1-s)ν-α-1Γ(ν),0≤t≤s≤1
is the Green function for this problem.Lemma 3 (see [24]).
LetG(t,s) be as given in the statement of Lemma 2. Then one finds that(i)
G(t,s) is a continuous function on the unit square [0,1]×[0,1];(ii)
G(t,s)≥0 for each (t,s)∈[0,1]×[0,1];(iii)
max⁡t∈[0,1]G(t,s)=G(1,s), for each s∈[0,1].Lemma 4 (see [24]).
LetG(t,s) be as given in the statement of Lemma 2. Then there exists a constant γ∈(0,1) such that
(9)min⁡t∈[(1/2),1]⁡G(t,s)≥γmax⁡t∈[0,1]⁡G(t,s)=γG(1,s).
To prove our results, we need the following Krasnoselskii's fixed point theorem which can be seen in Guo and Lakshmikantham [25].Lemma 5 (see [25]).
LetE be a Banach space, and let P be a cone. Assume that Ω1,Ω2 are open bounded subsets of E with 0∈Ω1, Ω1¯⊂Ω2, and let T:P⋂‍(Ω2¯∖Ω1)→P be a completely continuous operator such that(i)
∥Tu∥≤∥u∥,∀u∈P⋂‍∂Ω1, and ∥Tu∥≥∥u∥,∀u∈P⋂‍∂Ω2; or(ii)
∥Tu∥≥∥u∥,∀u∈P⋂‍∂Ω1, and ∥Tu∥≤∥u∥,∀u∈P⋂‍∂Ω2.ThenT has a fixed point in P⋂‍(Ω2¯∖Ω1).


## 3. Main Results

In this section, we apply Lemma5 to study problems (1)-(2) and (1), (3), and we obtain some new results on the existence of multiple positive solutions.


### 3.1. Problem (1)-(2) in the General Case

In our considerations, letE represent the Banach space of C([0,1]) when equipped with the usual supremum norm, ∥·∥. Then put X:=E×E, where X is equipped with the norm ∥(y1,y2)∥:=∥y1∥+∥y2∥ for (y1,y2)∈X. Observe that X is also a Banach space (see [26]). In addition, we define two operators T1,T2:X→E by
(10)T1(y1,y2)(t):=λ1∫01G1(t,s)a1(s)f(y1(s),y2(s))ds,T2(y1,y2)(t):=λ2∫01G2(t,s)a2(s)g(y1(s),y2(s))ds,
where G1(t,s) is the Green function of Lemma 2 with ν replaced by ν1 and, likewise, G2(t,s) is the Green function of Lemma 2 with ν replaced by ν2. Now, we define an operator S:X→X by
(11)S(y1,y2)(t)∶=(T1(y1,y2)(t),T2(y1,y2)(t))=(λ1∫01G1(t,s)a1(s)f(y1(s),y2(s))ds,‍hh.hλ2∫01G2(t,s)a2(s)g(y1(s),y2(s))ds).
We claim that whenever (y1,y2)∈X is a fixed point of the operator defined in (11), it follows that y1(t) and y2(t) solve problems (1)-(2). That is, a pair of functions y1,y2∈X is a solution of problems (1)-(2) if and only if y1,y2 is a fixed point of the operator S defined in (11) (see [26]).In the following, we will look for fixed points of the operatorS, because these fixed points coincide with solutions of problems (1)-(2). For use in the sequel, let γ1 and γ2 be the constants given by Lemma 4 associated, respectively, with the Green functions G1 and G2, and define γ~ by γ~:=min⁡{γ1,γ2}, and notice that γ~∈(0,1).For the sake of convenience, we set(12)f0=lim⁡(y1,y2)→(0+,0+)f(y1,y2)y1+y2,g0=lim⁡(y1,y2)→(0+,0+)g(y1,y2)y1+y2,f∞=lim⁡(y1,y2)→(∞,∞)f(y1,y2)y1+y2,g∞=lim⁡(y1,y2)→(∞,∞)g(y1,y2)y1+y2.Now we list some assumptions:(F1)
f0,g0∈(0,+∞);(F2)
f∞,g∞∈(0,+∞);(F3)
there are numbersΦ1,Φ2, where
(13)Φ1≔max⁡⁡{12[∫1/21γ~G1(1,s)a1(s)f∞ds‍]-1,ggggl12[∫1/21γ~G2(1,s)a2(s)g∞ds‍]-1},Φ2≔min⁡⁡{12[∫01G1(1,s)a1(s)f0ds‍]-1,ggggl12[∫01G2(1,s)a2(s)g0ds‍]-1},such that Φ1<λ1, λ2<Φ2.Next, we define the coneK by
(14)K≔{(y1,y2)∈X:y1,y2≥0,min⁡t∈[(1/2),1]hgmin⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ~∥(y1,y2)∥}.Lemma 6 (see [21]).
LetS be the operator defined by (11). Then S:K→K.Lemma 7.
S is a completely continuous operator.Proof.
The operatorT1:K→E is continuous in view of nonnegativeness and continuity of G1(t,s), f(y1,y2) and a1(t).
LetΩ⊆K be bounded; that is, there exists a positive constant M>0 such that ∥(y1,y2)∥≤M, for all (y1,y2)∈Ω. Let L=max⁡0≤t≤1,0≤∥(y1,y2)∥≤M|a1(t)f(y1(t),y2(t))|+1; then, for (y1,y2)∈Ω, we have
(15)|T1(y1,y2)(t)|≤λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≤λ1∫01G1(1,s)Lds‍≤LΦ2∫01G1(1,s)ds<+∞.‍
Hence, T1(Ω) is bounded.
On the other hand, givenε>0, setting δ=min⁡{(1/2)(Γ(ν1)ε/LΦ2)1/(ν1-1),εΓ(ν1)/(ν1-1)LΦ2}, then, for each (y1,y2)∈Ω, t1,t2∈[0,1], t1<t2, and t2-t1<δ, one has |T1(y1,y2)(t2)-T1(y1,y2)(t1)|<ε. That is to say, T(Ω) is equicontinuity. In fact,
(16)|T1(y1,y2)(t2)-T1(y1,y2)(t1)|=|λ1∫01G1(t2,s)a1(s)f(y1(s),y2(s))ds‍g-λ1∫01G1(t1,s)a1(s)f(y1(s),y2(s))ds‍|=|λ1∫0t1[G1(t2,s)-G1(t1,s)]a1(s)f(y1(s),y2(s))ds‍g+λ1∫t1t2[G1(t2,s)-G1(t1,s)]a1(s)‍hhhhhhhhhhh×f(y1(s),y2(s))dsg+λ1∫t21[G1(t2,s)-G1(t1,s)]a1(s)‍hhhhhhhhhhh∫t21‍×f(y1(s),y2(s))ds|≤Φ2LΓ(ν1)[(1-s)ν1-α-1(t2ν1-1-t1ν1-1)ds∫0t1(1-s)ν1-α-1(t2ν1-1-t1ν1-1)ds‍hhhhhhhhhh+∫t21(1-s)ν1-α-1(t2ν1-1-t1ν1-1)ds‍hhhhhhhhhh+∫t1t2(1-s)ν1-α-1(t2ν1-1-t1ν1-1)ds‍]=Φ2LΓ(ν1)1ν1-α(t2ν1-1-t1ν1-1)≤Φ2LΓ(ν1)(t2ν1-1-t1ν1-1).
In the following, we divide the proof into two cases.
Case
1. If δ≤t1<t2<1, then we have
(17)|T1(y1,y2)(t2)-T1(y1,y2)(t1)|≤Φ2LΓ(ν1)(t2ν1-1-t1ν1-1)=Φ2LΓ(ν1)(ν1-1)(t2-t1)tξν1-2<Φ2LΓ(ν1)(ν1-1)δ≤ε,
where tξ∈(t1,t2).
Case
2. If 0≤t1<δ,t2<2δ, then we have
(18)|T1(y1,y2)(t2)-T1(y1,y2)(t1)|≤Φ2LΓ(ν1)(t2ν1-1-t1ν1-1)≤Φ2LΓ(ν1)t2ν1-1<Φ2LΓ(ν1)(2δ)ν1-1≤ε.
By the means of the Arzela-Ascoli theorem, we have that T1 is completely continuous. Similarly, T2 is completely continuous. Consequently, S:K→K is a completely continuous operator. This completes the proof.In [21], Goodrich established the following result.Theorem 8 (see Theorem 3.3 in [21]).
Suppose that(F1)–(F3) are satisfied. Then problem (1)-(2) has at least one positive solution.From Theorem8, the following problem is natural: whether we can obtain some conclusions or not, if f0=f∞=g0=g∞=0  or f0=f∞=g0=g∞=∞? In the rest of this paper, we give some answers to this problem.For the sake of convenience, we make some assumptions:(H1)
there exist constantsρ1,A1>0, such that
(19)f(y1,y2),g(y1,y2)<A1-1ρ1for0≤∥(y1,y2)∥≤ρ1;(H2)
there exist constantsρ2,A2>0, such that
(20)f(y1,y2),g(y1,y2)≥A2-1ρ2forγ~ρ2≤∥(y1,y2)∥≤ρ2;(P1)
there are numbersΛ1,Λ2, where
(21)Λ1≔max⁡{12[∫1/21γ~G1(1,s)a1(s)ds‍]-1,ggggh12[∫1/21γ~G2(1,s)a2(s)ds‍]-1},Λ2≔min⁡{A12[∫01G1(1,s)a1(s)ds‍]-1,gggghA12[∫01G2(1,s)a2(s)ds‍]-1},such thatΛ1<λ1,λ2<Λ2;(P2)
there are numbersΛ3,Λ4, where
(22)Λ3≔max⁡⁡{A22[∫1/21G1(1,s)a1(s)ds‍]-1,ggghgA22[∫1/21G2(1,s)a2(s)ds‍]-1},Λ4≔min⁡⁡{12[∫01G1(1,s)a1(s)ds‍]-1,ggghg12[∫01‍G2(1,s)a2(s)ds]-1},such thatΛ3<λ1,λ2<Λ4.Theorem 9.
Suppose thatf0=f∞=g0=g∞=∞ and (H1),(P1) are satisfied. Then problem (1)-(2) has at least two positive solutions (y10,y20), (y¯1,y¯2), such that 0<∥(y10,y20)∥<ρ1<∥(y¯1,y¯2)∥.Proof.
From Lemma7, S is a completely continuous operator. At first, in view of f0=g0=∞, we have f(y1,y2)≥M(y1+y2), for 0<∥(y1,y2)∥<r1*<ρ1; g(y1,y2)≥M(y1+y2), for 0<∥(y1,y2)∥<r2*<ρ1, where M satisfies M≥1. Set ρ0:=min⁡{r1*,r2*}. So we define Ωρ0 by Ωρ0:={(y1,y2)∈X:∥(y1,y2)∥<ρ0}. Then for each (y1,y2)∈K⋂‍∂Ωρ0, we find that(23)T1(y1,y2)(1)=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)M(y1(s)+y2(s))ds‍≥λ1∫1/21γ~G1(1,s)a1(s)M∥(y1,y2)∥ds‍≥M2∥(y1,y2)∥≥12∥(y1,y2)∥.
So ∥T1(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ0.
Similarly, we find that∥T2(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ0. Consequently,
(24)∥S(y1,y2)∥=∥(T1(y1,y2),T2(y1,y2))∥=∥T1(y1,y2)∥+∥T2(y1,y2)∥≥∥(y1,y2)∥,
whenever (y1,y2)∈K⋂‍∂Ωρ0. Thus, S is cone expansion on K⋂‍∂Ωρ0.
Next, sincef∞=g∞=∞, we have f(y1,y2)≥M1(y1+y2) for y1+y2≥r1**>ρ1;g(y1,y2)≥M1(y1+y2) for y1+y2≥r2**>ρ1, where M1 satisfies M1≥1. Set ρ10:=max⁡{r1**,r2**}. Let ρ0*=max⁡{2ρ1,(ρ10/γ~)} and Ωρ0*:={(y1,y2)∈X:∥(y1,y2)∥<ρ0*}. Then (y1,y2)∈K⋂‍∂Ωρ0* implies
(25)y1(t)+y2(t)≥min⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ~∥(y1,y2)∥=γ~ρ0*≥ρ10.
So we obtain
(26)T1(y1,y2)(1)=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)M1(y1(s)+y2(s))ds‍≥λ1∫1/21γ~G1(1,s)a1(s)M1∥(y1,y2)∥ds‍≥M12∥(y1,y2)∥≥12∥(y1,y2)∥.
So ∥T1(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ0*.
Similarly, we find that∥T2(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ0*.
Consequently,∥S2(y1,y2)∥≥∥(y1,y2)∥, whenever (y1,y2)∈K⋂‍∂Ωρ0*. Thus, S is cone expansion on K⋂‍∂Ωρ0*.
Finally, letΩρ1:={(y1,y2)∈X:∥(y1,y2)∥<ρ1}. For (y1,y2)∈K⋂‍∂Ωρ1, from (H1), (P1), we have
(27)∥T1(y1,y2)∥=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≤A12[∫01G1(1,s)a1(s)ds‍]-1×∫01G1(1,s)a1(s)dsA1-1ρ1‍=ρ12=12∥(y1,y2)∥.
Similarly, we find that ∥T2(y1,y2)∥≤(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ1.
Consequently,∥S(y1,y2)∥≤∥(y1,y2)∥, whenever (y1,y2)∈K⋂‍∂Ωρ1. Thus, S is cone compression on K⋂‍∂Ωρ1.
So, from Lemma5, S has a fixed point (y10,y20)∈K⋂‍(Ω¯ρ1∖Ωρ0) and a fixed point (y¯1,y¯2)∈K⋂‍(Ω¯ρ0*∖Ωρ1). Both are positive solutions of BVP (1)-(2) with
(28)0<∥(y10,y20)∥<ρ1<∥(y¯1,y¯2)∥.
The proof is complete.Theorem 10.
Suppose thatf0=f∞=g0=g∞=0 and (H2), (P2) are satisfied. Then problem (1)-(2) has at least two positive solutions (y10,y20), (y¯1,y¯2), such that 0<∥(y10,y20)∥<ρ2<∥(y¯1,y¯2)∥.Proof.
At first, in view off0=g0=0, we have f(y1,y2)<ε(y1+y2), g(y1,y2)<ε(y1+y2), for 0<∥(y1,y2)∥≤ρ<ρ2, where ε satisfies ε≤1. Let Ωρ:={(y1,y2)∈X:∥(y1,y2)∥<ρ}.
Then for each(y1,y2)∈K⋂‍∂Ωρ, we find that
(29)∥T1(y1,y2)∥=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≤λ1∫01G1(1,s)a1(s)ε(y1(s)+y2(s))ds‍≤λ1∫01G1(1,s)a1(s)ε∥(y1,y2)∥ds‍≤ε2∥(y1,y2)∥≤12∥(y1,y2)∥.
Like Theorem 9, we get ∥S(y1,y2)∥≤∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ.
Next, in view off∞=g∞=0, we have f(y1,y2)<ε1(y1+y2), g(y1,y2)<ε1(y1+y2), for y1+y2≥ρ′>ρ2, where ε1 satisfies ε1≤1. We consider two cases.
Case
1. Suppose that f is unbounded; there exists ρ*>ρ′ such that
(30)f(y1,y2)≤f(y1*,y2*)for0≤∥(y1,y2)∥≤ρ*,hhhhhhhhhhhhhhhhh∥(y1*,y2*)∥=ρ*.
Since ρ*>ρ′, one has f(y1,y2)≤f(y1*,y2*)<ε1(y1*+y2*) for 0≤∥(y1,y2)∥≤ρ*. Then, for (y1,y2)∈K and ∥(y1,y2)∥=ρ*, we obtain
(31)∥T1(y1,y2)∥=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≤λ1∫01G1(1,s)a1(s)ε1(y1(s)+y2(s))ds‍≤λ1∫01G1(1,s)a1(s)ε1∥(y1,y2)∥ds‍≤ε12∥(y1,y2)∥≤12∥(y1,y2)∥.
Case
2. Suppose that f is bounded; there exists L1 such that f(y1,y2)≤L1 for all (y1,y2)∈K. Taking ρ*≥max⁡{2ρ2,L1}, for (y1,y2)∈K and ∥(y1,y2)∥=ρ*, we obtain
(32)∥T1(y1,y2)∥=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≤λ1∫01G1(1,s)a1(s)L1ds‍≤L12≤ρ*2=12∥(y1,y2)∥.
Hence, in either case, we always may set Ωρ*:={(y1,y2)∈X:∥(y1,y2)∥<ρ*} such that ∥T1(y1,y2)∥≤(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ*. Like Theorem 9, we get ∥S(y1,y2)∥≤∥(y1,y2)∥, for (y1,y2)∈K⋂‍∂Ωρ*.
Finally, setΩρ2:={(y1,y2)∈X:∥(y1,y2)∥<ρ2}. Then (y1,y2)∈K⋂‍∂Ωρ2 implies
(33)y1(t)+y2(t)≥min⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ~∥(y1,y2)∥=γ~ρ2.
Hence we have
(34)T1(y1,y2)(1)=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)A2-1ρ2ds‍≥A22A2-1ρ2=ρ22=12∥(y1,y2)∥.
Consequently, ∥T1(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ2. Like Theorem 9, we get ∥S(y1,y2)∥≥∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ2.
So, from Lemma5, S has a fixed point (y10,y20)∈K⋂‍(Ω¯ρ2∖Ωρ) and a fixed point (y¯1,y¯2)∈K⋂‍(Ω¯ρ*∖Ωρ2). Both are positive solutions of BVP (1)-(2) with
(35)0<∥(y10,y20)∥<ρ2<∥(y¯1,y¯2)∥,
which complete the proof.


### 3.2. Problem (1)–(3) in Case λ1=λ2=1

In the following, for the sake of convenience, set(36)B1:=max⁡{2∫01G1(1,s)a1(s)ds‍,2∫01G2(1,s)a2(s)ds‍},B2:=min⁡{2∫1/21G1(1,s)a1(s)ds‍,2∫1/21G2(1,s)a2(s)ds‍}.
Assume that there exist two positive constants ρ1≠ρ2 such that(H3)
f(y1,y2),g(y1,y2)≤B1-1ρ1, for 0≤∥(y1,y2)∥≤ρ1;(H4)
f(y1,y2),g(y1,y2)≥B2-1ρ2, for γ~ρ2≤∥(y1,y2)∥≤ρ2.Theorem 11.
Suppose that(H3) and (H4) are satisfied. Then problem (1)-(2), in the case where λ1=λ2=1, has at least one positive solution (y10,y20) such that ∥(y10,y20)∥ between ρ1 and ρ2.Proof.
With loss of generality, we may assume thatρ1<ρ2.
LetΩρ1:={(y1,y2)∈X:∥(y1,y2)∥<ρ1}. For (y1,y2)∈K⋂‍∂Ωρ1, one has
(37)∥T1(y1,y2)∥=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≤B12B1-1ρ1=ρ12=12∥(y1,y2)∥.
Like Theorem 9, we get ∥S(y1,y2)∥≤∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ1.
Now, setΩρ2:={(y1,y2)∈X:∥(y1,y2)∥<ρ2}. Then for (y1,y2)∈K⋂‍∂Ωρ2, one has
(38)y1(t)+y2(t)≥min⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ~∥(y1,y2)∥=γ~ρ2.
Thus, we get
(39)T1(y1,y2)(1)=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)B2-1ρ2ds‍≥B22B2-1ρ2=ρ22=12∥(y1,y2)∥.
Like Theorem 9, we get ∥S(y1,y2)∥≥∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ2. Hence, from Lemma 5, we complete the proof.Remark 12.
In [21], problem (1)-(2) with λ1=λ2=1 is not considered.


### 3.3. Problem (1), (3) in the General Case

Consider the following.Lemma 13 (see [21]).
A pair of functions(y1,y2)∈X is a solution of (1), (3) if and only if (y1,y2) is a fixed point of the operator U:X→X defined by(40)U(y1,y2)(t):=(U1(y1,y2)(t),U2(y1,y2)(t))=(β1(t)ϕ1(y1)+λ1∫01G1(t,s)a1(s)‍fhhhhhhhhhhhhhhhhhhh×(y1(s),y2(s))ds,hhhhβ2(t)ϕ2(y2)+λ2∫01G2(t,s)a2(s)g‍hhhhhhhhhhhhhhhhhh∫01‍×(y1(s),y2(s))ds),
where β1,β2:[0,1]→[0,1] are defined by
(41)β1(t):=Γ(ν1-α)Γ(ν1)tν1-1,β2(t):=Γ(ν2-α)Γ(ν2)tν2-1.Lemma 14 (see [21]).
Each ofβ1(t) and β2(t) is strictly increasing in t and satisfies β1(0)=β2(0)=0 and β1(1),β2(1)∈(0,1). Moreover, there exist constants Mβ1 and Mβ2 satisfying Mβ1,Mβ2∈(0,1) such that min⁡t∈[(1/2),1]β1(t)≥Mβ1∥β1∥ and min⁡t∈[(1/2),1]β2(t)≥Mβ2∥β2∥.
Let one define a new coneK1 by
(42)K1≔{(y1,y2)∈X:y1,y2≥0,min⁡t∈[(1/2),1]hlmin⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ0∥(y1,y2)∥},
where γ0:=min⁡{γ~,Mβ1,Mβ2}. It is obvious that γ0∈(0,1).Lemma 15 (see [21]).
U
:
K
1
→
K
1 is a completely continuous operator.
Now, one assumes(D1)
ϕ1(y1)≤∥y1∥/4, ϕ2(y2)≤∥y2∥/4 for each (y1,y2)∈K1;(P3)
There are numbersΛ5,Λ6, where
(43)Λ5≔max⁡⁡{12[∫1/21γ0G1(1,s)a1(s)ds‍]-1,hhhhhl12[∫1/21γ0G2(1,s)a2(s)ds‍]-1},Λ6≔min⁡⁡{A14[∫01G1(1,s)a1(s)ds‍]-1,hhhhhlA14[∫01G2(1,s)a2(s)ds‍]-1},such thatΛ5<λ1,λ2<Λ6.(P4)
There are numbersΛ7,Λ8, where
(44)Λ7≔max⁡⁡{A22[∫1/21G1(1,s)a1(s)ds‍]-1,hhhhhA22[∫1/21G2(1,s)a2(s)ds‍]-1},Λ8≔min⁡⁡{14[∫01G1(1,s)a1(s)ds‍]-1,hhhhh14[∫01G2(1,s)a2(s)ds‍]-1},such thatΛ7<λ1,λ2<Λ8.Theorem 16.
Suppose thatf0=f∞=g0=g∞=∞ and (H1), (D1), (P3) are satisfied. Then problem (1), (3) has at least two positive solutions (y10,y20),(y¯1,y¯2), such that
(45)0<∥(y10,y20)∥<ρ1<∥(y¯1,y¯2)∥.Proof.
At first, in view off0=g0=∞, we have f(y1,y2)≥M(y1+y2), g(y1,y2)≥M(y1+y2), for 0<∥(y1,y2)∥≤ρ0<ρ1, where M satisfies M≥1.
LetΩρ0:={(y1,y2)∈X:∥(y1,y2)∥<ρ0}. Then for each (y1,y2)∈K1⋂‍∂Ωρ0, we find that
(46)U1(y1,y2)(1)≥λ1∫1/21G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)M(y1(s)+y2(s))ds‍≥λ1∫1/21γ0G1(1,s)a1(s)M∥(y1,y2)∥ds‍≥M2∥(y1,y2)∥≥12∥(y1,y2)∥.
So ∥U1(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ0.
Similarly, we find that∥U2(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ0. Consequently,
(47)∥U(y1,y2)∥=∥(U1(y1,y2),U2(y1,y2))∥=∥U1(y1,y2)∥+∥U2(y1,y2)∥≥∥(y1,y2)∥,
whenever (y1,y2)∈K1⋂‍∂Ωρ0. Thus, U is cone expansion on K1⋂‍∂Ωρ0.
Next, sincef∞=g∞=∞, we get f(y1,y2)≥M1(y1+y2), g(y1,y2)≥M1(y1+y2), for y1+y2≥ρ10>ρ1, where M1 satisfies M1≥1. Let ρ0*=max⁡{2ρ1,(ρ10/γ0)} and Ωρ0*:={(y1,y2)∈X:∥(y1,y2)∥<ρ0*}; then, (y1,y2)∈K⋂‍∂Ωρ0* implies (48)y1(t)+y2(t)≥min⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ0∥(y1,y2)∥=γ0ρ0*≥ρ10.
So for (y1,y2)∈K1⋂‍∂Ωρ0*, we obtain
(49)U1(y1,y2)(1)≥λ1∫1/21G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)M1(y1(s)+y2(s))ds‍≥λ1∫1/21γ0G1(1,s)a1(s)M1∥(y1,y2)∥ds‍≥M12∥(y1,y2)∥≥12∥(y1,y2)∥.
That is, ∥U1(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ0*.
Similarly, we find that∥U2(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ0*. Consequently, ∥U(y1,y2)∥≥∥(y1,y2)∥, whenever (y1,y2)∈K1⋂‍∂Ωρ0*. Thus, U is cone expansion on K1⋂‍∂Ωρ0*.
Finally, letΩρ1:={(y1,y2)∈X:∥(y1,y2)∥<ρ1}. For (y1,y2)∈K⋂‍∂Ωρ1, from (H1),(D1), and (P3), we have
(50)∥U1(y1,y2)∥≤ϕ1(y1)+λ1∫01G1(t,s)a1(s)‍hhhhhhhhhhhhhh×f(y1(s),y2(s))ds≤∥y1∥4+A14A1-1ρ1≤ρ12=12∥(y1,y2)∥.
Similarly, we find that ∥U2(y1,y2)∥≤(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ1. Consequently, ∥U(y1,y2)∥≤∥(y1,y2)∥, whenever (y1,y2)∈K1⋂‍∂Ωρ1. Thus, U is cone compression on K1⋂‍∂Ωρ1.
So, from Lemma5, U has a fixed point (y10,y20)∈K1⋂‍(Ω¯ρ1∖Ωρ0) and a fixed point (y¯1,y¯2)∈K1⋂‍(Ω¯ρ0*∖Ωρ1). Both are positive solutions of BVP (1), (3) with
(51)0<∥(y10,y20)∥<ρ1<∥(y¯1,y¯2)∥.
The proof is complete.Theorem 17.
Suppose thatf0=f∞=g0=g∞=0 and (H2), (D1), (P4) are satisfied. Then problem (1), (3) has at least two positive solutions (y10,y20), (y¯1,y¯2), such that
(52)0<∥(y10,y20)∥<ρ2<∥(y¯1,y¯2)∥.Proof.
At first, in view off0=g0=0, we have f(y1,y2)<ε(y1+y2), g(y1,y2)<ε(y1+y2) for ∥(y1,y2)∥≤ρ<ρ2, where ε satisfies ε≤1. Let Ωρ:={(y1,y2)∈X:∥(y1,y2)∥<ρ}. Then for each (y1,y2)∈K1⋂‍∂Ωρ, we find that
(53)∥U1(y1,y2)∥≤ϕ1(y1)+λ1∫01G1(t,s)a1(s)f(y1(s),y2(s))ds‍≤∥y1∥4+ε4∥(y1,y2)∥≤12∥(y1,y2)∥.
Like Theorem 16, we get ∥U(y1,y2)∥≤∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ.
Next, in view off∞=g∞=0, we have f(y1,y2)<ε1(y1+y2), g(y1,y2)<ε1(y1+y2), for y1+y2≥ρ′>ρ2, where ε1 satisfies ε1≤1. We consider two cases.
Case
1. Suppose that f is unbounded and there exists ρ*>ρ′ such that
(54)f(y1,y2)≤f(y1*,y2*)for0≤∥(y1,y2)∥≤ρ*,hhhhhhhhhhhhhhhhh∥(y1*,y2*)∥=ρ*.
Since ρ*>ρ′, one has f(y1,y2)≤f(y1*,y2*)<ε1(y1*+y2*) for 0≤∥(y1,y2)∥≤ρ*.
Then, for(y1,y2)∈K1 and ∥(y1,y2)∥=ρ*, we obtain
(55)∥U1(y1,y2)∥≤ϕ1(y1)+λ1∫01G1(t,s)a1(s)f(y1(s),y2(s))ds‍≤∥y1∥4+ε14∥(y1,y2)∥≤12∥(y1,y2)∥.
Case
2. Suppose that f is bounded; there, exists L1 such that f(y1,y2)≤L1 for all (y1,y2)∈K1. Taking ρ*≥max⁡{2ρ2,L1}, for (y1,y2)∈K1 and ∥(y1,y2)∥=ρ*, we obtain
(56)∥U1(y1,y2)∥≤ϕ1(y1)+λ1∫01G1(t,s)a1(s)f(y1(s),y2(s))ds‍≤ϕ1(y1)+λ1∫01G1(t,s)a1(s)L1ds‍≤∥y1∥4+L14≤ρ*2=12∥(y1,y2)∥.
Hence, in either case, we always may set Ωρ*:={(y1,y2)∈X:∥(y1,y2)∥<ρ*} such that ∥U1(y1,y2)∥≤(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ*.
Like Theorem16, we get ∥U(y1,y2)∥≤∥(y1,y2)∥, for (y1,y2)∈K1⋂‍∂Ωρ*.
Finally, setΩρ2:={(y1,y2)∈X:∥(y1,y2)∥<ρ2}. Then (y1,y2)∈K1⋂‍∂Ωρ2 implies
(57)y1(t)+y2(t)≥min⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ0∥(y1,y2)∥=γ0ρ2.
Hence we have
(58)U1(y1,y2)(1)≥λ1∫1/21G1(1,s)a1(s)A2-1ρ2ds‍≥A22A2-1ρ2=ρ22=12∥(y1,y2)∥.
So, ∥U1(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ2.
Like Theorem16, we get ∥U(y1,y2)∥≥∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ2. So, from Lemma 5, U has a fixed point (y10,y20)∈K1⋂‍(Ω¯ρ2∖Ωρ) and a fixed point (y¯1,y¯2)∈K1⋂‍(Ω¯ρ*∖Ωρ2). Both are positive solutions of BVP (1), (3) with 0<∥(y10,y20)∥<ρ2<∥(y¯1,y¯2)∥, which complete the proof.


### 3.4. Problem (1), (3) in Case λ1=λ2=1

In [21], the author obtained that problem (1), (3) with λ1=λ2=1 having at least one positive solution. In the following, we also establish the existence of one positive solution to problem (1), (3) with λ1=λ2=1 under different conditions.For the sake of convenience, set(59)B3:=max⁡{4∫01G1(1,s)a1(s)ds‍,4∫01G2(1,s)a2(s)ds‍}.
Assume that there exist two positive constants ρ1≠ρ2 such that(H5)
f(y1,y2),g(y1,y2)≤B3-1ρ1 for 0≤∥(y1,y2)∥≤ρ1;(H6)
f(y1,y2),g(y1,y2)≥B2-1ρ2 for γ0ρ2≤∥(y1,y2)∥≤ρ2.Theorem 18.
Suppose that(H5),(H6), and (D1) are satisfied. Then problem (1), (3), in the case where λ1=λ2=1, has at least one positive solution (y10,y20) such that ∥(y1,y2)∥ between ρ1 and ρ2.Proof.
With loss of generality, we may assume thatρ1<ρ2. Let Ωρ1:={(y1,y2)∈X:∥(y1,y2)∥<ρ1}. For (y1,y2)∈K1⋂‍∂Ωρ1, from (H7), (D1), one has
(60)∥U1(y1,y2)∥≤ϕ1(y1)+∫01G1(t,s)a1(s)‍hhhhhhhhhhhh×f(y1(s),y2(s))ds≤ρ14+B34B3-1ρ1=ρ12=12∥(y1,y2)∥.
Like Theorem 16, we get ∥U(y1,y2)∥≤∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ1.
Now, setΩρ2:={(y1,y2)∈X:∥(y1,y2)∥<ρ2}. For (y1,y2)∈K1⋂‍∂Ωρ2, one has
(61)y1(t)+y2(t)≥min⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ0∥(y1,y2)∥=γ0ρ2.
Thus, from (H8), we get
(62)U1(y1,y2)(1)≥λ1∫1/21G1(1,s)a1(s)B2-1ρ2ds‍≥B22B2-1ρ2=ρ22=12∥(y1,y2)∥.
Like Theorem 16, we get ∥U(y1,y2)∥≥∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ2. Hence, from Lemma 5, we complete the proof.


## 3.1. Problem (1)-(2) in the General Case

In our considerations, letE represent the Banach space of C([0,1]) when equipped with the usual supremum norm, ∥·∥. Then put X:=E×E, where X is equipped with the norm ∥(y1,y2)∥:=∥y1∥+∥y2∥ for (y1,y2)∈X. Observe that X is also a Banach space (see [26]). In addition, we define two operators T1,T2:X→E by
(10)T1(y1,y2)(t):=λ1∫01G1(t,s)a1(s)f(y1(s),y2(s))ds,T2(y1,y2)(t):=λ2∫01G2(t,s)a2(s)g(y1(s),y2(s))ds,
where G1(t,s) is the Green function of Lemma 2 with ν replaced by ν1 and, likewise, G2(t,s) is the Green function of Lemma 2 with ν replaced by ν2. Now, we define an operator S:X→X by
(11)S(y1,y2)(t)∶=(T1(y1,y2)(t),T2(y1,y2)(t))=(λ1∫01G1(t,s)a1(s)f(y1(s),y2(s))ds,‍hh.hλ2∫01G2(t,s)a2(s)g(y1(s),y2(s))ds).
We claim that whenever (y1,y2)∈X is a fixed point of the operator defined in (11), it follows that y1(t) and y2(t) solve problems (1)-(2). That is, a pair of functions y1,y2∈X is a solution of problems (1)-(2) if and only if y1,y2 is a fixed point of the operator S defined in (11) (see [26]).In the following, we will look for fixed points of the operatorS, because these fixed points coincide with solutions of problems (1)-(2). For use in the sequel, let γ1 and γ2 be the constants given by Lemma 4 associated, respectively, with the Green functions G1 and G2, and define γ~ by γ~:=min⁡{γ1,γ2}, and notice that γ~∈(0,1).For the sake of convenience, we set(12)f0=lim⁡(y1,y2)→(0+,0+)f(y1,y2)y1+y2,g0=lim⁡(y1,y2)→(0+,0+)g(y1,y2)y1+y2,f∞=lim⁡(y1,y2)→(∞,∞)f(y1,y2)y1+y2,g∞=lim⁡(y1,y2)→(∞,∞)g(y1,y2)y1+y2.Now we list some assumptions:(F1)
f0,g0∈(0,+∞);(F2)
f∞,g∞∈(0,+∞);(F3)
there are numbersΦ1,Φ2, where
(13)Φ1≔max⁡⁡{12[∫1/21γ~G1(1,s)a1(s)f∞ds‍]-1,ggggl12[∫1/21γ~G2(1,s)a2(s)g∞ds‍]-1},Φ2≔min⁡⁡{12[∫01G1(1,s)a1(s)f0ds‍]-1,ggggl12[∫01G2(1,s)a2(s)g0ds‍]-1},such that Φ1<λ1, λ2<Φ2.Next, we define the coneK by
(14)K≔{(y1,y2)∈X:y1,y2≥0,min⁡t∈[(1/2),1]hgmin⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ~∥(y1,y2)∥}.Lemma 6 (see [21]).
LetS be the operator defined by (11). Then S:K→K.Lemma 7.
S is a completely continuous operator.Proof.
The operatorT1:K→E is continuous in view of nonnegativeness and continuity of G1(t,s), f(y1,y2) and a1(t).
LetΩ⊆K be bounded; that is, there exists a positive constant M>0 such that ∥(y1,y2)∥≤M, for all (y1,y2)∈Ω. Let L=max⁡0≤t≤1,0≤∥(y1,y2)∥≤M|a1(t)f(y1(t),y2(t))|+1; then, for (y1,y2)∈Ω, we have
(15)|T1(y1,y2)(t)|≤λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≤λ1∫01G1(1,s)Lds‍≤LΦ2∫01G1(1,s)ds<+∞.‍
Hence, T1(Ω) is bounded.
On the other hand, givenε>0, setting δ=min⁡{(1/2)(Γ(ν1)ε/LΦ2)1/(ν1-1),εΓ(ν1)/(ν1-1)LΦ2}, then, for each (y1,y2)∈Ω, t1,t2∈[0,1], t1<t2, and t2-t1<δ, one has |T1(y1,y2)(t2)-T1(y1,y2)(t1)|<ε. That is to say, T(Ω) is equicontinuity. In fact,
(16)|T1(y1,y2)(t2)-T1(y1,y2)(t1)|=|λ1∫01G1(t2,s)a1(s)f(y1(s),y2(s))ds‍g-λ1∫01G1(t1,s)a1(s)f(y1(s),y2(s))ds‍|=|λ1∫0t1[G1(t2,s)-G1(t1,s)]a1(s)f(y1(s),y2(s))ds‍g+λ1∫t1t2[G1(t2,s)-G1(t1,s)]a1(s)‍hhhhhhhhhhh×f(y1(s),y2(s))dsg+λ1∫t21[G1(t2,s)-G1(t1,s)]a1(s)‍hhhhhhhhhhh∫t21‍×f(y1(s),y2(s))ds|≤Φ2LΓ(ν1)[(1-s)ν1-α-1(t2ν1-1-t1ν1-1)ds∫0t1(1-s)ν1-α-1(t2ν1-1-t1ν1-1)ds‍hhhhhhhhhh+∫t21(1-s)ν1-α-1(t2ν1-1-t1ν1-1)ds‍hhhhhhhhhh+∫t1t2(1-s)ν1-α-1(t2ν1-1-t1ν1-1)ds‍]=Φ2LΓ(ν1)1ν1-α(t2ν1-1-t1ν1-1)≤Φ2LΓ(ν1)(t2ν1-1-t1ν1-1).
In the following, we divide the proof into two cases.
Case
1. If δ≤t1<t2<1, then we have
(17)|T1(y1,y2)(t2)-T1(y1,y2)(t1)|≤Φ2LΓ(ν1)(t2ν1-1-t1ν1-1)=Φ2LΓ(ν1)(ν1-1)(t2-t1)tξν1-2<Φ2LΓ(ν1)(ν1-1)δ≤ε,
where tξ∈(t1,t2).
Case
2. If 0≤t1<δ,t2<2δ, then we have
(18)|T1(y1,y2)(t2)-T1(y1,y2)(t1)|≤Φ2LΓ(ν1)(t2ν1-1-t1ν1-1)≤Φ2LΓ(ν1)t2ν1-1<Φ2LΓ(ν1)(2δ)ν1-1≤ε.
By the means of the Arzela-Ascoli theorem, we have that T1 is completely continuous. Similarly, T2 is completely continuous. Consequently, S:K→K is a completely continuous operator. This completes the proof.In [21], Goodrich established the following result.Theorem 8 (see Theorem 3.3 in [21]).
Suppose that(F1)–(F3) are satisfied. Then problem (1)-(2) has at least one positive solution.From Theorem8, the following problem is natural: whether we can obtain some conclusions or not, if f0=f∞=g0=g∞=0  or f0=f∞=g0=g∞=∞? In the rest of this paper, we give some answers to this problem.For the sake of convenience, we make some assumptions:(H1)
there exist constantsρ1,A1>0, such that
(19)f(y1,y2),g(y1,y2)<A1-1ρ1for0≤∥(y1,y2)∥≤ρ1;(H2)
there exist constantsρ2,A2>0, such that
(20)f(y1,y2),g(y1,y2)≥A2-1ρ2forγ~ρ2≤∥(y1,y2)∥≤ρ2;(P1)
there are numbersΛ1,Λ2, where
(21)Λ1≔max⁡{12[∫1/21γ~G1(1,s)a1(s)ds‍]-1,ggggh12[∫1/21γ~G2(1,s)a2(s)ds‍]-1},Λ2≔min⁡{A12[∫01G1(1,s)a1(s)ds‍]-1,gggghA12[∫01G2(1,s)a2(s)ds‍]-1},such thatΛ1<λ1,λ2<Λ2;(P2)
there are numbersΛ3,Λ4, where
(22)Λ3≔max⁡⁡{A22[∫1/21G1(1,s)a1(s)ds‍]-1,ggghgA22[∫1/21G2(1,s)a2(s)ds‍]-1},Λ4≔min⁡⁡{12[∫01G1(1,s)a1(s)ds‍]-1,ggghg12[∫01‍G2(1,s)a2(s)ds]-1},such thatΛ3<λ1,λ2<Λ4.Theorem 9.
Suppose thatf0=f∞=g0=g∞=∞ and (H1),(P1) are satisfied. Then problem (1)-(2) has at least two positive solutions (y10,y20), (y¯1,y¯2), such that 0<∥(y10,y20)∥<ρ1<∥(y¯1,y¯2)∥.Proof.
From Lemma7, S is a completely continuous operator. At first, in view of f0=g0=∞, we have f(y1,y2)≥M(y1+y2), for 0<∥(y1,y2)∥<r1*<ρ1; g(y1,y2)≥M(y1+y2), for 0<∥(y1,y2)∥<r2*<ρ1, where M satisfies M≥1. Set ρ0:=min⁡{r1*,r2*}. So we define Ωρ0 by Ωρ0:={(y1,y2)∈X:∥(y1,y2)∥<ρ0}. Then for each (y1,y2)∈K⋂‍∂Ωρ0, we find that(23)T1(y1,y2)(1)=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)M(y1(s)+y2(s))ds‍≥λ1∫1/21γ~G1(1,s)a1(s)M∥(y1,y2)∥ds‍≥M2∥(y1,y2)∥≥12∥(y1,y2)∥.
So ∥T1(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ0.
Similarly, we find that∥T2(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ0. Consequently,
(24)∥S(y1,y2)∥=∥(T1(y1,y2),T2(y1,y2))∥=∥T1(y1,y2)∥+∥T2(y1,y2)∥≥∥(y1,y2)∥,
whenever (y1,y2)∈K⋂‍∂Ωρ0. Thus, S is cone expansion on K⋂‍∂Ωρ0.
Next, sincef∞=g∞=∞, we have f(y1,y2)≥M1(y1+y2) for y1+y2≥r1**>ρ1;g(y1,y2)≥M1(y1+y2) for y1+y2≥r2**>ρ1, where M1 satisfies M1≥1. Set ρ10:=max⁡{r1**,r2**}. Let ρ0*=max⁡{2ρ1,(ρ10/γ~)} and Ωρ0*:={(y1,y2)∈X:∥(y1,y2)∥<ρ0*}. Then (y1,y2)∈K⋂‍∂Ωρ0* implies
(25)y1(t)+y2(t)≥min⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ~∥(y1,y2)∥=γ~ρ0*≥ρ10.
So we obtain
(26)T1(y1,y2)(1)=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)M1(y1(s)+y2(s))ds‍≥λ1∫1/21γ~G1(1,s)a1(s)M1∥(y1,y2)∥ds‍≥M12∥(y1,y2)∥≥12∥(y1,y2)∥.
So ∥T1(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ0*.
Similarly, we find that∥T2(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ0*.
Consequently,∥S2(y1,y2)∥≥∥(y1,y2)∥, whenever (y1,y2)∈K⋂‍∂Ωρ0*. Thus, S is cone expansion on K⋂‍∂Ωρ0*.
Finally, letΩρ1:={(y1,y2)∈X:∥(y1,y2)∥<ρ1}. For (y1,y2)∈K⋂‍∂Ωρ1, from (H1), (P1), we have
(27)∥T1(y1,y2)∥=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≤A12[∫01G1(1,s)a1(s)ds‍]-1×∫01G1(1,s)a1(s)dsA1-1ρ1‍=ρ12=12∥(y1,y2)∥.
Similarly, we find that ∥T2(y1,y2)∥≤(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ1.
Consequently,∥S(y1,y2)∥≤∥(y1,y2)∥, whenever (y1,y2)∈K⋂‍∂Ωρ1. Thus, S is cone compression on K⋂‍∂Ωρ1.
So, from Lemma5, S has a fixed point (y10,y20)∈K⋂‍(Ω¯ρ1∖Ωρ0) and a fixed point (y¯1,y¯2)∈K⋂‍(Ω¯ρ0*∖Ωρ1). Both are positive solutions of BVP (1)-(2) with
(28)0<∥(y10,y20)∥<ρ1<∥(y¯1,y¯2)∥.
The proof is complete.Theorem 10.
Suppose thatf0=f∞=g0=g∞=0 and (H2), (P2) are satisfied. Then problem (1)-(2) has at least two positive solutions (y10,y20), (y¯1,y¯2), such that 0<∥(y10,y20)∥<ρ2<∥(y¯1,y¯2)∥.Proof.
At first, in view off0=g0=0, we have f(y1,y2)<ε(y1+y2), g(y1,y2)<ε(y1+y2), for 0<∥(y1,y2)∥≤ρ<ρ2, where ε satisfies ε≤1. Let Ωρ:={(y1,y2)∈X:∥(y1,y2)∥<ρ}.
Then for each(y1,y2)∈K⋂‍∂Ωρ, we find that
(29)∥T1(y1,y2)∥=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≤λ1∫01G1(1,s)a1(s)ε(y1(s)+y2(s))ds‍≤λ1∫01G1(1,s)a1(s)ε∥(y1,y2)∥ds‍≤ε2∥(y1,y2)∥≤12∥(y1,y2)∥.
Like Theorem 9, we get ∥S(y1,y2)∥≤∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ.
Next, in view off∞=g∞=0, we have f(y1,y2)<ε1(y1+y2), g(y1,y2)<ε1(y1+y2), for y1+y2≥ρ′>ρ2, where ε1 satisfies ε1≤1. We consider two cases.
Case
1. Suppose that f is unbounded; there exists ρ*>ρ′ such that
(30)f(y1,y2)≤f(y1*,y2*)for0≤∥(y1,y2)∥≤ρ*,hhhhhhhhhhhhhhhhh∥(y1*,y2*)∥=ρ*.
Since ρ*>ρ′, one has f(y1,y2)≤f(y1*,y2*)<ε1(y1*+y2*) for 0≤∥(y1,y2)∥≤ρ*. Then, for (y1,y2)∈K and ∥(y1,y2)∥=ρ*, we obtain
(31)∥T1(y1,y2)∥=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≤λ1∫01G1(1,s)a1(s)ε1(y1(s)+y2(s))ds‍≤λ1∫01G1(1,s)a1(s)ε1∥(y1,y2)∥ds‍≤ε12∥(y1,y2)∥≤12∥(y1,y2)∥.
Case
2. Suppose that f is bounded; there exists L1 such that f(y1,y2)≤L1 for all (y1,y2)∈K. Taking ρ*≥max⁡{2ρ2,L1}, for (y1,y2)∈K and ∥(y1,y2)∥=ρ*, we obtain
(32)∥T1(y1,y2)∥=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≤λ1∫01G1(1,s)a1(s)L1ds‍≤L12≤ρ*2=12∥(y1,y2)∥.
Hence, in either case, we always may set Ωρ*:={(y1,y2)∈X:∥(y1,y2)∥<ρ*} such that ∥T1(y1,y2)∥≤(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ*. Like Theorem 9, we get ∥S(y1,y2)∥≤∥(y1,y2)∥, for (y1,y2)∈K⋂‍∂Ωρ*.
Finally, setΩρ2:={(y1,y2)∈X:∥(y1,y2)∥<ρ2}. Then (y1,y2)∈K⋂‍∂Ωρ2 implies
(33)y1(t)+y2(t)≥min⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ~∥(y1,y2)∥=γ~ρ2.
Hence we have
(34)T1(y1,y2)(1)=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)A2-1ρ2ds‍≥A22A2-1ρ2=ρ22=12∥(y1,y2)∥.
Consequently, ∥T1(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ2. Like Theorem 9, we get ∥S(y1,y2)∥≥∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ2.
So, from Lemma5, S has a fixed point (y10,y20)∈K⋂‍(Ω¯ρ2∖Ωρ) and a fixed point (y¯1,y¯2)∈K⋂‍(Ω¯ρ*∖Ωρ2). Both are positive solutions of BVP (1)-(2) with
(35)0<∥(y10,y20)∥<ρ2<∥(y¯1,y¯2)∥,
which complete the proof.


## 3.2. Problem (1)–(3) in Case λ1=λ2=1

In the following, for the sake of convenience, set(36)B1:=max⁡{2∫01G1(1,s)a1(s)ds‍,2∫01G2(1,s)a2(s)ds‍},B2:=min⁡{2∫1/21G1(1,s)a1(s)ds‍,2∫1/21G2(1,s)a2(s)ds‍}.
Assume that there exist two positive constants ρ1≠ρ2 such that(H3)
f(y1,y2),g(y1,y2)≤B1-1ρ1, for 0≤∥(y1,y2)∥≤ρ1;(H4)
f(y1,y2),g(y1,y2)≥B2-1ρ2, for γ~ρ2≤∥(y1,y2)∥≤ρ2.Theorem 11.
Suppose that(H3) and (H4) are satisfied. Then problem (1)-(2), in the case where λ1=λ2=1, has at least one positive solution (y10,y20) such that ∥(y10,y20)∥ between ρ1 and ρ2.Proof.
With loss of generality, we may assume thatρ1<ρ2.
LetΩρ1:={(y1,y2)∈X:∥(y1,y2)∥<ρ1}. For (y1,y2)∈K⋂‍∂Ωρ1, one has
(37)∥T1(y1,y2)∥=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≤B12B1-1ρ1=ρ12=12∥(y1,y2)∥.
Like Theorem 9, we get ∥S(y1,y2)∥≤∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ1.
Now, setΩρ2:={(y1,y2)∈X:∥(y1,y2)∥<ρ2}. Then for (y1,y2)∈K⋂‍∂Ωρ2, one has
(38)y1(t)+y2(t)≥min⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ~∥(y1,y2)∥=γ~ρ2.
Thus, we get
(39)T1(y1,y2)(1)=λ1∫01G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)B2-1ρ2ds‍≥B22B2-1ρ2=ρ22=12∥(y1,y2)∥.
Like Theorem 9, we get ∥S(y1,y2)∥≥∥(y1,y2)∥ for (y1,y2)∈K⋂‍∂Ωρ2. Hence, from Lemma 5, we complete the proof.Remark 12.
In [21], problem (1)-(2) with λ1=λ2=1 is not considered.


## 3.3. Problem (1), (3) in the General Case

Consider the following.Lemma 13 (see [21]).
A pair of functions(y1,y2)∈X is a solution of (1), (3) if and only if (y1,y2) is a fixed point of the operator U:X→X defined by(40)U(y1,y2)(t):=(U1(y1,y2)(t),U2(y1,y2)(t))=(β1(t)ϕ1(y1)+λ1∫01G1(t,s)a1(s)‍fhhhhhhhhhhhhhhhhhhh×(y1(s),y2(s))ds,hhhhβ2(t)ϕ2(y2)+λ2∫01G2(t,s)a2(s)g‍hhhhhhhhhhhhhhhhhh∫01‍×(y1(s),y2(s))ds),
where β1,β2:[0,1]→[0,1] are defined by
(41)β1(t):=Γ(ν1-α)Γ(ν1)tν1-1,β2(t):=Γ(ν2-α)Γ(ν2)tν2-1.Lemma 14 (see [21]).
Each ofβ1(t) and β2(t) is strictly increasing in t and satisfies β1(0)=β2(0)=0 and β1(1),β2(1)∈(0,1). Moreover, there exist constants Mβ1 and Mβ2 satisfying Mβ1,Mβ2∈(0,1) such that min⁡t∈[(1/2),1]β1(t)≥Mβ1∥β1∥ and min⁡t∈[(1/2),1]β2(t)≥Mβ2∥β2∥.
Let one define a new coneK1 by
(42)K1≔{(y1,y2)∈X:y1,y2≥0,min⁡t∈[(1/2),1]hlmin⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ0∥(y1,y2)∥},
where γ0:=min⁡{γ~,Mβ1,Mβ2}. It is obvious that γ0∈(0,1).Lemma 15 (see [21]).
U
:
K
1
→
K
1 is a completely continuous operator.
Now, one assumes(D1)
ϕ1(y1)≤∥y1∥/4, ϕ2(y2)≤∥y2∥/4 for each (y1,y2)∈K1;(P3)
There are numbersΛ5,Λ6, where
(43)Λ5≔max⁡⁡{12[∫1/21γ0G1(1,s)a1(s)ds‍]-1,hhhhhl12[∫1/21γ0G2(1,s)a2(s)ds‍]-1},Λ6≔min⁡⁡{A14[∫01G1(1,s)a1(s)ds‍]-1,hhhhhlA14[∫01G2(1,s)a2(s)ds‍]-1},such thatΛ5<λ1,λ2<Λ6.(P4)
There are numbersΛ7,Λ8, where
(44)Λ7≔max⁡⁡{A22[∫1/21G1(1,s)a1(s)ds‍]-1,hhhhhA22[∫1/21G2(1,s)a2(s)ds‍]-1},Λ8≔min⁡⁡{14[∫01G1(1,s)a1(s)ds‍]-1,hhhhh14[∫01G2(1,s)a2(s)ds‍]-1},such thatΛ7<λ1,λ2<Λ8.Theorem 16.
Suppose thatf0=f∞=g0=g∞=∞ and (H1), (D1), (P3) are satisfied. Then problem (1), (3) has at least two positive solutions (y10,y20),(y¯1,y¯2), such that
(45)0<∥(y10,y20)∥<ρ1<∥(y¯1,y¯2)∥.Proof.
At first, in view off0=g0=∞, we have f(y1,y2)≥M(y1+y2), g(y1,y2)≥M(y1+y2), for 0<∥(y1,y2)∥≤ρ0<ρ1, where M satisfies M≥1.
LetΩρ0:={(y1,y2)∈X:∥(y1,y2)∥<ρ0}. Then for each (y1,y2)∈K1⋂‍∂Ωρ0, we find that
(46)U1(y1,y2)(1)≥λ1∫1/21G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)M(y1(s)+y2(s))ds‍≥λ1∫1/21γ0G1(1,s)a1(s)M∥(y1,y2)∥ds‍≥M2∥(y1,y2)∥≥12∥(y1,y2)∥.
So ∥U1(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ0.
Similarly, we find that∥U2(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ0. Consequently,
(47)∥U(y1,y2)∥=∥(U1(y1,y2),U2(y1,y2))∥=∥U1(y1,y2)∥+∥U2(y1,y2)∥≥∥(y1,y2)∥,
whenever (y1,y2)∈K1⋂‍∂Ωρ0. Thus, U is cone expansion on K1⋂‍∂Ωρ0.
Next, sincef∞=g∞=∞, we get f(y1,y2)≥M1(y1+y2), g(y1,y2)≥M1(y1+y2), for y1+y2≥ρ10>ρ1, where M1 satisfies M1≥1. Let ρ0*=max⁡{2ρ1,(ρ10/γ0)} and Ωρ0*:={(y1,y2)∈X:∥(y1,y2)∥<ρ0*}; then, (y1,y2)∈K⋂‍∂Ωρ0* implies (48)y1(t)+y2(t)≥min⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ0∥(y1,y2)∥=γ0ρ0*≥ρ10.
So for (y1,y2)∈K1⋂‍∂Ωρ0*, we obtain
(49)U1(y1,y2)(1)≥λ1∫1/21G1(1,s)a1(s)f(y1(s),y2(s))ds‍≥λ1∫1/21G1(1,s)a1(s)M1(y1(s)+y2(s))ds‍≥λ1∫1/21γ0G1(1,s)a1(s)M1∥(y1,y2)∥ds‍≥M12∥(y1,y2)∥≥12∥(y1,y2)∥.
That is, ∥U1(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ0*.
Similarly, we find that∥U2(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ0*. Consequently, ∥U(y1,y2)∥≥∥(y1,y2)∥, whenever (y1,y2)∈K1⋂‍∂Ωρ0*. Thus, U is cone expansion on K1⋂‍∂Ωρ0*.
Finally, letΩρ1:={(y1,y2)∈X:∥(y1,y2)∥<ρ1}. For (y1,y2)∈K⋂‍∂Ωρ1, from (H1),(D1), and (P3), we have
(50)∥U1(y1,y2)∥≤ϕ1(y1)+λ1∫01G1(t,s)a1(s)‍hhhhhhhhhhhhhh×f(y1(s),y2(s))ds≤∥y1∥4+A14A1-1ρ1≤ρ12=12∥(y1,y2)∥.
Similarly, we find that ∥U2(y1,y2)∥≤(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ1. Consequently, ∥U(y1,y2)∥≤∥(y1,y2)∥, whenever (y1,y2)∈K1⋂‍∂Ωρ1. Thus, U is cone compression on K1⋂‍∂Ωρ1.
So, from Lemma5, U has a fixed point (y10,y20)∈K1⋂‍(Ω¯ρ1∖Ωρ0) and a fixed point (y¯1,y¯2)∈K1⋂‍(Ω¯ρ0*∖Ωρ1). Both are positive solutions of BVP (1), (3) with
(51)0<∥(y10,y20)∥<ρ1<∥(y¯1,y¯2)∥.
The proof is complete.Theorem 17.
Suppose thatf0=f∞=g0=g∞=0 and (H2), (D1), (P4) are satisfied. Then problem (1), (3) has at least two positive solutions (y10,y20), (y¯1,y¯2), such that
(52)0<∥(y10,y20)∥<ρ2<∥(y¯1,y¯2)∥.Proof.
At first, in view off0=g0=0, we have f(y1,y2)<ε(y1+y2), g(y1,y2)<ε(y1+y2) for ∥(y1,y2)∥≤ρ<ρ2, where ε satisfies ε≤1. Let Ωρ:={(y1,y2)∈X:∥(y1,y2)∥<ρ}. Then for each (y1,y2)∈K1⋂‍∂Ωρ, we find that
(53)∥U1(y1,y2)∥≤ϕ1(y1)+λ1∫01G1(t,s)a1(s)f(y1(s),y2(s))ds‍≤∥y1∥4+ε4∥(y1,y2)∥≤12∥(y1,y2)∥.
Like Theorem 16, we get ∥U(y1,y2)∥≤∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ.
Next, in view off∞=g∞=0, we have f(y1,y2)<ε1(y1+y2), g(y1,y2)<ε1(y1+y2), for y1+y2≥ρ′>ρ2, where ε1 satisfies ε1≤1. We consider two cases.
Case
1. Suppose that f is unbounded and there exists ρ*>ρ′ such that
(54)f(y1,y2)≤f(y1*,y2*)for0≤∥(y1,y2)∥≤ρ*,hhhhhhhhhhhhhhhhh∥(y1*,y2*)∥=ρ*.
Since ρ*>ρ′, one has f(y1,y2)≤f(y1*,y2*)<ε1(y1*+y2*) for 0≤∥(y1,y2)∥≤ρ*.
Then, for(y1,y2)∈K1 and ∥(y1,y2)∥=ρ*, we obtain
(55)∥U1(y1,y2)∥≤ϕ1(y1)+λ1∫01G1(t,s)a1(s)f(y1(s),y2(s))ds‍≤∥y1∥4+ε14∥(y1,y2)∥≤12∥(y1,y2)∥.
Case
2. Suppose that f is bounded; there, exists L1 such that f(y1,y2)≤L1 for all (y1,y2)∈K1. Taking ρ*≥max⁡{2ρ2,L1}, for (y1,y2)∈K1 and ∥(y1,y2)∥=ρ*, we obtain
(56)∥U1(y1,y2)∥≤ϕ1(y1)+λ1∫01G1(t,s)a1(s)f(y1(s),y2(s))ds‍≤ϕ1(y1)+λ1∫01G1(t,s)a1(s)L1ds‍≤∥y1∥4+L14≤ρ*2=12∥(y1,y2)∥.
Hence, in either case, we always may set Ωρ*:={(y1,y2)∈X:∥(y1,y2)∥<ρ*} such that ∥U1(y1,y2)∥≤(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ*.
Like Theorem16, we get ∥U(y1,y2)∥≤∥(y1,y2)∥, for (y1,y2)∈K1⋂‍∂Ωρ*.
Finally, setΩρ2:={(y1,y2)∈X:∥(y1,y2)∥<ρ2}. Then (y1,y2)∈K1⋂‍∂Ωρ2 implies
(57)y1(t)+y2(t)≥min⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ0∥(y1,y2)∥=γ0ρ2.
Hence we have
(58)U1(y1,y2)(1)≥λ1∫1/21G1(1,s)a1(s)A2-1ρ2ds‍≥A22A2-1ρ2=ρ22=12∥(y1,y2)∥.
So, ∥U1(y1,y2)∥≥(1/2)∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ2.
Like Theorem16, we get ∥U(y1,y2)∥≥∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ2. So, from Lemma 5, U has a fixed point (y10,y20)∈K1⋂‍(Ω¯ρ2∖Ωρ) and a fixed point (y¯1,y¯2)∈K1⋂‍(Ω¯ρ*∖Ωρ2). Both are positive solutions of BVP (1), (3) with 0<∥(y10,y20)∥<ρ2<∥(y¯1,y¯2)∥, which complete the proof.


## 3.4. Problem (1), (3) in Case λ1=λ2=1

In [21], the author obtained that problem (1), (3) with λ1=λ2=1 having at least one positive solution. In the following, we also establish the existence of one positive solution to problem (1), (3) with λ1=λ2=1 under different conditions.For the sake of convenience, set(59)B3:=max⁡{4∫01G1(1,s)a1(s)ds‍,4∫01G2(1,s)a2(s)ds‍}.
Assume that there exist two positive constants ρ1≠ρ2 such that(H5)
f(y1,y2),g(y1,y2)≤B3-1ρ1 for 0≤∥(y1,y2)∥≤ρ1;(H6)
f(y1,y2),g(y1,y2)≥B2-1ρ2 for γ0ρ2≤∥(y1,y2)∥≤ρ2.Theorem 18.
Suppose that(H5),(H6), and (D1) are satisfied. Then problem (1), (3), in the case where λ1=λ2=1, has at least one positive solution (y10,y20) such that ∥(y1,y2)∥ between ρ1 and ρ2.Proof.
With loss of generality, we may assume thatρ1<ρ2. Let Ωρ1:={(y1,y2)∈X:∥(y1,y2)∥<ρ1}. For (y1,y2)∈K1⋂‍∂Ωρ1, from (H7), (D1), one has
(60)∥U1(y1,y2)∥≤ϕ1(y1)+∫01G1(t,s)a1(s)‍hhhhhhhhhhhh×f(y1(s),y2(s))ds≤ρ14+B34B3-1ρ1=ρ12=12∥(y1,y2)∥.
Like Theorem 16, we get ∥U(y1,y2)∥≤∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ1.
Now, setΩρ2:={(y1,y2)∈X:∥(y1,y2)∥<ρ2}. For (y1,y2)∈K1⋂‍∂Ωρ2, one has
(61)y1(t)+y2(t)≥min⁡t∈[(1/2),1][y1(t)+y2(t)]≥γ0∥(y1,y2)∥=γ0ρ2.
Thus, from (H8), we get
(62)U1(y1,y2)(1)≥λ1∫1/21G1(1,s)a1(s)B2-1ρ2ds‍≥B22B2-1ρ2=ρ22=12∥(y1,y2)∥.
Like Theorem 16, we get ∥U(y1,y2)∥≥∥(y1,y2)∥ for (y1,y2)∈K1⋂‍∂Ωρ2. Hence, from Lemma 5, we complete the proof.


## 4. An Example

To illustrate how our main results can be used in practice, we present one example.Example 1.
Consider the following BVP, fort∈(0,1):
(63)-D0+5.2y1(t)=164500e-2t[(y1(t)+y2(t))1/2hhhhhhhhhhhhhhhhhhhh.(y1(t)+y2(t))1/2+(y1(t)+y2(t))2],-D0+5.95y2(t)=164000e-3t[(y1(t)+y2(t))1/3hhhhhhhhhhhhhhhhhhhhh.(y1(t)+y2(t))1/2+(y1(t)+y2(t))3],
subject to the boundary conditions
(64)y1(i)(0)=y2(i)(0)=0,0≤i≤4,(65)D0+1.5[y1(t)]t=1=D0+1.5[y2(t)]t=1=0.
Obviously, problem (63)–(65) fits the framework of problem (1)-(2) with
(66)ν1:=5.2,ν2:=5.95,α=1.5,λ1=164500,λ2=164000,n=6.
In addition, we have set
(67)f(y1,y2)≔(y1+y2)1/2+(y1+y2)2,g(y1,y2)≔(y1+y2)1/3+(y1+y2)3,a1(t):=e-2t,a2(t):=e-3t.We can see that f,g:[0,+∞)×[0,+∞)→[0,+∞) and are continuous. The functions a1(t) and a2(t) are obviously nonnegative.
Now, observe thatf0=f∞=g0=g∞=∞ holds. Again set A1=1/850, because f(y1,y2), g(y1,y2) is monotone increasing function for (y1,y2)≥0, taking ρ1=4; then, when ∥(y1,y2)∥∈[0,ρ1], we get
(68)f(y1,y2)≤∥(y1,y2)∥1/2+∥(y1,y2)∥2≤2+16=18<4×850=A1-1ρ1g(y1,y2)≤∥(y1,y2)∥1/3+∥(y1,y2)∥3≤41/3+64<4×850=A1-1ρ1,
which implies that (H1) holds.
On the other hand, to calculate the admissible range of the eigenvaluesλ1, λ2, as given by condition (P1), observe by numerical approximation, we find that
(69)Λ1≈163530,Λ2≈164547.25.
Thus, for any λ1, λ2 satisfying 163530<λ1, λ2<164547.25, condition (P1) will be satisfied.
Consequently, by Theorem9, problem (63)–(65) has at least two positive solutions.



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*Source: 817542-2014-01-23.xml*