We investigate the existence of positive solutions for the system of fourth-order p-Laplacian boundary value problems (|u′′|p-1u′′)′′=f1(t,u,v), (|v′′|q-1v′′)′′=f2(t,u,v), u(2i)(0)=u(2i)(1)=0, i=0,1, v(2i)(0)=v(2i)(1)=0, i=0,1, where p,q>0 and f1,f2∈C([0,1]×ℝ+2,ℝ+) (ℝ+:=[0,∞)). Based on a priori estimates achieved by utilizing Jensen’s integral inequalities and nonnegative matrices, we use fixed point index theory to establish our main results.

1. Introduction

This paper is concerned with the existence of positive solutions for the system of fourth-order p-Laplacian boundary value problems
(1)(|u′′|p-1u′′)′′=f1(t,u,v),(|v′′|q-1v′′)′′=f2(t,u,v),u(2i)(0)=u(2i)(1)=0,i=0,1,v(2i)(0)=v(2i)(1)=0,i=0,1,
where p,q>0 and f1,f2∈C([0,1]×ℝ+2,ℝ+)(ℝ+:=[0,∞)).

In recent years, boundary value problems for fourth-order nonlinear ordinary differential equations have been extensively studied, with many papers on this direction published. See [1–4] and the references cited therein. The so-called p-Laplacian boundary value problems arise in non-Newtonian mechanics, nonlinear elasticity, glaciology, population biology, combustion theory, and nonlinear flow laws; see [5, 6]. That explains why many authors have extensively studied the existence of positive solutions for p-Laplacian boundary value problems, by using topological degree theory, monotone iterative techniques, coincidence degree theory [7], and the Leggett-Williams fixed point theorem [8] or its variants; see [9–18] and the references therein.

To the best of our knowledge, problem (1) is a new topic in the existing literature. Closely related to our work here is [19] that studies the existence and multiplicity of positive solutions for the system of p-Laplacian boundary value problems
(2)-((ui′)pi-1)′=fi(t,u1,…,un),ui(0)=ui′(1)=0,i=1,…,n,
where n⩾2,pi>1,fi∈C([0,1]×ℝ+n,ℝ+)(i=1,…,n). Based on a priori estimates achieved by utilizing the Jensen integral inequalities and ℝ+n-monotone matrices, the author used fixed point index theory to establish the existence and multiplicity of positive solutions for the previous problem. For more details of the recent progress in the boundary value problems for systems of nonlinear ordinary differential equations, we refer the reader to [19–27] and the references cited therein.

We observe that a close link exists between (1) and the problem below
(3)u(4)=f1(t,u,v),v(4)=f2(t,u,v),u(2i)(0)=u(2i)(1)=0,i=0,1,v(2i)(0)=v(2i)(1)=0,i=0,1,
and this link can be established by Jensen's integral inequalities for concave functions and convex functions. In other words, (1) may be regarded as a perturbation of (3). With this perspective, a priori estimates of positive solutions for some problems associated with (1) can be derived by utilizing Jensen's integral inequalities. It is the a priori estimates that permit us to use fixed point index theory to establish our main results.

This paper is organized as follows. In Section 2, we provide some preliminary results. Our main results, namely, Theorems 6 and 7, are stated and proved in Section 3.

2. Preliminaries

We first make the following hypothesis throughout this paper:

f1,f2∈C([0,1]×ℝ+2,ℝ+).

Let E:=C[0,1] and
(4)P:={u∈E:u(t)⩾0,∀t∈[0,1]},∥u∥∶=maxt∈[0,1]|u(t)|;
then (E,∥·∥) is a real Banach space and P is a cone in E. It is easy to see that (1) is equivalent to the system of nonlinear integral equations
(5)u(t)=∫01k(t,s)(∫01k(s,τ)f1(τ,u(τ),v(τ))dτ)1/pds,v(t)=∫01k(t,s)(∫01k(s,τ)f2(τ,u(τ),v(τ))dτ)1/qds,
where
(6)k(t,s)∶={t(1-s),0⩽t⩽s⩽1,s(1-t),0⩽s⩽t⩽1.
Define the operators Ai(i=1,2):P→P and A:P×P→P×P by
(7)A1(u,v)(t)∶=∫01k(t,s)(∫01k(s,τ)f1(τ,u(τ),v(τ))dτ)1/pds,A2(u,v)(t)∶=∫01k(t,s)(∫01k(s,τ)f2(τ,u(τ),v(τ))dτ)1/qds,A(u,v)(t):=(A1,A2)(u,v)(t).
Obviously, under condition (H1), the operators Ai(i=1,2):P→P and A:P×P→P×P are completely continuous operators. In our setting, the existence of positive solutions for (1) is equivalent to that of positive fixed points of A.

Below are some elementary inequalities.

Lemma 1.

Let φ(t):=sinπt, then φ∈P∖{0} and
(8)∫01k(t,s)φ(t)dt=1π2φ(s),∫01k(t,s)φ(s)ds=1π2φ(t).

Lemma 2.

Suppose that ψ∈P is concave on [0,1]. Then
(9)∫01ψ(t)sinπtdt⩾2π2∥ψ∥.

Lemma 3.

Let a∈ℝ+,b∈ℝ+. If σ∈(0,1], then
(10)(a+b)σ⩾2σ-1(aσ+bσ).
If σ∈[1,+∞), then
(11)(a+b)σ⩽2σ-1(aσ+bσ).

Lemma 4 (see [<xref ref-type="bibr" rid="B28">28</xref>]).

Let E be a real Banach space, and K is a cone in E. Suppose that Ω⊂E is a bounded open set and T:Ω¯∩K→K is a completely continuous operator. If there exists u0∈K∖{0} such that
(12)u-Tu≠λu0,∀λ>0,u∈∂Ω∩K,
then i(T,Ω∩K,K)=0, where i indicates the fixed point index on K.

Lemma 5 (see [<xref ref-type="bibr" rid="B28">28</xref>]).

Let E be a real Banach space, and K is a cone in E. Suppose that Ω⊂E is a bounded open set, 0∈Ω, and T:Ω¯∩K→K is a completely continuous operator. If
(13)u-λTu≠0,∀λ∈[0,1],u∈∂Ω∩K,
then i(T,Ω∩K,K)=1.

3. Main Results

Let p*:=max{1,p}, p*:=min{1,p}, q*:=max{1,q}, q*:=min{1,q}. Then the previous constants satisfy the simple relations
(14)0<p*⩽1⩽p*,0<q*⩽1⩽q*,p*p*=p,q*q*=q.
We now list our hypotheses on f1 and f2.

(H2) There exist five constants ai1, bi1⩾0, c>0 such that
(15)f1(t,x,y)⩾a11xp*q*+b11yp*q*-c,f2(t,x,y)⩾a21xq*p*+b21yq*q*-c
for all x,y∈ℝ+, t∈[0,1], and the matrix A1:=(a1-1b1a2b2-1) is invertible with A1-1 nonnegative, where
(16)a1:=2p*/p-1a11p*/pπ-4,b1:=2p*/p-1b11p*/pπ-4,a2:=2q*/q-1a21q*/qπ-4,b2:=2q*/q-1b21q*/qπ-4.

(H3) There exist five constants ci1,di1⩾0, r>0 such that
(17)f1(t,x,y)⩽c11xp*p*+d11yp*q*,f2(t,x,y)⩽c21xq*p*+c21yq*q*
for all x,y∈[0,r], t∈[0,1], and the matrix A2:=(1-c1-d1-c21-d2) is invertible with A2-1 nonnegative, where
(18)c1:=2p*/p-1c11p*/pπ-4,d1:=2p*/p-1d11p*/pπ-4,c2:=2q*/q-1c21q*/qπ-4,d2:=2q*/q-1d21q*/qπ-4.

(H4) There exist five constants li1,mi1⩾0 and, a real number r>0 such that
(19)f1(t,x,y)⩾l11xp*p*+m11yp*q*,f2(t,x,y)⩾l21xq*p*+m21yq*q*
for all x,y∈[0,r], t∈[0,1], and the matrix A3:=(1-l1-m1-l21-m2) is invertible with A3-1 nonnegative, where
(20)l1:=2p*/p-1l11p*/pπ-4,m1:=2p*/p-1m11p*/pπ-4,l2:=2q*/q-1l21q*/qπ-4,m2:=2q*/q-1m21q*/qπ-4.

(H5) There exist five constants pi1,qi1⩾0, c>0 such that
(21)f1(t,x,y)⩽p11xp*p*+q11yp*q*+c,f2(t,x,y)⩽p21xq*p*+q21yq*q*+c
for all x,y∈ℝ+, t∈[0,1], and the matrix A4:=(p1-1q1p2q2-1) is invertible with A4-1 nonnegative, where
(22)p1:=4p*/p-1p11p*/pπ-4,q1:=4p*/p-1q11p*/pπ-4,p2:=4q*/q-1p21q*/qπ-4,q2:=4q*/q-1q21q*/qπ-4.
In the sequel, we set Ωρ:={(u,v)∈E2:∥(u,v)∥<ρ} for all ρ>0.

Theorem 6.

If (H1)–(H3) hold, then (1) has at least one positive solution.

Proof.

Let
(23)ℳ1:={(u,v)∈P×P:(u,v)=A(u,v)+λ(φ,φ),λ⩾0},
where φ∈P∖{0} is defined by φ(t):=sinπt. We want to show that ℳ1 is bounded. Indeed, if (u,v)∈ℳ1, then there exists λ⩾0 such that
(24)(u,v)=A(u,v)+λ(φ,φ),
and thus (u,v)⩾A(u,v). Note that u,v∈ℳ1 implies that u,v are concave. So are up*,vq*∈P with p*,q*∈(0,1]. Now Lemma 2 implies that
(25)∥up*∥⩽π22∫01up*(t)sinπtdt,∥vq*∥⩽π22∫01vq*(t)sinπtdt
hold for every (u,v)∈ℳ1. Note that k(t,s)∈[0,1] for all t,s∈[0,1] and p*,p*/p∈(0,1]. By (H2), Jensen's integral inequalities, and Lemma 3, we obtain
(26)up*(t)⩾(∫01k(t,s)(∫01k(s,τ)f1(τ,u(τ),v(τ))dτ)1/pds)p*⩾∬01k(t,s)k(s,τ)f1p*/p(τ,u(τ),v(τ))dτds⩾∬01k(t,s)k(s,τ)×((a11up*p*(τ)+b11vp*q*(τ))p*/p-cp*/p(a11up*p*(τ)+b11vp*q*(τ))p*/p)dτds⩾2p*/p-1∬01k(t,s)k(s,τ)×(a11p*/pup*(τ)+b11p*/pvq*(τ)-cp*/pa11p*/p)dτds
for all u,v∈ℳ1,t∈[0,1]. Note q*,q*/q∈(0,1]. The same argument for the previous inequality can be used to obtain
(27)vq*(t)⩾2q*/q-1∬01k(t,s)k(s,τ)×(a21q*/qup*(τ)+b21q*/qvq*(τ)-cq*/qb21q*/q)dτds
for all u,v∈ℳ1,t∈[0,1]. Multiply the last two inequalities by φ(t), integrate over [0,1], and use Lemma 1 and (H2) to obtain
(28)∫01up*(t)sinπtdt⩾a1∫01up*(t)sinπtdt+b1∫01vq*(t)sinπtdt-h1,∫01vq*(t)sinπtdt⩾a2∫01up*(t)sinπtdt+b2∫01vq*(t)sinπtdt-h2,
for every (u,v)∈ℳ1, where h1:=2p*/pcp*/pπ-5,h2:=2q*/qcq*/qπ-5. The last two inequalities can be written as
(29)(a1-1b1a2b2-1)(∫01up*(t)sinπtdt∫01vq*(t)sinπtdt)=A1(∫01up*(t)sinπtdt∫01vq*(t)sinπtdt)⩽(h1h2).
Now (H2) implies
(30)(∫01up*(t)sinπtdt∫01vq*(t)sinπtdt)⩽A1-1(h1h2).
Consequently, there exists a constant M>0 such that
(31)∫01up*(t)sinπtdt⩽M,∫01vq*(t)sinπtdt⩽M.
Recalling (25), we obtain
(32)∥up*∥⩽π2M2,∥vq*∥⩽π2M2,∀(u,v)∈ℳ1.
Therefore, ℳ1 is bounded, as required. Taking R>sup{∥(u,v)∥:(u,v)∈ℳ1}, we have
(33)(u,v)≠A(u,v)+λ(φ,φ),∀(u,v)∈∂QR∩(P×P),λ⩾0.
Now Lemma 4 yields
(34)i(A,QR∩(P×P),P×P)=0.
Let
(35)ℳ2:={(u,v)∈Qr¯∩(P×P):(u,v)=λA(u,v),λ∈[0,1]}.
Now we want to show that ℳ2={0}. Indeed, if (u,v)∈ℳ2, then there exists λ∈[0,1] such that (u,v)=λA(u,v), and thus (u,v)⩽A(u,v) for all (u,v)∈ℳ2. Note that k(t,s)∈[0,1] for all t,s∈[0,1] and p*,p*/p⩾1. By (H3), Jensen’s integral inequalities, and Lemma 3, we obtain
(36)up*(t)⩽(×f1(τ,u(τ),v(τ))dτ∫01)1/p∫01k(t,s)(∫01k(s,τ)×f1(τ,u(τ),v(τ))dτ∫01)1/pds)p*⩽∬01k(t,s)k(s,τ)f1p*/p(τ,u(τ),v(τ))dτds⩽∬01k(t,s)k(s,τ)×(c11up*p*(τ)+d11vp*q*(τ))p*/pdτds⩽2p*/p-1∬01k(t,s)k(s,τ)×(c11p*/pup*(τ)+d11p*/pvq*(τ))dτds
for all (u,v)∈ℳ2,t∈[0,1]. Note that q*,q*/q⩾1. The same argument for the previous inequality can be used to obtain
(37)vq*(t)⩽2q*/q-1∬01k(t,s)k(s,τ)×(c21q*/qup*(τ)+d21q*/qvq*(τ))dτds
for all (u,v)∈ℳ2,t∈[0,1]. Multiply the last two inequalities by φ(t), integrate over [0,1], and use Lemma 1 and (H3) to obtain
(38)∫01up*(t)sinπtdt⩽c1∫01up*(t)sinπtdt+d1∫01vq*(t)sinπtdt,∫01vq*(t)sinπtdt⩽c2∫01up*(t)sinπtdt+d2∫01vq*(t)sinπtdt
for all (u,v)∈ℳ2. The last two inequalities can be written as
(39)(1-c1-d1-c21-d2)(∫01up*(t)sinπtdt∫01vq*(t)sinπtdt)=A2(∫01up*(t)sinπtdt∫01vq*(t)sinπtdt)⩽(00).
Now (H3) implies
(40)(∫01up*(t)sinπtdt∫01vq*(t)sinπtdt)⩽A2-1(00)=(00),
so that
(41)∫01up*(t)sinπtdt=0,∫01vq*(t)sinπtdt=0,
whence u≡0,v≡0, and ℳ2={0}, as required. Thus we have
(42)(u,v)≠λA(u,v),∀(u,v)∈∂Qr∩(P×P),λ∈[0,1].
Now Lemma 5 yields
(43)i(A,Qr¯∩(P×P),P×P)=1.
Combining (34) and (43) gives
(44)i(A,(QR∖Qr¯)∩(P×P),P×P)=0-1=-1.
Consequently, A has at least one fixed point on (QR∖Qr¯)∩(P×P). Thus (1) has at least one positive solution. This completes the proof.

Theorem 7.

If (H1), (H4), and (H5) hold, then (1) has at least one positive solution.

Proof.

Let
(45)ℳ3:={(u,v)∈Qr¯∩(P×P):(u,v)=A(u,v)+λ(φ,φ),λ⩾0}.
We now assert that ℳ3⊂{0}. Indeed, if (u,v)∈ℳ3, then there exists λ∈[0,1] such that (u,v)=A(u,v)+λ(φ,φ). Thus (u,v)⩾A(u,v). Note that k(t,s)∈[0,1] for all t,s∈[0,1] and p*,p*/p∈(0,1]. By (H4), Jensen's integral inequalities, and Lemma 3, we obtain
(46)up*(t)⩾(×f1(τ,u(τ),v(τ))dτ∫01)1/p∫01k(t,s)(∫01k(s,τ)×f1(τ,u(τ),v(τ))dτ∫01)1/pds)p*⩾∬01k(t,s)k(s,τ)f1p*/p(τ,u(τ),v(τ))dτds⩾∬01k(t,s)k(s,τ)×(l11up*p*(τ)+m11vp*q*(τ))p*/pdτds⩾2p*/p-1∬01k(t,s)k(s,τ)×(l11p*/pup*(τ)+m11p*/pvq*(τ))dτds
for all (u,v)∈ℳ3,t∈[0,1]. Note that q*,q*/q∈(0,1]. The same argument for the previous inequality can be used to obtain
(47)vq*(t)⩾2q*/q-1∬01k(t,s)k(s,τ)×(l21q*/qup*(τ)+m21q*/qvq*(τ))dτds
for all (u,v)∈ℳ3,t∈[0,1]. Multiply the last two inequalities by φ(t), integrate over [0,1], and use Lemma 1 and (H4) to obtain
(48)∫01up*(t)sinπtdt⩾l1∫01up*(t)sinπtdt+m1∫01vq*(t)sinπtdt,∫01vq*(t)sinπtdt⩾l2∫01up*(t)sinπtdt+m2∫01vq*(t)sinπtdt
for all (u,v)∈ℳ3. The last two inequalities can be written as
(49)(l1-1m1l2m2-1)(∫01up*(t)sinπtdt∫01vq*(t)sinπtdt)=A3(∫01up*(t)sinπtdt∫01vq*(t)sinπtdt)⩽(00).
Now (H4) implies
(50)(∫01up*(t)sinπtdt∫01vq*(t)sinπtdt)⩽A3-1(00)=(00),
so that
(51)∫01up*(t)sinπtdt=0,∫01vq*(t)sinπtdt=0,
whence u≡0,v≡0, and ℳ3⊂{0}, as asserted. As a result of this, we obtain
(52)(u,v)≠A(u,v)+λ(φ,φ),∀(u,v)∈∂Qr∩(P×P),λ⩾0.
By Lemma 4, we have
(53)i(A,Qr∩(P×P),P×P)=0.
Let
(54)ℳ4:={(u,v)∈P×P:(u,v)=λA(u,v),λ∈[0,1]}.
Now we want to show that ℳ4 is bounded. Indeed, if (u,v)∈ℳ4, then there exists λ∈[0,1] such that (u,v)=λA(u,v). This implies (u,v)⩽A(u,v). Note that k(t,s)∈[0,1] for all t,s∈[0,1] and p*,p*/p⩾1. By (H5), Jensen's integral inequalities, and Lemma 3, we obtain
(55)up*(t)⩽∬01k(t,s)k(s,τ)f1p*/p×(τ,u(τ),v(τ))dτds⩽∬01k(t,s)k(s,τ)(p11up*p*(τ)+q11vp*q*(τ)+cp11up*p*)p*/pdτds⩽2p*/p-1∬01k(t,s)k(s,τ)×((p11up*p*(τ)+q11vp*q*(τ))p*/p+cp*/pp11up*p*)dτds⩽2p*/p-1∬01k(t,s)k(s,τ)×(2p*/p-1(p11p*/pup*(τ)+q11p*/pvq*(τ))+cp*/pp11up*p*)dτds=4p*/p-1∬01k(t,s)k(s,τ)×(p11p*/pup*+q11p*/pvq*)dτds+2p*/p-1∬01k(t,s)k(s,τ)cp*/pdτds
for all (u,v)∈ℳ4,t∈[0,1]. Note that q*,q*/q⩾1. The same argument for the previous inequality can be used to obtain
(56)vq*(t)⩽4q*/q-1∬01k(t,s)k(s,τ)×(p21q*/qup*+q21q*/qvq*)dτds+2q*/q-1∬01k(t,s)k(s,τ)cq*/qdτds
for all (u,v)∈ℳ4,t∈[0,1]. Multiply the last two inequalities by φ(t), integrate over [0,1], and use Lemma 1 and (H5) to obtain
(57)∫01up*(t)sinπtdt⩽p1∫01up*(t)sinπtdt+q1∫01vq*(t)sinπtdt+h3,∫01vq*(t)sinπtdt⩽p2∫01up*(t)sinπtdt+q2∫01vq*(t)sinπtdt+h4
for all (u,v)∈ℳ4, where h3:=2p*/pcp*/pπ-5,h4:=2q*/qcq*/qπ-5. The last two inequalities can be written as
(58)(1-p1-q1-p21-q2)(∫01up*(t)sinπtdt∫01vq*(t)sinπtdt)=A4(∫01up*(t)sinπtdt∫01vq*(t)sinπtdt)⩽(h3h4).
Now (H3) implies
(59)(∫01up*(t)sinπtdt∫01vq*(t)sinπtdt)⩽A4-1(h3h4),
and thus there exists a constant M>0 such that
(60)∫01up*(t)sinπtdt⩽M,∫01vq*(t)sinπtdt⩽M
for every (u,v)∈ℳ4. The concavity of u,v and Lemma 2 imply
(61)∥u∥⩽π22∫01u(t)sinπtdt,∥v∥⩽π22∫01v(t)sinπtdt.
The fact that p*,q*⩾1, together with Jensen's inequalities, implies
(62)(∫01u(t)sinπtdt)p*⩽∫01up*(t)(sinπt)p*dt⩽∫01up*(t)sinπtdt⩽M,(∫01v(t)sinπtdt)q*⩽∫01vq*(t)(sinπt)q*dt⩽∫01vq*(t)sinπtdt⩽M,
and, in turn,
(63)∫01u(t)sinπtdt⩽M1/p*,∫01v(t)sinπtdt⩽M1/q*.
Recalling (61), we obtain
(64)∥u∥⩽π2M1/p*2,∥v∥⩽π2M1/q*2
for all (u,v)∈ℳ4. This proves the boundedness of ℳ4. Taking R>sup{∥(u,v)∥:(u,v)∈ℳ4}, we have
(65)(u,v)≠λA(u,v),∀(u,v)∈∂QR∩(P×P),λ∈[0,1].
Now Lemma 3 yields
(66)i(A,QR∩(P×P),P×P)=1.
Combining (53) with (66) gives
(67)i(A,(QR∖Qr¯)∩(P×P),P×P)=1-0=1.
Hence, A has at least one fixed point on (QR∖Qr¯)∩(P×P). Thus (1) has at least one positive solution. This completes the proof.

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