Symmetric positive solutions for fourth-order n-dimensional m-Laplace systems

R E S E A R C H

Open Access

Symmetric positive solutions

for fourth-order

n

-dimensional

m

-Laplace systems

Meiqiang Feng

1*

_{, Ping Li}

1

_{and Sujing Sun}

2

*_{Correspondence:}

meiqiangfeng@sina.com

1_{School of Applied Science, Beijing}

Information Science & Technology University, Beijing, People’s Republic of China

Full list of author information is available at the end of the article

Abstract

This paper investigates the existence, multiplicity, and nonexistence of symmetric positive solutions for the fourth-ordern-dimensionalm-Laplace system

⎧ ⎨ ⎩

φ

m(x(t)))=

(t)f(t,x(t)), 0 <t< 1,

x(0) =x(1) =₀1g(s)x(s)ds,

φ

m(x(0)) =

φ

m(x(1)) = 1

0 h(s)

φ

m(x(s))ds.

The vector-valued functionxis defined byx= [x1,x2,. . .,xn],

(t) = diag[

ψ

1(t),. . .,

ψ

i(t),. . .,

ψ

n(t)], where

ψ

i∈Lp[0, 1] for somep≥1. Our methods

employ the fixed point theorem in a cone and the inequality technique. Finally, an example illustrates our main results.

Keywords: Symmetric positive solutions;n-dimensional system;m-Laplace operator; Matrix theory; Fixed point technique

1 Introduction

Consider the fourth-ordern-dimensionalm-Laplace system

φm

x(t)=(t)ft, x(t), 0 <t< 1, (1.1) subject to the following boundary conditions:

x(0) = x(1) =₀1g(s)x(s)ds, φm(x(0)) =φm(x(1)) =

1

0 h(s)φm(x(s))ds,

(1.2)

where

x(t) =x1(t),x2(t), . . . ,xn(t)

T ,

(t) =diag ψ1(t),ψ2(t), . . . ,ψn(t)

,

f(t, x) =f1(t, x), . . . ,fi(t, x), . . . ,fn(t, x)

T ,

φm

x(t)=φm

x1(t)

,φm

x2(t)

, . . . ,φm

xn(t)

T ,

g(s) =diag g1(s),g2(s), . . . ,gn(s)

,

h(s) =diag h1(s),h2(s), . . . ,hn(s)

.

Here, we understand thatfi(t, x) means thatfi(t,x1,x2, . . . ,xn),i= 1, 2, . . . ,n. Therefore, system (1.1) means that

⎛ ⎜ ⎜ ⎜ ⎜ ⎝

(φm(x1(t))) (φm(x2(t)))

.. . (φm(xn(t)))

⎞ ⎟ ⎟ ⎟ ⎟ ⎠= ⎛ ⎜ ⎜ ⎜ ⎜ ⎝

ψ1(t) 0 · · · 0 0 ψ2(t) · · · 0

..

. ... . .. ... 0 0 · · · ψn(t)

⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⎛ ⎜ ⎜ ⎜ ⎜ ⎝

f1(t, x)

f2(t, x) .. .

fn(t, x)

⎞ ⎟ ⎟ ⎟ ⎟

⎠. (1.3)

Similarly, (1.2) means that

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎛ ⎜ ⎝

x1(0)

x2(0)

.. . xn(0)

⎞ ⎟ ⎠= ⎛ ⎜ ⎝

x1(1)

x2(1)

.. . xn(1)

⎞ ⎟ ⎠=₀1

⎛ ⎜ ⎝

g1(s) 0 ··· 0

0 g2(s)··· 0

..

. ... ... ... 0 0 ··· gn(s)

⎞ ⎟ ⎠ ⎛ ⎜ ⎝

x1(s)

x2(s)

.. . xn(s)

⎞ ⎟ ⎠ds,

⎛ ⎜ ⎝

φm(x1(0)) φm(x2(0))

.. .

φm(xn(0)) ⎞ ⎟ ⎠= ⎛ ⎜ ⎝

φm(x1(1)) φm(x2(1))

.. .

φm(xn(1)) ⎞ ⎟ ⎠=₀1

⎛ ⎜ ⎝

h1(s) 0 ··· 0

0 h2(s)··· 0

..

. ... ... ... 0 0 ···hn(s)

⎞ ⎟ ⎠ ⎛ ⎜ ⎝

φm(x1(s)) φm(x2(s))

.. .

φm(xn(s)) ⎞ ⎟ ⎠ds.

(1.4)

And then it follows respectively from (1.3) and (1.4) that

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩

(φm(x1(t)))=ψ1(t)f1(t,x1(t), . . . ,xn(t)), 0 <t< 1, (φm(x2(t)))=ψ2(t)f2(t,x1(t), . . . ,xn(t)), 0 <t< 1, ..

.

(φm(xn(t)))=ψn(t)fn(t,x1(t), . . . ,xn(t)), 0 <t< 1,

(1.5) ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

x1(0) =x1(1) =₀1g1(s)x1(s)ds, φm(x1(0)) =φm(x1(1)) =

1

0h1(s)φm(x1(s))ds,

x2(0) =x2(1) =₀1g2(s)x2(s)ds, φm(x2(0)) =φm(x2(1)) =

1

0h2(s)φm(x2(s))ds, ..

.

xn(0) =xn(1) =

1

0 gn(s)xn(s)ds, φm(xn(0)) =φm(xn(1)) =

1

0 hn(s)φm(xn(s))ds.

(1.6)

From above, we know that system (1.1)–(1.2) is equivalent to system (1.3)–(1.4), and system (1.3)–(1.4) is equivalent to system (1.5)–(1.6); thus, system (1.1)–(1.2) is equivalent to (1.5)–(1.6).

A vector-valued function x is called a solution of (1.1)–(1.2) if x∈C2([0, 1],Rn) with φm(x)∈C2((0, 1),Rn), and satisfies (1.1) and (1.2). If, for eachi= 1, 2, . . . ,n,xi(t)≥0 for allt∈(0, 1) and there is at least one nontrivial component of x, then we say that x(t) = (x1(t),x2(t), . . . ,xn(t))Tis positive onJ.

positive solutions for the above problem. For the case ofn= 1,m= 2,g(t)≡0,h(t)≡0 for

t∈Jand∈C[0, 1] not∈Lp_{[0, 1], system (1.1)–(1.2) reduces to the problem studied} by Graef et al. in [2]. By using Krasnosel’skii’s fixed-point theorem, the authors obtained some existence and nonexistence results. For other related results on system (1.1)–(1.2), we refer the reader to the references [3–27]. Moreover, for the latest development direc-tion of the fourth order differential equadirec-tions, see the references [28–31].

At the same time, we notice that a class of boundary value problems with integral bound-ary conditions has attracted many authors (see [20, 32–42]). It is an important and inter-esting problem, which contains two-point, three-point, and multi-point boundary value problems as special cases; for instance, see [43–58] and the references cited therein.

Here we point out that our problem is new in the sense of fourth-ordern-dimensional

m-Laplace systems with integral boundary conditions introduced here. To the best of our knowledge, the existence of single or multiple positive solutions for fourth-ordern -dimensionalm-Laplace systems (1.1)–(1.2) has not yet been studied, especially for the case

(t) =diag ψ1(t), . . . ,ψi(t), . . . ,ψn(t)

,

whereψi∈Lp[0, 1] for somep≥1. In consequence, our main results of the present work will be a useful contribution to the existing literature on the topic of fourth-order n -dimensionalm-Laplace systems with integral boundary conditions. The existence, multi-plicity, and nonexistence of symmetric positive solutions for the given problem are new, though they are proved by applying the well-known method based on the fixed point the-orem of cone expansion and compression of norm type.

Throughout this paper, we usei= 1, 2, . . . ,n, unless otherwise stated. LetJ= [0, 1],R+= [0, +∞),Rn

+=R+ ×R+× · · · ×R+ n

, and

x= [x1,x2, . . . ,xn]∈Rn+.

In addition, let the components of, f, h, and g satisfy the following conditions: (H1) ψi(t)∈Lp(J)for some1≤p≤+∞, andψi(t)is nonnegative, symmetric onJ, and

there existsN> 0such thatψi(t)≥Na.e. onJ;

(H2) fi:J×Rn+→R+is continuous, and for allx∈Rn+,fi(t, x)is symmetric onJ; (H3) gi,hi∈L1(J)are nonnegative, symmetric onJwith

μi:=

1

0

gi(s)ds∈[0, 1), νi:=

1

0

hi(s)ds∈[0, 1). (1.7)

2 Preliminaries

In this part, we give some properties of Green’s function associated with system (1.1)– (1.2), and we present some definitions and lemmas which are needed throughout this pa-per.

Definition 2.1(see [59]) LetEbe a real Banach space overR. A nonempty closed set

P⊂Eis said to be a cone provided that

(i) au+bv∈Pfor allu,v∈Pand alla≥0,b≥0and (ii) u, –u∈Pimpliesu= 0.

Every coneP⊂Einduces a semi-ordering inEgiven byu≤vif and only ifv–u∈P.

Definition 2.2 Ifx(t) =x(1 –t),t∈J, thenxis said to be symmetric inJ.

In our discussion, x = (x1,x2, . . . ,xn)T is symmetric onJ if and only ifxiis symmetric onJ.

Next, we reduce system (1.1)–(1.2) to an integral system. It follows from system (1.5)– (1.6) that system (1.1)–(1.2) can be written as follows:

⎧ ⎪ ⎨ ⎪ ⎩

(φm(xi(t)))=ψi(t)fi(t,x1(t), . . . ,xn(t)), 0 <t< 1,

xi(0) =xi(1) =

1

0 gi(s)xi(s)ds, φm(xi(0)) =φm(xi(1)) =

1

0 hi(s)φm(xi(s))ds,

(2.1)

whereφm(s) =|s|m–2s,m> 1,φm∗=φ–1m,m1 + 1 m∗= 1. Firstly, by means of the transformation

φmxi(t)

= –yi(t), (2.2)

we can convert system (2.1) into

yi(t) = –ψi(t)fi(t, x(t)), 0 <t< 1,

yi(0) =yi(1) =

1

0hi(s)yi(s)ds,

(2.3)

and

xi(t) = –φm∗(yi(t)), 0 <t< 1,

xi(0) =xi(1) =

1

0 gi(s)xi(s)ds.

(2.4)

Lemma 2.1 Assume that(H3)holds.Then system(2.4)has a unique solution xi(t)and xi(t)

can be expressed in the form

xi(t) = –

1

0

Hi(t,s)φm∗

yi(s)

ds, (2.5)

where

Hi(t,s) =G(t,s) + 1 1 –μi

1

0

G(τ,s)gi(τ)dτ, (2.6)

G(t,s) =

t(1 –s), 0≤t≤s≤1,

Proof The proof is similar to that of Lemma 2.1 in [60]. By (2.6) and (2.7), we can show thatHi_{(t,s) andG(t,s) have the following properties.}

Proposition 2.1 Assume that(H3)holds.Then we have Hi(t,s) > 0, G(t,s) > 0, ∀t,s∈(0, 1);

Hi(t,s)≥0, G(t,s)≥0, ∀t,s∈J.

(2.8)

Proposition 2.2 For all t,s∈J,we have

e(t)e(s)≤G(t,s)≤G(t,t) =t(1 –t) =e(t)≤e=max

t∈J e(t) = 1

4, (2.9)

G(1 –t, 1 –s) =G(t,s). (2.10)

Proposition 2.3 Assume that(H3)holds.Then,for all t,s∈J,we have

ρie(s)≤Hi(t,s)≤γis(1 –s) =γie(s)≤1 4γ

i_{, (2.11)}

where

γi= 1

1 –μi

, ρi=

1

0 e(τ)gi(τ)dτ 1 –μi

. (2.12)

Proof By (2.6) and (2.9), we have

Hi(t,s) =G(t,s) + 1 1 –μi

1

0

G(s,τ)gi(τ)dτ

≥ 1

1 –μi

1

0

G(s,τ)gi(τ)dτ

≥

1

0 e(τ)gi(τ)dτ 1 –μi

s(1 –s)

=ρie(s), t∈J. (2.13)

In addition, noticing thatG(t,s)≤s(1 –s), we have

Hi(t,s) =G(t,s) + 1 1 –μi

1

0

G(s,τ)gi(τ)dτ

≤s(1 –s) + 1 1 –μi

1

0

s(1 –s)gi(τ)dτ

≤s(1 –s)

1 + 1 1 –μi

1

0

gi(τ)dτ

=s(1 –s) 1 1 –μi

=γie(s), t∈J. (2.14)

Proposition 2.4 Assume that(H3)holds.Then,for all t,s∈J,we have

Hi(1 –t, 1 –s) =Hi(t,s). (2.15)

Proof The proof is similar to that of Proposition 2.1 of [60].

Lemma 2.2 Assume that(H1)–(H3)hold.Then system(2.3)has a unique solution

yi(t) = –

1

0

H₁i(t,s)ψi(s)fi

s, x(s)ds, (2.16)

where

H₁i(t,s) =G(t,s) + 1 1 –νi

1

0

G(v,s)hi(v)dv. (2.17)

Proof The proof is similar to Lemma 2.1 of [60].

Remark2.1 Assume that (H3) holds. Then, for allt,s∈J, it follows from (2.17) that

Hi

1(t,s)≥0, ρ1ie(s)≤H1i(t,s)≤γ1is(1 –s)≤ 1 4γ

i

1, H1i(1 –t, 1 –s) =H1i(t,s), where

ρ₁i=

1

0 e(τ)hi(τ)dτ 1 –νi

, γ₁i= 1 1 –νi

.

Assume thatxiis a solution of system (2.1). Then it follows from Lemma 2.1 that

xi(t) = –

1

0

Hi(t,s)φm∗

yi(s)

ds, (2.18)

and then, it follows from Lemma 2.2 that

xi(t) =

1

0

Hi(t,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, x(τ)dτ

ds. (2.19)

LetE=C[0, 1],X=E×E× · · · × E

n

, and for all x = (x1,x2, . . . ,xn)T∈X, the norm inXis defined as

x= n

i=1

sup

t∈J |

xi|.

Then (X, · ) is a real Banach space. Define a coneKinXby

K=

x= (x1,x2, . . . ,xn)T∈X:xi≥0,xi(t) is symmetric and concave onJ,

min

t∈J n

i=1

xi(t)≥δx

where

δ= min

1≤i≤nδi, δi= ρi_(ρi

1)m ∗_–1

γi_(γi 1)m

∗_–1.

We also define two setsKr,Kr,Rby

Kr={x∈K:x<r}, Kr,R={x∈K:r<x<R}, where 0 <r<R.

To make our research significant, letgi(t)≡0,hi(t)≡0 for anyt∈J,i= 1, 2, . . . ,n.

Remark2.2 By the definition ofρi,ρ₁i,γi,γ₁i, we have 0 <δi< 1, and then 0 <δ< 1. Let T :K→Xbe a map with components (T1, . . . ,Ti, . . . ,Tn). Here, we understand Tx = (T1x, . . . ,Tix, . . . ,Tnx)T, where

(Tix)(t) =

1

0

Hi(t,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, x(τ)dτ

ds. (2.21)

From the proof of Lemma 2.1 and Lemma 2.2, we have the following remark.

Remark2.3 From (2.21), we know that x∈Xis a solution of system (1.1)–(1.2) if and only if x is a fixed point of the map T.

Lemma 2.3 Assume that(H1)–(H3)hold.Then we have T(K)⊂K,andT:K→K is

completely continuous.

Proof For all x∈K, from (2.21), we know that

(Tix)= –φm∗

1

0

Hi₁(t,s)ψi(s)fi

s, x(s)ds

≤0, (2.22)

which implies thatTixis concave onJ. In addition, it follows from (2.21) that

(Tix)(0) = (Tix)(1)≥0.

Thus, for allt∈J, we have (Tix)(t)≥0. Noticing thatψi(t) is symmetric on (0, 1),xi(t) is symmetric onJ, andfi(·, x) is symmetric onJ, we have

(Tix)(1 –t) =

1

0

Hi(1 –t,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, x(τ)dτ

ds

=

0

1

Hi(1 –t, 1 –s)φm∗

1

0

H₁i(1 –s,τ)ψi(τ)fi

τ, x(τ)dτ

d(1 –s)

=

1

0

Hi(t,s)φm∗

1

0

H₁i(1 –s,τ)ψi(τ)fi

τ, x(τ)dτ

ds

=

1

0

×

0

1

H₁i(1 –s, 1 –τ)ψi(1 –τ)fi

1 –τ, x(1 –τ)d(1 –τ)

ds

=

1

0

Hi(t,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, x(τ)dτ

ds

= (Tix)(t),

which shows that (Tix)(1 –t) = (Tix)(t),t∈J. And hence (Tix)(t) is symmetric onJ. In addition, according to (2.14), we know that

(Tix)(t) =

1

0

Hi(t,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, x(τ)dτ

ds

≤γiγ₁im∗–1

1

0

e(s)φm∗

1

0

e(τ)ψi(τ)fi

τ, x(τ)dτ

ds, ∀t∈J.

Then

Tx= n

i=1

sup

t∈J

(Tix)(t)≤ n

i=1

γiγ₁im∗–1

1

0

e(s)φm∗

1

0

e(τ)ψi(τ)fi

τ, x(τ)dτ

ds.

Similarly, according to (2.13), we know that

min

t∈J n

i=1

(Tix)(t) =min t∈J

n

i=1

1

0

Hi(t,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, x(τ)dτ

ds ≥ n i=1

ρiρ₁im∗–1

1

0

e(s)φm∗

1

0

e(τ)ψi(τ)fi

τ, x(τ)dτ

ds = n i=1 δiγi

γ₁im∗–1

1

0

e(s)φm∗

1

0

e(τ)ψi(τ)fi

τ, x(τ)dτ

ds

≥δTx.

Thus, we haveTix∈K, then there is T(K)⊂K.

Next, we show T is completely continuous, and we need to showTiis completely con-tinuous.

Letl> 0 and define

f_li=sup

t∈J

fi

t, x(t): x∈Rn₊,x ≤l> 0.

We show thatTiis compact.

For eachl> 0, letBl={x∈K:x ≤l}. ThenBl is a bounded closed convex set inK. ∀(xm)m∈N∈K, it follows from (2.21) that

|Tixm|=

₀1Hi(t,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, xm(τ)

dτ ds ≤1 4γ i 1 4γ i 1

m∗–1

1

0 φm∗

1

0 ψi(τ)fi

τ, xm(τ)

dτ

≤1 4γ

i

1 4γ

i 1

m∗–1

f_lim∗–1

1

0 φm∗

1

0

ψi(τ)dτ

ds

≤1 4γ

i

1 4γ

i 1

m∗–1

_fi l

m∗–1

1

0 φm∗

ψi1

ds

=1 4γ

i

1 4γ

i 1

m∗–1

f_lim∗–1ψi1

m∗–1

=

1 4γ

i 1

m∗

fi l

m∗–1 ψi1

m∗–1 .

Therefore, (Ti(Bl)) is uniformly bounded.

Next we show the equicontinuity of (Tixm)m∈N. Due toHi(t,s) is continuous onJ×J, thenHi_{(t,s) is uniformly continuous. Thus, for anyε> 0, there existl}

1> 0,t1,t2∈J, if |t1–t2|<l1, we have

(Tixm)(t2) – (Tixm)(t1) =

₀1Hi(t2,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, xm(τ)

dτ

ds

–

1

0

Hi(t1,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, xm(τ)

dτ

ds

=

1

0

Hi(t2,s) –Hi(t1,s)

×φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, xm(τ)

dτ

ds

≤

1 4γ

i 1

m∗–1 ψi1

m∗–1

f_lim∗–1

1

0

Hi(t1,s) –Hi(t2,s) ds

≤ε,

which shows that (Tixm)m∈Nis equicontinuous onJ. Therefore, it follows from theArzelà–

Ascolitheorem that there exist a functionT_i1∈C[0, 1] and a subsequence of (Tixm)m∈N converging uniformly toT1

i onJ.

We prove the continuity ofTi. Let (xm)m∈N be any sequence converging onKto x∈K, and letL> 0 be such thatxm ≤Lfor allm∈N. Note thatfi(t, x) is continuous onJ×KL. It is not difficult to see that the dominated convergence theorem guarantees that

lim

m→∞(Tixm)(t) = (Tix)(t) (2.23)

for eacht∈J. Moreover, the compactness ofTiimplies that (Tixm)(t) converges uniformly to (Tix)(t) onJ. If not, then there existε0> 0 and a subsequence (xmj)j∈Nof (xm)m∈N such

that

sup

t∈J

(Tixmj)(t) – (Tix)(t) ≥ε0, j∈N. (2.24)

Now, it follows from the compactness ofTithat there exists a subsequence of (xmj)j∈N

converges uniformly toy0∈C[0, 1]. Thus, from (2.24), we easily see that

sup

t∈J

y0(t) – (Tix)(t) ≥ε0, j∈N. (2.25)

On the other hand, from the pointwise convergence (2.23) we obtain

y0(t) = (Tix)(t), t∈J.

This is a contradiction to (2.25). ThereforeTiis continuous.

Therefore Ti : K → K is completely continuous. This completes the proof of

Lemma 2.3.

In the following lemma, we employ Hölder’s inequality to obtain some of the norm in-equalities in our main results.

Lemma 2.4(Hölder) Let e∈Lp_{[a,b]with p> 1,h∈L}q_{[a,b]with q> 1,and}1 p+

1

q= 1.Then

eh∈L1_[a,b],and eh1≤ ephq.

Let e∈L1_{[a,b]and h∈L}∞_{[a,b].Then eh∈L}1_[a,b],and eh1≤ e1h∞.

Finally, we state the well-known fixed point theorem of cone expansion and compression of norm type.

Lemma 2.5(see [59]) Let P be a cone in a real Banach space E.Assume1,2are bounded

open sets in E with0∈1⊂2,A:P∩(2\1)→P is completely continuous such that

either

(i) Ax ≤ x,∀x∈P∩∂1;Ax ≥ x,∀x∈P∩∂2or (ii) Ax ≥ x,∀x∈P∩∂1;Ax ≤ x,∀x∈P∩∂2.

Then A has at least one fixed point in P∩(2\1).

Remark2.4 To make it clear for the reader what1,2,∂1,∂2, and¯2\1mean, we give typical examples of1and2.

1=

x∈C[a,b] :x<r, 2=

x∈C[a,b] :x<R,

2\1=

x∈C[a,b] :r≤ x ≤R, where 0 <r<R,x=maxt∈[a,b]|x(t)|.

3 Main results

For convenience’s sake, we introduce the notations:

f_iβ=lim sup

x→β

max

t∈J

fi(t, x) φm(x)

, fiβ=lim inf x→βmint∈J

fi(t, x) φm(x)

,

Di= 1 6nγ

i_γi

1eqωip

m∗–1

, Di₁=

1 6

m∗

nγiγ₁iNm∗–1,

whereβis 0 or∞.

The first existence theorem deals with the case 1 <p<∞.

Theorem 3.1 Assume that(H1)–(H3)hold.Furthermore,assume that one of the following

conditions is satisfied.

(C1) There exist two constantsr,Rwith0 <r≤δRsuch thatfi(t, x)≤φm(_Dr_i)fort∈J, 0≤ x ≤r,andfi(t, x)≥φm(_δR

iDi1

)fort∈J,δR≤ x ≤R; (C2) fi0>φm(_δ1

iDi1)

andf_i∞<φm(_D1_i)(particularly,fi0=∞andfi∞= 0).

Then,system(1.1)–(1.2)has at least one symmetric positive solution.

Proof Case (1). Considering the condition (C1), for all x∈∂Kr, we have x=r and

fi(t, x(t))≤φm(_Dr_i),i= 1, 2, . . . ,n. Thus, for allt∈J, we have

Tx= n

i=1

sup

t∈J (Tix)(t)

= n

i=1

sup

t∈J

1

0

Hi(t,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, x(τ)dτ

ds ≤ n i=1

γiγ₁im∗–1

1

0

e(s)φm∗

1

0

e(τ)ψi(τ)φm

r Di dτ ds ≤ n i=1

γiγ₁im∗–1 r

Di

1

0

e(s)φm∗

1

0

eqψipdτ

ds = n i=1 1 6γ

i_γi

1eqψip

m∗–1 r

Di = n i=1 r

n=x. (3.1)

In addition, for all x∈∂KR, we havex=R, and then it follows from (2.20) and (C1) that

Tx= n

i=1

sup

t∈J (Tix)(t)

= n

i=1

sup

t∈J

1

0

Hi_(t,s)φ m∗

1

0

Hi

1(s,τ)ψi(τ)fi

τ, x(τ)dτ

ds ≥ n i=1

ρiρ₁iNm∗–1

1

0

e(s)φm∗

1

0

e(τ)φm

R

δiDi1

dτ

= n

i=1

1 6

m∗ δiγi

γ₁iNm∗–1 R

δiDi1

= n

i=1

R

n=x. (3.2)

Case (2). Considering condition (C2), it follows from the definition of fi0 and fi0 > φm(_δ1

iDi1

) that there existsr1> 0 such that

fi(t, x)≥

fi0–εi1

φm

x, ∀t∈J, 0≤ x ≤r1,

whereε₁i > 0 satisfiesDi

1δi(fi0–εi1)m ∗_–1

≥1,i= 1, 2, . . . ,n. Then, for allt∈J, x∈∂Kr1, we

have

Tx= n

i=1

sup

t∈J (Tix)(t)

= n

i=1

sup

t∈J

1

0

Hi(t,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, x(τ)dτ

ds

≥ n

i=1

ρiρ₁iNm∗–1

1

0

e(s)φm∗

1

0

e(τ)fi0–ε1i

φm

xdτ

ds

= n

i=1

1 6

m∗ δiγi

γ₁iNm∗–1fi0–εi1

m∗–1 x

≥ n

i=1 x

n =x. (3.3)

Next, turning tof_i∞<φm(_D1_i),i= 1, 2, . . . ,n, and we know that there existsR1> 0 such that

fi(t, x)≤

f_i∞+εi₂φm

x, ∀t∈J,x ≥R1,

whereεi₂> 0 satisfiesDi(fi∞+εi2)m ∗_–1

≤1,i= 1, 2, . . . ,n. Let

Mi= max

0≤x≤R1,t∈J

fi(t, x), i= 1, 2, . . . ,n.

Thus, for allt∈J, x∈K, we havefi(t, x)≤Mi+ (fi∞+ε2i)φm(x),i= 1, 2, . . . ,n. Letting

R1

n >max

r1,R1,Mim∗–1Di

1 –Di

then, for allt∈J, x∈∂KR1, we have

Tx= n

i=1

sup

t∈J (Tix)(t)

= n

i=1

sup

t∈J

1

0

Hi(t,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, x(τ)dτ

ds

≤ n

i=1

γiγ₁ieqψipm∗–1

1

0

e(s)φm∗

1

0

Mi+f_i∞+εi₂φm

xdτ

ds

≤ n

i=1

Mim∗–1Di

n +

f_i∞+εi₂m∗–1xDi

n

< n

i=1

R1

n – Di

nR1

f_i∞+εi₂m∗–1+f_i∞+εi₂m∗–1xDi

n

= n

i=1

R1

n =x.

(3.4)

Applying Lemma 2.5 to (3.1) and (3.2), or (3.3) and (3.4) yields that T has at least one fixed point x∗∈Kr,R, or x∗∈Kr1,R1. Thus it follows from Remark 2.3 that system (1.1)–(1.2)

has at least one symmetric positive solution. This finishes the proof of Theorem 3.1.

The following theorem deals with the casep=∞.

Theorem 3.2 Assume that(H1)–(H3), (C1)or(H1)–(H3), (C2)hold.Then system(1.1)–

(1.2)has at least one symmetric positive solution.

Proof Lete1ψi∞replaceeqψipand repeat the argument above. Finally, we consider the case ofp= 1.

Let

Di=nγi

γ₁im∗–1ψi1

m∗–11 4

m∗–1 .

Theorem 3.3 Assume that(H1)–(H3), (C1)or(H1)–(H3), (C2)hold.Then system(1.1)–

(1.2)has at least one symmetric positive solution.

Proof Similar to the proof of Theorem 3.1. For allt∈J, x∈∂Kr, we have

Tx= n

i=1

sup

t∈J (Tix)(t)

= n

i=1

sup

t∈J

1

0

Hi(t,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, x(τ)dτ

≤ n

i=1

γiγ₁im∗–1

1

0

e(s)φm∗

1

0

e(τ)ψi(τ)φm

r Di

dτ

ds

= n

i=1

γiγ₁im∗–1 r

Di

ψi1

m∗–11 4

m∗–1

= n

i=1

r

n=x. (3.5)

Next, similar to the proof of Theorem 3.1, we can finish the proof.

In the following theorems we only consider the case of 1 <p<∞.

Theorem 3.4 Assume that(H1)–(H3)hold.Furthermore,assume that one of the following conditions is satisfied.

(C3) There exists two constantsr,R,which satisfy0 <r≤δR,such thatfi(t, x)≥φm(_δr

iDi1

)

fort∈J,0≤ x ≤r,andfi(t, x)≤φm(_DR_i)fort∈J,δR≤ x ≤R; (C4) f0

i <φm(_D1_i)andfi∞>φm(_δi1_Di 1

)(particularly,f0

i = 0andfi∞=∞),

where i= 1, 2, . . . ,n.Thus,system(1.1)–(1.2)has at least one symmetric positive solutionx∗.

Proof The proof is similar to that of Theorem 3.1, so we omit it here. Next, we discuss the multiplicity of system (1.1)–(1.2).

Theorem 3.5 Assume that(H1)–(H3)and the following conditions hold. (C5) fi0>φm(_δ1

iDi1

)andfi∞>φm(_δ1

iDi1

)(particularly,fi0=fi∞=∞);

(C6) there exists a constantb> 0such thatmaxt∈J,x=bfi(t, x) <φm(_Db_i).

Then system(1.1)–(1.2)has at least two symmetric positive solutionsx∗, x∗∗with

0 <!!x∗∗!!<b<!!x∗!!. (3.6)

Proof Choose two constantsr,Rwith 0 <r<b<R. It follows from (C5) that iffi0>φm(_δ1

iDi1), then by means of the proof of (3.3), we obtain that

Tx>x, ∀x∈∂Kr; (3.7)

iffi∞>φm( 1

δiDi₁), then by means of the proof of (3.3), we obtain that

Tx>x, ∀x∈∂KR. (3.8)

On the other hand, it follows from (C6) that

Tx= n

i=1

sup

t∈J

1

0

Hi_(t,s)φ m∗

1

0

Hi

1(s,τ)ψi(τ)fi

τ, x(τ)dτ

ds

≤ n

i=1

γiγ₁im∗–1

1

0

e(s)φm∗

1

0

eqψipfi

τ, x(τ)dτ

< n

i=1

γiγ₁im∗–1

1

0

e(s)φm∗

1

0

eqψipφm

b Di

dτ

ds

= n

i=1 1 6γ

i_γi

1eqψip

m∗–1 b

Di

=b=x, ∀x∈∂Kb. (3.9)

Applying Lemma 2.5 to (3.7), (3.8), and (3.9) yields that T has a fixed point x∗∗∈Kr,b and a fixed point x∗∈Kb,R. Then it follows from Remark 2.3 that system (1.1)–(1.2) has at least two symmetric positive solutions x∗and x∗∗. Noticing (3.9), we obtainx∗ =band x∗∗ =b. Therefore (3.6) holds, and the proof of Theorem 3.5 is complete.

Similarly, the following Theorem 3.6 can be obtained.

Theorem 3.6 Assume that(H1)–(H3)and the following conditions hold. (C7) fi0<φm(_D1_i)andfi∞<φm(_D1_i);

(C8) There exists a constantB> 0such thatmint∈J,x=Bfi(t, x) >φm(_δB

iDi1

).

Then system(1.1)–(1.2)has at least two symmetric positive solutionsx∗andx∗∗with

0 <!!x∗∗!!<B<!!x∗!!.

Theorem 3.7 Assume that (H1)–(H3) hold. If there exist 2n positive numbers bk, dk,

k= 1, 2, . . . ,n,with b1<δd1<d1<b2<δd2<d2<· · ·<bn<δdn<dn,such that (C9) fi(t, x)≤φm(b_Dk_i)fort∈J,δbk≤ x ≤bk andfi(t, x)≥φm(_δid_Dki

1

)fort∈J,δdk≤ x ≤dk,k= 1, 2, . . . ,n;or

(C10) fi(t, x)≥φm(_δbk

iDi1

)fort∈J,δbk≤ x ≤bk andfi(t, x)≤φm(d_Dk_i)fort∈J,δdk≤ x ≤dk,k= 1, 2, . . . ,n.

Then system(1.1)–(1.2)has at least n symmetric positive solutionsxk,andxksatisfy

bk≤ xk ≤dk, k= 1, 2, . . . ,n.

Finally, we discuss the existence result corresponding to the case when system (1.1)– (1.2) has no symmetric positive solutions.

Theorem 3.8 Assume that conditions(H1)–(H3)hold and fi(t, x) <φm(_Dx_i),∀t∈J,x> 0.

Then system(1.1)–(1.2)has no positive solution.

Proof Assume to the contrary that x is a positive solution of system (1.1)–(1.2), then for any 0 <t< 1, we have x∈K,xi(t) > 0, and

x= n

i=1

sup

t∈J

xi(t)

= n

i=1

sup

t∈J

1

0

Hi_(t,s)φ m∗

1

0

Hi

1(s,τ)ψi(τ)fi

τ, x(τ)dτ

ds

≤ n

i=1

γiγ₁im∗–1

1

0

e(s)φm∗

1

0

e(τ)ψi(τ)fi

τ, x(τ)dτ

< n

i=1

γiγ₁im∗–1

1

0

e(s)φm∗

1

0

e(τ)ψi(τ)φm

x Di dτ ds ≤ n i=1

γiγ₁im∗–1x

Di

1

0

e(s)φm∗

1

0

eqψipdτ

ds = n i=1 1 6γ

i_γi

1eqψip

m∗–1x

Di =x.

This is a contradiction, and this completes the proof.

Similarly, we have the following results.

Theorem 3.9 Assume that (H1)–(H3) hold and fi(t, x) > φm(_δix_Di 1

), ∀t ∈ J, x > 0,

i= 1, 2, . . . ,n.Then system(1.1)–(1.2)has no positive solution.

Proof Assume that x is a positive solution of system (1.1)–(1.2). Then, for any 0 <t< 1, we have x∈K,xi(t) > 0, and

x= n

i=1

sup

t∈J

xi(t)

= n

i=1

sup

t∈J

1

0

Hi(t,s)φm∗

1

0

H₁i(s,τ)ψi(τ)fi

τ, x(τ)dτ

ds ≥ n i=1

ρiρ₁iNm∗–1

1

0

e(s)φm∗

1

0

e(τ)fi

τ, x(τ)dτ

ds = n i=1 δiγi

γ₁iNm∗–1

1

0

e(s)φm∗

1

0

e(τ)fi

τ, x(τ)dτ

ds > n i=1 δiγi

γ₁iNm∗–1

1

0

e(s)φm∗

1

0

e(τ)φm

x

δiDi1

dτ ds ≥ n i=1 1 6 m∗ δiγi

γ₁iNm∗–1x

δiDi1 =x.

This leads to a contradiction, and this finishes the proof.

4 An example

In the following example, we selectn= 2,m= 2,p= 2, andN= 1.

Example4.1 Consider the following system:

⎧ ⎪ ⎨ ⎪ ⎩

(φ2(x(t)))=(t)f(t, x(t)), 0 <t< 1,

x(0) = x(1) =₀1g(s)x(s)ds,

φ2(x(0)) =φ2(x(1)) =₀1h(s)φ2(x(s))ds,

where

g(t) =

"

1

2 0

0 1₂

#

, h(t) =

"

1

2 0

0 1₃

#

,

(t) =

"

6 0 0 3

#

, f(t, x) =

"

(1 +sinπt)x6₁ x4

2

#

.

Then, by calculations, we obtain thatm∗= 2,μ1=μ2=ν1=1₂,ν2=1₃,γ1=γ2=γ11= 2, γ₁2=3₂,ρ1=ρ2=ρ₁1=1₆,ρ₁2=₁₂1,δ1= ₁₄₄1 ,δ2=₂₁₆1 ,δ=₂₁₆1 ,D1=√8₃₀,D2= √3₃₀,D11=29,

D2

1=16, and

H1(t,s) =H₁1(t,s) =H2(t,s) =G(t,s) +

1

0

G(s,τ)dτ,

H₁2(t,s) =G(t,s) +1 2

1

0

G(s,τ)dτ,

where

G(t,s) =

t(1 –s), 0≤t≤s≤1,

s(1 –t), 0≤s≤t≤1.

Clearly, conditions (H1)–(H3) hold. Next, we show that the condition (C1) of Theo-rem 3.1 holds. Choosingr=1₂,R= 65_×√6

3, we obtain that

φ2

r D1

=1 2×

√ 30 8 =

√ 30 16 , φ2

R

δ1D11

= 144×9 2×6

5_×√6

3 = 3×68×√63,

φ2

r D2

=1 2×

√ 30 3 =

√ 30 6 , φ2

R

δ2D21

= 216×6×65×√63 =×69×√63,

f1(t, x) = (1 +sinπt)x61≤2×r6= 2×

1 2

6 = 1

32, fort∈J, 0≤ x ≤r,

f1(t, x) = (1 +sinπt)x6₁≥(δ1R)6=

1 144×6

5_×√6

3

6

= 3×(54)6,

fort∈J,δR≤ x ≤R,

f2(t, x) =x4₂≤r4=

1 2

4 = 1

16, fort∈J, 0≤ x ≤r,

f1(t, x) =x4₂≥(δ2R)4=

1 216×6

5_×√6

3

4

= 323 ×₆8_{, fort}∈_J,δR≤ _x ≤_R.

Therefore,fi(t, x)≤φ2(_Dr_i), for allt∈J, 0≤ x ≤r, andfi(t, x)≥φ2(_δiR_Di 1

), for allt∈J, δR≤ x ≤R,i= 1, 2.

Therefore, it follows from Theorem 3.1 that system (4.1) has at least one symmetric positive solution.

5 Conclusion

-Laplace system with integral boundary conditions. Our results will be a useful contribu-tion to the existing literature on the topic of fourth-ordern-dimensionalm-Laplace sys-tems.

Acknowledgements

The authors are grateful to anonymous referees for their constructive comments and suggestions which have greatly improved this paper.

Funding

This work is sponsored by the National Natural Science Foundation of China (11401031), the Beijing Natural Science Foundation (1163007), the Scientific Research Project of Construction for Scientific and Technological Innovation Service Capacity (KM201611232017), and the teaching reform project of Beijing Information Science & Technology University (2018JGZD41).

Availability of data and materials Not applicable.

Ethics approval and consent to participate Not applicable.

Competing interests

The authors declare that there is no conflict of interest regarding the publication of this manuscript. The authors declare that they have no competing interests.

Consent for publication Not applicable.

Authors’ contributions

The authors contributed equally in this article. They have all read and approved the final manuscript.

Author details

1_{School of Applied Science, Beijing Information Science & Technology University, Beijing, People’s Republic of China.} 2_{College of Mathematics and System Science, Shandong University of Science and Technology, Qingdao, People’s}

Republic of China.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 15 January 2018 Accepted: 14 April 2018

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