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Nontrivial solutions for a boundary value problem with integral boundary conditions

Abstract

This paper concerns the existence of nontrivial solutions for a boundary value problem with integral boundary conditions by topological degree theory. Here the nonlinear term is a sign-changing continuous function and may be unbounded from below.

1 Introduction

Consider the following Sturm-Liouville problem with integral boundary conditions

{ ( L u ) ( t ) + h ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , ( cos γ 0 ) u ( 0 ) ( sin γ 0 ) u ( 0 ) = 0 1 u ( τ ) d α ( τ ) , ( cos γ 1 ) u ( 1 ) + ( sin γ 1 ) u ( 1 ) = 0 1 u ( τ ) d β ( τ ) ,
(1.1)

where (Lu)(t)= ( p ˜ ( t ) u ( t ) ) +q(t)u(t), p ˜ (t) C 1 [0,1], p ˜ (t)>0, q(t)C[0,1], q(t)<0, α and β are right continuous on [0,1), left continuous at t=1 and nondecreasing on [0,1] with α(0)=β(0)=0; γ 0 , γ 1 [0,π/2], 0 1 u(τ)dα(τ) and 0 1 u(τ)dβ(τ) denote the Riemann-Stieltjes integral of u with respect to α and β, respectively. Here the nonlinear term f:[0,1]×(,+)(,+) is a continuous sign-changing function and f may be unbounded from below, h:(0,1)[0,+) with 0< 0 1 h(s)ds<+ is continuous and is allowed to be singular at t=0,1.

Problems with integral boundary conditions arise naturally in thermal conduction problems [1], semiconductor problems [2], hydrodynamic problems [3]. Integral BCs (BCs denotes boundary conditions) cover multi-point BCs and nonlocal BCs as special cases and have attracted great attention, see [414] and the references therein. For more information about the general theory of integral equations and their relation with boundary value problems, we refer to the book of Corduneanu [4], Agarwal and O’Regan [5]. Yang [6], Boucherif [8], Chamberlain et al. [10], Feng [11], Jiang et al. [14] focused on the existence of positive solutions for the cases in which the nonlinear term is nonnegative. Although many papers investigated two-point and multi-point boundary value problems with sign-changing nonlinear terms, for example, [1520], results for boundary value problems with integral boundary conditions when the nonlinear term is sign-changing are rarely seen except for a few special cases [7, 12, 13].

Inspired by the above papers, the aim of this paper is to establish the existence of nontrivial solutions to BVP (1.1) under weaker conditions. Our findings presented in this paper have the following new features. Firstly, the nonlinear term f of BVP (1.1) is allowed to be sign-changing and unbounded from below. Secondly, the boundary conditions in BVP (1.1) are the Riemann-Stieltjes integral, which includes multi-point boundary conditions in BVPs as special cases. Finally, the main technique used here is the topological degree theory, the first eigenvalue and its positive eigenfunction corresponding to a linear operator. This paper employs different conditions and different methods to solve the same BVP (1.1) as [7]; meanwhile, this paper generalizes the result in [17] to boundary value problems with integral boundary conditions. What we obtain here is different from [620].

2 Preliminaries and lemmas

Let E=C[0,1] be a Banach space with the maximum norm u= max 0 t 1 |u(t)| for uE. Define P={uEu(t)0,t[0,1]} and B r ={uEu<r}. Then P is a total cone in E, that is, E= P P ¯ . P denotes the dual cone of P, namely, P ={g E g(u)0, for all uP}. Let E denote the dual space of E, then by Riesz representation theorem, E is given by

E = { v v  is right continuous on  [ 0 , 1 )  and is bounded variation on  [ 0 , 1 ] with  v ( 0 ) = 0 } .

We assume that the following condition holds throughout this paper.

(H1) u(t)0 is the unique C 2 solution of the linear boundary value problem

{ ( L u ) ( t ) = 0 , 0 < t < 1 , ( cos γ 0 ) u ( 0 ) ( sin γ 0 ) u ( 0 ) = 0 , ( cos γ 1 ) u ( 1 ) + ( sin γ 1 ) u ( 1 ) = 0 .

Let φ,ψ C 2 ([0,1], R + ) solve the following inhomogeneous boundary value problems, respectively:

{ ( L φ ) ( t ) = 0 , 0 < t < 1 , ( cos γ 0 ) φ ( 0 ) ( sin γ 0 ) φ ( 0 ) = 1 , ( cos γ 1 ) φ ( 1 ) + ( sin γ 1 ) φ ( 1 ) = 0 and { ( L ψ ) ( t ) = 0 , 0 < t < 1 , ( cos γ 0 ) ψ ( 0 ) ( sin γ 0 ) ψ ( 0 ) = 0 , ( cos γ 1 ) ψ ( 1 ) + ( sin γ 1 ) ψ ( 1 ) = 1 .

Let κ 1 =1 0 1 φ(τ)dα(τ), κ 2 = 0 1 ψ(τ)dα(τ), κ 3 = 0 1 φ(τ)dβ(τ), κ 4 =1 0 1 ψ(τ)dβ(τ).

(H2) κ 1 >0, κ 4 >0, k= κ 1 κ 4 κ 2 κ 3 >0.

Lemma 2.1 ([7])

If (H1) and (H2) hold, then BVP (1.1) is equivalent to

u(t)= 0 1 G(t,s)h(s)f ( s , u ( s ) ) ds,

where G(t,s)C([0,1]×[0,1], R + ) is the Green function for (1.1).

Define an operator A:EE as follows:

(Au)(t)= 0 1 G(t,s)h(s)f ( s , u ( s ) ) ds,uE.
(2.1)

It is easy to show that A:EE is a completely continuous nonlinear operator, and if uE is a fixed point of A, then u is a solution of BVP (1.1) by Lemma 2.1.

For any uE, define a linear operator K:EE as follows:

(Ku)(t)= 0 1 G(t,s)h(s)u(s)ds,uE.
(2.2)

It is easy to show that K:EE is a completely continuous nonlinear operator and K(P)P holds. By [7], the spectral radius r(K) of K is positive. The Krein-Rutman theorem [21] asserts that there are ϕP{0} and ω P {0} corresponding to the first eigenvalue λ 1 =1/r(K) of K such that

λ 1 Kϕ=ϕ
(2.3)

and

λ 1 K ω=ω,ω(1)=1.
(2.4)

Here K : E E is the dual operator of K given by:

( K v ) (s)= 0 s 0 1 G(t,τ)h(τ)dv(t)dτ,v E .

The representation of K , the continuity of G and the integrability of h imply that ω C 1 [0,1]. Let e(t):= ω (t). Then eP{0}, and (2.4) can be rewritten equivalently as

r(K)e(s)= 0 1 G(t,s)h(s)e(t)dt, 0 1 e(t)dt=1.
(2.5)

Lemma 2.2 ([7])

If (H1) holds, then there is δ>0 such that P 0 ={uP 0 1 u(t)e(t)dtδu} is a subcone of P and K(P) P 0 .

Lemma 2.3 ([22])

Let E be a real Banach space and ΩE be a bounded open set with 0Ω. Suppose that A: Ω ¯ E is a completely continuous operator. (1) If there is y 0 E with y 0 0 such that uAu+μ y 0 for all uΩ and μ0, then deg(IA,Ω,0)=0. (2) If Auμu for all uΩ and μ1, then deg(IA,Ω,0)=1. Here deg stands for the Leray-Schauder topological degree in E.

Lemma 2.4 Assume that (H1), (H2) and the following assumptions are satisfied:

(C1) There exist ϕP{0}, ω P {0} and δ>0 such that (2.3), (2.4) hold and K maps P into P 0 .

(C2) There exists a continuous operator H:EP such that

lim u + H u u =0.

(C3) There exist a bounded continuous operator F:EE and u 0 E such that Fu+ u 0 +HuP for all uE.

(C4) There exist v 0 E and ζ>0 such that KFu λ 1 (1+ζ)KuKHu v 0 for all uE.

Let A=KF, then there exists R>0 such that

deg(IA, B R ,0)=0,

where B R ={uEu<R}.

Proof Choose a constant L 0 = ( δ λ 1 ) 1 (1+ ζ 1 )+K>0. From (C2), for 0< ε 0 < L 0 1 , there exists R 1 >0 such that u> R 1 implies

Hu< ε 0 u.
(2.6)

Now we shall show

uKFu+μϕfor any u B R  and μ0,
(2.7)

provided that R is sufficiently large.

In fact, if (2.7) is not true, then there exist u 1 B R and μ 1 0 satisfying

u 1 =KF u 1 + μ 1 ϕ.
(2.8)

Since ϕP{0}, e(t)P{0}, 0 1 ϕ(t)e(t)dt>0. Multiply (2.8) by e(t) on both sides and integrate on [0,1]. Then, by (C4), (2.5), we get

0 1 u 1 ( t ) e ( t ) d t = 0 1 ( K F u 1 ) ( t ) e ( t ) d t + μ 1 0 1 ϕ ( t ) e ( t ) d t λ 1 ( 1 + ζ ) 0 1 0 1 G ( t , s ) h ( s ) u 1 ( s ) d s e ( t ) d t 0 1 ( K H u 1 ) ( t ) e ( t ) d t 0 1 v 0 ( t ) e ( t ) d t = λ 1 ( 1 + ζ ) 0 1 0 1 G ( t , s ) h ( s ) u 1 ( s ) e ( t ) d s d t 0 1 0 1 G ( t , s ) h ( s ) ( H u 1 ) ( s ) e ( t ) d s d t 0 1 v 0 ( t ) e ( t ) d t = λ 1 ( 1 + ζ ) 0 1 [ 0 1 G ( t , s ) h ( s ) e ( t ) d t ] u 1 ( s ) d s 0 1 [ 0 1 G ( t , s ) h ( s ) e ( t ) d t ] ( H u 1 ) ( s ) d s 0 1 v 0 ( t ) e ( t ) d t = λ 1 ( 1 + ζ ) r ( K ) 0 1 e ( s ) u 1 ( s ) d s r ( K ) 0 1 ( H u 1 ) ( s ) e ( s ) d s 0 1 v 0 ( t ) e ( t ) d t = ( 1 + ζ ) 0 1 u 1 ( t ) e ( t ) d t r ( K ) 0 1 ( H u 1 ) ( t ) e ( t ) d t 0 1 v 0 ( t ) e ( t ) d t .
(2.9)

Thus,

0 1 u 1 (t)e(t)dt ζ 1 ( r ( K ) 0 1 ( H u 1 ) ( t ) e ( t ) d t + 0 1 v 0 ( t ) e ( t ) d t ) .
(2.10)

By (2.9), 0 1 (KH u 1 )(t)e(t)dt=r(K) 0 1 (H u 1 )(t)e(t)dt holds. Then (2.3), (2.6) and (2.10) imply

0 1 ( u 1 ( t ) + ( K H u 1 ) ( t ) + ( K u 0 ) ( t ) ) e ( t ) d t ζ 1 ( r ( K ) 0 1 ( H u 1 ) ( t ) e ( t ) d t + 0 1 v 0 ( t ) e ( t ) d t ) + r ( K ) 0 1 ( H u 1 ) ( t ) e ( t ) d t + 0 1 ( K u 0 ) ( t ) e ( t ) d t ζ 1 ( 1 + ζ ) r ( K ) 0 1 ( H u 1 ) ( t ) e ( t ) d t + ζ 1 0 1 v 0 ( t ) e ( t ) d t + 0 1 ( K u 0 ) ( t ) e ( t ) d t ζ 1 ( 1 + ζ ) r ( K ) ε 0 u 1 + L 1 ,
(2.11)

where L 1 = ζ 1 0 1 v 0 (t)e(t)dt+ 0 1 (K u 0 )(t)e(t)dt is a constant.

(C3) shows F u 1 + u 0 +H u 1 P and (C1) implies μ 1 ϕ= μ 1 λ 1 K φ 1 P 0 . Then (C1), (2.8) and Lemma 2.2 tell us that

u 1 +KH u 1 +K u 0 =KF u 1 + μ 1 ϕ+KH u 1 +K u 0 =K(F u 1 +H u 1 + u 0 )+ μ 1 ϕ P 0 .

The definition of P 0 yields

0 1 ( u 1 + K H u 1 + K u 0 ) ( t ) e ( t ) d t δ u 1 + K H u 1 + K u 0 δ u 1 δ K H u 1 δ K u 0 .
(2.12)

It follows from (2.6), (2.11) and (2.12) that

u 1 = δ 1 0 1 ( u 1 + K H u 1 + K u 0 ) ( t ) e ( t ) d t + K H u 1 + K u 0 ε 0 ( δ λ 1 ) 1 ( 1 + ζ 1 ) u 1 + L 1 δ 1 + ε 0 K u 1 + K u 0 = ε 0 L 0 u 1 + L 2 ,
(2.13)

where L 2 =K u 0 + L 1 δ 1 is a constant.

Since 0< ε 0 L 0 <1, then (2.13) deduces that (2.7) holds provided that R is sufficiently large such that R>max{ L 2 /(1 ε 0 L 0 ), R 1 }. By (2.13) and Lemma 2.3, we have

deg(IA, B R ,0)=0.

 □

3 Main results

Theorem 3.1 Assume that (H1), (H2) hold and the following conditions are satisfied:

(A1) There exist two nonnegative functions b(t),c(t)C[0,1] with c(t)0 and one continuous even function B:R R + such that f(t,x)b(t)c(t)B(x) for all xR. Moreover, B is nondecreasing on R + and satisfies lim x + B ( x ) x =0.

(A2) f:[0,1]×RR is continuous.

(A3) lim inf x + f ( t , x ) x > λ 1 uniformly on t[0,1].

(A4) lim sup x 0 | f ( t , x ) x |< λ 1 uniformly on t[0,1].

Here λ 1 is the first eigenvalue of the operator K defined by (2.2).

Then BVP (1.1) has at least one nontrivial solution.

Proof We first show that all the conditions in Lemma 2.4 are satisfied. By Lemma 2.2, condition (C1) of Lemma 2.4 is satisfied. Obviously, B:EP is a continuous operator. By (A1), for any ε>0, there is L>0 such that when x>L, B(x)<εx holds. Thus, for uE with u>L, B(u)<εu holds. The fact that B is nondecreasing on R + yields (Bu)(t)B(u) for any uP, t[0,1]. Since B:R R + is an even function, for any uE and t[0,1], (Bu)(t)B(u) holds, which implies BuB(u) for uE. Therefore,

BuB ( u ) <εu,uE with u>L,

that is, lim u + B u u =0. Take Hu= c 0 Bu, for any uE, where c 0 = max t [ 0 , 1 ] c(t)>0. Obviously, lim u + H u u =0 holds. Therefore H satisfies condition (C2) in Lemma 2.4.

Take u 0 (t)b= max t [ 0 , 1 ] b(t)>0 and (Fu)(t)=f(t,u(t)) for t[0,1], uE, then it follows from (A1) that

Fu+ u 0 +HuPfor all uE,

which shows that condition (C3) in Lemma 2.4 holds.

By (A3), there exist ε 1 >0 and a sufficiently large number l 1 >0 such that

f(t,x) λ 1 (1+ ε 1 )x,x l 1 .
(3.1)

Combining (3.1) with (A1), there exists b 1 0 such that

f(t,x) λ 1 (1+ ε 1 )x b 1 c 0 B(x)for all xR,

and so

Fu λ 1 (1+ ε 1 )u b 1 Hufor all uE.
(3.2)

Since K is a positive linear operator, from (3.2) we have

(KFu)(t) λ 1 (1+ ε 1 )(Ku)(t)K b 1 (KHu)(t),t[0,1],uE.

So condition (C4) in Lemma 2.4 is satisfied.

According to Lemma 2.4, we derive that there exists a sufficiently large number R>0 such that

deg(IA, B R ,0)=0.
(3.3)

From (A4) it follows that there exist 0< ε 2 <1 and 0<r<R such that

| f ( t , x ) | (1 ε 2 ) λ 1 |x|,t[0,1],xR with |x|r.

Thus

| ( A u ) ( t ) | (1 ε 2 ) λ 1 ( K | u | ) (t),t[0,1],uE with ur.
(3.4)

Next we will prove that

uμAufor all u B r  and μ[0,1].
(3.5)

If there exist u 1 B r and μ 1 [0,1] such that u 1 = μ 1 A u 1 . Let z(t)=| u 1 (t)|. Then zP and by (3.4), z(1 ε 2 ) λ 1 Kz. The n th iteration of this inequality shows that z ( 1 ε 2 ) n λ 1 n K n z (n=1,2,), so z ( 1 ε 2 ) n λ 1 n K n z, that is, 1 ( 1 ε 2 ) n λ 1 n K n . This yields 1 ε 2 =(1 ε 2 ) λ 1 r(K)=(1 ε 2 ) λ 1 lim n K n n 1, which is a contradictory inequality. Hence, (3.5) holds.

It follows from (3.5) and Lemma 2.3 that

deg(IA, B r ,0)=1.
(3.6)

By (3.3), (3.6) and the additivity of Leray-Schauder degree, we obtain

deg(IA, B R B ¯ r ,0)=deg(IA, B R ,0)deg(IA, B r ,0)=1.

So A has at least one fixed point on B R B ¯ r , namely, BVP (1.1) has at least one nontrivial solution. □

Corollary 3.1 Using ( A 1 ) instead of (A1), the conclusion of Theorem  3.1 remains true.

( A 1 ) There exist three constants b>0, c>0 and α(0,1) such that

f(x)bc | x | α for any xR.

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Acknowledgements

The first two authors were supported financially by the National Natural Science Foundation of China (11201473, 11271364) and the Fundamental Research Funds for the Central Universities (2013QNA35, 2010LKSX09, 2010QNA42). The third author was supported financially by the National Natural Science Foundation of China (11371221, 11071141), the Specialized Research Foundation for the Doctoral Program of Higher Education of China (20123705110001) and the Program for Scientific Research Innovation Team in Colleges and Universities of Shandong Province.

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Correspondence to Bingmei Liu.

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Keywords

  • Nonlinear Term
  • Nontrivial Solution
  • Continuous Operator
  • Real Banach Space
  • Positive Linear Operator