positivesolutionsforasystemof 2nth-order...

Hao et al. Journal of Inequalities and Applications (2020) 2020:20 https://doi.org/10.1186/s13660-020-2296-z

R E S E A R C H Open Access

Positive solutions for a system of 2nth-orderboundary value problems involvingsemipositone nonlinearitiesXinan Hao1* , Donal O’Regan2 and Jiafa Xu1

*Correspondence:[email protected] of Mathematical Sciences,Qufu Normal University, Qufu,P.R. ChinaFull list of author information isavailable at the end of the article

AbstractIn this paper we use the fixed point index to study the existence of positive solutionsfor a system of 2nth-order boundary value problems involving semipositonenonlinearities.

MSC: 34B18; 45J05; 47H11

Keywords: 2nth-order boundary value problems; Fixed point index; Positive solution

1 IntroductionIn this paper we investigate the existence of positive solutions for the following system of2nth-order boundary value problems involving semipositone nonlinearities:

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

(–1)nx(2n) = f1(t, x, x′, . . . , (–1)n–2x(2n–4), (–1)n–2x(2n–3), (–1)n–1x(2n–2),

y, y′, . . . , (–1)n–2y(2n–4), (–1)n–2y(2n–3), (–1)n–1y(2n–2)),

(–1)ny(2n) = f2(t, x, x′, . . . , (–1)n–2x(2n–4), (–1)n–2x(2n–3), (–1)n–1x(2n–2),

y, y′, . . . , (–1)n–2y(2n–4), (–1)n–2y(2n–3), (–1)n–1y(2n–2)),

x(2i)(0) = x(2i+1)(1) = 0, y(2i)(0) = y(2i+1)(1) = 0, i = 0, 1, . . . , n – 1,

(1.1)

where n ∈ N with n ≥ 1, and fj ∈ C([0, 1] × R4n–2+ , R) (R+ := [0,∞), R := (–∞, +∞), j = 1, 2)satisfy the semipositone condition:

(H0) there is a positive constant M such that

fj(t, z1, z2, . . . , z4n–2) ≥ –M, t ∈ [0, 1], zi ∈ R+, i = 1, 2, . . . , 4n – 2, j = 1, 2.

In recent years, coupled systems of boundary value problems have been investigated bymany authors since such systems appear naturally in many real-world situations. Somerecent results on the topic can be found in a series of papers [1–26] and the referencestherein. In [1], Yang used nonnegative matrix theory to study the existence of positive

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Hao et al. Journal of Inequalities and Applications (2020) 2020:20 Page 2 of 17

solutions for the system of generalized Lidstone problems,

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

(–1)mu(2m) = f1(t, u, –u′′, . . . , (–1)m–1u(2m–2), v, –v′′, . . . , (–1)n–1v(2n–2)),

(–1)nv(2n) = f2(t, u, –u′′, . . . , (–1)m–1u(2m–2), v, –v′′, . . . , (–1)n–1v(2n–2)),

α0u(2i)(0) – β0u(2i+1)(0) = α1u(2i)(1) + β1u(2i+1)(1) = 0, i = 0, 1, . . . , m – 1,

α0v(2j)(0) – β0v(2j+1)(0) = α1v(2j)(1) + β1v(2j+1)(1) = 0, j = 0, 1, . . . , n – 1,

(1.2)

where f1, f2 ∈ C([0, 1] × Rm+n+ , R+), and in [2] Xu and Yang used some concave functions todepict the coupling behaviors for the nonlinearities fi (i = 1, 2), and they established theexistence of positive solutions for (1.2). In [3], Wang and Yang used similar methods asin [1] to study the existence of positive solutions for the system of higher-order boundaryvalue problems involving all derivatives of odd orders

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

(–1)mw(2m) = f (t, w, w′, –w′′′, . . . , (–1)m–1w(2m–1), z, z′, –z′′′, . . . , (–1)n–1z(2n–1)),

(–1)nz(2n) = g(t, w, w′, –w′′′, . . . , (–1)m–1w(2m–1), z, z′, –z′′′, . . . , (–1)n–1z(2n–1)),

w(2i)(0) = w(2i+1)(1) = 0, i = 0, 1, . . . , m – 1,

z(2j)(0) = z(2j+1)(1) = 0, j = 0, 1, . . . , n – 1,

where f , g ∈ C([0, 1] × Rm+n+2+ , R+). Moreover, they used a condition of Berstein–Nagumotype to obtain a priori estimates for w(2m–1) and z(2n–1). For related papers, we refer thereader to [27–33]. In [27] the authors used topological degree theory to study the existenceof nontrivial solutions for the higher-order nonlinear fractional boundary value probleminvolving Riemann–Liouville fractional derivatives:

⎧⎨

⎩

Dα0+u(t) = –f (t, u(t), Dβ10+u(t), D

β20+u(t), . . . , D

βn–10+ (t)), 0 < t < 1,

u(0) = u′(0) = · · · = u(n–2)(0) = Dβ0+u(1) = 0,

where Dα0+, Dβ0+, D

βi0+ are the Riemann–Liouville fractional derivatives, and f ∈ C([0, 1] ×

Rn, R).Motivated by the above work, in this paper we investigate the positive solutions for the

system of 2nth-order boundary value problems (1.1) involving semipositone nonlinear-ities. We first use the method of order reduction to transform (1.1) into an equivalentsystem of integro-integral equations, and then we establish a system of nonnegative oper-ator equations. Using the fixed point index and nonnegative matrix theory, we study theexistence of positive fixed points for the operator equations, and obtain positive solutionsfor (1.1).

2 PreliminariesLet E = C[0, 1], ‖z‖ = maxt∈[0,1] |z(t)|, P = {t ∈ [0, 1] : z(t) ≥ 0,∀t ∈ [0, 1]}. Then (E,‖ · ‖) is aBanach space, and P a cone on E. Let

k1(t, s) := min{t, s}, ki(t, s) :=∫ 1

0ki–1(t, τ )k1(τ , s) dτ , t, s ∈ [0, 1], i = 2, 3, . . . , n,


and

(Biz)(t) :=∫ 1

0ki(t, s)z(s) ds, hi(t, s) := ∂ki(t, s)/∂t, i = 1, 2, . . . , n – 1.

Note

((Biz)(t)

)′ :=∫ 1

0hi(t, s)z(s) ds, i = 1, 2, . . . ,

and Bi, B′i : E → E are completely continuous linear operators, Bi, B′i are also positive op-erators, i.e., they will map P into P.

Lemma 2.1 ([28]) Let κψ = 1 – 2/e, and ψ(t) = tet , t ∈ [0, 1]. Then we have

κψψ(s) ≤∫ 1

0k1(t, s)ψ(t) dt ≤ ψ(s).

Lemma 2.2 ([28]) Let z ∈ P. Then we have∫ 1

0

[

(Bn–1z)(t) + 2n–2∑

i=0

((Bn–1–iz)(t)

)′]

ψ(t) dt =∫ 1

0z(t)ψ(t) dt.

Lemma 2.3 ([34]) Let E be a real Banach space and P a cone on E. Suppose that Ω ⊂ Eis a bounded open set and that A : Ω ∩ P → P is a continuous compact operator. If thereexists a ω0 ∈ P \ {0} such that

ω – Aω �= λω0, ∀λ ≥ 0,ω ∈ ∂Ω ∩ P,

then i(A,Ω ∩ P, P) = 0, where i denotes the fixed point index on P.

Lemma 2.4 ([34]) Let E be a real Banach space and P a cone on E. Suppose that Ω ⊂ E isa bounded open set with 0 ∈ Ω and that A : Ω ∩ P → P is a continuous compact operator.If

ω – λAω �= 0, ∀λ ∈ [0, 1],ω ∈ ∂Ω ∩ P,

then i(A,Ω ∩ P, P) = 1.

Now, we consider the following auxiliary problem associated with (1.1):

⎧⎨

⎩

(–1)nx(2n) = f (t, x, x′, . . . , (–1)n–2x(2n–4), (–1)n–2x(2n–3), (–1)n–1x(2n–2)),

x(2i)(0) = x(2i+1)(1) = 0, i = 0, 1, . . . , n – 1,

where f ∈ C([0, 1] × R2n–1+ , R) satisfies the condition:(H0)′ there is a positive constant M such that

f (t, z1, z2, . . . , z2n–1) ≥ –M, t ∈ [0, 1], zi ∈ R+, i = 1, 2, . . . , 2n – 1.


Let (–1)n–1x(2n–2)(t) = z(t), t ∈ [0, 1]. Then we have⎧⎨

⎩

–z′′(t) = f (t, (Bn–1z)(t), ((Bn–1z)(t))′, . . . , (B1z)(t), ((B1z)(t))′, z(t)),

z(0) = z′(1) = 0,(2.1)

which can be expressed in the integral form

z(t) =∫ 1

0k1(t, s)f

(s, (Bn–1z)(s),

((Bn–1z)(s)

)′, . . . , (B1z)(s),((B1z)(s)

)′, z(s))

ds. (2.2)

For convenience, let

(Aiz)(t) =((Biz)(t)

)′, t ∈ [0, 1], i = 1, 2, . . . , n – 1.

As a result, we can also write (2.2) in the form

z(t) =∫ 1

0k1(t, s)f

(s, (Bn–1z)(s), (An–1z)(s), . . . , (B1z)(s), (A1z)(s), z(s)

)ds.

Let w(t) = M∫ 1

0 k1(t, s) ds = M(t – t2/2). We need to consider the following problem:

⎧⎪⎪⎨

⎪⎪⎩

–z′′(t) = f̃ (t, (Bn–1(z – w))(t), (An–1(z – w))(t), . . . ,

(B1(z – w))(t), (A1(z – w))(t), (z – w)(t)),

z(0) = z′(1) = 0,

(2.3)

where

f̃ (t, z1, . . . , z2n–1) =

⎧⎨

⎩

f (t, z1, . . . , z2n–1) + M, t ∈ [0, 1], zi ≥ 0, i = 1, 2, . . . , 2n – 1,f (t, 0, . . . , 0) + M, t ∈ [0, 1], for else cases.

Note that (2.3) can be expressed in the integral form

z(t) =∫ 1

0k1(t, s)̃f

(s,(Bn–1(z – w)

)(s),

(An–1(z – w)

)(s), . . . ,

(B1(z – w)

)(s),

(A1(z – w)

)(s), (z – w)(s)

)ds.

Using (H0)′, we see that f̃ ∈ C([0, 1] × R2n–1+ , R+).

Lemma 2.5(i) If z∗ is a positive solution of (2.1), then z∗ + w is a positive solution of (2.3).

(ii) If z∗∗ is a positive solution of (2.3), and greater than w, then z∗∗ – w is a positivesolution of (2.1).

Proof Substituting z∗ + w into (2.3), we have

⎧⎪⎪⎨

⎪⎪⎩

–z∗′′(t) – w′′(t) = f̃ (t, (Bn–1(z∗ + w – w))(t), (An–1(z∗ + w – w))(t), . . . ,

(B1(z∗ + w – w))(t), (A1(z∗ + w – w))(t), (z∗ + w – w)(t)),

(z∗ + w)(0) = (z∗ + w)′(1) = 0.

(2.4)


Note that w satisfies the boundary value problem

⎧⎨

⎩

–z′′(t) = M,

z(0) = z′(1) = 0.

By virtue of (2.4), we have

⎧⎨

⎩

–z∗′′(t) – w′′(t) = f (t, (Bn–1z∗)(t), (An–1z∗)(t), . . . , (B1z∗)(t), (A1z∗)(t), z∗(t)) + M,

z∗(0) = z∗′(1) = 0,

which is (2.1).On the other hand, we substitute z∗∗ – w into (2.1), and obtain

⎧⎪⎪⎨

⎪⎪⎩

–z∗∗′′(t) + w′′(t) = f (t, (Bn–1(z∗∗ – w))(t), (An–1(z∗∗ – w))(t), . . . ,

(B1(z∗∗ – w))(t), (A1(z∗∗ – w))(t), (z∗∗ – w)(t)),

(z∗∗ – w)(0) = (z∗∗ – w)′(1) = 0.

Note that, from the definitions of w and f̃ , we have

⎧⎪⎪⎨

⎪⎪⎩

–z∗∗′′(t) = f̃ (t, (Bn–1(z∗∗ – w))(t), (An–1(z∗∗ – w))(t), . . . ,

(B1(z∗∗ – w))(t), (A1(z∗∗ – w))(t), (z∗∗ – w)(t)),

z∗∗(0) = z∗∗′(1) = 0,

which is (2.3). This completes the proof. �

From Lemma 2.5, if we wish to seek the positive solutions for (2.1), we only need tostudy the positive solutions for (2.3), which are greater than w. Consequently, we definean operator T : P → E as follows:

(Tz)(t) =∫ 1

0k1(t, s)̃f

(s,(Bn–1(z – w)

)(s),

(An–1(z – w)

)(s), . . . ,

(B1(z – w)

)(s),

(A1(z – w)

)(s), (z – w)(s)

)ds.

Then T is a completely continuous operator, and if there exists a z ∈ P with z ≥ w suchthat Tz = z, we see that z – w is a positive solution of (2.1).

Let

P0 ={

z ∈ P : z(t) ≥ t‖z‖,∀t ∈ [0, 1]}.

Then P0 is also a cone on E, and we have the following lemma.

Lemma 2.6 T(P) ⊂ P0.

Note that, for t, s ∈ [0, 1], tk1(s, s) ≤ k1(t, s) ≤ k1(s, s) and k1(s, s) = s, so we can easily ob-tain this lemma (the details are omitted).


From Lemmas 2.5 and 2.6, we have z ∈ P0 if z is a fixed point of T . Consequently, if‖z‖ ≥ M we have

z(t) – w(t) ≥ t‖z‖ – M(t – t2/2)≥ t‖z‖ – tM ≥ 0.

Hence, we only need to seek T ’s positive fixed point z with ‖z‖ ≥ M, and then z – w is apositive solution of (2.1).

3 Main resultsIn (1.1), let (–1)n–1x(2n–2) = u and (–1)n–1y(2n–2) = v, then we obtain the following systemof boundary value problems:

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

–u′′(t) = f1(t, (Bn–1u)(t), (An–1u)(t), . . . , (B1u)(t), (A1u)(t), u(t),

(Bn–1v)(t), (An–1v)(t), . . . , (B1v)(t), (A1v)(t), v(t)),

–v′′(t) = f2(t, (Bn–1u)(t), (An–1u)(t), . . . , (B1u)(t), (A1u)(t), u(t),

(Bn–1v)(t), (An–1v)(t), . . . , (B1v)(t), (A1v)(t), v(t)),

u(0) = u′(1) = 0, v(0) = v′(1) = 0,

(3.1)

which has the integral form

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

u(t) =∫ 1

0 k1(t, s)f1(s, (Bn–1u)(s), (An–1u)(s), . . . , (B1u)(s), (A1u)(s), u(s),

(Bn–1v)(s), (An–1v)(s), . . . , (B1v)(s), (A1v)(s), v(s)) ds,

v(t) =∫ 1

0 k1(t, s)f2(s, (Bn–1u)(s), (An–1u)(s), . . . , (B1u)(s), (A1u)(s), u(s),

(Bn–1v)(s), (An–1v)(s), . . . , (B1v)(s), (A1v)(s), v(s)) ds.

For j = 1, 2, let

Fj(t, z1, z2, . . . , z4n–2)

=

⎧⎨

⎩

fj(t, z1, z2, . . . , z4n–2) + M, t ∈ [0, 1], zi ≥ 0, i = 1, 2, . . . , 4n – 2,fj(t, 0, 0, . . . , 0) + M, t ∈ [0, 1], for other cases.

Then we can define the operators Tj (j = 1, 2) : P4n–2 → P and T : P2 → P2 as follows:

Tj(u, v)(t) =∫ 1

0k1(t, s)Fj

(s,(Bn–1(u – w)

)(s),

(An–1(u – w)

)(s), . . . ,

(B1(u – w)

)(s),

(A1(u – w)

)(s), (u – w)(s),

(Bn–1(v – w)

)(s),

(An–1(v – w)

)(s), . . . ,

(B1(v – w)

)(s),

(A1(v – w)

)(s), (v – w)(s)

)ds

and

T(u, v)(t) = (T1, T2)(u, v)(t), t ∈ [0, 1].


Then, if we find the positive fixed point (u∗, v∗) of T with u∗, v∗ ≥ w, then (u∗ – w, v∗ – w)is a positive solution for (3.1). Let

x(t) =∫ 1

0kn–1(t, s)

(u∗(s) – w(s)

)ds, y(t) =

∫ 1

0kn–1(t, s)

(v∗(s) – w(s)

)ds, (3.2)

and we will obtain the positive solution for (1.1) (note from the discussion in Sect. 2, weneed the norms of u∗, v∗ to be greater than M).

Now, we list our assumptions for Fj (j = 1, 2):(H1) There exist aj1, bj1, cj1, dj1, lj > 0 (j = 1, 2) such that

κψ(b11 + a11(n – 1)

)< 1, κψ

(d21 + c21(n – 1)

)< 1,

�11 = det

(κψ (d11 + c11(n – 1)) κψ (b11 + a11(n – 1)) – 1

κψ (d21 + c21(n – 1)) – 1 κψ (b21 + a21(n – 1))

)

> 0,

and, for all t ∈ [0, 1], zi, z̃i ∈ R+, i = 1, 2, . . . , 2n – 1,

F1(t, z1, z2, . . . , z2n–3, z2n–2, z2n–1, z̃1, z̃2, . . . , z̃2n–3, z̃2n–2, z̃2n–1)

≥ a11(z1 + z3 + · · · + z2n–3) + a11(2z2 + 4z4 + · · · + 2(n – 1)z2n–2

)+ b11z2n–1

+ c11(̃z1 + z̃3 + · · · + z̃2n–3) + c11(2̃z2 + 4̃z4 + · · · + 2(n – 1)̃z2n–2

)+ d11̃z2n–1 – l1,


≥ a21(z1 + z3 + · · · + z2n–3) + a21(2z2 + 4z4 + · · · + 2(n – 1)z2n–2

)+ b21z2n–1

+ c21(̃z1 + z̃3 + · · · + z̃2n–3) + c21(2̃z2 + 4̃z4 + · · · + 2(n – 1)̃z2n–2

)+ d21̃z2n–1 – l2.

(H2) There exist Qj (j = 1, 2) : [0, 1] → R such that∫ 1

0k1(s, s)Qj(s) ds < M

and

Fj(t, z1, z2, . . . , z2n–3, z2n–2, z2n–1, z̃1, z̃2, . . . , z̃2n–3, z̃2n–2, z̃2n–1) ≤ Qj(t),

for all t ∈ [0, 1], zi, z̃i ∈ [0, M], i = 1, 2, . . . , 2n – 1, j = 1, 2.(H3) There exist ãj1, b̃j1, c̃j1, d̃j1,̃ lj > 0 (j = 1, 2) such that

b̃11 + ã11(n – 1) < 1, d̃21 + c̃21(n – 1) < 1,

�22 = det

(1 – [̃b11 + ã11(n – 1)] –[̃d11 + c̃11(n – 1)]–[̃b21 + ã21(n – 1)] 1 – [̃d21 + c̃21(n – 1)]

)

> 0,

and, for all t ∈ [0, 1], zi, z̃i ∈ R+, i = 1, 2, . . . , 2n – 1,


≤ ã11(z1 + z3 + · · · + z2n–3) + ã11(2z2 + 4z4 + · · · + 2(n – 1)z2n–2

)+ b̃11z2n–1


+ c̃11(̃z1 + z̃3 + · · · + z̃2n–3) + c̃11(2̃z2 + 4̃z4 + · · · + 2(n – 1)̃z2n–2

)+ d̃11̃z2n–1 + l̃1,


≤ ã21(z1 + z3 + · · · + z2n–3) + ã21(2z2 + 4z4 + · · · + 2(n – 1)z2n–2

)+ b̃21z2n–1

+ c̃21(̃z1 + z̃3 + · · · + z̃2n–3) + c̃21(2̃z2 + 4̃z4 + · · · + 2(n – 1)̃z2n–2

)+ d̃21̃z2n–1 + l̃2.

(H4) There exist Q̃j (j = 1, 2) : [0, 1] → R and t1, t2 ∈ (0, 1] such that

∫ 1

0k1(tj, s)Q̃j(s) ds > M

and

Fj(t, z1, z2, . . . , z2n–3, z2n–2, z2n–1, z̃1, z̃2, . . . , z̃2n–3, z̃2n–2, z̃2n–1) ≥ Q̃j(t),

for all t ∈ [0, 1], zi, z̃i ∈ [0, M], i = 1, 2, . . . , 2n – 1, j = 1, 2.Let Bρ = {u ∈ P : ‖u‖ < ρ} for ρ > 0 in the sequel. Then we easily have ∂Bρ = {u ∈ P :

‖u‖ = ρ}, Bρ = {u ∈ P : ‖u‖ ≤ ρ}.

Theorem 3.1 Suppose that (H0)–(H2) hold. Then (1.1) has at least one positive solu-tion.

Proof We first prove that there exists R1 > M such that

(u, v) �= T(u, v) + λ(φ1,φ2), for (u, v) ∈ ∂BR1 ∩ (P × P),λ ≥ 0, (3.3)

where φi (i = 1, 2) are given elements in the cone P0. We argue by contradiction. Supposethere exist (u, v) ∈ ∂BR1 ∩ (P × P) and λ0 ≥ 0 with

(u, v) = T(u, v) + λ0(φ1,φ2). (3.4)

This, together with Lemma 2.6, implies that u, v ∈ P0. Moreover, from (H1) we have(

u(t)v(t)

)

=

(T1(u, v)(t) + λ0φ1(t)T2(u, v)(t) + λ0φ2(t)

)

≥(

T1(u, v)(t)T2(u, v)(t)

)

≥

⎛

⎜⎜⎜⎜⎜⎜⎜⎝

∫ 10 k1(t, s)(a11

∑n–1i=1 [(Bi(u – w))(s) + 2(n – i)(Ai(u – w))(s)] + b11(u – w)(s)) ds

+∫ 1

0 k1(t, s)(c11∑n–1

i=1 [(Bi(v – w))(s) + 2(n – i)(Ai(v – w))(s)] + d11(v – w)(s)) ds– l1

∫ 10 k1(t, s) ds∫ 1

0 k1(t, s)(a21∑n–1

i=1 [(Bi(u – w))(s) + 2(n – i)(Ai(u – w))(s)] + b21(u – w)(s)) ds+∫ 1

0 k1(t, s)(c21∑n–1

i=1 [(Bi(v – w))(s) + 2(n – i)(Ai(v – w))(s)] + d21(v – w)(s)) ds– l2

∫ 10 k1(t, s) ds

⎞

⎟⎟⎟⎟⎟⎟⎟⎠

.


Multiply by ψ(t) on both sides, integrate over [0, 1], and use Lemma 2.1, and wehave

(∫ 10 u(t)ψ(t) dt∫ 10 v(t)ψ(t) dt

)

≥

⎛

⎜⎜⎜⎜⎜⎜⎜⎝

κψ∫ 1

0 ψ(t)(a11∑n–1

i=1 [(Bi(u – w))(t) + 2(n – i)(Ai(u – w))(t)] + b11(u – w)(t)) dt+ κψ

∫ 10 ψ(t)(c11

∑n–1i=1 [(Bi(v – w))(t) + 2(n – i)(Ai(v – w))(t)] + d11(v – w)(t)) dt

– l1∫ 1

0 ψ(t) dtκψ

∫ 10 ψ(t)(a21

∑n–1i=1 [(Bi(u – w))(t) + 2(n – i)(Ai(u – w))(t)] + b21(u – w)(t)) dt

+ κψ∫ 1

0 ψ(t)(c21∑n–1

i=1 [(Bi(v – w))(t) + 2(n – i)(Ai(v – w))(t)] + d21(v – w)(t)) dt– l2

∫ 10 ψ(t) dt

⎞

⎟⎟⎟⎟⎟⎟⎟⎠

=

⎛

⎜⎜⎜⎝

κψ∫ 1

0 ψ(t)(a11∑n–1

i=1 [(Bi(u – w))(t) + 2(n – i)(Bi(u – w))′(t)] + b11(u – w)(t)) dt+ κψ

∫ 10 ψ(t)(c11

∑n–1i=1 [(Bi(v – w))(t) + 2(n – i)(Bi(v – w))′(t)] + d11(v – w)(t)) dt – l1

κψ∫ 1

0 ψ(t)(a21∑n–1

i=1 [(Bi(u – w))(t) + 2(n – i)(Bi(u – w))′(t)] + b21(u – w)(t)) dt+ κψ

∫ 10 ψ(t)(c21

∑n–1i=1 [(Bi(v – w))(t) + 2(n – i)(Bi(v – w))′(t)] + d21(v – w)(t)) dt – l2

⎞

⎟⎟⎟⎠

.

Using Lemma 2.2 we obtain

(∫ 10 u(t)ψ(t) dt∫ 10 v(t)ψ(t) dt

)

≥(

κψ (b11 + a11(n – 1))∫ 1

0 (u – w)(t)ψ(t) dt + κψ (d11 + c11(n – 1))∫ 1

0 (v – w)(t)ψ(t) dt – l1κψ (b21 + a21(n – 1))

∫ 10 (u – w)(t)ψ(t) dt + κψ (d21 + c21(n – 1))

∫ 10 (v – w)(t)ψ(t) dt – l2

)

.

Let

N1 = κψ[(

b11 + a11(n – 1))

+(d11 + c11(n – 1)

)]∫ 1

0w(t)ψ(t) dt + l1

= κψ[(

b11 + a11(n – 1))

+(d11 + c11(n – 1)

)](2e – 5)M + l1,

N2 = κψ[(

b21 + a21(n – 1))

+(d21 + c21(n – 1)

)]∫ 1

0w(t)ψ(t) dt + l2

= κψ[(

b21 + a21(n – 1))

+(d21 + c21(n – 1)

)](2e – 5)M + l2.

Therefore, we have

([κψ (b11 + a11(n – 1)) – 1]

∫ 10 u(t)ψ(t) dt + κψ (d11 + c11(n – 1))

∫ 10 v(t)ψ(t) dt

κψ (b21 + a21(n – 1))∫ 1

0 u(t)ψ(t) dt + [κψ (d21 + c21(n – 1)) – 1]∫ 1

0 v(t)ψ(t) dt

)

≤(N1N2

)

and

(κψ (d11 + c11(n – 1)) κψ (b11 + a11(n – 1)) – 1

κψ (d21 + c21(n – 1)) – 1 κψ (b21 + a21(n – 1))

)(∫ 10 v(t)ψ(t) dt∫ 10 u(t)ψ(t) dt

)

≤(N1N2

)

.


Solving this matrix inequality, we obtain

(∫ 10 v(t)ψ(t) dt∫ 10 u(t)ψ(t) dt

)

≤ 1�11

(κψ (b21 + a21(n – 1)) 1 – κψ (b11 + a11(n – 1))

1 – κψ (d21 + c21(n – 1)) κψ (d11 + c11(n – 1))

)(N1N2

)

.

Consequently, there exist Ñ1, Ñ2 > 0 such that

(∫ 10 v(t)ψ(t) dt∫ 10 u(t)ψ(t) dt

)

≤(Ñ1Ñ2

)

.

Note that u, v ∈ P0, and we have(

‖v‖‖u‖

)

≤(

Ñ1e–2Ñ2e–2

)

.

Therefore, we can choose R1 > max{M, Ñ1e–2 , Ñ2e–2 } such that (3.4) is false, and thus (3.3) holds.From Lemma 2.3 we have

i(T , BR1 ∩ (P × P), P × P

)= 0. (3.5)

Next we prove that

(u, v) �= λT(u, v), for (u, v) ∈ ∂BM ∩ (P × P),∀λ ∈ [0, 1]. (3.6)

If not, there exist (u, v) ∈ ∂BM ∩ (P × P) and λ1 ∈ [0, 1] such that

(u, v) = λ1T(u, v).

This, combining with (H2), implies that

(MM

)

=

(‖u‖‖v‖

)

≤(

‖T1(u, v)‖‖T2(u, v)‖

)

≤(∫ 1

0 k1(s, s)Q1(s) ds∫ 10 k1(s, s)Q2(s) ds

)

<

(MM

)

.

This is a contradiction, and thus (3.6) is true. From Lemma 2.4 we have

(T , BM ∩ (P × P), P × P

)= 1. (3.7)

From (3.5) and (3.7) we have

i(T , (BR1 \ BM) ∩ (P × P), P × P

)

= i(T , BR1 ∩ (P × P), P × P

)– i(T , BM ∩ (P × P), P × P

)= 0 – 1 = –1.

Therefore the operator T has at least one fixed point (u∗, v∗) on (BR1 \ BM) ∩ (P × P) with‖u∗‖ ≥ M, ‖v∗‖ ≥ M, and note from (3.2) we see that (1.1) has at least one positive solution.This completes the proof. �


Theorem 3.2 Suppose that (H0), and (H3)–(H4) hold. Then (1.1) has at least one positivesolution.

Proof We first prove that

(u, v) �= T(u, v) + λ(φ̃1, φ̃2), for (u, v) ∈ ∂BM ∩ (P × P),λ ≥ 0, (3.8)

where φ̃i (i = 1, 2) ∈ P are fixed elements. If this claim is false, there exist (u, v) ∈ ∂BM ∩(P × P) and λ2 ≥ 0 such that

(u, v) = T(u, v) + λ2(φ̃1, φ̃2).

This, together with (H4), gives

(‖u‖‖v‖

)

≥(

u(t1)v(t2)

)

≥(

T1(u, v)(t1)T2(u, v)(t2)

)

≥(∫ 1

0 k1(t1, s)Q̃1(s) ds∫ 10 k1(t2, s)Q̃2(s) ds

)

>

(MM

)

.

This is a contradiction, and thus (3.8) holds. From Lemma 2.3 we have

i(T , BM ∩ (P × P), P × P

)= 0. (3.9)

Next we show that there is a large number R2 > M such that

(u, v) �= λT(u, v), for (u, v) ∈ ∂BR2 ∩ (P × P),∀λ ∈ [0, 1]. (3.10)

We argue by contradiction, so we assume there exist (u, v) ∈ ∂BR2 ∩ (P × P) and λ3 ∈ [0, 1]such that

(u, v) = λ3T(u, v).

Lemma 2.6 implies that u, v ∈ P0, and from (H3) we obtain(

u(t)v(t)

)

≤(

T1(u, v)(t)T2(u, v)(t)

)

≤

⎛

⎜⎜⎜⎜⎜⎜⎜⎝

∫ 10 k1(t, s)(̃a11

∑n–1i=1 [(Bi(u – w))(s) + 2(n – i)(Ai(u – w))(s)] + b̃11(u – w)(s)) ds

+∫ 1

0 k1(t, s)(̃c11∑n–1

i=1 [(Bi(v – w))(s) + 2(n – i)(Ai(v – w))(s)] + d̃11(v – w)(s)) ds+ l̃1

∫ 10 k1(t, s) ds∫ 1

0 k1(t, s)(̃a21∑n–1

i=1 [(Bi(u – w))(s) + 2(n – i)(Ai(u – w))(s)] + b̃21(u – w)(s)) ds+∫ 1

0 k1(t, s)(̃c21∑n–1

i=1 [(Bi(v – w))(s) + 2(n – i)(Ai(v – w))(s)] + d̃21(v – w)(s)) ds+ l̃2

∫ 10 k1(t, s) ds

⎞

⎟⎟⎟⎟⎟⎟⎟⎠

.


Multiply by ψ(t) on both sides, integrate over [0, 1], and use Lemma 2.1, we have

(∫ 10 u(t)ψ(t) dt∫ 10 v(t)ψ(t) dt

)

≤

⎛

⎜⎜⎜⎝

∫ 10 ψ(t)(̃a11

∑n–1i=1 [(Bi(u – w))(t) + 2(n – i)(Ai(u – w))(t)] + b̃11(u – w)(t)) dt

+∫ 1

0 ψ(t)(̃c11∑n–1

i=1 [(Bi(v – w))(t) + 2(n – i)(Ai(v – w))(t)] + d̃11(v – w)(t)) dt + l̃1∫ 10 ψ(t)(̃a21

∑n–1i=1 [(Bi(u – w))(t) + 2(n – i)(Ai(u – w))(t)] + b̃21(u – w)(t)) dt

+∫ 1

0 ψ(t)(̃c21∑n–1

i=1 [(Bi(v – w))(t) + 2(n – i)(Ai(v – w))(t)] + d̃21(v – w)(t)) dt + l̃2

⎞

⎟⎟⎟⎠

.

This, combining with Lemma 2.2, implies that

(∫ 10 u(t)ψ(t) dt∫ 10 v(t)ψ(t) dt

)

≤

⎛

⎜⎜⎜⎝

∫ 10 ψ(t)(̃a11

∑n–1i=1 [(Biu)(t) + 2(n – i)(Biu)′(t)] + b̃11u(t)) dt

+∫ 1

0 ψ(t)(̃c11∑n–1

i=1 [(Biv)(t) + 2(n – i)(Biv)′(t)] + d̃11v(t)) dt + l̃1∫ 10 ψ(t)(̃a21

∑n–1i=1 [(Biu)(t) + 2(n – i)(Biu)′(t)] + b̃21u(t)) dt

+∫ 1

0 ψ(t)(̃c21∑n–1

i=1 [(Biv)(t) + 2(n – i)(Biv)′(t)] + d̃21v(t)) dt + l̃2

⎞

⎟⎟⎟⎠

=

([̃b11 + ã11(n – 1)]

∫ 10 u(t)ψ(t) dt + [̃d11 + c̃11(n – 1)]

∫ 10 v(t)ψ(t) dt + l̃1

[̃b21 + ã21(n – 1)]∫ 1

0 u(t)ψ(t) dt + [̃d21 + c̃21(n – 1)]∫ 1

0 v(t)ψ(t) dt + l̃2

)

.

Consequently, we have

(1 – [̃b11 + ã11(n – 1)] –[̃d11 + c̃11(n – 1)]–[̃b21 + ã21(n – 1)] 1 – [̃d21 + c̃21(n – 1)]

)(∫ 10 u(t)ψ(t) dt∫ 10 v(t)ψ(t) dt

)

≤(

l̃1l̃2

)

.

Solving this matrix inequality, we obtain

(∫ 10 u(t)ψ(t) dt∫ 10 v(t)ψ(t) dt

)

≤ 1�22

(1 – [̃d21 + c̃21(n – 1)] d̃11 + c̃11(n – 1)

b̃21 + ã21(n – 1) 1 – [̃b11 + ã11(n – 1)]

)(l̃1l̃2

)

.

Therefore, there exist Ñ3, Ñ4 > 0 such that

(∫ 10 u(t)ψ(t) dt∫ 10 v(t)ψ(t) dt

)

≤(Ñ3Ñ4

)

.

Note that u, v ∈ P0, and then we obtain(

‖u‖‖v‖

)

≤(

Ñ3e–2Ñ4e–2

)

.

If we choose R2 > max{M, Ñ3e–2 , Ñ4e–2 } then (3.10) holds. From Lemma 2.4 we have

i(T , BR2 ∩ (P × P), P × P

)= 1. (3.11)


From (3.9) and (3.11) we have

i(T , (BR2 \ BM) ∩ (P × P), P × P

)

= i(T , BR2 ∩ (P × P), P × P

)– i(T , BM ∩ (P × P), P × P

)= 1 – 0 = 1.

Therefore the operator T has at least one fixed point (u∗, v∗) on (BR2 \ BM) ∩ (P × P) with‖u∗‖ ≥ M, ‖v∗‖ ≥ M, and note from (3.2) we see that (1.1) has at least one positive solution.This completes the proof. �

Example 3.3 Let

a11 =e

4(e – 2)(n – 1), b11 =

e2(e – 2)

,

c11 =e

(e – 2)(n – 1), d11 =

ee – 2

,

a21 =e

2(e – 2)(n – 1), b21 =

e5(e – 2)

,

c21 =e

3(e – 2)(n – 1), d21 =

e2(e – 2)

.

Then

κψ(b11 + a11(n – 1)

)=

e – 2e

(e

2(e – 2)+

e4(e – 2)(n – 1)

(n – 1))

=34

< 1,

κψ(d21 + c21(n – 1)

)=

e – 2e

(e

2(e – 2)+

e3(e – 2)(n – 1)

(n – 1))

=56

< 1,

�11 =

∣∣∣∣∣

e–2e (

ee–2 +

e(e–2)(n–1) (n – 1)) –

14

– 16e–2

e (e

5(e–2) +e

2(e–2)(n–1) (n – 1))

∣∣∣∣∣

=163120

> 0.

Consider


=95

M[

eMe – 2

(32

+54

(1 + n))]–δ1

× [a11(z1 + z3 + · · · + z2n–3) + a11(2z2 + 4z4 + · · · + 2(n – 1)z2n–2

)

+ b11z2n–1 + c11(̃z1 + z̃3 + · · · + z̃2n–3)+ c11

(2̃z2 + 4̃z4 + · · · + 2(n – 1)̃z2n–2

)+ d11̃z2n–1

]δ1 ,


=1910

M[

eMe – 2

(7

10+

56

(1 + n))]–δ2

× [a21(z1 + z3 + · · · + z2n–3) + a21(2z2 + 4z4 + · · · + 2(n – 1)z2n–2

)

+ b21z2n–1 + c21(̃z1 + z̃3 + · · · + z̃2n–3)+ c21

(2̃z2 + 4̃z4 + · · · + 2(n – 1)̃z2n–2

)+ d21̃z2n–1

]δ2 ,

for all t ∈ [0, 1], zi, z̃i ∈ R+, i = 1, 2, . . . , 2n – 1, and δ1, δ2 > 1.


For all t ∈ [0, 1], zi, z̃i ∈ [0, M], i = 1, 2, . . . , 2n – 1, j = 1, 2, we have

F1(t, z1, z2, . . . , z2n–3, z2n–2, z2n–1, z̃1, z̃2, . . . , z̃2n–3, z̃2n–2, z̃2n–1) ≤ 95 M,

F2(t, z1, z2, . . . , z2n–3, z2n–2, z2n–1, z̃1, z̃2, . . . , z̃2n–3, z̃2n–2, z̃2n–1) ≤ 1910 M.

Consequently, if let Q1(t) ≡ 95 M, Q2(t) ≡ 1910 M for t ∈ [0, 1], then (H2) holds.On the other hand, for all t ∈ [0, 1] we note that

lim infa11

∑n–1i=1 (z2i–1+2iz2i)+b11z2n–1+c11

∑n–1i=1 (̃z2i–1+2ĩz2i)+d11̃z2n–1→+∞

F1(t, z1, z2, . . . , z2n–3, z2n–2, z2n–1, z̃1, z̃2, . . . , z̃2n–3, z̃2n–2, z̃2n–1)a11

∑n–1i=1 (z2i–1 + 2iz2i) + b11z2n–1 + c11

∑n–1i=1 (̃z2i–1 + 2ĩz2i) + d11̃z2n–1

= lim infa11

∑n–1i=1 (z2i–1+2iz2i)+b11z2n–1+c11

∑n–1i=1 (̃z2i–1+2ĩz2i)+d11̃z2n–1→+∞

95 M[

eMe–2 (

32 +

54 (1 + n))]

–δ1 [a11∑n–1

i=1 (z2i–1 + 2iz2i) + b11z2n–1 + c11∑n–1

i=1 (̃z2i–1 + 2ĩz2i) + d11̃z2n–1]δ1

a11∑n–1

i=1 (z2i–1 + 2iz2i) + b11z2n–1 + c11∑n–1

i=1 (̃z2i–1 + 2ĩz2i) + d11̃z2n–1

= +∞

and

lim infa21

∑n–1i=1 (z2i–1+2iz2i)+b21z2n–1+c21

∑n–1i=1 (̃z2i–1+2ĩz2i)+d21̃z2n–1→+∞

F2(t, z1, z2, . . . , z2n–3, z2n–2, z2n–1, z̃1, z̃2, . . . , z̃2n–3, z̃2n–2, z̃2n–1)a21

∑n–1i=1 (z2i–1 + 2iz2i) + b21z2n–1 + c21

∑n–1i=1 (̃z2i–1 + 2ĩz2i) + d21̃z2n–1

= lim infa21

∑n–1i=1 (z2i–1+2iz2i)+b21z2n–1+c21

∑n–1i=1 (̃z2i–1+2ĩz2i)+d21̃z2n–1→+∞

1910 M[

eMe–2 (

710 +

56 (1 + n))]

–δ2 [a21∑n–1

i=1 (z2i–1 + 2iz2i) + b21z2n–1 + c21∑n–1

i=1 (̃z2i–1 + 2ĩz2i) + d21̃z2n–1]δ2

a21∑n–1

i=1 (z2i–1 + 2iz2i) + b21z2n–1 + c21∑n–1

i=1 (̃z2i–1 + 2ĩz2i) + d21̃z2n–1

= +∞.

Therefore, (H1) holds.

Example 3.4 Let t1 = 12 , t2 = 1, and note that∫ 1

0 k1(t, s) ds = t –12 t

2 for t ∈ [0, 1], and if weconsider the case Q̃j ≡ constant, we have

∫ 1

0k1(t1, s)Q̃1(s) ds =

38

Q̃1,∫ 1

0k1(t2, s)Q̃2(s) ds =

12

Q̃2.

To obtain the first inequality in (H4), we can take Q̃1 = 3M, Q̃2 = 52 M.Let

(ã11 b̃11 c̃11 d̃11ã21 b̃21 c̃21 d̃21

)

=

(1

20(n–1)1

101

80(n–1)1

401

40(n–1)1

201

100(n–1)1

50

)

.


Then we have

b̃11 + ã11(n – 1) =1

10+

120

< 1, d̃21 + c̃21(n – 1) =1

50+

1100

< 1,

�22 =

∣∣∣∣∣

0.85 –[ 140 +1

80 ]–[ 120 +

140 ] 0.97

∣∣∣∣∣≈ 0.82 > 0.

Let


= 3M exp{

M(0.125 + 0.0625(1 + n)

)}

× exp{–̃a11(z1 + z3 + · · · + z2n–3) – ã11(2z2 + 4z4 + · · · + 2(n – 1)z2n–2

)

– b̃11z2n–1 – c̃11(̃z1 + z̃3 + · · · + z̃2n–3)– c̃11

(2̃z2 + 4̃z4 + · · · + 2(n – 1)̃z2n–2

)– d̃11̃z2n–1

},


= 2.5M exp{

M(0.07 + 0.035(1 + n)

)}

× exp{–̃a21(z1 + z3 + · · · + z2n–3) – ã21(2z2 + 4z4 + · · · + 2(n – 1)z2n–2

)

– b̃21z2n–1 – c̃21(̃z1 + z̃3 + · · · + z̃2n–3)– c̃21

(2̃z2 + 4̃z4 + · · · + 2(n – 1)̃z2n–2

)– d̃21̃z2n–1

},

for all t ∈ [0, 1], zi, z̃i ∈ R+, i = 1, 2, . . . , 2n – 1.For all t ∈ [0, 1], zi, z̃i ∈ [0, M], i = 1, 2, . . . , 2n – 1, j = 1, 2, we have

F1(t, z1, z2, . . . , z2n–3, z2n–2, z2n–1, z̃1, z̃2, . . . , z̃2n–3, z̃2n–2, z̃2n–1) ≥ 3M,F2(t, z1, z2, . . . , z2n–3, z2n–2, z2n–1, z̃1, z̃2, . . . , z̃2n–3, z̃2n–2, z̃2n–1) ≥ 2.5M,

and thus (H4) holds. On the other hand, for all t ∈ [0, 1] we also have

lim supã11

∑n–1i=1 (z2i–1+2iz2i)+̃b11z2n–1+̃c11

∑n–1i=1 (̃z2i–1+2ĩz2i)+d̃11̃z2n–1→+∞

F1(t, z1, z2, . . . , z2n–3, z2n–2, z2n–1, z̃1, z̃2, . . . , z̃2n–3, z̃2n–2, z̃2n–1)ã11

∑n–1i=1 (z2i–1 + 2iz2i) + b̃11z2n–1 + c̃11

∑n–1i=1 (̃z2i–1 + 2ĩz2i) + d̃11̃z2n–1

= lim supã11

∑n–1i=1 (z2i–1+2iz2i)+̃b11z2n–1+̃c11

∑n–1i=1 (̃z2i–1+2ĩz2i)+d̃11̃z2n–1→+∞

(

3M exp{

M(0.125 + 0.0625(1 + n)

)}

× exp{

–

[

ã11n–1∑

i=1

(z2i–1 + 2iz2i) + b̃11z2n–1 + c̃11n–1∑

i=1

(̃z2i–1 + 2ĩz2i) + d̃11̃z2n–1

]}

/(

ã11n–1∑

i=1

(z2i–1 + 2iz2i) + b̃11z2n–1 + c̃11n–1∑

i=1

(̃z2i–1 + 2ĩz2i) + d̃11̃z2n–1

))

= 0


and

lim supã21

∑n–1i=1 (z2i–1+2iz2i)+̃b21z2n–1+̃c21

∑n–1i=1 (̃z2i–1+2ĩz2i)+d̃21̃z2n–1→+∞

F2(t, z1, z2, . . . , z2n–3, z2n–2, z2n–1, z̃1, z̃2, . . . , z̃2n–3, z̃2n–2, z̃2n–1)ã21

∑n–1i=1 (z2i–1 + 2iz2i) + b̃21z2n–1 + c̃21

∑n–1i=1 (̃z2i–1 + 2ĩz2i) + d̃21̃z2n–1

= lim supã21

∑n–1i=1 (z2i–1+2iz2i)+̃b21z2n–1+̃c21

∑n–1i=1 (̃z2i–1+2ĩz2i)+d̃21̃z2n–1→+∞

(

2.5M exp{

M(0.07 + 0.035(1 + n)

)}

× exp{

–

[

ã21n–1∑

i=1

(z2i–1 + 2iz2i) + b̃21z2n–1 + c̃21n–1∑

i=1

(̃z2i–1 + 2ĩz2i) + d̃21̃z2n–1

]}

/(

ã21n–1∑

i=1

(z2i–1 + 2iz2i) + b̃21z2n–1 + c̃21n–1∑

i=1

(̃z2i–1 + 2ĩz2i) + d̃21̃z2n–1

))

= 0.

Therefore, (H3) holds.

4 ConclusionIn this paper we use the fixed point index to study the existence of positive solutions forthe system of 2nth-order boundary value problems (1.1) involving semipositone nonlin-earities. Our nonlinearities not only depend on all derivatives of unknown functions, butthey also grow superlinearly and sublinearly at infinity.

AcknowledgementsThe authors would like to thank the referees for their pertinent comments and valuable suggestions.

FundingThis work is supported by the China Postdoctoral Science Foundation (Grant Nos. 2017M612230, 2019M652348), and theNatural Science Foundation of Chongqing Normal University (Grant No. 16XYY24).

Availability of data and materialsNot applicable.

Competing interestsThe authors declare that they have no competing interests.

Authors’ contributionsAll authors contributed equally to the manuscript. All authors read and approved the final manuscript.

Author details1School of Mathematical Sciences, Qufu Normal University, Qufu, P.R. China. 2School of Mathematics, Statistics andApplied Mathematics, National University of Ireland, Galway, Ireland.

Publisher’s NoteSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 8 November 2019 Accepted: 21 January 2020

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Positive solutions for a system of 2nth-order boundary value problems involving semipositone nonlinearitiesAbstractMSCKeywords

IntroductionPreliminariesMain resultsConclusionAcknowledgementsFundingAvailability of data and materialsCompeting interestsAuthors' contributionsAuthor detailsPublisher's NoteReferences

positivesolutionsforasystemof 2nth-order...

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