Positive Solutions for Nonlineary nth-order Boundary Value Problem＊

LI Lang

(Department of Mathematics，QiLu Normal University，Jinan 250013，China)

In this paper，we study the existence of positive solutions for the following nonlinear n-th-order boundary value problem

where h∈C(0，1)∩L(0，1)is nonnegative and may be singular at t=0 and t=1 ，f∈C(［0，1］× R+，that is，α(t)and β(t)have bounded variation.

The boundary value problems for nonlinear higher-order differential equations arise in a variety of areas of applied mathematics，physics，and variational problems of control theory.Many authors have discussed the existence of solutions of higher-order boundary value problems，see for example［1－13］and the references therein.Recently，J.R.L.Webb［1］studied the following n-th order nonlocal boundary value problem

where a(t)have singularity，and the nonlinearity f satisfies Caratheodory conditions.Under weak assumptions，J.R.L.Webb obtained sharp results on the existence of one positive solution and multiple positive solutions under suitable condition on f.

By means of fixed point index theory，Hao［2］discussed the existence of positive solutions for eigenvalue problem of(2).In［2］，f:［0，1］× (0，∞)→［0，∞)is sublinear.It should be remarked that our nonlinearity f in this paper，unlike the f in［2］，may be both sublinear and superlinear.Our second theorem involves the existence of at least one positive solution for(1)with f growing superlinearly，thereby complementing the results in［2］.In addition，it is interesting to note that，unlike most of the authors，especially in［2］，who discuss multi-point or integral boundary value problems，we do not endeavor to construct a new Green＇s function associated with(1).Thus our work here is meaningful and interesting and improves and extends the corresponding ones in［1，2］.

It is worth mentioning that the idea using a Riemann-Stieltjes integral with a signed measure is due to Webb and Infante in［3，4］.The papers［3 －5］contain several new ideas，and give a unified approach to many boundary value problems.It is wellknown that the Riemann-Stieltjes integral，as in the form of，where α(s)is of bounded variation，that is dα(s)can be a signed measure，includes as special cases the multi-point boundary value problems and integral boundary value problems.That is why many authors are particularly interested in Riemann-Stieltjes integral boundary value problems.

1 Preliminaries

We can easily see the problem(1)is equivalent to the following integral equation

Where

Lemma 1.1［6］G(t，s)has the following properties

(i)0 ≤ G(t，s)≤ K(s)，∀t，s ∈ ［0，1］，Where K(s)

(ii)G(t，s)≥ γ(t)K(s)，∀t，s∈［0，1］Where

Lemma 1.1implies G(t，s)≥γ(t)G(τ，s)，∀t，s∈［0，1］and γ(t)is almost everywhere positive in ［0，1］.

Now A:P→P is a completely continuous operator and L:P → P is a completely continuous，linear，positive operator.From now on we suppose that the spectral radius of L，denoted by r(L)，is positive.Now the well-known Krein-Rutman theorem［14］asserts that there exist two functions φ ∈ P{0}and ψ ∈L(0，1){0}with ψ(x)≥ 0 .for which

Let

on E.The following is a result obtained by Lemma 1.1.

Lemma 1.2L(P)⊂ P0.

Lemma 1.3［15］Suppose is a bounded open set and A:∩P→P is a completely continuous operator.If there exists u0∈P{0}such that u － Au≠μu0，∀u∈ ∂Ω ∩ P，then i(A，Ω ∩ P，P)=0.

Lemma 1.4［15］Let be a bounded open set with0 ∈Ω.Suppose A:∩P→P is a completely continuous operator，If u≠μAu，∀u∈∂Ω∩P，0≤u≤1.then i(A，Ω ∩ P，P)=1 .

Throughout the paper we assume that:

(H2)h ∈ L(0，1)∩ C(0，1)is nonnegative and does not vanish identically on any subinterval of(0，1)and

uniformly with respect to t∈［0，1］，where

uniformly with respect to A:Ω ∩ P → P，t∈［0，1］.

2 Main Results

Theorem 2.1If(H1)，(H2)and(H3)hold，the problem(1)has at least one positive solution.

Proof. By f0＞ λ1，there exist r＞0 and ε ＞ 0 such that f(t，u)≥ (λ1+ ε)u ，for all u ∈ ［0，r］and t∈ ［0，1］.Consequently，by definition of A.we find

Now we claim that the set.M1={u∈ P:u=Au+μφ，μ≥0}⊂{0}

Indeed，If u∈ M1，then we have u≥ Au，∀u∈ M1.Combining this and(7)，we get

Multiply by ψ(t)on both sides of the above and integrate over［0，1］and use(6)to obtain

0，then u(t)≡0.As a result，M1⊂{0}holds.

Therefore，u≠Au+μφ，∀u∈∂Br∩P，μ≥0.Lemma 1.3 gives

On the other hand，by f∞=0，there exist a sufficiently small ε ＞ 0 and b ＞0 such that f(t，u)≤ εu+b for all u ≥0 and t∈［0，1］.

Note that we may choose R sufficiently big to satisfy R ＞r.We shall claim

If the claim is false，there exist u1∈ ∂BR∩ P and u1∈［0，1］such that u1= μ1Au1.Therefore，

u1(t)≤(Au1)(t)

Multiply by dα(t)and dβ(t)respectively and integrate over［0，1］to obtain

Let

We make assumptions involving κi(i=1，2，3，4)and κ as follows.

and

Combining the above two inequalities and(10)，we find

where

Indeed，we can choose sufficiently small ε ＞ 0 such that ρQε ＜ 1，therefore，‖u‖Taking R ＞ ，a contradiction with u1∈∂BR∩P.As a result(9)is true.Now Lemma 1.4 implies

Now(8)and(12)combined imply

Hence the operator A has at least one fixed point on(BR)∩ P.Therefore(1)has at least one positive solution，which completes the proof.

Theorem 2.2If(H1)，(H2)，(H4)and(H5)hold，the problem(1)has at least one positive solution.

Proof. We find by definition of f∞，there exist ε＞0 and b＞0 such that

f(t，u)≥(λ1+ ε)u －b，∀0≤ u≤ ∞..

This implies

for all u∈ P.

Let

where φ ∈ P is determined by(6).We shall prove M2is bounded.Indeed，if u∈M2，then we have u≥ Au by definition.This together with(13)leads to

Multiply by ψ(t)on both sides of the above and integrate over［0，1］and use(6)to obtain

Note we have M2⊂P0by Lemma 1.2.This together with the preceding inequality implies‖u‖≤(εω)－1b for all u∈M2，which establishes the boundedness of M2，as required.Taking R ＞(εω)－1b，we obtain u≠ Au+ μφ，∀u∈ ∂BR∩ P，μ ≥ 0.Now Lemma 1.3 yields

On the other hand，by definition of f0，there existr∈(0，r)and a sufficiently small ε ＞ 0 ，such that f(t，u)≤ εu for all u∈［0，r］and t∈［0，1］.This implies

for all u∈Br∩P.

Now we claim

Indeed，if there exist u2∈∂Br∩P and μ2∈［0，1］ for which u2= μ2Au2，then this together with(15)leads to

By(H5)，similarly with(11)，we obtain

We have also assumed eQε ＜1 as Theorem 2.1.Thus ‖u2‖ =0，a contradiction to u2∈∂Br∩P.

As a result，(16)is true and we have by Lemma 1.4

Now(14)and(17)combined imply

Hence the operator A has at least one fixed point on(BR)∩ P.Therefore(1)has at least one positive solution，which completes the proof.

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