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Multiplicity of positive solutions to a singular ( p 1 , p 2 ) -Laplacian system with coupled integral

boundary conditions

Jeongmi Jeong

1

, Chan-Gyun Kim

B2

and Eun Kyoung Lee

2

1Department of mathematics, Pusan National University, Busan, 609-735, Korea

2Department of Mathematics Education, Pusan National University, Busan, 609-735, Korea

Received 14 July 2015, appeared 1 June 2016 Communicated by Gennaro Infante

Abstract. In this work, we investigate the existence and multiplicity results for positive solutions to a singular(p1,p2)-Laplacian system with coupled integral boundary condi- tions and a parameter(µ,λ)∈R3+. Using sub-super solutions method and fixed point index theorems, it is shown that there exists a continuous surface C which separates R2+×(0,)into two regionsO1andO2such that the problem under consideration has two positive solutions for(µ,λ)∈ O1, at least one positive solution for(µ,λ)∈ C, and no positive solutions for(µ,λ)∈ O2.

Keywords: nonlocal boundary condition, multiplicity, positive solution, singular weight function.

2010 Mathematics Subject Classification: 34B10, 35J57, 35J92.

1 Introduction

Nonlocal boundary conditions appear when the information on the boundary are connected to values inside the domain. Various types of boundary value problems involving nonlocal conditions have been extensively studied by various methods such as fixed point theorems on cones and the Leray–Schauder alternative, etc. We refer the reader to [11,14,15,24,30–33,37]

and the references therein.

For example, Ma [24] considered





u00(t) =g1(t,u(t),u0(t)) +e(t), a.e.t ∈(0, 1), u0(0) =0, u(1) =

m i=1

αiu(ξi),

where g1 : [0, 1]×R2Ris continuous,e ∈ C[0, 1], m∈N, ξi ∈ (0,∞)with 0 < ξ1 <· · · <

ξm < 1, and αi ∈ (0,∞). Using the Leray–Schauder alternative, the existence of at least one

BCorresponding author. Email: cgkim75@gmail.com

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solution is obtained for two cases:

m i=1

αi 6=1 (Nonresonance) and

m i=1

αi =1 (Resonance).

Webb and Infante [31,32] studied the existence of multiple positive solutions of nonlinear differential equations of the form

−u00(t) =w1(t)g2(t,u(t)), a.e.t ∈(0, 1)

with various nonlocal boundary conditions involving linear functionals on C[0, 1] including the following conditions: either

u(0) =α1[u], u(1) =α2[u] or u(0) =α1[u], u0(1) =α2[u].

Here,w1 ∈ L1((0, 1),R+),R+:= [0,∞), g2 :[0, 1]×RR+ is continuous, and fori∈ {1, 2}, αi is bounded linear functionals onC[0, 1]involving Stieltjes integrals with signed measures.

Recently, Zhang and Feng [37] studied the following one-dimensional singular p-Laplacian problems of the form

(

λ(ϕp(u0(t)))0+w2(t)g3(t,u(t)) =0, t ∈(0, 1), au(0)−bu0(0) =R1

0 w3(t)u(t)dt, u0(1) =0,

whereλ is a positive parameter, ϕp(s) =|s|p2s, p> 1, a,b> 0, andw2,w3 ∈ L1((0, 1),R+). Using fixed point index theory on cones of Banach spaces, they obtained several results about the existence, multiplicity, and nonexistence of positive solutions under various assumptions on the nonlinearityg3(t,s)which satisfiesL1-Carathéodory condition.

The systems of differential equations equipped with a variety of boundary conditions have been extensively studied by many authors, see, e.g., [2–7,10,19,21,23,25,26,34]. For example, in [25], do Ó et al. considered a class of system of second-order differential equations of the

form 





−u00= g4(t,u,v,a,b), in(0, 1),

−v00 =g5(t,u,v,a,b), in(0, 1), u(0) =u(1) =v(0) =v(1) =0,

(1.1) where the nonlinearities g4 and g5 are superlinear at the origin as well as at infinity, and a,b∈R+. Using fixed point theorems of cone expansion/compression type, the upper-lower solutions method and degree argument, it was shown that there exists a continuous curve Γ which splits the positive quadrant of the(a-b)-plane into disjoint setsS1andS2such that (1.1) has at least two positive solutions in S1, has at least one positive solution on the boundary of S1, and has no positive solutions in S2. The result was applied to establish the existence and multiplicity of positive radial solutions for a certain class of semilinear elliptic systems in annular domains.

We are concerned with the existence of positive solutions to the following singular(p1,p2)- Laplacian system with coupled integral boundary conditions

((Φ(u0(t)))0+λh(t)•f(t,u(t)) =θ, t∈(0, 1),

u(0) =α[u], u(1) =µ, (Pλ,µ)

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where Φ(s1,s2) = (ϕp1(s1),ϕp2(s2)), ϕpi(s) := |s|pi2s with pi > 1 for i ∈ {1, 2}, θ is the origin of R2, α : C[0, 1]×C[0, 1] → R2 is a linear transformation which is defined by, for u∈C[0, 1]×C[0, 1],

α[u]:=

Z 1

0 u(s)k(s)ds

and k = (kij)2×2 with kij ∈ L1((0, 1),R+) for i,j ∈ {1, 2}, (µ,λ) = (µ1,µ2,λ) ∈ R3+ is a parameter, and•denotes the entrywise product, i.e.,(a1,a2)•(b1,b2):= (a1b1,a2b2).

Throughout this paper, we assume the following hypotheses are satisfied unless otherwise stated:

(H1) f = (f1,f2) : [0, 1]×R2 → (0,∞)2 with fi ∈ C([0, 1]×R2,(0,∞)) and h = (h1,h2) : (0, 1)→(0,∞)2 withhi ∈C((0, 1),(0,∞))satisfying hi ∈ Ai fori∈ {1, 2}, where

Ai = (

hˆ : Z 12

0 ϕpi1 Z 12

s

hˆ(τ)dτ

! ds+

Z 1

1 2

ϕpi1 Z s

1 2

hˆ(τ)dτ

ds< )

;

(H2) fori∈ {1, 2},R1

0(1−s)kii(s)ds∈[0, 1); (H3) detK>0, where

K:= I− Z 1

0

(1−s)k(s)ds

and I is the identity matrix of size 2.

For convenience, we identify (a,b) ∈ R2 with the 1-by-2 matrix a b

if necessary. Con- sequently,α[u]is well defined for u∈C[0, 1]×C[0, 1].

The main purpose of this paper is to study the existence and multiplicity results for posi- tive solutions to problem (Pλ,µ) using sub-super solutions method and fixed point index theo- rems. For sub-super solutions method concerning semilinear problems with nonlocal bound- ary conditions, we refer to [27–29]. It seems not obvious that sub-super solutions method can be applicable to our problem with (p1,p2)-Laplacian due to the coupled integral boundary condition in (Pλ,µ). Thus we prove a theorem for sub-super solutions (see Theorem 2.12), and it is shown that there exists a continuous surface C which separates R2+×(0,∞) into two regions O1 and O2 such that (Pλ,µ) has two positive solutions for (µ,λ) ∈ O1, at least one positive solution for (µ,λ)∈ C, and no positive solutions for(µ,λ)∈ O2 (see Theorem3.10).

Deng and Li [9] considered a semilinear elliptic problem of the form





∆u+A(x)uq=0 in Ω1,

u>0 inΩ1, u∈ Hloc1 (1)∩C(¯1), u|∂Ω1 =0, u→µ1>0 as|x| →,

(1.2)

where Ω1 =RN\ω is an exterior domain inRN, ωRN is a bounded domain with smooth boundary, N > 2, andq> 1. Among other results, whenq = (N+2)/(N−2)and 0 ≤ A∈ L1(1)satisfies certain additional conditions, it was shown that there existsµ >0 such that (1.2) has at least two positive solutions for µ1 ∈ (0,µ), exactly one positive solution forµ1 = µ, and no positive solutions forµ1∈ (µ,∞). The existence, multiplicity and nonexistence of positive radial solutions to p-Laplacian problems similar to (1.2) were studied in [16–18].

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As applications, we study existence results for positive radial solutions to p-Laplacian systems defined in an exterior domain as follows:

(div |∇z1|p2∇z1

+λK1(|x|)fˆ1(|x|,z1,z2) =0 inΩ, div |∇z2|p2∇z2

+λK2(|x|)fˆ2(|x|,z1,z2) =0 inΩ, (1.3) subject to coupled integral boundary conditions

(z1(x) =R

[l11(|y|)z1(y) +l21(|y|)z2(y)]dy on|x|=r0, z2(x) =R

[l12(|y|)z1(y) +l22(|y|)z2(y)]dy on|x|=r0, (1.4) or

(z1(x) =R

[l11(|y|)z1(y) +l21(|y|)z2(y)]dy as|x| →, z2(x) =R

[l12(|y|)z1(y) +l22(|y|)z2(y)]dy as|x| →∞, (1.5) where Ω = {x ∈ RN : |x| > r0}, r0 > 0, N > p > 1, Ki ∈ C((r0,∞),(0,∞)), lij ∈ L1((r0,∞),R+), and ˆfi ∈C([r0,∞)×R2+,(0,))fori,j∈ {1, 2}.

Using the main result (Theorem3.10), we investigate the existence, multiplicity and nonex- istence of positive solutionsz = (z1,z2)to (1.3)+(1.4) (resp. (1.3)+(1.5)) satisfyingz(x)→µas

|x| →(resp.z(x) =µon |x|=r0) for givenµR2+ andλ∈(0,)(see Corollary4.1).

For u,v ∈ R2, u ≤ v (resp. u < v) meansui ≤ vi (resp. ui < vi)for all i ∈ {1, 2}, where ui andvi arei-th coordinates ofu andv, respectively. For functions w1,w2 : [0, 1]→ Rn with n ∈ {1, 2}, w1 ≤ w2 (resp. w1 < w2) also means w1(t) ≤ w2(t) (resp. w1(t) < w2(t)) for t∈ [0, 1]. We also denoteθ the zero function from[0, 1]toR2 as well as the origin ofR2.

This paper is organized as follows. In Section 2, well-known theorems such as generalized Picone identity and a fixed point index theorem are recalled, and a solution operator and a theorem for sub-super solutions related to problem (Pλ,µ) are also introduced. In Section 3, the main result in this paper is given (see Theorem3.10). Finally, in Section 4, applications for problem (1.3)+(1.4) or (1.3)+(1.5) are given (see Corollary4.1).

2 Preliminaries

For semilinear problems, we usually use integration by parts twice in order to obtain the useful information for solutions such as a block for parameters λ and a priori estimates for solutions. However, it is not effective for thep-Laplacian problem. The following generalized Picone identity can be used to overcome the difficulty (see Lemma3.1and Lemma 3.3). The identity can be verified by straightforward differentiation, but for completeness, we give the proof of it.

Theorem 2.1(generalized Picone identity, see, e.g., [13,20]). Let us define lp[y] = (ϕp(y0))0+b1(t)ϕp(y),

Lp[z] = (ϕp(z0))0+b2(t)ϕp(z),

whereϕp(s) =|s|p2s, s ∈R, p> 1and b1,b2are continuous functions on an interval I. Let y and z be functions such that y, z, ϕp(y0), ϕp(z0)are differentiable on I and z(t)6= 0for t ∈ I.Then the

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generalized Picone identity can be written as d

dt

|y|pϕp(z0)

ϕp(z) −yϕp(y0)

= (b1−b2)|y|p

|y0|p+ (p−1)

yz0 z

p

−py0ϕp yz0

z

−ylp[y] + |y|p

ϕp(z)Lp[z]. (2.1) Proof. By straightforward differentiation,

d dt

|y|pϕp(z0) ϕp(z)

= (|y|pϕp(z0))0ϕp(z)− |y|pϕp(z0)(ϕp(z))0 (ϕp(z))2

= py

0ϕp(y)ϕp(z0) +|y|p(Lp[z]−b2(t)ϕp(z))

ϕp(z) −(p−1)

yz0 z

p

= py0ϕp yz0

z

+ |y|p

ϕp(z)Lp[z]−b2(t)|y|p−(p−1)

yz0 z

p

(2.2)

and d

dt(yϕp(y0)) =|y0|p+y(lp[y]−b1(t)ϕp(y)). (2.3) Then subtracting (2.3) from (2.2) yields the identity (2.1).

Remark 2.2. By Young’s inequality, y0ϕp

yz0 z

≤ |y0|p

p +

1− 1

p

yz0 z

p

,

and the equality holds if and only if sgny0 =sgn(yz0/z)and|y0|p =|yz0/z|p. Thus,

|y0|p+ (p−1)

yz0 z

p

−py0ϕp yz0

z

≥0, which implies, by (2.1),

d dt

|y|pϕp(z0)

ϕp(z) −yϕp(y0)

≤(b1−b2)|y|p−ylp[y] + |y|p

ϕp(z)Lp[z]. (2.4) Now we recall a well-known theorem for the existence of a global continuum of solutions by Leray and Schauder [22] and a fixed point index theorem:

Theorem 2.3(see, e.g., [35, Corollary 14.12]). Let X be a Banach space with X 6={0}and letP be an order cone in X.Consider

x= H(λ,x), (2.5)

where λR+ and x ∈ P.If H : R+× P → P is completely continuous and H(0,x) = 0 for all x∈ P,thenC+(P),the component of the solution set of (2.5)containing(0, 0), is unbounded.

Theorem 2.4(see, e.g., [12]). Let X be a Banach space,Pbe a cone in X andObe a bounded open set containingθ in X,whereθ is the origin of X. Let A:P ∩ O → P be completely continuous. Suppose that Ax6= νx for all x∈ P ∩Oand allν ≥1.Then i(A,O ∩ P,P) =1.

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2.1 Solution operator

In this subsection, we define an operator related to problem (Pλ,µ) and prove the complete continuity of it.

Denote X := C[0, 1]×C[0, 1] with norm k(u1,u2)kX := ku1k+ku2k, and P := {u = (u1,u2)∈X:u1,u2∈ K}. Then(X,k · kX)is a Banach space andPis a cone inX. Here,C[0, 1] denotes the Banach space of continuous functions u defined on [0, 1] with usual maximum normkuk:=maxt∈[0,1]|u(t)|andK:={u∈C[0, 1]: uis a nonnegative concave function}.

By(H3), detK>0 and

K1 = 1 detK

 1−

Z 1

0

(1−s)k22(s)ds

Z 1

0

(1−s)k12(s)ds Z 1

0

(1−s)k21(s)ds 1−

Z 1

0

(1−s)k11(s)ds

 .

Then all entries ofK1are nonnegative by(H2)and nonnegativity ofkij fori,j∈ {1, 2}. Defineβ:R2+×X→R2 by, for(µ,v)∈R2+×X,

β[µ,v]:= (β1[µ,v],β2[µ,v]):=

Z 1

0

(v(s) +µs)k(s)K1ds.

Then β[µn,vn] → β[µ0,v0] in R2 as (µn,vn) → (µ0,v0) in R2+×X, and β[µ,v] ∈ R2+ for all (µ,v)∈R2+× P.

Consider the following problem

((Φ(v0(t)−β[µ,v] +µ))0+λh(t)•F(µ,v)(t) =θ, t ∈(0, 1),

v(0) =v(1) =θ, ( ˆPλ,µ)

whereF:= (F1,F2):R2+×X→Xis defined by, for(µ,v)∈R2+×X,

F(µ,v)(t):= f(t,v(t) + (1−t)β[µ,v] +tµ), t∈[0, 1]. (2.6) ThenF(µ,v)>θ, since f(t,s)>θ for all(t,s)∈ [0, 1]×R2.

For i = 1, 2, define continuous transformations Li : R2+×X → X by, for t ∈ [0, 1] and (µ,u),(µ,v)∈R2+×X,

L1(µ,u)(t):= (L11(µ,u),L12(µ,u))(t):=u(t)−((1−t)α[u] +tµ) (2.7) and

L2(µ,v)(t):= (L21(µ,v),L22(µ,v))(t):=v(t) + ((1−t)β[µ,v] +tµ). (2.8) With the above transformations (2.7) and (2.8), we have the following lemma.

Lemma 2.5. Assume that(H3)holds. Thenα[u] =β[µ,L1(µ,u)]for(µ,u)∈ R2+×X andβ[µ,v] = α[L2(µ,v)]for(µ,v)∈ R2+×X.

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Proof. We only show thatα[u] = β[µ,L1(µ,u)], since the other case can be proved in a similar manner. For (µ,u)∈R2+×X,

β[µ,L1(µ,u)] =

Z 1

0

(L1(µ,u)(s) +µs)k(s)K1ds

=

Z 1

0

(u(s)−(1−s)α[u])k(s)K1ds

=

Z 1

0

u(s)k(s)K1ds−α[u]

Z 1

0

(1−s)k(s)K1ds

=α[u]

I−

Z 1

0

(1−s)k(s)ds

K1=α[u], and the proof is complete.

By a non-negative solution (resp. a positive solution)uto problem (Pλ,µ) or (Pˆλ,µ), we mean u= (u1,u2)is a solution to problem (Pλ,µ) or (Pˆλ,µ) which satisfies ui(t) ≥0 (resp.ui(t)> 0) for all t∈(0, 1)and alli∈ {1, 2}.

Remark 2.6. (1) Let v = (v1,v2) be a solution to problem (Pˆλ,µ). Then for i ∈ {1, 2}, v0i is decreasing in (0, 1), and vi is concave on(0, 1). Since v(0) = v(1) = θ, v ∈ P. Moreover, v is a positive solution to problem (Pˆλ,µ) if λ > 0, and v = θ only if λ = 0. Similarly, let u be a non-negative solution to problem (Pλ,µ). Then u∈ P, andu is a positive solution to problem (Pλ,µ) if λ>0.

(2)By Lemma2.5, for fixed(λ,µ)∈R3+, ifvis a (non-negative) solution to problem (Pˆλ,µ), thenu= L2(µ,v)is a non-negative solution to problem (Pλ,µ), and conversely ifuis a solution to problem (Pλ,µ), thenv= L1(µ,u)is a (non-negative) solution to problem (Pˆλ,µ).

Lemma 2.7. Assume that(H1)–(H3)hold. For fixed(λ,µ,v)∈(0,∞)×R2+×X withµ= (µ1,µ2) and i∈ {1, 2}, there exists a unique constant Mi =Mi(λ,µ,v)∈(0, 1)satisfying

(βi[µ,v]−µi)Mi+

Z Mi

0 ϕpi1

ϕpi(µiβi[µ,v]) +λ Z Mi

s hi(τ)Fi(µ,v)(τ)dτ

ds

= −(βi[µ,v]−µi)(1−Mi) +

Z 1

Mi

ϕpi1

ϕpi(−µi+βi[µ,v]) +λ Z s

Mi

hi(τ)Fi(µ,v)(τ)dτ

ds.

(2.9)

Proof. Let (λ,µ,v) ∈ (0,∞)×R2+×X with µ = (µ1,µ2) and i ∈ {1, 2} be fixed. Define a continuous function xi = xi(λ,µ,v):(0, 1)→Rby

xi(t) =βi[µ,v]−µi+x1i(t) +x2i(t), t ∈(0, 1), where

x1i(t) =

Z t

0 ϕpi1

ϕpi(µiβi[µ,v]) +λ Z t

s hi(τ)Fi(µ,v)(τ)dτ

ds and

x2i(t) =

Z 1

t ϕpi1

ϕpi(µiβi[µ,v])−λ Z s

t hi(τ)Fi(µ,v)(τ)dτ

ds.

We claim thatxiis strictly increasing in(0, 1). Indeed, by(H1), limt0+x1i(t) =limt1x2i(t) = 0, limt1 x1i(t)> µiβi[µ,v], and limt0+x2i(t)<µiβi[µ,v]. Consequently,

tlim0+xi(t)<0 and lim

t1xi(t)>0.

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For 0<t1<t2<1, one has xi1(t2)−x1i(t1)>

Z t2

t1 ϕpi1

ϕpi(µiβi[µ,v]) +λ Z t2

s hi(τ)Fi(µ,v)(τ)dτ

ds and

x2i(t2)−x2i(t1)>−

Z t2

t1

ϕpi1

ϕpi(µiβi[µ,v]) +λ Z t1

s hi(τ)Fi(µ,v)(τ)dτ

ds

>−

Z t2

t1

ϕpi1

ϕpi(µiβi[µ,v]) +λ Z t2

s hi(τ)Fi(µ,v)(τ)dτ

ds.

Thus,xi(t2)−xi(t1)>0 for 0<t1< t2 <1, and there exists a uniqueMi = Mi(λ,µ,v)∈(0, 1) such thatxi(Mi) =0. Consequently,Mi satisfies (2.9).

Fori=1, 2, defineTi :R3+×X→C[0, 1]by, for(λ,µ,v)∈R3+×X,

Ti(λ,µ,v)(t):=













(βi[µ,v]−µi)t+Rt 0 ϕpi1

ϕpi(µiβi[µ,v]) +λRMi

s hi(τ)Fi(µ,v)(τ)dτ

ds, 0≤t ≤Mi,

−(βi[µ,v]−µi)(1−t) +R1 t ϕpi1

ϕpi(−µi+βi[µ,v]) +λRs

Mihi(τ)Fi(µ,v)(τ)dτ

ds, Mi ≤t ≤1,

where Mi = Mi(λ,µ,v) ∈ (0, 1)is a constant satisfying (2.9) forλ > 0 and Mi may be taken arbitrary number in(0, 1)forλ=0, since Ti(0,µ,v) =0 for all (µ,v)∈ R2+×X.

For i ∈ {1, 2}, by Lemma 2.7, we can see that Ti is well-defined, kTi(λ,µ,v)k = Ti(λ,µ,v)(Mi), and (Ti(λ,µ,v))0(Mi) = 0. Moreover, since hi(t)fi(t,s) ≥ 0 for all (t,s) ∈ [0, 1]×R2,(Ti(λ,µ,v))0 is decreasing in(0, 1)for all(λ,µ,v)∈R3+×X. SinceTi(λ,µ,v)(0) = Ti(λ,µ,v)(1) =0,Ti(R3+×X)⊆ K.

Define T : R3+×X → P by T(λ,µ,v) := (T1(λ,µ,v),T2(λ,µ,v)) for (λ,µ,v) ∈ R3+×X.

ThenTis well defined, and by standard argument we have the following lemma.

Lemma 2.8. Assume that (H1)–(H3) hold. Let (λ,µ) ∈ R3+ be fixed. Then problem (Pˆλ,µ) has a (non-negative) solution v if and only if T(λ,µ,·)has a fixed point v inP. Moreover, ifθ is a solution to problem(Pˆλ,µ), thenλ=0,and ifλ=0,thenθ is a unique solution to problem(Pˆλ,µ).

To show the complete continuity ofT, we first prove the following lemma.

Lemma 2.9. Assume that(H1)–(H3)hold, let{(λn,µn,vn)}be a bounded sequence inR3+×X and let i∈ {1, 2}be fixed. If Min = Mi(λn,µn,vn) →0 (or1)as n → ∞,thenkTi(λn,µn,vn)k → 0 as n→∞.

Proof. We only prove the caseMni →0, since the other case can be proved in a similar manner.

Since {(λn,µn,vn)} is a bounded sequence in R3+×X, there exists C > 0 such that µni + λnkFi(µn,vn)k+|βi[µn,vn]| ≤Cfor alln∈N. Here,µni is thei-th component ofµn. Then

kTi(λn,µn,vn)k = Ti(λn,µn,vn)(Mni)

= (βi[µn,vn]−µni)Mni +

Z Mn

i

0 ϕpi1

ϕpi(µniβi[µn,vn]) +λn Z Mn

i

s hi(τ)Fi(µn,vn)(τ)dτ

ds

≤CMni +γ 1 pi1

Z Mn

i

0

C+ϕpi1(C)ϕpi1 Z Mn

i

s hi(τ)dτ

ds,

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whereγq=max{1, 2q1}forq>0. It follows fromhi ∈ Athat, asn→∞,kTi(λn,µn,vn)k →0, and thus the proof is complete.

Combining Lemma 2.9 with the standard arguments (see, e.g., [1, Lemma 3] and [14, Lemma 2.4]), we have the following lemma. We omit the proof of it.

Lemma 2.10. Assume that(H1)–(H3)hold. Then T:R3+×X→ P is completely continuous.

2.2 Sub-super solutions theorem

In this subsection, we give a theorem for sub-super solutions to problem (Pλ,µ).

Definition 2.11. Letλ >0 andµR2+be given. We say that ζ = (ζ1,ζ2)is asupersolutionto problem (Pλ,µ) if ζi ∈C1(0, 1)with ϕpi(ζi0)absolutely continuous fori=1, 2, and

((Φ(ζ0(t)))0+λh(t)•f(t,ζ(t))≤θ, t ∈(0, 1), ζ(0)≥α[ζ], ζ(1)≥ µ.

We also say thatψ= (ψ1,ψ2)is asubsolutionto problem (Pλ,µ) ifψi ∈C1(0, 1)with ϕpi(ψ0i) absolutely continuous fori=1, 2, and it satisfies the reverse of the above inequalities.

To get a theorem for sub-super solutions to problem (Pλ,µ), we make the following hy- potheses:

(H20) max

i∈{1,2}

Z 1

0

(1−s)[k1i(s) +k2i(s)]ds

=:Ck ∈ [0, 1);

(F1) For fixed (t,u) ∈ [0, 1]×R+, f1 = f1(t,u,v) is quasi-monotone nondecreasing with respect to v, i.e., f1(t,u,v1) ≤ f1(t,u,v2) whenever 0 ≤ v1 ≤ v2. For fixed (t,v) ∈ [0, 1]×R+, f2= f2(t,u,v)is quasi-monotone nondecreasing with respect tou.

Note that(H20)implies(H2).

Now, a theorem for sub-super solutions for the problem (Pλ,µ) is given as follows.

Theorem 2.12. Letλ > 0 andµR2+ be fixed. Assume that(H1),(H20),(H3)and(F1)hold, and that there exist ψ andζ, respectively, a subsolution and a supersolution to problem (Pλ,µ) such that ψζ. Then problem(Pλ,µ)has at least one solution u such thatψ≤u≤ ζ.

Proof. Defineγ:[0, 1]×R2R2+by, fort ∈[0, 1]andu= (u1,u2)∈R2, γ(t,u):= (γ1(t,u1),γ2(t,u2)),

where, fori∈ {1, 2},γi :[0, 1]×RR+by

γi(t,s):=





ζi(t), s≥ζi(t),

s, ψi(t)≤s ≤ζi(t), ψi(t), s≤ψi(t).

Letλ>0 andµR2+ be fixed, and consider the following modified problem ((Φ(u0(t)))0+λh(t)•f(t,γ(t,u(t))) =θ, t∈ (0, 1),

u(0) =αγ[u], u(1) =µ, (2.10)

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whereαγ[u]:=R1

0 γ(s,u(s))k(s)ds.

For givenv∈ X, definegv:R2R2by gv(x):=

Z 1

0 γ(s,v(s) + (1−s)x+sµ)k(s)ds.

Then gv is a contraction mapping on (R2,| · |R2). Here, for (x1,x2) ∈ R2, |(x1,x2)|R2 := max(|x1|,|x2|). Indeed, since |γi(t,s)−γi(t,τ)| ≤ |s−τ|for any s,τR and t ∈ [0, 1], for v¯ = (v¯1, ¯v2), ˜v = (v˜1, ˜v2)∈R2,

|γ(t, ¯v)−γ(t, ˜v)|R2 =max{|γ1(t, ¯v1)−γ1(t, ˜v1)|,|γ2(t, ¯v2)−γ2(t, ˜v2)|}

≤max{|v¯1−v˜1|,|v¯2−v˜2|}=|v¯−v˜|R2. Then, fors ∈[0, 1]andx,y∈ R2,

|γ(s,v(s) + (1−s)x+sµ)−γ(s,v(s) + (1−s)y+sµ)|R2 ≤(1−s)|x−y|R2,

and|gv(x)−gv(y)|R2 ≤Ck|x−y|R2. Thengv is a contraction mapping on(R2,| · |R2)by(H20). Thus, for given v ∈ X, there is a unique solution βγ[v] ∈ R2 of the equation x = gv(x), in other words, it is the unique element ofR2which satisfies that

βγ[v] =

Z 1

0 γ(s,v(s) + (1−s)βγ[v] +sµ)k(s)ds.

From this fact, it follows that

αγ[u] =βγ[v] under the transformations

u(t):=v(t) + (1−t)βγ[v] +tµ, v(t):= u(t)−((1−t)αγ[u] +tµ). (2.11) Thus (2.10) can be equivalently rewritten as follows:

((Φ(v0(t)−βγ[v] +µ))0+λh(t)•Fγ(v)(t) =θ, t∈(0, 1),

v(0) =v(1) =θ, (2.12)

whereFγ := (F1γ,F2γ):X→X is defined by, forv ∈X,

Fγ(v)(t):= f(t,γ(t,v(t) + (1−t)βγ[v] +tµ)), t∈[0, 1].

Consequently,vis a solution to problem (2.12) if and only ifu is a solution to problem (2.10) under the transformations (2.11), respectively.

Now, defineTγ = (T1γ,T2γ): X→ P by, for eachi=1, 2 andv∈ X,

Tiγ(v)(t):=













(βγi[v]−µi)t+Rt 0 ϕpi1

ϕpi(µiβγi[v]) +λRMγi

s hi(τ)Fiγ(v)(τ)dτ

ds, 0≤t≤ Mγi,

−(βγi[v]−µi)(1−t) +R1 t ϕpi1

ϕpi(−µi+βγi[v]) +λRs

Mγi hi(τ)Fiγ(v)(τ)dτ

ds, Mγi ≤ t≤1,

(11)

where Mγi = Miγ(v)is the constant satisfying (βγi[v]−µi)Miγ+

Z Mγ

i

0 ϕpi1 ϕpi(µiβγi [v]) +λ Z Mγ

i

s hi(τ)Fiγ(v)(τ)dτ

! ds

=−(βγi[v]−µi)(1−Mγi ) +

Z 1

Mγi

ϕpi1

ϕpi(−µi+βγi[v]) +λ Z s

Mγi hi(τ)Fiγ(v)(τ)dτ

ds.

Thenvis a fixed point ofTγinXif and only ifvis a solution to problem (2.12). It follows that Tγ is completely continuous onXandTγ(X)is bounded inX. Then, by Theorem2.4,Tγ has a fixed point v, and consequently (2.10) has a solutionuunder the first transformation (2.11).

Now if we prove thatψ≤u≤ζ, then, by the definition ofγ, (Pλ,µ) has a solutionusuch that ψ ≤ u ≤ ζ and the proof is complete. In order to show u ≤ ζ, assume on the contrary that u16≤ζ1or u26≤ζ2. We only consider the caseu1 6≤ζ1, since the case u26≤ζ2can be dealt in a similar manner. Set X1(t) := u1(t)−ζ1(t)for t ∈ [0, 1]. Then, since X1(0) =u1(0)−ζ1(0)≤ R1

02i=1[γi(s,ui(s))−ζi(s)]ki1(s)ds ≤0 and X1(1) =u1(1)−ζ1(1)≤ 0, there exists σ ∈ (0, 1) such that X1(σ) = maxt∈[0,1]X1(t) > 0. Then X01(σ) = 0 and we may assume that there is a∈(σ, 1)such thatX10(t)<0 andX1(t)>0 fort∈(σ,a], which imply that

u01(σ) =ζ10(σ), u01(t)<ζ10(t), and u1(t)>ζ1(t) fort ∈(σ,a]. (2.13) By the quasi-monotonicity of f1 and (2.13), fort ∈(σ,a],

−(ϕp1(u01(t)))0 = λh1(t)f1(t,γ1(t,u1(t)),γ2(t,u2(t)))

= λh1(t)f1(t,ζ1(t),γ2(t,u2(t)))

λh1(t)f1(t,ζ1(t),ζ2(t))≤ −(ϕp1(ζ01(t)))0. For t∈(σ,a], integrating this inequality fromσtot, we get

ϕp1(u01(t))≥ ϕp1(ζ10(t)), fort∈ (σ,a].

Since ϕp1 is monotone increasing, u01(t) ≥ ζ10(t) fort ∈ (σ,a], and it contradicts (2.13). Thus u1ζ1, and we can show that u ≤ ζ. In a similar manner, it is shown thatψ ≤ u, and thus the proof is complete.

3 Main results

First, we give a hypothesis which will be used in this section:

(F) For eachi∈ {1, 2}, there exists an interval Ii := [θ1i,θ2i]⊂(0, 1)withθi1<θi2such that fi:= lim

|s1|+|s2|→

fi(t,s1,s2) sipi1

= uniformly int ∈ Ii. Set

mh := min

tI1I2

{h1(t),h2(t)}>0. (3.1) Now we givea prioriestimates for solutions as follows.

Lemma 3.1. Assume that (H1)–(H3) and (F) hold. Let α ∈ (0,∞) be given. Then there exists M = M(α) > 0 such that kukX < M for any non-negative solutions u to problem (Pλ,µ) for all λ∈ [α,)andµR2+.

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