We consider the existence of positive solutions for the Neumann boundary value problem x′′(t)+m2(t)x(t)=f(t,x(t))+e(t),t∈(0,1),x′(0)=0,x′(1)=0, where m∈C([0,1],(0,+∞)),e∈C[0,1], and f:[0,1]×(0,+∞)→[0,+∞) is continuous. The theorem obtained is very general and complements previous known results.

1. Introduction

The existence of solutions of Neumann boundary value problem of second-order ordinary differential equations has been studied by many authors; see Sun et al. [1], Cabada and Pouso [2], Cabada et al. [3], Canada et al. [4], Chu et al. [5], Jiang, and Liu [6], Yazidi [7], Sun and Li [8] and the references therein.

Recently, Chu et al. [5] have studied the existence of positive solution to the Neumann boundary value problem
x′′(t)+m2x(t)=f(t,x(t))+e(t),t∈(0,1),x′(0)=0,x′(1)=0,
where m∈(0,π/2) is a constant, e∈C[0,1] and nonlinearity f(t,x) may be singular at x=0. Their approach was based upon the nonlinear alternative principle of Leray-Schauder and Green's function, G1(t,s), of the associated linear problem
x′′(t)+m2x(t)=0,t∈(0,1),x′(0)=0,x′(1)=0.
Notice that Green's function G1(t,s) can be explicitly expressed by
G1(t,s)={cosm(1-s)cosmtmsinm,0≤t≤s≤1,cosm(1-t)cosmsmsinm,0≤s≤t≤1.

In this paper, we will consider the more general problem
x′′(t)+m2(t)x(t)=f(t,x(t))+e(t),t∈(0,1),x′(0)=0,x′(1)=0,
where m∈C([0,1],(0,+∞)),e∈C[0,1], and f:[0,1]×(0,+∞)→[0,+∞) is continuous.

Of course, the natural question is what would happen when the constant m in (1.1) is replaced with a function m(t)? Obviously, Green's function of the associated linear problem
x′′(t)+m2(t)x(t)=0,t∈(0,1),x′(0)=0,x′(1)=0
cannot be explicitly expressed by elementary functions! The primary contribution of this paper is to construct Green's function associated with the Neumann boundary value problem with a variable coefficient (1.5) and study the properties of the Green's function. We apply the Krasnoselskii and Guo fixed point theorem as an application. This application was first made by Erbe and Wang [9] to ordinary differential equations. Since that time, there has been a tremendous amount of work to study the existence of positive solutions to BVPs for ordinary differential equations. Once we obtain Theorem 2.2, many of those applications would work here as well.

The rest of the paper is organized as follows: Section 2 is devoted to constructing Green's function and proves some preliminary results. In Section 3, we state and prove our main results. In Section 4, an example illustrates the applicability of the main existence result.

2. Preliminaries and Lemmas

Let us fix some notation to be used. Given φ∈L1[0,1], we write φ≻0 if φ≥0 for a.e. t∈[0,1], and it is positive in a set of positive measure. Let us denote by p* and p* the essential supremum and infimum of a given function p∈L1[0,1] if they exist. To study the boundary value problem (1.4), we need restriction on m(t)

To rewrite (1.4) to an equivalent integral equation, we need to construct Green's function of the corresponding linear problem. To do this, we need the following.

Lemma 2.1.

Let (H0) hold. Suppose φ and ψ be the solution of the linear problems
φ′′(t)+m2(t)φ(t)=0,t∈[0,1],φ′(0)=0,φ(0)=1,ψ′′(t)+m2(t)ψ(t)=0,t∈[0,1],ψ′(1)=0,ψ(1)=1,
respectively. Then

φ(t)>0 on [0,1], and φ′(t)<0 on (0,1];

(ii) ψ(t)>0 on [0,1], and ψ′(t)>0 on [0,1).

Proof.

We will give a proof for (i) only. The proof of (ii) follows in a similar manner.

It is easy to see that the problem
x′′(t)+m¯2x(t)=0,t∈[0,1],x′(0)=0,x(0)=1
has the unique solution x(t)=cosm¯t and t∈[0,1]. From (H0), we know that
cosm¯t>0,t∈[0,1].
On the other hand, for all t∈[0,1], we have
(cosm¯t)′′+m2(t)cosm¯t=-m¯2cosm¯t+m2(t)cosm¯t=(m2(t)-m¯2)cosm¯t≤0.
By using comparison theorem (see [10]), we obtain
cosm¯t≤φ(t),t∈[0,1].
Therefore, we have from (2.3) and (2.5) that
0<cosm¯t≤φ(t),t∈[0,1].
Thus
φ′′(t)=-m2(t)φ(t)<0,t∈[0,1].
From the fact φ′(0)=0 and (2.7), we obtain φ′(t)<0 on (0,1].

Now, let
G(t,s)=1ψ′(0){ψ(t)φ(s),0≤s≤t≤1,ψ(s)φ(t),0≤t≤s≤1.

Theorem 2.2.

Let (H0) hold. Then for any y∈C[0,1], the problem
x′′(t)+m2(t)x(t)=y(t),t∈(0,1),x′(0)=0,x′(1)=0
is equivalent to the integral equation
x(t)=∫01G(t,s)y(s)ds.

Proof.

First we show that the unique solution of (2.9) can be represented by (2.10).

In fact, we know that the equationx′′(t)+m2(t)x(t)=0,t∈[0,1]
has known two linear independent solutions φ and ψ since |φ(0)ψ(0)φ′(0)ψ′(0)|=ψ′(0)≠0.

Now by the method of variation of constants, we can obtain that the unique solution of the problem (2.9) can be represented byx(t)=∫01G(t,s)y(s)ds,
where G(t,s) is as (2.8).

Next we check that the function defined by (2.10) is a solution of (2.9).

From (2.10), we know thatx(t)=∫0tψ(t)φ(s)ψ′(0)y(s)ds+∫t1ψ(s)φ(t)ψ′(0)y(s)ds,x′(t)=ψ′(t)∫0tφ(s)ψ′(0)y(s)ds+φ′(t)∫t1ψ(s)ψ′(0)y(s)ds,x′′(t)=ψ′′(t)∫0tφ(s)ψ′(0)y(s)ds+φ′′(t)∫t1ψ(s)ψ′(0)y(s)ds+ψ′(t)φ(t)ψ′(0)y(t)-φ′(t)ψ(t)ψ′(0)y(t).
So that
x′′(t)+m2(t)x(t)=1ψ′(0)|φ(t)ψ(t)φ′(t)ψ′(t)|y(t)=1ψ′(0)|φ(0)ψ(0)φ′(0)ψ′(0)|y(t)=y(t).
Finally, it is easy to see that
x′(0)=φ′(0)∫01ψ(s)ψ′(0)y(s)ds=0,x′(1)=ψ′(1)∫01φ(s)ψ′(0)y(s)ds=0.

From Lemma 2.1, we know that
G(t,s)>0,∀t,s∈[0,1].
Let A=min0≤t,s≤1G(t,s),B=max0≤t,s≤1G(t,s),σ=A/B. Then B>A>0 and 0<σ<1.

In order to prove the main result of this paper, we need the following fixed-point theorem of cone expansion-compression type due to Krasnoselskii's (see [11]).

Theorem 2.3.

Let E be a Banach space, and K⊂E is a cone in E. Assume that Ω1 and Ω2 are open subsets of E with θ∈Ω1 and Ω¯1⊂Ω2. Let T:K∩(Ω¯2∖Ω1)→K be a completely continuous operator. In addition, suppose that either

∥Tu∥≤∥u∥,∀u∈K∩∂Ω1 and ∥Tu∥≥∥u∥,∀u∈K∩∂Ω2 or

∥Tu∥≥∥u∥,∀u∈K∩∂Ω1 and ∥Tu∥≤∥u∥,∀u∈K∩∂Ω2 holds.

Then T has a fixed point in K∩(Ω¯2∖Ω1).3. Main Results

In this section, we state and prove the main results of this paper.

Let us define the function
γ(t)=∫01G(t,s)e(s)ds,
which is just the unique solution of the linear problem (2.9) with y(t)=e(t). For our constructions, let E=C[0,1], with norm ∥x∥=sup0≤t≤1|x(t)|. Define a cone P, by
P={x∈E∣x(t)≥0on[0,1],andmin0≤t≤1x(t)≥σ‖x‖}.

Theorem 3.1.

Let (H0) hold. Suppose that there exist a constant r>0 such that

there exist continuous, nonnegative functions g,h, and k, such that
0≤f(t,x)≤k(t)[g(x)+h(x)]∀(t,x)∈[0,1]×(0,r],

g(x)>0 is nonincreasing, and h(x)/g(x)is nondecreasing in x∈(0,r];

r-γ*/(g(σr)(1+h(r)/g(r)))>K*, here K(t)=∫01G(t,s)k(s)ds;

there exists a continuous function ϕr≻0 such that
f(t,x)≥ϕr(t)∀(t,x)∈[0,1]×(0,r];

ϕr(t)+e(t)≻0forallt∈[0,1].

Then problem (1.4) has at least one positive solution x with 0<∥x∥<r.Remark 3.2.

When m(t)≡m,t∈[0,1], then (1.4) reduces to (1.1), (H0) reduce to m∈(0,π/2). So Theorem 3.1 is more extensive than [5, Theorem 3.1].

Proof of Theorem <xref ref-type="statement" rid="thm3.1">3.1</xref>.

Let δ=min0≤t≤1∫01G(t,s)ϕr(s)ds+γ*. Choose n0∈{1,2,…} such that 1/n0<σr1, where 0<r1<min{δ,r} is a constant. Let N0={n0+1,n0+2,…}. Fix n∈N0. Consider the boundary value problem
x′′(t)+m2(t)x(t)=fn(t,x(t))+e(t),t∈(0,1),x′(0)=0,x′(1)=0,
where
fn(t,x)={f(t,x),ifx≥1n,f(t,1n),if0≤x≤1n.
We note that x is a solution of (3.1n) if and only if
x(t)=∫01G(t,s)[fn(s,x(s))+e(s)]ds,0≤t≤1.
Define an integral operator Tn:P→E by
(Tnx)(t)=∫01G(t,s)[fn(s,x(s))+e(s)]ds,0≤t≤1,x∈P.
Then, (3.1n) is equivalent to the fixed point equation x=Tnx. We seek a fixed point of Tn in the cone P.

Set Ω1={x∈E∣∥x∥<r1},Ω2={x∈E∣∥x∥<r}. If x∈P∩(Ω¯2∖Ω1), thenr≥x(t)≥σ∥x∥≥σr1>1n0>1n>0on[0,1].
Notice from (2.16), (H3), and (H4) that, for x∈P∩(Ω¯2∖Ω1),(Tnx)(t)≥0 on [0,1]. Also, for x∈P∩(Ω¯2∖Ω1), we have
(Tnx)(t)=∫01G(t,s)[fn(s,x(s))+e(s)]ds≤max0≤t,s≤1G(t,s)∫01[fn(s,x(s))+e(s)]ds,t∈[0,1],
so that
‖Tnx‖≤max0≤t,s≤1G(t,s)∫01[fn(s,x(s))+e(s)]ds.
And next, if x∈P∩(Ω¯2∖Ω1), we have by (3.10),
min0≤t≤1(Tnx)(t)=min0≤t≤1∫01G(t,s)[fn(s,x(s))+e(s)]ds≥min0≤t,s≤1G(t,s)∫01[fn(s,x(s))+e(s)]ds=σmax0≤t,s≤1G(t,s)∫01[fn(s,x(s))+e(s)]ds≥σ‖Tnx‖.
As a consequence, Tn:P∩(Ω¯2∖Ω1)→P. In addition, standard arguments show that Tn is completely continuous.

If x∈P with ∥x∥=r, thenr≥x(t)≥σ‖x‖=σr>σr1>1n0>1n>0on[0,1],
and we have by (H1),(H2), and (H3)(Tnx)(t)=∫01G(t,s)[fn(s,x(s))+e(s)]ds≤∫01G(t,s)k(s)[g(x(s))+h(x(s))]ds+γ*≤g(σr)(1+h(r)g(r))K*+γ*<r=‖x‖,t∈[0,1].
Thus, ∥Tnx∥≤∥x∥. Hence,
‖Tnx‖≤‖x‖,forx∈P∩∂Ω2.

If x∈P with ∥x∥=r1, thenr>r1≥x(t)≥σ‖x‖=σr1>1n0>1n>0on[0,1],
and we have by (H3) and (H4)(Tnx)(12)=∫01G(12,s)[fn(s,x(s))+e(s)]ds≥∫01G(12,s)[ϕr(s)+e(s)]ds≥min0≤t≤1∫01G(t,s)[ϕr(s)+e(s)]ds≥min0≤t≤1∫01G(t,s)ϕr(s)ds+γ*=δ>r1=‖x‖.
Thus, ∥Tnx∥≤∥x∥. Hence,
‖Tnx‖≤‖x‖,forx∈P∩∂Ω1.

Applying (ii) of Theorem 2.3 to (3.14) and (3.17) yields that Tn has a fixed point xn∈P∩(Ω¯2∖Ω1), and r1≤∥xn∥≤r. As such, xn is a solution of (3.1n), andr≥xn(t)≥σ‖xn‖≥σr1>1n0>1n>0,t∈[0,1].

Next we prove the fact‖xn′‖≤H
for some constant H>0 and for all n>n0. To this end, integrating the first equation of (3.1n) from 0 to 1, we obtain
∫01m2(t)xn(t)dt=∫01[fn(t,xn(t))+e(t)]dt.
Then
‖xn′‖=max0≤t≤1|xn′(t)|=max0≤t≤1|∫0txn′′(s)ds|=max0≤t≤1|∫0t[fn(s,xn(s))+e(s)-m2(s)xn(s)]ds|≤∫01[fn(s,xn(s))+e(s)]ds+∫01m2(s)xn(s)ds=2∫01m2(s)xn(s)ds≤2m¯2r=:H.

The fact ∥xn∥≤r and (3.19) show that {xn}n∈N0 is a bounded and equicontinuous family on [0,1]. Now the Arzela-Ascoli Theorem guarantees that {xn}n∈N0 has a subsequence, {xnk}k∈ℕ, converging uniformly on [0,1] to a function x∈C[0,1]. From the fact ∥xn∥≤r and (3.18), x satisfies σr1≤x(t)≤r for all t∈[0,1]. Moreover, xnk satisfies the integral equationxnk(t)=∫01G(t,s)[fn(s,xnk(s))+e(s)]ds.
Let k→∞, and we arrive at
x(t)=∫01G(t,s)[f(s,x(s))+e(s)]ds,
where the uniform continuity of f(t,x) on [0,1]×[σr1,r] is used. Therefore, x is a positive solution of boundary value problem (1.4). Finally, it is not difficult to show that, ∥x∥<r.

By Theorem 3.1, we have the following Corollary.

Corollary 3.3.

Let (H0) hold. Assume that there exist continuous functions b¯,b≻0 and λ>0 such that

0≤b¯(t)/xλ≤f(t,x)≤b(t)/xλ, for all x>0 and t∈[0,1].

Then problem (1.4) has at least one positive solution if one of the following two conditions holds:

e*≥0;

e*<0,b¯*+(B*/σλ)λ/λ+1e*>0, where B(t)=∫01G(t,s)b(s)ds.

Remark 3.4.

When m(t)≡m,t∈[0,1], then (1.4) reduces to (1.1), (H0) reduce to m∈(0,π/2). So Corollary 3.3 is more extensive than [5, Corollary 3.1].

4. Example

Consider second-order Neumann boundary value problem
x′′(t)+[π12(3-t)]2x(t)=√2¯t15[x-1(t)+1],t∈(0,1),x′(0)=0,x′(1)=0.
Here ,f(t,x)=√2¯t15[x-1+1],(t,x)∈[0,1]×(0,+∞),e(t)≡0,m(t)=(π/12)(3-t),t∈[0,1]. Obviously, (H0) is satisfied. Let k(t)=√2¯t15,g(x)=1/x,h(x)≡1,ϕr(t)=√2¯t15/2,r=2, then we can check that (H1),(H3), and (H4) are satisfied. In addition, for r=2, we have
r-γ*g(σr)(1+h(r)/g(r))=4σ3=43min0≤t,s≤1G(t,s)max0≤t,s≤1G(t,s)≥4cos2m¯3=23.
On the other hand, by Lemma 2.1, we have
ψ′(0)=∫01m2(t)ψ(t)dt≥m̲2cosm¯=√2¯π272.
By (4.3), we have
K*=max0≤t≤1∫01G(t,s)(2s15)ds≤√2¯16ψ′(0)≤92π2<23.
Hence, r-γ*/(g(σr)(1+h(r)/g(r)))>K*. So that (H2) is satisfied. According to Theorem 3.1, the boundary value problem (4.1) has at least one positive solution x with 0<∥x∥<2.

For boundary value problem (4.1), however, we cannot obtain the above conclusion by Theorem 3.1 of paper [5] since m(t)=(π/12)(3-t),t∈[0,1] is not a constant. These imply that Theorem 3.1 in this paper complement and improve those obtained in [5].

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