On positive solutions for nonhomogeneous m-point boundary value problems with two parameters

Abstract

This paper is concerned with the existence, multiplicity, and nonexistence of positive solutions for nonhomogeneous m-point boundary value problems with two parameters. The proof is based on the fixed-point theorem, the upper-lower solutions method, and the fixed-point index.

MSC: 34B10, 34B18.[PUBLICATION ABSTRACT]

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R E S E A R C H Open Access

On positive solutions for nonhomogeneous m-point boundary value problems with two parameters

Fanglei Wang1* and Yukun An2

*Correspondence: mailto:[email protected]

Web End [email protected]

1College of Science, Hohai University, Nanjing, 210098, P.R. ChinaFull list of author information is available at the end of the article

Abstract

This paper is concerned with the existence, multiplicity, and nonexistence of positive

solutions for nonhomogeneous m-point boundary value problems with two parameters. The proof is based on the xed-point theorem, the upper-lower solutions

method, and the xed-point index.

MSC: 34B10; 34B18Keywords: nonhomogeneous BVP; positive solutions; upper-lower solutions;

xed-point theorem; xed point index

1 Introduction

Many authors have studied the existence, nonexistence, and multiplicity of positive solutions for multipoint boundary value problems by using the xed-point theorem, the xed point index theory, and the lower and upper solutions method. We refer the readers to the references []. Recently, Hao, Liu and Wu [] studied the existence, nonexistence, and multiplicity of positive solutions for the following nonhomogeneous boundary value problems:

u (t) = a(t)f (t, u(t)), u() = , u()

mi= kiu(i) = b,

where b > , ki > (i = , , . . . , m ), < < < < m < ,

mi= kii < , a(t) may be singular at t = and/or t = . They showed that there exists a positive number b* >

such that the problem has at least two positive solutions for < b < b*, at least one positive solution for b = b* and no solution for b > b* by using the Krasnoselskii-Guo xed-point theorem, the upper-lower solutions method, and the topological degree theory.

Inspired by the above references, the purpose of this paper is to study the following more general nonhomogeneous boundary value problems:

u (t) = h(t)f (u(t)), u() = , u()

mi= kiu(i) =

()

where , are positive parameters, ki > , < < < < m < . The main result of

the present paper is summarized as follows.

2012 Wang and An; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0

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g(u(s)) ds,

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Theorem . Assume the following conditions hold:(H) (, ) R+\{(, )} are nonnegative parameters;

(H) h : [, ] [, +) is continuous, h(t) does not vanish identically on any

subinterval of [, ] and

G(s, s)h(s) ds < +, where G(s, s) is given in Sect. ; (H) f , g C(R+, R+) is nondecreasing with respect to u, respectively, that is,

f (u) f (u) if u u,

g(u) g(u) if u u.

And either f () > or g() > ;(H) There exist constants m, m > such that f (u) mu and g(u) mu,

respectively, for all u ;

(H) lim|u|+ f(u)u = +, lim|u|+ g(u)u = +.

If <

mi= ki < , then there exists a bounded and continuous curve separating R+\{(, )} into two disjoint subsets and such that () has at least two positive so

lutions for (, ) , one positive solution for (, ) , and no solution for (, ) .

Moreover, let + be the parametric representation of , where

+ : = () > , : = () = .

Then on +, the function = () is continuous and nonincreasing, that is, if , we

have () ( ).

For the proof of Theorem ., we also need the following lemmas.

Lemma . [] Let E be a Banach space, K a cone in E and bounded open in E. Let and T : K K be condensing. Suppose that Tx = x for all x K and all

. Then

i(T, K , K) = .

Lemma . [] Let E be a Banach space and K a cone in E. For r > , dene Kr = {x K : x < r}. Assume that T : Kr K is a compact map such that Tx = x for x Kr. If x Tx for all x Kr, then

i(T, Kr, K) = .

2 Preliminaries

Lemma . [] Assume that <

then the Green function for the homogeneous BVP

u (t) = y(t),

u() = , u()

mi= kii < . If y(t) C(, ) with

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

mi= kiu(i) =

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is given by

G(t, s) =

mi= kii

s( t)

mi= ki(i t)s, s t, s ,

t[( s)

mi= ki(i s)], t s ,

s( t) +

mi=j+ ki(i t)s, j s j+, s t, j = , , . . . , m ,t[( s)

mi=j+ ki(i s)],j s j+, t s, j = , , . . . , m , s( t) +

ji= kii(t s)

mi= kii(t s), m s t,

t( s), m s, t s.

Moreover, the Green function satises the following properties:(i) G(t, s) > for t, s (, ), and G(t, s) is continuous on [, ] [, ];

(ii) G(t, s) G(s, s) for all t, s [, ].

Lemma . Assume that (H)-(H) hold. If <

mi= kii < , then u C[, ] is a solution of () if and only if u C[, ] satises the following nonlinear integral equation:

u(t) =

G(t, s)h(s)f u(s)

ds +

g(u(s)) ds

mi= kii

t.

Proof Integrating both sides of () from to t twice and applying the boundary conditions, then we can obtain

u(t) =

g(u(s)) ds

t

mi= kii

t

(t s)h(s)f u(s)

ds

+ t

mi= kii

( s)h(s)f u(s)

ds

m

i=ki

i

(

i s)h(s)f

u(s) ds

.

Furthermore, by Lemma ., we can obtain

u(t) =

G(t, s)h(s)f

u(s)

ds +

g(u(s)) ds

t.

Let E denote the Banach space C[, ] with the norm u = maxt[,] |u(t)|. A function u(t) is said to be a solution of () if u C[, ] C(, ) satises (). Moreover, from

Lemma ., it is clear to see that u(t) is a solution of () is equivalent to the xed point of the operator T dened as

Tu(t) =

mi= kii

G(t, s)h(s)f u(s)

ds +

g(u(s)) ds

mi= kii

t.

In addition, dene a cone K E as

K =

u E : u(t) , t [, ], inft[,m] u(t) u

,

where = k min[ , ]. Then we have

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Lemma . If (H)-(H) hold, then T : K K is completely continuous.

The proof procedure of Lemma . is standard, so we omit it.

Now, we will establish the classical lower and upper solutions method for our problem. As usual, we say that x(t) is a lower solution for () if

x (t) + h(t)f (x(t)) , x() , x()

mi= kix(i)

g(x(s)) ds.

Similarly, we dene the upper solution y(t) of the problem ():

y (t) + h(t)f (y(t)) , y() , y()

mi= kiy(i)

g(y(s)) ds.

Lemma . Let x(t), y(t) be lower and upper solutions, respectively, of () such that

x(t) y(t). Then () has a nonnegative solution u(t) satisfying x(t) u(t) y(t) for t [, ].

Proof Dene

Dyx =

u R : x(t) u(t) y(t), t [, ]

.

It is clear to see that Dyx is a bounded, convex and closed subset in Banach space E. Now we can prove that T : Dyx Dyx.

For any u(t) Dyx, from (H), we have

Tu(t) =

G(t, s)h(s)f u(s)

ds +

g(u(s)) ds

mi= kii

t

G(t, s)h(s)f y(s)

ds +

g(y(s)) ds

mi= kii

t

= y(t).

On the other hand, we also have

Tu(t) =

G(t, s)h(s)f u(s)

ds +

g(u(s)) ds

mi= kii

t

G(t, s)h(s)f x(s)

ds +

g(x(s)) ds

mi= kii

t

= x(t).

From above inequalities, we obtain that T : Dyx Dyx.

Therefore, by Schauders xed theorem, the operator T has a xed point u(t) Dyx, which

is the solution of ().

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3 Proof of Theorem 1.1

Lemma . Assume (H)-(H) hold and be a compact subset of R+\{(, )}. Then there

exists a constant C > such that for all (, ) and all possible positive solutions u(t)

of () at (, ), one has u C .

Proof Suppose on the contrary that there exists a sequence {un} of positive solutions of

Eq. () at (n, n) such that (n, n) for all n N and

un .

Then un(t) K, and thus

inf

t[,m] un(t)

un . ()

Since is compact, the sequence {(n, n)}n= has a convergent subsequence which we

denote without loss of generality still by {(n, n)}n= such that

lim

n

n = *, lim

n

n = *

and at least * > or * > .Case (I). If * > , we have n */ > for n sucient large. Then by (H), there exists

a R > such that

f (u) Lu, u R,

where L satises

*

L

min

t[,]

m

G(t, s)h(s) ds > .

Since un , for n sucient large, we

un(t) = n

G(t, s)h(s)f u

n(s)

ds + n

g(un(s)) ds

mi= kii

t

n

G(t, s)h(s)f u

n(s)

ds

G(t, s)h(s)Lun(s) ds

n

m

G(t, s)h(s)Lun(s) ds

*

*

L

un min

t[,]

m

G(t, s)h(s) ds

> un .

This is a contradiction.

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Case (II). If * > , then we have n */ > for n sucient large. Since lim|u|+ g(u)u =

+, there exists a R > such that

g(u) Mu, u R,

where M satises

*

M(m )

> .

Since un , then for n sucient large, we have

un() = n

mi= kii

, s)h(s)f

un(s) ds + n

g(un(s)) ds

mi= kii

n

g(un(s)) ds

mi= kii

n

Mun(s) ds

mi= kii

*

M u (m )

mi= kii

> un .

This is a contradiction.

Lemma . Assume (H)-(H) hold. If () has a positive solution at (, ), then Eq. () has a positive solution at (, ) R+\{(, )} for all (, ) (, ).

Proof Let u(t) be the solution of Eq. () at (, ), then u(t) be the upper solution of () at (, ) R+\{(, )} with (, ) (, ). Since f () > or g() > , u = is not a solution

of (), but it is the lower solution of () at (, ). Therefore, by Lemma ., we obtain the result.

Lemma . Assume (H)-(H) hold. Then there exists (*, *) > (, ) such that Eq. () has a positive solution for all (, ) (*, *).

Proof Let (t) be the unique solution of

u (t) = h(t),

u() = , u()

()

It is clear to see that (t) is a positive solution of (). Let Mf = maxt[,] f ((t)), Mg = maxt[,] g((t)), then by (H), we know that Mf > and Mg > . Set (*, *) =

(/Mf , /Mg), we have

mi= kiu(i) = .

(t) + *h(t)f ((t)) = h(t) + *h(t)f ((t)) = h(t)(*f ((t)) ) ,

() = , ()

mi= ki(i) *

g((s)) ds =

*g((s)) ds ,

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which implies that (t) is an upper solution of () at (*, *). On the other hand, is a lower solution of () and (t). By (H), is not a solution of (). Hence, () has a

positive solution at (*, *), Lemma . now implies the conclusion of Lemma ..

Dene a set S by

S = (, ) R+\

(, ) : () has a positive solution at (, )

.

Then it follows from Lemma . that S = and (S, ) is a partially ordered set.

Lemma . Assume (H)-(H) hold. Then (S, ) is bounded above.

Proof Let (, ) S and u(t) be a positive solution of () at (, ), then we have

u u() =

, s)h(s)f

u(s) ds +

g(u(s)) ds

mi= kii

, s)h(s)mu(s) ds +

mu(s) ds

mi= kii

m

G(

, s)h(s)mu(s) ds + m

u (m )

mi= kii

m u

m

G(

, s)h(s) ds + m

u (m )

mi= kii

by (H). Furthermore, we can obtain that

m

m

G(

, s)h(s) ds + m

(m )

mi= kii

.

Lemma . Assume (H)-(H) hold. Then every chain in S has a unique supremum in S.

Lemma . Assume (H)-(H) hold. Then there exists a

[*, ] such () has a positive

solution at (, ) for all <

, no solution at (, ) for all >

. Similarly, there exists a

[*, ] such that () has a positive solution at (, ) for all <

, and no solution

at (, ) for all >

.

Lemma . Assume (H)-(H) hold. Then there exists a continuous curve separating R+\{(, )} into two disjoint subsets and such that is bounded and is un

bounded, Eq. () has at least one solution for (, ) , and no solution for (, ) .

The function = () is nonincreasing, that is, if

,

then

()

.

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Lemma . Let (, ) . Then there exists > such that (u* + , v* + ) is an upper

solution of () at (, ) for all < , where (u*, v*) is the positive solution of Eq. ()

corresponding to some (*, *) satisfying

(, )

*, *

.

Proof From (H), there exists constant M > such that

f

u*(t)

M > , g

u*(t) M > , for all t [, ].

Then by the uniform continuity of f and g on a compact set, there exist > such that

f u*(t) +

f

u*(t)

< M(

* )

,

g u*(t) +

g

u*(t)

< M(

* ) ,

for all t [, ] and < . Let u* = u* + , then we have

u* (t) + h(t)f

u* (t) = *h(t)f

u* (t) + h(t)f

u* (t) h(t)

* M f

u*

,

u* () = > ,

and

u* ()

m

i=kiu* (i)

g u*

ds

= u*() +

m

i=ki

u*(i) +

g u*

+

ds

= u*()

m

i=ki

u*(i)

+

m

i=ki

u* +

ds

= *

g u*

ds

g u*

+

ds +

m

i=ki

=

*

g u*

ds +

g u*

g

u* +

ds

+

m

i=ki

m

i=ki

> .

From above inequalities, it is clear to see that u* , is an upper solution of () at (, ) for all < .

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Proof of Theorem . From above lemmas, we need only to show the existence of the second positive solution of () for (, ) . Let (, ) , then there exists (*, *)

such that

(, )

*, *

.

Let (u*, v*) be the positive solution of () at (*, *). Then for > given by Lemma . and for all : < , denote

u* = u* + ,

v* = v* + .

Dene the set

D = u E : < u <

u*

.

Then D is bounded open set in E and D. The map T satises K D K and is

condensing, since it is completely continuous. Now let (u, v) K D, then there exists

[, ] such that either u() =

u*(). Then by (H) and Lemma ., we obtain

Tu() =

, s)h(s)f

u(s) ds +

g(u(s)) ds

mi= kii

g(

u*(s)) ds

, s)h(s)f

u*(s)

ds +

mi= kii

< u*() = u() u()

for all . Thus, T(u) = u for all u K D and , Lemma . now implies that

i(T, K D, K) = .

Now for some xed and , it follows from assumption (H) that there exists a R > such that

f (u) Lu, and g(u) Lu, u R, ()

where L satises

L

m

G(

, s)h(s) ds +

(m )

> .

mi= kii

v*) } where C is given by Lemma . with a compact set in R+\{(, )} containing (, ). Let

KR* =

u K : u < R*

.

Let R* = max{C , R, (

u*,

Then it follows from Lemma .,

T(u) = u u KR

* .

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Moreover, for u KR

* , we have

t[,m] u(t)

u R.

inf

Furthermore, we have

Tu() =

, s)h(s)f

u(s) ds +

g(u(s)) ds

mi= kii

, s)h(s)Lu(s) ds +

Lu(s) ds

mi= kii

L u

m

G(

, s)h(s) ds + L

u (m )

mi= kii

= L

m

G(

, s)h(s) ds +

(m )

mi= kii

u

> u .

Thus, Tu > u and it follows from Lemma . that

i(T, KR* , K) = .

By the additivity of the xed-point index,

= i(T, KR* , K) = i(T, K D, K) + i(T, KR

* \K D, K)

= + i(T, KR* \K D, K),

which yields

i(T, KR* \K D, K) = .

Hence, T has at least one xed point in K D and another one in KR

* \K D; this shows

that in , () has at least two positive solution.

Example Consider the following boundary value problem:

u (t) = (u + ), u() = , u()

()

mi= kiu(i) =

(u(s) + ) ds,

where f (u) = (u + ), g(u) = (u + ), and h(t) = .

Competing interests

The authors declare that they have no competing interests.

Authors contributions

In this paper, the author studies the existence, multiplicity, and nonexistence of positive solutions for nonhomogeneous m-point boundary value problems with two parameters. The proof is based on the upper-lower solutions method and xed-point index. All authors typed, read, and approved the nal manuscript.

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Author details

1College of Science, Hohai University, Nanjing, 210098, P.R. China. 2Department of Mathematics, Nanjing University of Aeronautics and Astronautics, Nanjing, 210016, P.R. China.

Acknowledgements

The authors would like to thank the referees for valuable comments and suggestions for improving this paper. The rst author is supported nancially by the Fundamental Research Funds for the Central Universities.

Received: 14 May 2012 Accepted: 27 July 2012 Published: 6 August 2012

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doi:10.1186/1687-2770-2012-87Cite this article as: Wang and An: On positive solutions for nonhomogeneous m-point boundary value problems with two parameters. Boundary Value Problems 2012 2012:87.