Open Access

Positive solutions for the third-order boundary value problems with the second derivatives

Boundary Value Problems20122012:34

DOI: 10.1186/1687-2770-2012-34

Received: 3 November 2011

Accepted: 26 March 2012

Published: 26 March 2012

Abstract

By using the fixed-point index theory in a cone and defining a linear operator, we obtain the existence of at least one positive solution for the third-order boundary value problem with integral boundary conditions

u ( t ) + f ( t , u ( t ) , u ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = 0 , u ( 0 ) = 0 , u ( 1 ) = 0 1 g ( t ) u ( t ) d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equa_HTML.gif

where f : [0, 1] × R+ × R-→ R+ is a nonnegative function. The associated Green's function for the above problem is also used, and a new reproducing cone also used.

Keywords

fixed-point index theory Green's function positive solution boundary value problem

1 Introduction

By eigenvalue criteria, Webb [1] obtained the existence of multiple positive solutions of a Hammerstein integral equation of the form
u ( t ) = 0 1 k ( t , s ) g ( s ) f ( s , u ( s ) ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equb_HTML.gif
where k can have discontinuities and g L1. Then, some articles have studied different BVPs by this way (see [25]). Webb [4] introduced an unified method to study existence of at least one nonzero solution for higher order boundary value problems
u ( n ) ( t ) + g ( t ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( n ) ( 0 ) = 0 , 0 k n - 2 , u ( 1 ) = 0 1 u ( s ) d A ( s ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equc_HTML.gif
In 2010, Hao [5] considered the existence of positive solutions of the n th-order BVP
u ( n ) ( t ) + λ a ( t ) f ( t , u ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = u ( 0 ) = = u ( n - 2 ) ( 0 ) = 0 , u ( 1 ) = 0 1 u ( s ) d A ( s ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equd_HTML.gif
Guo [6] studied the existence of positive solutions for the there-point boundary problem with the first-order derivative.
x + f ( t , x , x ) = 0 , 0 < t < 1 , x ( 0 ) = 0 , x ( 1 ) = α x ( η ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Eque_HTML.gif
where f is a nonnegative continuous function. In 2011, Zhao [7] studied third-order differential equations:
x + f ( t , x ( t ) ) = θ , t [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equf_HTML.gif
subject to integral boundary condition of the form
x ( 0 ) = θ , x ( 0 ) = θ , x ( 1 ) = 0 1 g ( t ) x ( t ) d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equg_HTML.gif

where f C([0, 1] × P, P).

In this article, we study the existence of positive solutions for the following boundary value problem
u ( t ) + f ( t , u ( t ) , u ( t ) ) = 0 , t ( 0 , 1 ) , u ( 0 ) = 0 , u ( 0 ) = 0 , u ( 1 ) = 0 1 g ( t ) u ( t ) d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ1_HTML.gif
(1.1)

The results are proved by applying the fixed point index theory in a cone and spectral radius of a linear operator. Unlike reference [7], the nonlinear part f involves the second-order derivative and just satisfies Caratheodory conditions.

The following conditions are satisfied throughout this article:

(H1) f : [0, 1] × R+ × R-R+ satisfies Caratheodory conditions, that is, f(·,u, v) is measurable for each fixed u R+, v R-, and f(t, ·,·) is continuous for a.e. t [0, 1]. For any r, r' > 0, there exists φ r , r ( t ) L [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq1_HTML.gif, such that 0 f ( t , u , v ) φ r , r ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq2_HTML.gif, where (u, v) [0, r] × [-r', 0], a.e. t [0, 1];

(H2) g L[0, 1] is nonnegative, b [0, 1), where b = 0 1 s g ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq3_HTML.gif.

2 Preliminaries

Lemma 2.1[7]. Let y L1[0, 1] and y ≥ 0, the problem
u ( t ) + y ( t ) = 0 , t ( 0 , 1 ) , u ( 0 ) = 0 , u ( 0 ) = 0 , u ( 1 ) = 0 1 g ( t ) u ( t ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ2_HTML.gif
(2.1)
has a unique solution
u ( t ) = 0 1 H ( t , s ) y ( t ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equh_HTML.gif
where H ( t , s ) = G ( t , s ) + t 1 - b 0 1 G ( τ , s ) y ( τ ) d τ , b = 0 1 s g ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq4_HTML.gif,
G ( t , s ) = 1 2 t ( 1 - s ) 2 - 1 2 ( t - s ) 2 , 0 s t 1 , 1 2 t ( 1 - s ) 2 , 0 t s 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equi_HTML.gif

Lemma 2.2. Let y L1[0, 1] and y ≥ 0, the unique solution of the boundary value problem (2.1) satisfies the following conditions: u(t) ≥ 0, u"(t) ≤ 0, for t [0, 1].

Proof. By Lemma 2.1, u(t) ≥ 0. By differential equations u'"(t) + y(t) = 0, t (0, 1), we get
u ( t ) - u ( 0 ) = - 0 1 y ( s ) d s , u ( t ) = - 0 1 y ( s ) d s 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equj_HTML.gif
Let X = C2[0, 1] with u = max 0 t 1 u ( t ) + max 0 t 1 u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq5_HTML.gif. Obviously, (X, ||·||) is a Banach space. Define the cone P X by
P = u X u ( t ) 0 , u ( t ) 0 , P r = u P u < r , r > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equk_HTML.gif

Obviously P is a cone in X, and P r is a bounded open subset in P.

Definition 2.1[1]. Let P be a cone in a Banach space X. If for any x X and x+, x- P, writing x = x+ + x- shows that P is a reproducing cone.

Lemma 2.3. P is a reproducing cone in X.

Proof. Suppose u X, so u" C[0, 1] and
u = u - - u + , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ3_HTML.gif
(2.2)
where u- = min{u"(t), 0}, u+ = min{-u"(t), 0}. Obviously u+,u- C[0, 1] and u+ ≤ 0,u- ≤ 0. For (2.2), we get
u ( t ) = 0 t u - ( s ) d s - 0 t u + ( s ) d s + u ( 0 ) , u ( t ) = 0 t d s 0 s u - ( τ ) d τ - 0 t d s 0 s u + ( τ ) d τ + u ( 0 ) t + u ( 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equl_HTML.gif
If u(0) ≥ 0, u'(0)t ≥ 0, let
u 1 = - 0 t d s 0 s u + ( τ ) d τ + u ( 0 ) t + u ( 0 ) , u 2 = - 0 t d s 0 s u - ( τ ) d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equm_HTML.gif

So u1 ≥ 0, u2 ≥ 0, then u1, u2 P and u = u1 - u2.

If u(0) ≤ 0, u'(0)t ≤ 0, let
u 1 = - 0 t d s 0 s u + ( τ ) d τ , u 2 = - 0 t d s 0 s u - ( τ ) d τ - u ( 0 ) t - u ( 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equn_HTML.gif

So u1 ≥ 0, u2 ≥ 0, then u1, u2 P and u = u1 - u2.

If u(0) ≥ 0, u'(0)t ≤ 0, let
u 1 = - 0 t d s 0 s u + ( τ ) d τ + u ( 0 ) , u 2 = - 0 t d s 0 s u - ( τ ) d τ - u ( 0 ) t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equo_HTML.gif

So u1 ≥ 0, u2 ≥ 0, then u1, u2 P and u = u1 - u2.

If u(0) ≤ 0, u'(0)t ≥ 0, let
u 1 = - 0 t d s 0 s u + ( τ ) d τ + u ( 0 ) t , u 2 = - 0 t d s 0 s u - ( τ ) d τ - u ( 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equp_HTML.gif

So u1 ≥ 0, u2 ≥ 0, then u1, u2 P and u = u1 - u2.

Then P is a reproducing cone in X.

Lemma 2.4 (Krein-Rutman) [8]. Let K be a reproducing cone in a real Banach space X and let L : KK be a compact linear operator with L(K) K. r(L) is the spectral radius of L. If r(L) > 0, then there is φ1 K\{0} such that 1 = r(L)φ1.

Lemma 2.5[9]. Let X be a Banach space, P be a cone in X and Ω(P) be a bounded open subset in P. Suppose that A : Ω ( P ) ¯ P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq6_HTML.gif is a completely continuous operator. Then the following results hold
  1. (1)

    If there exists u 0 P\{0} such that uAu + λu 0, for any u Ω(P), λ ≥ 0, then the fixed-point index i(A, Ω(P), P) = 0.

     
  2. (2)

    If 0 Ω(P), Au ≠ λu, for any u Ω(P), λ ≥ 1, then the fixed-point index i(A, Ω(P), P) = 1.

     
Define the operator A: XX, L: XX, by
A u ( t ) = 0 1 H ( t , s ) f ( s , u ( s ) , u ( s ) ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equq_HTML.gif
L u ( t ) = 0 1 H ( t , s ) ( u ( s ) - u ( s ) ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equr_HTML.gif

So A : PP is completely continuous operator; L : P → P is a compact linear operator.

Lemma 2.6[7]. Assume that (H2) holds, then choose δ 0 , 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq7_HTML.gif, for all t [δ, 1 - δ],v, s [0, 1], we have
G ( t , s ) ρ G ( v , s ) , H ( t , s ) ρ H ( v , s ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equs_HTML.gif

where ρ = 4δ2(1 - δ).

Note: r(L) is the spectral radius of L. h = min t [ δ , 1 - δ ] δ 1 - δ H ( t , s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq8_HTML.gif, where δ 0 , 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq7_HTML.gif. By Lemma 2.6, obviously h > 0.

Lemma 2.7. Suppose conditions (H1), (H2) hold, then r(L) > 0.

Proof. Take u(t) ≡ 1, then u"(t) = 0, for any t [δ, 1 - δ] we get
L u ( t ) δ 1 - δ H ( t , s ) d s h > ( 0 ) . L 2 u ( t ) δ 1 - δ H ( t , s ) L u ( s ) d s h δ 1 - δ H ( t , s ) d s h 2 > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equt_HTML.gif
Repeating the process gives
L k u ( t ) h k . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equu_HTML.gif

So, we get L k h k , r ( L ) = lim k L k 1 k h > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq9_HTML.gif. The proof is completed.

By Lemma 2.4, then there is φ1 P\{0} such that 1 = r(L)φ1.

3 Main results

In the following, we use the notation:
f ¯ ( u , v ) = sup t [ 0 , 1 ] \ E f ( t , u , v ) , f ( u , v ) = inf t [ 0 , 1 ] \ E f ( t , u , v ) , f = max { lim u sup { sup v R f ¯ ( u , v ) u v } , lim v sup { sup u R + f ¯ ( u , v ) u v } } , f 0 d = max { lim u 0 + inf { inf v [ d , 0 ] f ( u , v ) u v } , lim v 0 inf { inf u [ 0 , d ] f ( u , v ) u v } } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equv_HTML.gif

where E is a fixed subset of [0, 1] of measure zero, d > 0.

Lemma 3.1. Suppose
0 f < μ , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ4_HTML.gif
(3.1)

where μ = 1/r(L), then there exists R0 > 0 such that i(A, P r , P) = 1 for each r > R0.

Proof. Let ε > 0 satisfy fμ - ε, then there exist r1 > 0 such that
f ( t , u , v ) ( μ - ε ) ( u - v ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equw_HTML.gif

for all u > r1 or v < -r1 and a.e. t [0, 1].

By (H1), there exists φ r 1 L [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq10_HTML.gif such that
0 f ( t , u , v ) φ r 1 ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equx_HTML.gif
for all (u, v) [0, r1] × [-r1, 0] and a.e. t [0, 1]. Hence, we have
f ( t , u , v ) ( μ - ε ) ( u - v ) + φ r 1 ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ5_HTML.gif
(3.2)

for all u R+, v R- and a.e. t [0, 1].

Since 1 μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq11_HTML.gif is the spectrum radius of L. It follows from 1 μ - ε I - L - 1 = n = 0 ( μ - ε ) n + 1 L n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq12_HTML.gif, (I/(μ - ε) - L)-l exists, let
C = 0 1 H ( t , s ) φ r 1 ( s ) d s , R 0 = 1 μ - ε I - L - 1 C μ - ε 3 2 - 1 2 t 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equy_HTML.gif
For r > R0, by Lemma 2.5 we will prove
A u λ u , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equz_HTML.gif

for each u P r and λ ≥ 1.

In fact, if not, there exist u0 ∂P r and λ0 ≥ 1 such that Au0 = λ0u0.

Together with (3.2) implies
u 0 ( t ) A u 0 ( t ) 0 1 H ( t , s ) [ ( μ ε ) u 0 ( s ) + φ r 1 ( t ) ] d s 0 1 H ( t , s ) ( μ ε ) [ u 0 ( s ) v 0 ( s ) ) + φ r 1 ( s ) ] d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equaa_HTML.gif
So
u 0 ( t ) ( μ - ε ) L u 0 ( t ) + C , u 0 ( t ) λ 0 u 0 ( t ) = ( A u 0 ( t ) ) ( μ - ε ) ( L u 0 ( t ) ) - C . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equab_HTML.gif
Then
1 μ - ε I - L u 0 ( t ) C μ - ε 3 2 - 1 2 t 2 , 1 μ - ε I - L u 0 ( t ) C μ - ε 3 2 - 1 2 t 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equac_HTML.gif
So
C μ - ε 3 2 - 1 2 t 2 - 1 μ - ε I - L u 0 ( t ) P . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equad_HTML.gif
Then
u 0 ( t ) ( I μ ε L ) 1 C μ ε ( 3 2 1 2 t 2 ) , u 0 ' ' ( t ) [ ( I μ ε L ) 1 C μ ε ( 3 2 1 2 t 2 ) ] ' ' , u 0 ( t ) R 0 < r . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equae_HTML.gif

This is a contradiction. By Lemma 2.5 (2), we get that i(A, P r , P) = 1 for each r > R0. The proof is completed.

Lemma 3.2. Suppose there exists d > 0 such that
μ < f 0 d . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ6_HTML.gif
(3.3)

Then there exists ρ0 > 0 and dρ0 such that for each ρ (0, ρ0], if uAu for u , then i(A, P ρ , P) = 0.

Proof. Let ε > 0 satisfy f 0 d μ + ε https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq13_HTML.gif, there exist dρ0 > 0 such that
f ( t , u , v ) ( μ + ε ) ( u - v ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ7_HTML.gif
(3.4)

for u [0, ρ0],v [-ρ0,0] and a.e. t [0, 1].

Let ρ (0,ρ0], by Lemma 2.5 (1), we prove that: uAu + λφ1 for all u ∂Pρ, λ > 0, where φ1 P\{0} is the eigenfunction of L corresponding to the eigenvalue 1 μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq11_HTML.gif. In fact, if not, there exist u0 ∂P ρ , λ0 > 0 such that u0 = Au0 + λ0φ1. This implies
u 0 λ 0 φ 1 a n d u 0 λ 0 φ 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equaf_HTML.gif

Let: λ * = sup λ | u 0 λ φ 1 , u 0 " λ φ 1 " https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq14_HTML.gif.

So 0 < λ0 < λ* < ∞ and u 0 λ * φ 1 , u 0 λ * φ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq15_HTML.gif. Then, u0- λ*φ1 P.

For L(P) P, we get
μ L u 0 λ * μ L φ 1 = λ * φ 1 , μ ( L u 0 ) λ * μ ( L φ 1 ) = λ * φ 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equag_HTML.gif
By (3.4), we get
A u 0 = 0 1 H ( t , s ) f ( s , u 0 ( s ) , u 0 ' ' ( s ) ) d s ( μ + ε ) L u 0 . ( A u 0 ) ( μ + ε ) ( L u 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equah_HTML.gif
So, we know
u 0 = A u 0 + λ 0 φ 1 ( μ + ε ) L u 0 + λ 0 φ 1 ( λ * + λ 0 ) φ 1 . ( u 0 ) = ( A u 0 ) + λ 0 φ 1 ( μ + ε ) ( L u 0 ) + λ 0 φ 1 ( λ * + λ 0 ) φ 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equai_HTML.gif

which contradicts the definition of λ*.

Lemma 3.3. Suppose there is ρ1 > 0 such that
f ( t , u , v ) d 1 ρ 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ8_HTML.gif
(3.5)

for u [0, ρ1] and v [-ρ1, 0] a.e. t [0, 1], where d 1 = 1 0 1 H ( t , s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq16_HTML.gif, if Auu for u P ρ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq17_HTML.gif, then i ( A , P ρ 1 , P ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq18_HTML.gif.

Proof. Suppose u P ρ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq17_HTML.gif, by Lemma 2.2, we get
A u = max 0 t 1 A u ( t ) - min 0 t 1 ( A u ( t ) ) = max 0 t 1 0 1 H ( t , s ) f ( t , u ( t ) , u ( t ) ) d s + max 0 t 1 0 1 H ( t , s ) f ( t , u ( t ) , u ( t ) ) d s d 1 ρ 1 max 0 t 1 0 1 H ( t , s ) d s + max 0 t 1 0 1 H ( t , s ) d s ρ 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equaj_HTML.gif

That is Au ≠ λu for each u P ρ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq17_HTML.gif, λ > 1. If Auu for u P ρ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq17_HTML.gif, by Lemma 2.5, then i ( A , P ρ 1 , P ) = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq18_HTML.gif.

Lemma 3.4. Suppose there is ρ2 > 0 such that
f ( t , u , v ) d 2 ρ 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equ9_HTML.gif
(3.6)

for u [0, ρ 2] and v [-ρ 2, 0] a.e. t [0, 1], where d 2 = 1 min t [ δ , 1 - δ ] 0 1 H ( t , s ) d s - max t [ δ , 1 - δ ] 0 1 H ( t , s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq19_HTML.gif. If Auu for u P ρ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq20_HTML.gif, then i ( A , P ρ 2 , P ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq21_HTML.gif.

Proof. For u P ρ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq20_HTML.gif, t [δ, 1 - δ], by Lemma 2.2, we get
A u + A u = 0 1 H ( t , s ) f ( t , u ( t ) , u ( t ) ) d s + 0 1 H ( t , s ) f ( t , u ( t ) , u ( t ) ) d s d 2 ρ 2 0 1 H ( t , s ) d s + 0 1 H ( t , s ) d s ρ 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_Equak_HTML.gif

This implies that uAu + λφ for each u P ρ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq20_HTML.gif, λ > 0, where φ P\{0} is the eigenfunction of L corresponding to r(L). Suppose uAu for u P ρ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq20_HTML.gif, by Lemma 2.5, then i ( A , P ρ 2 , P ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq21_HTML.gif.

Theorem 3.1. The boundary value problem (1.1) has at least one positive solution if one of the following conditions holds.

(C1) There exists d > 0 such that (3.3) and (3.1) hold.

(C2) There exists d > 0, ρ1> 0 such that (3.3) and (3.5) hold.

(C3) There exists ρ2 > 0 such that (3.6) and (3.1) hold.

(C4) There exists ρ1, ρ2 > 0 with 0 < ρ2 < ρ1d1/d2 such that (3.5) and (3.6) hold.

Proof. When condition (C1) holds, by Lemma 3.1 and 0 ≤ f < μ, we get that there exists r > 0 such that i(A, P r , P) = 1. It follows from Lemma 3.2 and μ < f 0 b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-34/MediaObjects/13661_2011_Article_160_IEq22_HTML.gif, then there exists 0 < ρ < min{r, d} such that either there exists u ∂P ρ that i(A, P ρ , P) = 0 or u = Au. So BVP (1.1) has at least one positive solution u P with ρ ≤ ||u|| < r.

When one of other conditions holds, the results can be proved similarly.

Declarations

Acknowledgements

The project is supported by the Natural Science Foundation of China (10971045) and the Natural Science Foundation of Hebei Province (A2009000664, A2011208012). The research item financed by the talent training project funds of Hebei Province. The authors would like to thank the referee for helpful comments and suggestions.

Authors’ Affiliations

(1)
College of Electrical Engineering and Information, Hebei University of Science and Technology
(2)
College of Sciences, Hebei University of Science and Technology

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© Guo et al; licensee Springer. 2012

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