Monotone iterative technique for ( k, n − k ) conjugate boundary value problems

Monotone iterative technique for ( k, n − _k ) conjugate boundary value problems

Yujun Cui

and Yumei Zou

1Key Laboratory of Mining Disaster Prevention and Control Co-founded by Shandong Province and the Ministry of Science and Technology, Shandong University of Science and Technology, Qingdao 266590, China

2Department of Statistics and Finance, Shandong University of Science and Technology, Qingdao 266590, P. R. China

Received 10 July 2015, appeared 23 October 2015 Communicated by Jeff R. L. Webb

Abstract. In this paper, a comparison result for (k,n−k) conjugate boundary value problems is established. By using the monotone iterative technique and the method of upper and lower solutions, we investigate the existence of extremal solutions for a nonlinear differential equation with(k,n−k)conjugate boundary value problems. As an application, an example is presented to illustrate the main results.

Keywords: (k,n−k) conjugate boundary value problems, monotone iterative tech- nique, comparison result.

2010 Mathematics Subject Classification: 34B15.

1 Introduction

We consider the existence of solution of the following (k,n−k) conjugate boundary value problems for nonlinear ordinary differential equations, using the method of upper and lower solutions and its associated monotone iterative technique







(−1)^n−kx⁽ⁿ⁾(t) = f(t,x(t)), 0<t <1, n≥2, 1≤k ≤n−1, x⁽ⁱ⁾(0) =x^(j)(1) =0, 0≤i≤k−1, 0≤ j≤ n−k−1,

(1.1)

wheren ≥2 andk≥1 are fixed integers.

The subject of (k,n−k) conjugate boundary value problems for nonlinear ordinary dif- ferential equations derives from its theoretical challenge, and have close relationship with oscillation theory (see [4] for more details). Recently, many people paid attention to existence result of solution of(k,n−k)conjugate boundary value problems, such as [1,2,5–7,9,10,12–20], by means of some fixed point theorems.

BCorresponding author. Email: cyj720201@163.com

The method of upper and lower solutions coupled with the monotone iterative technique plays a very important role in investigating the existence of solutions to ordinary differential equation problems, for example [3,8,11]. However, as far as we know, there are no papers dealing with the existence of solutions for (k,n−k)conjugate boundary value problems, by means of the lower and upper solutions method.

The aims of this paper are to establish comparison result for(k,n−k)conjugate boundary value problems and to investigate the existence of extremal solutions of problem (1.1).

The rest of this paper is organized as follows: in Section 2, we present some useful pre- liminaries and lemmas. The main results are given in Section 3. In Section 4, examples are presented to illustrate the main results.

2 Preliminaries and lemmas

Let C[0, 1] denote the Banach space of real-valued continuous function with norm kxk = max_t∈[0,1]|x(t)|_.

Throughout this paper, we shall use the following notation:

G(t,s) =











1

(k−1)!(n−k−1)!

Z _t(₁−s)

0 u^k−1(u+s−t)^n−k−1du, 0≤t ≤s≤1, 1

(k−1)!(n−k−1)!

Z _s(1−t)

0 u^n−k−1(u+t−s)^k−1du, 0≤s ≤t≤1.

It is well known from the papers [10,17] that G(t,s) is the Green’s function of the following homogeneous boundary value problem:







(−1)^n−kx⁽ⁿ⁾(t) =0, 0< t<1, n≥2, 1≤ k≤n−1, x⁽ⁱ⁾(0) =x^(j)(1) =0, 0≤i≤k−1, 0≤ j≤n−k−1.

Lemma 2.1([14,19]). The function G(t,s)defined as above has the following properties:

G(t,s)≤ βs^n−k(1−s)^k, 0≤t,s≤1, β

n−1g(t)s^n−k(1−s)^k ≤ G(t,s)≤αg(t)s^n−k−1(1−s)^k−1, 0≤t,s≤1, where

β= ¹

(k−1)_!(n−k−1)_!^{, g}(t) =t^k(1−t)^n−k,

α= ¹

min{k,n−k}(k−₁)_!(n−k−₁)_!^.

In the rest of this paper, we also make the following assumptions:

(H₁) ∅ 6= I⁺∪I⁻ ⊂ {0, 1, . . . ,k−₁}_{, where} i ∈ I⁺ (ori ∈ I⁻) means that the following (k,n−k)conjugate boundary value problem











(−1)^n−kx⁽ⁿ⁾(t) =0, 0<t<1, n≥2, 1≤ k≤n−1,

x(₀) =x⁰(₀) =· · ·= x⁽ⁱ⁻¹⁾(₀) =x⁽ⁱ⁺¹⁾(₀) =· · ·= x^(k−1)(₀) =_0, x⁽ⁱ⁾(0) =1, x^(j)(1) =0, 0≤ j≤ n−k−1

has a unique nonnegative (or nonpositive) solution I_i(t)with|I_i(t)| ≥ ^tk(1−_n!^t)n−k,t∈ [0, 1]. (H₂) ∅6= J⁺∪J⁻ ⊂ {0, 1, . . . ,n−k−₁}_{, where}j∈ J⁺(orj∈ J⁻) means that the following (k,n−k)conjugate boundary value problem











(−1)^n−kx⁽ⁿ⁾(t) =0, 0<t<1, n≥2, 1≤ k≤n−1, x⁽ⁱ⁾(0) =0, 0≤i≤k−1, x^(j)(1) =1,

x(1) =x⁰(1) =· · ·= x^(j−1)(1) =x^(j+1)(1) =· · ·= x^(n−k−1)(1) =0 has a unique nonnegative (nonpositive) solution J_j(_t)_with|J_j(_t)| ≥ ^tk(1−_n!^t)n−k,t∈ [_{0, 1}]_.

Remark 2.2. It follows from (H₁) and (H₂) that for any a_i,b_j ∈ _R (0 ≤ i ≤ k−1, 0 ≤ j ≤ n−k−1) such that

a_i =0, ifi6∈ I⁺∪I⁻ and

b_j =_{0, if}j6∈ J⁺∪J⁻, the(k,n−k)conjugate boundary value problem







(−1)^n−kx⁽ⁿ⁾(t) =0, 0<t <1, n ≥2, 1≤k≤ n−1, x⁽ⁱ⁾(0) =a_i, x^(j)(1) =b_j, 0≤i≤k−1, 0≤ j≤ n−k−1

has a unique solutionψ(t) =_∑^k_i=⁻₀¹a_iI_i(t) +_∑ⁿ_j=⁻₀^k−1b_jJ_j(t), in which we may takeI_i(t) =J_j(t)≡ 0 fori6∈I⁺∪I⁻andj6∈ J⁺∪J⁻. Moreover, if

a_i ≥0, ifi∈ I⁺; a_i ≤0, ifi∈ I⁻ and

b_j ≥0, if j∈ J⁺; b_j ≤0, if j∈ J⁻ hold, ψ(t)becomes a nonnegative function.

Remark 2.3. We point out from examples below that the assumptions(H₁)and(H₂)appear naturally in the study involving(k,n−k)conjugate boundary value problem.

Example 2.4. Whenn=3,k=1, the unique solution of

x⁰⁰⁰(t) =0, x(0) =a, x(1) =b, x⁰(1) =c can be explicitly given by

ψ(t) =aI0(t) +bJ0(t) +cJ₁(t), where

I₀(t) =1−t² ≥0, J₀(t) =−t²+2t≥0, J₁(t) =−t(1−t)≤0, t ∈[0, 1]. Example 2.5([15]). Whenn=4,k=2, the unique solution of

x⁽⁴⁾(t) =_0, x(₀) =a, x(₁) =b, x⁰(₀) =c, x⁰(₁) =d

can be explicitly given by

ψ(t) =aI₀(t) +bJ₀(t) +cI₁(t) +dJ₁(t), where

I₀(t) =2t³−3t²+1≥0, J₀(t) =−2t³+3t²≥0,

I₁(t) =t³−2t²+t≥0, J₁(t) =t³−t²≤0, t ∈[0, 1]. Example 2.6. Whenn=5,k =3, the unique solution of

x⁽⁵⁾(t) =0, x(0) =a, x(1) =b, x⁰(0) =c, x⁰(1) =d, x⁰⁰(0) =e can be explicitly given by

ψ(t) =aI₀(t) +bJ₀(t) +cI₁(t) +dJ₁(t) +eI₂(t), where

I₀(t) =3t⁴−4t³+1≥0, J₀(t) =−3t⁴+4t³ ≥0, I₁(t) =t(2t+1)(1−t)² ≥0, J₁(t) =t³−t⁴ ≤0, I2(t) = ¹

2t²(1−t)²≥0, t ∈[0, 1].

Remark 2.7. Under assumptions (H₁), (H₂), we give the definition of lower and upper solution for(k,n−k)conjugate boundary value problem.

Definition 2.8. u ∈ Cⁿ[0, 1] is called a lower solution of(k,n−k)conjugate boundary value problem if











(−1)^n−ku⁽ⁿ⁾(t)≤ f(t,u(t)), 0<t <1, n≥2, 1≤k≤ n−1,

u⁽ⁱ⁾(0)≤0, ifi∈ I⁺; u⁽ⁱ⁾(0)≥0, ifi∈ I⁻; u⁽ⁱ⁾(0) =0, ifi6∈ I⁺∪I⁻; u^(j)(1)≤0, ifj∈ J⁺; u^(j)(1)≥0, ifj∈ J⁻; u^(j)(1) =0, if j6∈ J⁺∪J⁻.

Analogously,v ∈ Cⁿ[0, 1]is called an upper solutions of(k,n−k)conjugate boundary value problem if the above inequalities are reversed.

For example,uis a lower solution of(3, 2)conjugate boundary value problem if











u⁽⁵⁾(t)≤ f(t,u(t)), 0<t <1, u(0)≤0, u⁰(0)≤0, u⁰⁰(0)≤0;

u(1)≤0, u⁰(1)≥0.

Now we consider the linear(k,n−k)conjugate boundary value problem







(−1)^n−kx⁽ⁿ⁾(t) =−Mx(t) +σ(t), 0<t<1, n≥2, 1≤k ≤n−1, x⁽ⁱ⁾(0) =a_i, x^(j)(1) =b_j, 0≤i≤k−1, 0≤ j≤n−k−1

(2.1) where Mis a nonnegative constant andσ∈C[_{0, 1}]_,a_i,b_j ∈_R.

Lemma 2.9. If

αMB(n,n)<1, (2.2)

whereαis given in Lemma2.1 and B(t,s)denotes the Beta function, then(2.1) has a unique solution x, which can be expressed by

x(t) =ψ(t) +

Z ₁

0 Q(t,s)ψ(s)ds+

Z ₁

0 H(t,s)σ(s)ds, (2.3) whereψ(t)is given in Remark2.2,

G₁(t,s) =−MG(t,s), Q(t,s) =

+_∞ m

∑

Gm(t,s), (2.4)

Gm(t,s) = (−M)^m

Z ₁

0

· · ·

Z ₁

0 G(t,r₁)G(r₁,r₂)· · ·G(r_m−1,s)dr₁· · ·dr_m−1, and

H(t,s) =G(t,s) +

Z ₁

0 Q(t,τ)G(τ,s)dτ.

All functions G_n(t,s), H(t,s), Q(t,s)are continuous on[_{0, 1}]×[_{0, 1}]and the series on the right-hand side of (2.4)converges uniformly on[0, 1]×[0, 1].

Proof. It follows from the paper [10] that x ∈ Cⁿ[0, 1] is a solution of (2.1) if and only if x∈C[_{0, 1}]is a solution of the following operator equation

x+Tx= ϕ (2.5)

with operatorT: C[0, 1]→C[0, 1]given by (Tx)(t) =M

Z ₁

0 G(t,s)x(s)ds, and

ϕ(t) =ψ(t) +

Z ₁

0 G(t,s)σ(s)ds. (2.6) We shall prover(T) < 1, wherer(T)denotes the spectral radius of operator T. Actually, forx ∈C[0, 1], by Lemma2.1, we have

|Tx(t)| ≤ M Z ₁

0 G(t,s)|x(s)|ds

≤αMt^k(1−t)^n−k

Z ₁

0 s^n−k−1(1−s)^k−1dskxk

=αMB(k,n−k)kxkt^k(1−t)^n−k. Hence, we have

|T²x(t)| ≤M Z ₁

0 G(t,s)|Tx(s)|ds

≤α²M²B(k,n−k)kxkt^k(1−t)^n−k

Z ₁

0 sⁿ⁻¹(1−s)ⁿ⁻¹ds

=α²M²B(k,n−k)B(n,n)kxkt^k(1−t)^n−k.

By the induction method, we have

|T^mx(t)| ≤α^mM^mB(k,n−k)B^m−1(n,n)kxkt^k(1−t)^n−k,

which implies thatkT^mk ≤α^mM^mB(k,n−k)B^m−1(n,n). It follows fromr(T) = lim

m→_∞kT^mk^1/m that

r(T)≤ αMB(n,n)<1.

This yields that the unique solution of operator equation (2.5) is given by x = (I+T)⁻¹ϕ= (I−T+T²+· · ·+ (−1)^mT^m+· · ·)ϕ.

Substituting (2.6) into the above equality, we get (2.3) and the proof is complete.

Lemma 2.10. Suppose that x∈Cⁿ[0, 1]satisfies











(−1)^n−kx⁽ⁿ⁾(t)≥ −Mx(t), 0<t <1, n≥2, 1≤k ≤n−1,

x⁽ⁱ⁾(₀)≥_0, if i ∈ I⁺; x⁽ⁱ⁾(₀)≤_0, if i∈ I⁻; x⁽ⁱ⁾(₀) =_0, if i6∈I⁺∪I⁻, x^(j)(1)≥0, if j∈ J⁺; x^(j)(1)≤0, if j∈ J⁻; x^(j)(1) =0, if j6∈ J⁺∪J⁻, where the nonnegative constant M satisfies(2.2),

B(k,n−k)

Mαβ+ ^M

3α²β²B(_n,n)_B(_k+_1,n−k+₁) 1−M²α²β²B²(n,n)

< ^β

n−1, (2.7)

MNα+ ^{N M}

3α²βB(n,n)B(k,n−k) 1−M²α²β²B²(n,n) < ¹

n!, (2.8)

in which N=max

_{Z 1}

0 s^n−k−1(1−s)^k−1y(s)ds: y∈ {|I_i|, i∈ I⁺∪I⁻} ∪ {|J_j|, j∈ J⁺∪J⁻}

. Then x(t)≥0for t ∈[0, 1].

Proof. Letσ(t) = (−₁)^n−kx⁽ⁿ⁾(t) +Mx(t)_and

a_i = x⁽ⁱ⁾(0), 0≤i≤k−1; b_j = x^(j)(1), 0≤j≤n−k−1.

Thenσ(_t)≥0 and







a_i ≥_{0, if}i∈ I⁺; a_i ≤_{0, if}i∈ I⁻; a_i =_{0, if}i6∈ I⁺∪I⁻; b_j ≥0, ifj∈ J⁺; b_j ≤0, ifj∈ J⁻; b_j =0, ifj6∈ J⁺∪J⁻.

By Lemma2.9, (2.3) holds in which ψ(t) ≥ 0 fort ∈ [0, 1]. It follows from the expression of G_m(t,s)_that G_m(t,s) ≤ _{0 when} mis odd and G_m(t,s) ≥ _{0 when} m is even. Thus, we obtain

form=3, 5, . . ., by using Lemma2.1, G_m(t,s) = −M^m

Z ₁

0

· · ·

Z ₁

0 G(t,r₁)G(r₁,r₂)· · ·G(r_m−2,r_m−1)G(r_m−1,s)dr₁· · ·dr_m−1

≥ −M^m Z ₁

0

· · ·

Z ₁

0

αg(t)rⁿ₁^−k−1(1−r₁)^k−1·αr^k₁(1−r₁)^n−krⁿ₂^−k−1(1−r₂)^k−1· · ·

×αr^k_m−2(1−r_m−2)^n−krⁿ_m⁻₋^k₁⁻¹(1−r_m−1)^k−1·βs^n−k(1−s)^kdr₁· · ·dr_m−1

= −M^mα^m−1βg(t)s^n−k(1−s)^k

Z ₁

0 rⁿ₁⁻¹(1−r₁)ⁿ⁻¹dr₁

×

Z ₁

0 rⁿ₂⁻¹(1−r₂)ⁿ⁻¹dr₂· · ·

Z ₁

0 rⁿ_m⁻₋¹₂(1−r_m−2)ⁿ⁻¹dr_m−2

×

Z ₁

0 rⁿ_m⁻₋^k₁⁻¹(1−r_m−1)^k−1dr_m−1

= −M^mα^m−1βg(t)s^n−k(1−s)^kB^m−2(n,n)B(k,n−k). Consequently, we have

H(t,s) =G(t,s) +

Z ₁

0 Q(t,τ)G(τ,s)dτ=G(t,s) +

+_∞ m

∑

Z ₁

0 G_m(t,τ)G(τ,s)dτ

≥ G(t,s)−M Z ₁

0 G(t,τ)G(τ,s)dτ+

+_∞ m

∑

Z ₁

0 G_2m+1(t,τ)G(τ,s)dτ

≥ ^β

n−1g(t)s^n−k(1−s)^k−Mαβg(t)s^n−k(1−s)^k

Z ₁

0 τ^n−k−1(1−τ)^k−1dτ

−

+_∞ m

∑

M^2m+1α^2mβ²g(t)s^n−k(1−s)^kB^2m−1(n,n)B(k,n−k)

Z ₁

0 τ^n−k(1−τ)^kdτ

= g(t)s^n−k(1−s)^k β

n−₁−MαβB(k,n−k)

−

+_∞ m

∑

M^2m+1α^2mβ²B^2m−1(n,n)B(k,n−k)B(k+1,n−k+1)

. and for y∈ {I_i, i∈ I⁺} ∪ {−I_i, i∈ I⁻} ∪ {J_j, j∈ J⁺} ∪ {−J_j,j∈ J⁻}_,

y(t) +

Z ₁

0 Q(t,s)y(s)ds

≥y(t)−M Z ₁

0

G(t,s)y(s)ds+

+_∞ m

∑

Z ₁

0

G_2m+1(t,s)y(s)ds

≥ ^g(t)

n! −Mαg(t)

Z ₁

0 s^n−k−1(1−s)^k−1y(s)ds+

+_∞ m

∑

Z ₁

0 G_2m+1(t,s)y(s)ds

≥ ^g(t)

n! −Mαg(t)

Z ₁

0 s^n−k−1(1−s)^k−1y(s)ds

−

+_∞ m

∑

M^2m+1α^2mβB^2m−1(n,n)B(k,n−k)g(t)

Z ₁

0 s^n−k(1−s)^ky(s)ds

≥ ^g(t)

n! −MNαg(t)−N

+_∞ m

∑

M^2m+1α^2mβB^2m−1(n,n)B(k,n−k)g(t)

= g(t)

"

1

n!−MNα−N

+_∞ m

∑

M^2m+1α^2mβB^2m−1(n,n)B(k,n−k)

# .

Thus, by (2.8), we have thatx(t)≥0 fort∈ [0, 1], and the lemma is proved.

3 Main results

In this section, we prove the existence of extremal solutions of differential equation (1.1).

Theorem 3.1. Let f ∈ C([0, 1]×_R,R); v₀,w₀ be lower and upper solutions of (1.1) such that v₀(t)≤w₀(t)on[_{0, 1}]. Suppose further that there exists M>₀such that

f(t,x)− f(t,y)≥ −M(x−y), (3.1) whenever v₀(t) ≤ y ≤ x ≤ w₀(t)and M satisfies (2.2), (2.7)and(2.8). Then there exist monotone sequences{v_m(t)}_,{w_m(t)}which converge uniformly on [_{0, 1}]to the extremal solutions of problem (1.1)in the order interval[v₀,w₀] ={u∈C[0, 1]:v₀(t)≤ u(t)≤w₀(t), t∈ [0, 1]}.

Proof. For anyη∈[v₀,w₀], we consider the linear differential equation







(−1)^n−kx⁽ⁿ⁾(t) =−Mx(t) + f(t,η(t)) +Mη(t), 0<t <1, n≥2, 1≤ k≤n−1, x⁽ⁱ⁾(0) = x^(j)(1) =0, 0≤i≤k−1, 0≤ j≤n−k−1.

(3.2)

By Lemma2.9, (3.2) has a unique solution x(t) = R1

0 H(t,s)[f(s,η(s)) +Mη(s)]ds in C[0, 1]. Define the mappingAbyAη= xwith operator A:[v₀,w₀]→C[0, 1]given by

(Aη)(t) =

Z ₁

0 H(t,s)[f(s,η(s)) +Mη(s)]ds and use it to construct the sequences{vm(t)},{wm(t)}. Let us prove that

(i) v₀ ≤ Av₀, Aw₀≤w₀;

(ii) Ais a monotone operator on[v₀,w₀].

To prove (i), set Av₀ = v₁, wherev₁ is the unique solution of (3.2) with η = v₀. Setting p=v₁−v0, we see that











(−1)^n−kp⁽ⁿ⁾(t)≥ −Mp(t), 0< t<1, n≥2, 1≤k ≤n−1,

p⁽ⁱ⁾(0)≥0, ifi∈ I⁺; p⁽ⁱ⁾(0)≤0, ifi∈ I⁻; p⁽ⁱ⁾(0) =0, ifi6∈ I⁺∪I⁻, p^(j)(1)≥0, ifj∈ J⁺; p^(j)(1)≤0, ifj∈ J⁻; p^(j)(1) =0, if j6∈ J⁺∪J⁻.

This shows, by Lemma2.10, that p(t) ≥ 0 on [0,1] and hence v0 ≤ Av0. Similarly, we can show thatAw₀≤w₀.

To prove (ii), letη₁,η₂ ∈ [v₀,w₀]such thatη₁ ≤ η₂. Suppose thatx₁ = Aη₁, and x₂ = Aη₂. Set p=x2−x₁ so that







(−1)^n−kp⁽ⁿ⁾(t)≥ −Mp(t), 0<t <1, n≥2, 1≤k≤ n−1, p⁽ⁱ⁾(0) = p^(j)(1) =0, 0≤i≤k−1, 0≤ j≤n−k−1,

(3.3) here we have used the condition (3.1). By Lemma2.10, (3.3) implies that Aη₁ ≤ Aη₂ proving (ii).

Now let v_m = Av_m−1, w_m = Aw_m−1, m = 1, 2, . . . . From the foregoing arguments, we conclude that

v₀≤ v₁ ≤ · · · ≤vm ≤ · · · ≤ · · ·wm ≤ · · · ≤w₁ ≤w₀. (3.4) Obviously the sequences {vm},{wm}are uniformly bounded on[0, 1], and by (3.1), we have

f(t,v₀(t)) +Mv₀(t)≤ f(t,v_m(t)) +Mv_m(t)

≤ f(t,w_m(t)) +Mw_m(t)≤ f(t,w₀(t)) +Mw₀(t), m∈ _N, t ∈[0, 1]. This together with the continuity of H(t,s)on [0, 1]×[0, 1]imply that {v_m}^∞_m=2 = {Av_m}^∞_m=1 and{w_m}^∞_m=2 = {Aw_m}^∞_m=1 are two sequentially compact sets. As a result, there exist subse- quences{vm_j},{wm_j}that converge uniformly on[0, 1]. In view of (3.4), it also follows that the entire sequences {v_m}, {w_m} converge uniformly and monotonically to their limit functions v^∗(t),w^∗(t)respectively, that is,

mlim→_∞v_m(t) =v^∗(t), lim

m→_∞w_m(t) =w^∗(t), uniformly on[0, 1].

It is now easy to show that v^∗,w^∗ are solutions of conjugate boundary value problem (1.1), using the corresponding integral equation

x(t) = (Aη)(t) =

Z ₁

0 H(t,s)[f(s,η(s)) +Mη(s)]ds for (3.2).

Next, we prove that v^∗,w^∗ are extremal solutions of (1.1) in [_v0,w0]. In fact, we assume that xis any solution of (1.1). That is,







(−1)^n−kx⁽ⁿ⁾(t) = f(t,x(t)), 0<t <1, n≥2, 1≤k≤n−1, x⁽ⁱ⁾(0) = x^(j)(1) =0, 0≤i≤ k−1, 0≤ j≤n−k−1.

By (3.1) and Lemma2.10, it is easy by induction to show that

vm ≤ x≤wm, m=1, 2, 3 . . . . (3.5) Now, letting m→_∞in (3.5), we have v^∗ ≤x ≤ w^∗. That is,v^∗ andw^∗ are extremal solutions of (1.1) in [v₀,w₀].

4 Examples

Consider the following(2, 2)conjugate boundary value problems:







x⁽⁴⁾(t) = ¹

5(t²−x(t))³−¹

5t⁹, 0<t <1, x(0) =x⁰(0) = x(1) =x⁰(1) =0.

(4.1)

Let f(t,x) = ¹₅(t²−x)³− ¹₅t⁹. Obviously, f ∈ C([0, 1]×_R,R). Take w₀(t) = t²−3t³/4, v0(t) =0, thenv0(t)≤w0(t)fort∈ [0, 1]and we have











w⁽₀⁴⁾(_t) =₀≥ − ³⁷

320t⁹ = ¹

5(_t²−w0(_t))]³− ¹

5t⁹, 0< _t<_1, w₀(₀) =w⁰₀(₀) =_0, w₀(₁) = ¹

4 ≥_0, w₀⁰(₁) =−¹ 4 ≤_0,







v₀⁽⁴⁾(t) =0≤ ^t6−t⁹

5 = ¹

5(t²−v₀(t))³− ¹

5t⁹, 0< t<1, v₀(0) =v⁰₀(0) =v₀(1) =v⁰₀(1) =0.

Consequently, by Definition2.8and Example2.5,v0,w0 are lower and upper solutions of (4.1) respectively. Ifv₀(t)≤v≤u ≤w₀(t), we have

f(t,u)− f(t,v) = ¹

5(t²−u)³−¹

5v(t²−v)³≥ −³

5(u−v). It is clear thatM = ³_5, α= ¹_2,β=_1,n=_4, k=_2,

N=max _{Z 1}

0 s(1−s)y(s)ds: y∈ {2t³−3t²+1,−2t³+3t²,t³−2t²+t,t²−t³}

= ¹ 12, and so, it is easy to show that inequalities (2.2), (2.7) and (2.8) are satisfied.

By Theorem3.1, problem (4.1) has extremal solutions in[v₀,w₀].

Acknowledgements

The authors sincerely thank the reviewer for careful reading and useful comments that have led to the present improved version of the original paper. The Project Supported by NNSF of China (11371221, 11571207), the Specialized Research Foundation for the Doctoral Program of Higher Education of China (20123705110001), the Program for Scientific Research Innovation Team in Colleges, a project of Shandong province Higher Educational Science and Technology Program (J15LI16) and SDNSF(ZR2015AL002),

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