Upper and lower solutions theories are established for a kind of m-point impulsive boundary value problems with p-Laplacian. By using such techniques and the Schauder fixed point theorem, the existence of solutions and positive solutions is obtained. Nagumo conditions play an important role in the nonlinear term involved with the first-order derivatives.

1. Introduction

In this paper, we study the following m-point impulsive boundary value problem with one-dimensional p-Laplacian:
(1)(ϕp(u′(t)))′+q(t)f(t,u(t),u′(t))=0,t∈J′,Δu|t=tk=I1k(u(tk)),k=1,2,…,n,Δϕp(u′)|t=tk=I2k(u(tk),u′(tk)),k=1,2,…,n,u(0)=∑i=1m-2aiu(ηi),u′(1)=0,
where ϕp(s)=|s|p-2s, p>1. J=[0,1], tk (k=1,2,…,n, where n is a fixed positive integer) are fixed points with 0<t1<t2<⋯<tn<1. J′=J∖{t1,t2,…,tn}. 0<η1<η2<⋯<ηm-2<1, ∑i=1m-2ai<1. Δu|t=tk=u(tk+)-u(tk-), k=1,2,…,n, where u(tk+) and u(tk-) represent the right-hand limit and left-hand limit of u(t) at t=tk, respectively.

The theory of impulsive differential equations describes processes which experience a sudden change of their state at certain moments. Processes with such a character arise naturally and often, especially in phenomena studied in physics, chemical technology, population dynamics, biotechnology, and economics. For an introduction of the basic theory of impulsive differential equations in Rn, see Lakshmikantham et al. [1], Baĭnov and Simeonov [2], Samoilenko and Perestyuk [3], and the references therein. The theory of impulsive differential equations has become an important area of investigation in recent years and is much richer than the corresponding theory of differential equations (see for instance [4–9] and the references therein).

At the same time, it is well known that the method of lower and upper solutions is a powerful tool for proving the existence results for a large class of boundary value problems. For a few of such works, we refer the readers to [10–14].

In [15], Cabada and Pouso considered
(2)(ϕp(u′(t)))′=f(t,u(t),u′(t)),fora.e.t∈I=[a,b],g(u(a),u′(a),u′(b))=0,u(b)=h(u(a)),
by using the upper and lower method, the authors get the existence of solution to the above BVP.

In [16], Lü et al. studied
(3)(ϕp(u′(t)))′=f(t,u(t),u′(t)),fora.e.t∈I=[0,1],u(0)=0,u(1)=0,
by giving conditions on f involving pairs of lower and upper solutions, they get the existence of at least three solutions to the above BVP.

In [12], Shen and Wang studied
(4)x′′(t)=f(t,x(t),x′(t)),fort∈J,t≠tk,δx(tk)=Ik(x(tk)),k=1,2,…,p,δx′(tk)=Jk(x(tk),x′(tk)),k=1,2,…,p,g(x(0),x′(0))=0,h(x(1),x′(1))=0,
they prove the existence of solutions to the problem under the assumption that there exist lower and upper solutions associated with the problem.

Motivated by the works mentioned above, in this paper, we considered BVP (1), the main tool is upper and lower method, and the Schäuder fixed point theorem. We not only get the existence of solutions, but also the existence of positive solutions.

The main structure of this paper is as follows. In Section 2, we give the preliminary and present some lemmas in order to prove our main results. Section 4 presents the main theorems of this paper, and at the end of Section 4, we give an example to illustrate our main results.

2. Preliminary

Define PC[0,1]={u∣u:[0,1]→R is continuous at t≠tk, u(tk)=u(tk-), u(tk+) exists, left continuous}, and PC1[0,1]={u∣u∈PC[0,1], u′:[0,1]→R is continuous at t≠tk, u′(tk+) and u′(tk-) exists, u′(tk)=u′(tk-), k=1,2,…,n}. Then, PC1[0,1] is a real Banach space with the norm ∥u∥=max{∥u∥∞,∥u′∥∞}, where ∥u∥∞=sup0≤t≤1|u(t)|.

By a solution of (1), we mean a function u∈E which satisfies (1), where E=PC[0,1]∩C2(J′).

Definition 1.

A function α∈E, with ϕp(α′)∈PC1(0,1), will be called a lower solution of (1) if
(5)(ϕp(α′(t)))′+q(t)f(t,α(t),α′(t))≥0,t∈J′,Δα|t=tk≤I1k(α(tk)),k=1,2,…,n,Δϕp(α′)|t=tk≤I2k(α(tk),α′(tk)),k=1,2,…,n,α(0)≤∑i=1m-2aiα(ηi),α′(1)<0.
A function β∈E, with ϕp(β′)∈PC1(0,1), satisfying the reversed inequalities is an upper solution of problem (1).

Next, we define the Nagumo condition we are going to use. Note that the condition does not depend on the boundary data of the problem.

Definition 2.

We say that f satisfies a Nagumo condition relative to the pair α and β, with α,β∈E, α≤β in [0,1], if there exists a function ψ:C([0,+∞),[0,+∞)) such that
(6)|f(t,x,y)|≤ψ(|y|),∀(t,x,y)∈P,
where P={(t,x,y)∈[0,1]×[0,+∞)×R; α(t)≤x(t)≤β(t)}, and also that
(7)∫ϕp(ν)∞dsψ(ϕp-1(s))>∫01q(s)ds,
here, ν=maxk=1,2,…,n{|(β(tk+1)-α(tk))/(tk+1-tk)|, |(α(tk+1)-β(tk))/(tk+1-tk)|}.

From Definition 2, we can find a real number L>ν>0 such that
(8)∫ϕp(ν)ϕp(L)dsψ(ϕp-1(s))>∫01q(t)dt.

In addition, we assume that the following three conditions hold:

q∈C(0,1) with q>0 on (0,1) and ∫01q(s)ds<∞;

f∈C([0,1]×R2,R);

I1k, I2k∈C(R,R). I2k(y,z) is nondecreasing in z∈[-L,L] for all 1≤k≤n.

Firstly, we define
(9)Pα,β(t,x)={β(t),x(t)≥β(t),x(t),α(t)≤x(t)≤β(t)β(t),x(t)≤α(t).∀x∈[0,+∞),

One can find the next result, with its proof, in [17].

Lemma 3.

For each u∈E, the next two properties hold:

dPα,β(t,u)/dt exists for a.e. t∈J′,

if u, um∈E and um→u in E, then
(10)ddtPα,β(t,um(t))⟶ddtPα,β(t,u(t))fora.e.t∈J′.

We consider the following modified problem:
(11)(ϕp(u′(t)))′(t)+q(t)f*(t,u(t),ddtPα,β(t,u))=0,t∈J′,Δu|t=tk=I1k(Pα,β(t,u)),k=1,2,…,n,Δϕp(u′)|t=tk=I2k(Pα,β(t,u),h(t,u′)),k=1,2,…,n,u(0)=∑i=1m-2aiu(ηi),u′(1)=0,
with
(12)f*(t,x,y)=f(t,Pα,β(t,x),h(t,y))+tanh(x-Pα,β(t,x)),
where h is defined by
(13)h(t,y)={L,y(t)>L,y(t),|y(t)|≤L,L,y(t)<-L.∀y∈R,

Thus f* is a continuous function on [0,1]×[0,∞)×R and satisfies
(14)|f*(t,x,y)|≤ψ(|y|)+1,∀|y|<L,|f*(t,x,y)|≤M,∀(t,x,y)∈P,
for some constant M. Moreover, we may choose M so that ∥α∥∞,∥β∥∞<M.

Lemma 4.

Suppose that ∑i=1m-2ai≠1, denote a=∑i=1m-2ai, then the following boundary value problem:
(15)(ϕp(u′(t)))′+v(t)=0,t∈J′,Δu|t=tk=I1k(u(tk)),k=1,2,…,n,Δϕp(u′)|t=tk=I2k(u(tk),u′(tk)),k=1,2,…,n,u(0)=∑i=1m-2aiu(ηi),u′(1)=0,
has a unique solution as follows:
(16)u(t)=11-a∑i=1m-2ai(∑tk<ηiI1k(u(tk))+∫0ηiϕp-1(∫s1v(τ)dτ∑s<tk-∑s<tkI2k(u(tk),u′(tk)))ds)+∫0tϕp-1(∫s1v(τ)dτ-∑s<tkI2k(u(tk),u′(tk)))ds+∑tk<tI1k(u(tk)).

Proof.

(17)(ϕp(u′(t)))′=-v(t).
For t∈(tn,1), integrating (17) from t to 1, we get
(18)u′(t)=ϕp-1(∫t1v(s)ds).
For t∈(tn-1,tn], integrating (17) from t to tn, we get
(19)u′(t)=ϕp-1[(∫ttnv(s)ds)+ϕp(u′(tn-))].
Combining (18), (19), and the boundary condition, for t∈(tn-1,tn], we have
(20)u′(t)=ϕp-1[(∫ttnv(s)ds)+ϕp(u′(tn+))(∫ttnv(s)ds)-I2n(u(tn),u′(tn))]=ϕp-1[(∫ttnv(s)ds)+∫tn1v(s)ds(∫ttnv(s)ds)-I2n(u(tn),u′(tn))]=ϕp-1[(∫t1v(s)ds)-I2n(u(tn),u′(tn))].
For t∈(tn-2,tn-1], integrating (17) from t to tn-1, and combining (20), we have
(21)u′(t)=ϕp-1[(∫ttn-1v(s)ds)+ϕp(u′(tn-1-))]=ϕp-1[(∫ttn-1v(s)ds)+ϕp(u′(tn-1+))∫t1-I2(n-1)(u(tn-1),u′(tn-1))]=ϕp-1[(∫ttn-1v(s)ds)+∫tn-11v(s)ds-I2n(u(tn),u′(tn))∫t1-I2(n-1)(u(tn-1),u′(tn-1))(∫ttn-1v(s)ds)]=ϕp-1[(∫t1v(s)ds)-I2n(u(tn),u′(tn))∫t1-I2(n-1)(u(tn-1),u′(tn-1))]=ϕp-1[(∫t1v(s)ds)-∑t<tkI2k(u(tk),u′(tk))].

By induction, for t∈(0,1), there holds
(22)u′(t)=ϕp-1[(∫t1v(s)ds)-∑t<tkI2k(u(tk),u′(tk))].
Integrating (22) from 0 to t1, we have
(23)u(t1)=u(0)+∫0t1ϕp-1(∫s1v(τ)dτ-∑s<tkI2k(u(tk),u′(tk)))ds.
For t∈(t1,t2], integrating (22) from t1 to t, we have
(24)u(t)=u(t1+)+∫t1tϕp-1(∫s1v(τ)dτ-∑s<tkI2k(u(tk),u′(tk)))ds.
By the boundary condition Δu|t=tk=I1k(u(tk)), (23), and (24), for t∈(t1,t2], we have
(25)u(t)=u(0)+I11(u(t1))+∫0tϕp-1(∫s1v(τ)dτ-∑s<tkI2k(u(tk),u′(tk)))ds.
Repeating in the above manner, for t∈[0,1], we have
(26)u(t)=u(0)+∑tk<tI1k(u(tk))+∫0tϕp-1(∫s1v(τ)dτ-∑s<tkI2k(u(tk),u′(tk)))ds.
By (26), we have
(27)u(ηi)=u(0)+∑tk<ηiI1k(u(tk))+∫0ηiϕp-1(∫s1v(τ)dτ-∑s<tkI2k(u(tk),u′(tk)))ds,i=0,1,…,m-2.
By the boundary condition u(0)=∑i=1m-2aiu(ηi), we obtain
(28)u(0)=11-a∑i=1m-2ai(∑tk<ηiI1k(u(tk))+∫0ηiϕp-1(∫s1v(τ)dτ∑s<tk-∑s<tkI2k(u(tk),u′(tk)))ds).
Substituting (26) into (28), we have (16) which holds. The proof is complete.

Lemma 5.

If u is a solution of BVP (11), α(t) and β(t) are lower and upper solutions of (1), respectively, α≤β, and
(29)I1k(α(tk))≤I1k(u)≤I1k(β(tk)),k=1,…,p,forα(tk)≤u≤β(tk),
then,
(30)α(t)≤u(t)≤β(t),t∈[0,1].

Proof.

Denote y(t)=u(t)-β(t), we will only see that u(t)≤β(t) for every t∈J. An analogous reasoning shows that u(t)≥α(t) for all t∈J.

Otherwise, if u(t)≤β(t), t∈[0,1] does not hold, then sup0≤t≤1(u(t)-β(t))>0. Since u′(1)-β′(1)<0, there are three cases.

Case 1. Denote t0=0, tn+1=1. There exists p∈{0,1,2,…,n}, t¯0∈(tp,tp+1) such that y(t¯0)=sup0≤t≤1y(t)>0. Then, sup0≤t≤1y(t)=max0≤t≤1y(t). Let t¯1=max{t∣y(t)≤0,t∈[tp,t¯0)}, t¯2=min{t∣y(t)≤0,t∈(t¯0,tp+1]}. If y(t)>0 for all t∈[tp,t¯0), let t¯1=tp, and if y(t)>0 for all t∈(t¯0,tp+1], let t¯2=tp+1. Thus,
(31)[ϕp(u′(t))]′-[ϕp(β′(t))]′≥f(t,β(t),β′(t))+tanh(u(t)-β(t))-f(t,β(t),β′(t))=tanh(u(t)-β(t))>0,∀t∈(t¯1,t¯2).
As a consequence, ϕp(u′(t))-ϕp(β′(t))>ϕp(u′(t¯0))-ϕp(β′(t¯0))=0 for all t∈(t¯0,t¯2) and since ϕp is increasing and, in particular, one to one, we have
(32)u′(t)>β′(t),∀t∈(t¯0,t¯2).
Thus, u(t¯2)-β(t¯2)>u(t¯0)-β(t¯0)>0, which is a contradiction.

Case 2. According to Case 1, if sup0≤t≤1y(t)>0, then sup0≤t≤1y(t)=y(tk-)=y(tk) or sup0≤t≤1y(t)=y(tk+), k=0,1,2,…,n. Since y′(1)<0, y(1) cannot be sup0≤t≤1y(t).

Next, we claim that for k=0,1,2,…,n-1, if supt∈[tk,tk+1]y(t)>0, then supt∈[tk,tk+1]y(t)=y(tk+). Otherwise, supt∈[tk,tk+1]y(t)=y(tk+1-)=y(tk+1). Suppose that sup0≤t≤t1y(t)=y(t1-)=y(t1)>0 and y′(t1)=y′(t1-)≥0.

From the boundary condition and (29), we have
(33)u(t1+)-u(t1)=u(t1+)-u(t1-)=I1k(Pα,β(t,u))=I1k(β(t1))=β(t1+)-β(t1-)=β(t1+)-β(t1),ϕp(u′(t1+))-ϕp(u′(t1))=I2k(Pα,β(t1,u),h(t1,u′))=I2k(β(t1),β′(t1))≥ϕp(β′(t1+))-ϕp(β′(t1)).
Hence,
(34)y(t1+)=u(t1+)-β(t1+)=u(t1)-β(t1)>0,y′(t1+)=u′(t1+)-β′(t1+)≥u′(t1)-β′(t1)≥0.
Suppose that y′(t1+)=0 and y is nonincreasing on some interval (t1,t1+γ)⊂(t1,t2), where γ>0 is sufficiently small such that y(t)>0 on t∈(t1,t1+γ). For t∈(t1,t1+γ),
(35)[ϕp(u′(t))]′-[ϕp(β′(t))]′≥f(t,β(t),ddtPα,β(t,u))+tanh(u(t)-β(t))-f(t,β(t),β′(t))=tanh(u(t)-β(t))>0.
Note that f(t,β(t),(d/dt)Pα,β(t,u))-f(t,β(t),β′(t))→0 as γ→0+. Hence, when γ>0 is sufficiently small, [ϕp(u′(t))]′-[ϕp(β′(t))]′>0, t∈(t1,t1+γ), which contradicts the assumption of monotonicity of y. Thus, we obtain
(36)0<y(t1+)<y(t2)=y(t2-),y′(t2)=y′(t2-)≥0.
We use the preceding procedure and deduce by induction that
(37)y(tk)>0,y′(tk)≥0,k=1,2,…,n+1,
which contradicts that y′(1)<0. Essentially, by the same analysis, we can get that y(tk)>0, k=2,3,…,n cannot hold.

Case 3. u(0)-β(0)=sup0≤t≤1(u(t)-β(t))=max0≤t≤1(u(t)-β(t))>0. Easily, it holds that u′(0+)-β′(0+)≤0. While by the boundary condition and ∑i=1m-2ai<1, we have
(38)u(0)-β(0)≤∑i=1m-2ai(u(ηi)-β(ηi))<maxi=1,2,…,m-2{u(ηi)-β(ηi)},
which is also a contradiction.

Consequently, u(t)≤β(t) holds for all t∈[0,1]. The proof is complete.

Lemma 6.

If u is a solution of (11), then -L<u′(t)<L for all t∈[0,1].

Proof.

Let u∈E be a solution of (11). From Lemma 5, we have α(t)≤u(t)≤β(t), and so
(39)(ϕp(u′(t)))′+q(t)f(t,u(t),u′(t))=0fort∈J′.
By the mean value theorem, there exists ξ0∈(tk,tk+1) with u′(ξ0)=(u(tk+1)-u(tk))/(tk+1-tk), k=1,2,…,n, and as a result,
(40)-L<-ν≤α(tk+1)-β(tk)tk+1-tk≤u′(ξ0)≤β(tk+1)-α(tk)tk+1-tk≤ν<L.
Let v0=|u′(ξ0)|. Suppose that there exists a point in the interval [0,1] for which u′>L or u′<-L. From the continuity of u′, we can choose ξ1∈[0,1], such that one of the following situations holds:

u′(ξ0)=ν0, u′(ξ1)=L and ν0≤u′(t)≤L for all t∈(ξ0,ξ1);

u′(ξ1)=L, u′(ξ0)=ν0 and ν0≤u′(t)≤L for all t∈(ξ1,ξ0);

u′(ξ0)=-ν0, u′(ξ1)=-L and -L≤u′(t)≤-ν0 for all t∈(ξ0,ξ1);

u′(ξ1)=-L, u′(ξ0)=-ν0 and -L≤u′(t)≤-ν0 for all t∈(ξ1,ξ0).

Without loss of generality, suppose that -L≤ν0≤u′(t)≤L for all t∈(ξ0,ξ1). Then,
(41)(ϕp(u′(t)))′n+q(t)f(t,u(t),u′(t))=0fort∈(ξ0,ξ1),
and so
(42)|(ϕp(u′(t)))′|=|q(t)f(t,u(t),u′(t))|≤q(t)|ψ(|u′(t)|)|,fort∈(ξ0,ξ1).
As a result,
(43)∫ϕp(ν0)ϕp(L)dsψ(ϕp-1(s))=∫ξ0ξ1|(ϕp(u′(t)))′|ψ(|u′(t)|)dt≤∫ξ0ξ1q(t)dt<∫01q(t)dt.
Note also that ϕp-1(s)≥0 for s∈[ϕp(ν0),ϕp(L)], so we have ν0≤ν and thus ϕp(ν0)≤ϕp(ν), which leads to the following:
(44)∫ϕp(ν0)ϕp(L)duψ(ϕp-1(u))≥∫ϕp(ν)ϕp(L)1ψ(ϕp-1(s))ds>∫01q(t)dt,
a contradiction. The proof is complete.Theorem 7 (Schäuder fixed point theorem).

Let P be a convex subset of a normed Linear space E. Each continuous, compact map T:P→P has a fixed point.

3. The Main ResultsTheorem 8.

Suppose that conditions (H1)–(H3) hold. Then, BVP (1) has at least one solution u∈E∩C2(J′) such that
(45)α(t)≤u(t)≤β(t),-L<u′(t)<L,t∈[0,1].

Proof.

Solving (11) is equivalent to finding a u∈E which satisfies
(46)u(t)=11-a∑i=1m-2ai(∑tk<ηiI1k*(u(tk))+∫0ηiϕp-1(∫s1Fu*(τ)dτ∑s<tk-∑s<tkI2k*(u(tk),u′(tk)))ds)+(∫0tϕp-1(∫s1Fu*dτ)ds-∑s<tkI2k*(u(tk),u′(tk)))+∑tk<tI1k*(u(tk)),
where Fu*(t)=q(t)f*(t,u(t),u′(t)), I1k*(u(tk))=I1k(Pα,β(t,u)), I2k*(u(tk),u′(tk))=I2k(Pα,β(t,u),h(t,u′)).

Now, define the following operator T:E→E by
(47)(Tu)(t)=11-a∑i=1m-2ai(∑tk<ηiI1k*(u(tk))+∫0ηiϕp-1(∫s1Fu*(τ)dτ∑s<tk-∑s<tkI2k*(u(tk),u′(tk)))ds)+∫0tϕp-1(∫s1Fu*(τ)dτ-∑s<tkI2k*(u(tk),u′(tk)))ds+∑tk<tI1k*(u(tk)).
It is obvious that T:E→E is completely continuous.

By the Schäuder fixed point Theorem 7, we can easily obtain that T has a fixed point u∈E, which is a solution of BVP (11). And by Lemmas 5 and 6, we know that α(t)≤u(t)≤β(t), -L<u′(t)<L, then BVP (11) becomes BVP (1), therefore u(t) is a solution of BVP (1). The proof is complete.

Theorem 9.

Suppose that conditions (H1)–(H3) hold. Assume that there exist two lower solutions α1 and α2 and two upper solutions β1 and β2 for problem (1), satisfying the following:

α1≤α2≤β2;

α1≤β1≤β2;

α2≰β1, which means that there exists t∈[0,1] such that α2(t)>β1(t);

if u is a solution of (1) with u≥α2, then u>α2 on (0,1);

if u is a solution of (1) with u≤β1, then u<β1 on (0,1).

If f satisfies the Nagumo condition with respect to α1, β2, then problem (1) has at least three solutions u1, u2, and u3 satisfying
(48)α1≤u1≤β1,α2≤u2≤β2,u3≰β1,u3≱α2.Proof.

We consider the following modified problem:
(49)(ϕp(u′(t)))′+q(t)f*(t,u(t),ddtPα1,β2(t,u))=0,t∈J′,Δu|t=tk=I¯1k(u(tk)),k=1,2,…,n,Δϕp(u′)|t=tk=I¯2k(u(tk),u′(tk)),k=1,2,…,n,u(0)=∑i=1m-2aiu(ηi),u′(1)=0.
Now, define the following operator T¯:E→E by
(50)(T¯u)(t)=11-a∑i=1m-2ai(∑tk<ηiI¯1k(u(tk))+∫0ηiϕp-1(∫s1F¯u(τ)dτ∑s<tk-∑s<tkI¯2k(u(tk),u′(tk)))ds)+∫0tϕp-1(∫s1F¯u(τ)dτ-∑s<tkI¯2k(u(tk),u′(tk)))ds+∑tk<tI¯1k(u(tk)),
where F¯u=q(t)f*(t,u(t),(d/dt)Pα1,β2(t,u)), I¯1k(u(tk))=I1k(Pα1,β2(tk,u)), and I¯2k=I2k(Pα1,β2(t,u),h(t,u′)).

It is standard that T¯:E→E is completely continuous.

Let N>(4/(1-a))((MQ)1/(p-1)+nI), where M=max{f(t,x,y)∣(t,x,y)∈[0,1]×[α1,β2]×[-L,L]}, Q=∫01q(t)dt, I=max{maxtk<1(I1k(Pα1,β2(tk,u)), I2k(Pα,β(t,u),h(t,u′))}.

Let Ω={u∈E:∥u∥<N}.

It is immediate from the argument above that T¯(Ω¯)⊂Ω. Thus,
(51)deg(I-T¯,Ω,0)=1.
Let
(52)Ωα2={u∈Ω:u>α2on(0,1)},Ωβ1={u∈Ω:u<β1on(0,1)}.
Since α2≰β1, α2>-M¯, β1<M¯ (i.e., choose M¯ such that ∥α2∥∞, ∥β1∥∞<M). It follows that Ωβ1≠∅≠Ωα2, Ωβ1∩Ωα2=∅, and Ω∖(Ωβ1∪Ωα2¯)≠∅.

By assumptions (iv) and (v), there are no solutions in ∂Ωβ1∪∂Ωα2. Thus,
(53)deg(I-T¯,Ω,0)=deg(I-T¯,Ω∖(Ωβ1∪Ωα2¯),0)+deg(I-T¯,Ωβ1,0)+deg(I-T¯,Ωα2,0).
We show that deg(I-T¯,Ωβ1,0)=deg(I-T¯,Ωα2,0)=1, then
(54)deg(I-T¯,Ω∖(Ωβ1∪Ωα2¯),0)=-1,
and there are solutions in Ωβ1∪Ωα2¯, Ωβ1, Ωα2 as required.

We now show deg(I-T¯,Ωα2,0)=1. The proof that deg(I-T¯,Ωβ1,0)=1 is similar and hence omitted. We define I-W, the extension to Ω¯ of the restriction of I-T¯ to Ω¯α2 as follows. Let
(55)w(t,x,y)=f(t,Pα2,β2(t,x),h(t,y))+tanh(x-Pα2,β2(t,x)),
where Pα2,β2(t,x) (in Pα2,β2(t,x) replace α1 by α2) and h are previously defined. Thus, w is a continuous function on [0,1]×R2 and satisfies
(56)|w(t,x,y)|≤ψ(|y|)+π2,for|y|<L,|w(t,x,y)|≤M1,for(t,x,y)∈[0,1]×R2,
for some constants M1. Moreover, we may choose M1 so that ∥α2∥∞, ∥β2∥∞<M1.

Consider the following problem:
(57)(ϕp(u′(t)))′+q(t)w(t,u(t),ddtPα2,β2(t,u))=0,t∈J′,Δu|t=tk=I~1k(u(tk)),k=1,2,…,n.Δϕp(u′)|t=tk=I~2k(u(tk),u′(tk)),k=1,2,…,n.u(0)=∑i=1m-2aiu(ηi),u′(1)=0.
Now, define the following operator:
(58)(Wu)(t)=11-a∑i=1m-2ai[∫0ηiϕp-1(∫s1q(τ)Wu*(τ)dτ)ds∑tk<ηi+∑tk<ηiI¯1k(u(tk))]+∫0tϕp-1(∫s1q(τ)Wu*(τ)dτ-∑s<tkI~2k(u(tk),u′(tk)))ds+∑tk<tI~1k(u(tk)),
where Wu*(t)=q(t)w(t,u(t),(d/dt)Pα2,β2(t,u)), I~1k(u(tk))=I1k(Pα2,β2(tk,u)), and I~2k=I2k(Pα2,β2(t,u),h(t,u′)). Again, it is easy to check (from a previous argument and (v)) that u is a solution of (57) if u∈Ωα2 and Wu=u (note that W:E→E is compact).

Thus, deg(I-T¯,Ω∖Ω¯α2,0)=0. Moreover, it is easy to see that W(Ω¯)⊂Ω. By assumptions (iv) and (v), there are no solutions in ∂Ωα2∩∂Ωβ1. Thus,
(59)deg(I-T¯,Ωα2,0)=deg(I-W,Ω∖Ω¯α2,0)+deg(I-W,Ωα2,0)=deg(I-W,Ω,0)=1.
Thus there are three solutions, as required. The proof is complete.

A slight modification of the argument in Theorem 9 yields the next result.

Theorem 10.

Suppose that conditions (H1)–(H3) hold. Assume that there exist two lower solutions α1 and α2 and two upper solutions β1 and β2 for problem (1), satisfying

α1≤α2≤β2;

α1≤β1≤β2;

α2≰β1;

there exists 0<ɛ¯<mint∈[0,1]{α2(t)-α1(t),β2(t)-β1(t)} such that all ɛ∈(0,ɛ¯], the function α2-ɛ, and β1+ɛ are, respectively, lower and upper solution of (1);

α2-ɛ¯≰β1+ɛ¯.

If f satisfies the Nagumo condition with respect to α1, β2, then problem (1) has at least three solutions u1, u2, and u3 satisfying
(60)α1≤u1≤β1,α2≤u2≤β2,u3≰β1,u3≱α2.Proof.

In the proof of Theorem 9, define
(61)Ωα2={u∈Ω:u>α2-ɛ¯on(0,1)},Ωβ1={u∈Ω:u<β1+ɛ¯on(0,1)},
where Ω is defined in Theorem 9.

Theorem 11.

Let f∈C([0,1]×(0,+∞)×(0,+∞),[0,+∞)) and q satisfies (H1). Furthermore, the following conditions hold.

BVP (1) has a pair of positive upper and lower solutions β, α∈X satisfying
(62)α(t)≤β(t),t∈[0,1].

f satisfies a Nagumo condition relative to the pair α and β.

Then, BVP (1) has at least one positive solution such that
(63)α(t)≤u(t)≤β(t),t∈[0,1].

Proof.

It can be proved easily, we omit it here.

4. ExamplesExample 1.

Consider the following BVP:
(64)(ϕp(u′(t)))′+q(t)f(t,u(t),u′(t))=0,t∈J′,Δu|t=tk=u(tk)14,k=1,2,…,n,Δϕp(u′)|t=tk=-24u′(tk)+u(tk),k=1,2,…,n,u(0)=15u(13)+16u(23),u′(1)=0,
where q(t)=1, f(t,u(t),u′(t))=4t2+sin2u(t)+20u′(t)3. Let p=3; then, q=3/2, which means that ϕp-1=ϕq.

It is clear that
(65)α(t)={-t2+4,t∈[0,12],-(t-12)2+3,t∈(12,34],-(t-34)2+2,t∈(34,1],β(t)={-(t-34)2+8916,t∈[0,12],-(t-1)2+254,t∈(12,34],-(t-2)2+8,t∈(34,1]
are lower and upper solutions of BVP (64), respectively. The figures of α(t) and β(t) are as shown in Figure 1.

Clearly, α(t)≤β(t). After substituting α(t), β(t) into (ϕp(u′(t)))′+q(t)f(t,u(t),u′(t))=0, t∈J′, denote upper=(ϕp(β′(t)))′+q(t)f(t,β(t),β′(t)) and lower=(ϕp(α′(t)))′+q(t)f(t,α(t),α′(t)), we can get Figure 2. From Figure 2, we can see that when t∈J′, (ϕp(β′(t)))′+q(t)f(t,β(t),β′(t))≤0 and (ϕp(β′(t)))′+q(t)f(t,β(t),β′(t))≥0. Other inequalities can be verified easily. So, according to Definition 1, we can get that β(t) and α(t) are upper and lower solutions of BVP (64), respectively.

Let ψ(y)=20y3+5, and it is clear that f satisfies Nagumo condition relative to α and β. Obviously, all the conditions of Theorem 11 hold. Hence, (64) has at least one positive solution α(t)≤u(t)≤β(t).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This research is supported by the National Natural Science Foundation (no. 11226133), the Fundamental Research Funds for the Central Universities (no. 2652012141), and the Young Talents Programme of Beijing.

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