Open Access

Infinitely many solutions to superlinear second order m-point boundary value problems

Boundary Value Problems20112011:14

DOI: 10.1186/1687-2770-2011-14

Received: 28 April 2011

Accepted: 15 August 2011

Published: 15 August 2011

Abstract

We consider the boundary value problem

u ( x ) + g ( u ( x ) ) + p ( x , u ( x ) , u ( x ) ) = 0 , x ( 0 , 1 ) , u ( 0 ) = 0 , u ( 1 ) = i = 1 m - 2 α i u ( η i ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equa_HTML.gif

where:

(1) m ≥ 3, η i (0, 1) and α i > 0 with A : = i = 1 m - 2 α i < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq1_HTML.gif;

(2) g : is continuous and satisfies

g ( s ) s > 0 , s 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equb_HTML.gif

and

lim s g ( s ) s = ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equc_HTML.gif

(3) p : [0, 1] × 2 is continuous and satisfies

| p ( x , u , v ) | C + β | u | , x [ 0 , 1 ] ( u , v ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equd_HTML.gif

for some C > 0 and β (0, 1/2).

We obtain infinitely many solutions having specified nodal properties by the bifurcation techniques.

MSC(2000). 34B15, 58E05, 47J10

Keywords

Nodal solutions Second order equations Multi-point boundary value problems Bifurcation

1 Introduction

We consider the nonlinear boundary value problem
u ( x ) + g ( u ( x ) ) + p ( x , u ( x ) , u ( x ) ) = 0 , x ( 0 , 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ1_HTML.gif
(1.1)
u ( 0 ) = 0 , u ( 1 ) = i = 1 m - 2 α i u ( η i ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ2_HTML.gif
(1.2)

where

(H1) m ≥ 3, η i (0, 1) and α i > 0 with
A : = i = 1 m - 2 α i < 1 ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Eque_HTML.gif
(H2) g : is continuous and satisfies
g ( s ) s > 0 , s 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ3_HTML.gif
(1.3)
and
lim s g ( s ) s = ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ4_HTML.gif
(1.4)
(H3) p : [0, 1] × 2 is continuous and satisfies
| p ( x , u , v ) | C + β | u | , x [ 0 , 1 ] , ( u , v ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ5_HTML.gif
(1.5)

for some C > 0 and β (0, 1/2).

In order to state our results, we first recall some standard notations to describe the nodal properties of solutions. For any integer, n ≥ 0, C n [0, 1] will denote the usual Banach space of n-times continuously differentiable functions on [0, 1], with the usual sup-type norm, denoted by || · ||n. Let X := {u C2[0, 1]: u satisfies (1.2)}, Y := C0[0, 1], with the norms | · |2 and | · |0, respectively. Let
E : = { u C 1 [ 0 , 1 ] : u satisfies ( 1 . 2 ) } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equf_HTML.gif

with the norms | · | E .

We define a linear operator L : XY by
L u : = - u , u X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ6_HTML.gif
(1.6)

In addition, for any continuous function g : and any u Y, let g(u) Y denote the function g(u(x)), x [0, 1].

Next, we state some notations to describe the nodal properties of solutions of (1.1), see [1] for the details. For any C1 function u, if u(x0) = 0, then x0 is a simple zero of u, if u'(x0) ≠ 0. Now, for any integer k ≥ 1 and any ν {+, -}, we define sets S k ν , Γ k ν C 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq2_HTML.gif consisting of the set of functions u C2[0, 1] satisfying the following conditions:
S k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equg_HTML.gif
  1. (i)
    u(0) = 0, νu'(0) > 0; (ii) u has only simple zeros in [0, 1] and has exactly k - 1 zeros in (0, 1).
    Γ k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equh_HTML.gif
     
  2. (i)

    u(0) = 0, νu'(0) > 0; (ii) u' has only simple zeros in (0, 1) and has exactly k such zeros; (iii) u has a zero strictly between each two consecutive zeros of u'.

     

Remark 1.1 If we add the restriction u' (1) ≠ 0 on the functions in Γ k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq3_HTML.gif then Γ k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq3_HTML.gif becomes the set T k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq4_HTML.gif, which used in [1]. The reason we use Γ k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq3_HTML.gif rather than T k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq4_HTML.gif is that the Equation (1.1) is not autonomous anymore.

In [1, Remarks 2.1 and 2.2], Rynne pointed out that
  1. a.

    If u T k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq5_HTML.gif, then u has exactly one zero between each two consecutive zeros of u', and all zeros of u are simple. Thus, u has at least k - 1 zeros in (0, 1), and at most k zeros in (0, 1];

     
  2. b.

    The sets T k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq4_HTML.gif are open in X and disjoint;

     
  3. c.

    When considering the multi-point boundary condition (1.2), the sets T k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq4_HTML.gif are in fact more appropriate than the sets S k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq6_HTML.gif.

     

The main result of this paper is the following

Theorem 1.1 Let (H1)-(H3) hold. Then there exists an integer k0 ≥ 1 such that for all integers kk0 and each ν {+, -} the problem (1.1), (1.2) has at least one solution u k ν Γ k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq7_HTML.gif.

Superlinear problems with classical boundary value conditions have been considered in many papers, particularly in the second and fourth order cases, with either periodic or separated boundary conditions, see for example [211] and the references therein. Specifically, the second order periodic problem is considered in [2, 3], while [47] consider problems with separated boundary conditions, and results similar to Theorem 1.1 were obtained in each of these papers. The fourth order periodic problem is considered in [810]. Rynne [11] and De Coster [12] consider some general higher order problems with separated boundary conditions also.

Calvert and Gupta [13] studied the superlinear three-point boundary value problem
u ( x ) + g ( u ( x ) ) + p ( x , u ( x ) , u ( x ) ) = 0 , x ( 0 , 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ7_HTML.gif
(1.7)
u ( 0 ) = 0 , u ( 1 ) = β u ( η ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ8_HTML.gif
(1.8)

(which is a nonlocal boundary value problem), under the assumptions:

(A0) β (0, 1) (1, ∞);

(A1) g : is continuous and satisfies g(s)s > 0, s ≠ 0, g ( s ) s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq8_HTML.gifis increasing and
lim | s | g ( s ) s = ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equi_HTML.gif
(A2) p : [0, 1] × 2 is a function satisfying the Carathéodory conditions and satisfies
| p ( x , u , v ) | M 1 ( t , max ( | u | , | v | ) ) , x [ 0 , 1 ] , ( u , v ) 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equj_HTML.gif

where M1 : [0, 1] × [0, ∞) → [0, ∞) satisfies the condition: for each s [0, ∞), M1(·, s) is integrable on [0, 1] and for each t [0, 1], M1(t, ·) is increasing on [0, ∞) with s - 1 0 1 M 1 ( t , s ) d s 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq9_HTML.gif as s → ∞.

Calvert and Gupta used Leray-Schauder degree and some ideas from Henrard [14] and Cappieto et al. [5] to prove the existence of infinity many solutions for (1.7), (1.8). Their results extend the main results in [14].

It is the purpose of this paper to use the global bifurcation theorem, see [15] and [1], to obtain infinity many nodal solutions to m-point boundary value problems (1.1), (1.2) under the assumptions (H1)-(H3). Obviously, our conditions (H2) and (H3) are much weaker than the corresponding restrictions imposed in [13]. Our paper uses some of ideas of Rynne [10], which deals with fourth order two-point boundary value problems. By the way, the proof [10, Lemma 2.8] contains a small error (since ||u″|0ζ4(0) |u″|0ζ4(R) there). So, we introduce a new function χ (see (3.7)) with
χ ( 0 ) ζ 2 ( R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equk_HTML.gif

which are required in applying Lemma 3.4.

2 Eigenvalues of the linear problem

First, we state some preliminary results related to the linear eigenvalue problem
L u = λ u , u X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ9_HTML.gif
(2.1)

Denote the spectrum of L by σ(L). The following spectrum results on (2.1) were established by Rynne [1], which extend the main result of Ma and O'Regan [16].

Lemma 2.1. [1, Theorem 3.1] The spectrum σ(L) consists of a strictly increasing sequence of eigenvalues λ k > 0, k = 1, 2, ..., with corresponding eigenfunctions ϕ k ( x ) = sin ( λ k 1 2 x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq10_HTML.gif. In addition,
  1. (i)

    limk→∞ λ k = ∞;

     
  2. (ii)

    ϕ T k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq11_HTML.gif, for each k ≥ 1, and ϕ 1 is strictly positive on (0, 1).

     

Lemma 2.2 [1, Theorem 3.8] For each k ≥ 1, the algebraic multiplicity of the characteristic value λ k of L-1 : YY is equal to 1.

3 Proof of the main results

For any u X, we define e(u)(·): [0, 1] → by
e ( u ) ( x ) = p ( x , u ( x ) , u ( x ) ) , x [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equl_HTML.gif
It follows from (1.5) that
| e ( u ) ( x ) | C + β | u ( x ) | , x [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ10_HTML.gif
(3.1)
For any s , let
G ( s ) = 0 s g ( τ ) d τ 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equm_HTML.gif
and for any s ≥ 0, let
γ ( s ) = max { | g ( r ) | : | r | s } , Γ ( s ) = max { G ( r ) : | r | s } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equn_HTML.gif
We now consider the boundary value problem
u + λ u + α ( g ( u ) + e ( u ) ) = 0 , u X , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ11_HTML.gif
(3.2)

where α [0, 1] is an arbitrary fixed number and λ . In the following lemma (λ, u) × X will be an arbitrary solution of (3.2).

By (H2), we can choose b1 ≥ 1 such that
| s | b 1 | g ( s ) | C + β | s | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ12_HTML.gif
(3.3)

By (1.2), we have the following

Lemma 3.1. Let (H1) hold and let u X. Then
| u | 0 | u | 0 | u | 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ13_HTML.gif
(3.4)
Lemma 3.2. Let u be a solution of (3.2). Then for any x0, x1 [0, 1],
u ( x 1 ) 2 + λ u ( x 1 ) 2 + 2 α G ( u ( x 1 ) ) = u ( x 0 ) 2 + λ u ( x 0 ) 2 + 2 α G ( u ( x 0 ) ) - 2 α x 0 x 1 e ( u ) ( s ) u ( s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equo_HTML.gif

Proof. Multiply (3.2) by u' and integrate from x0 to x1, then we get the desired result. ■

In the following, let us fix R (0, ∞) so large that Rb1 and
g ( r ) + p ( t , r , v ) > 0 , t [ 0 , 1 ] , v , r > R , g ( r ) + p ( t , r , v ) < 0 , t [ 0 , 1 ] , v , r < - R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ14_HTML.gif
(3.5)
Lemma 3.3. There exists an increasing function ζ1 : [0, ∞) → [0, ∞), such that for any solution u of (3.2) with 0 ≤ λR and |u(x 0)| + |u'(x0)| ≤ R for some x0 [0, 1], we have
| u | 0 ζ 1 ( R ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equp_HTML.gif
Proof. Choose x1 [0, 1] such that |u'|0 = |u'(x1)|. We obtain from Lemma 3.2 that
| u | 0 2 = u ( x 1 ) 2 (1) u ( x 1 ) 2 + λ u ( x 1 ) 2 + 2 α G ( u ( x 1 ) ) (2) = u ( x 0 ) 2 + λ u ( x 0 ) 2 + 2 α G ( u ( x 0 ) ) - 2 α x 0 x 1 e ( u ) ( ξ ) u ( ξ ) d ξ . (3) (4) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equq_HTML.gif
Combining this with (3.1), (3.4), it concludes that
| u | 0 2 R 2 + R 3 + 2 Γ ( R ) + 2 C + β | u | 0 | u | 0 K ( R ) + 2 C | u | 0 + 2 β | u | 0 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equr_HTML.gif
with
K ( R ) = R 2 + R 3 + 2 Γ ( R ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equs_HTML.gif
This implies
| u | 0 ζ 1 ( R ) : = 2 C + 4 C 2 + 4 ( 1 - 2 β ) K ( R ) 2 ( 1 - 2 β ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equt_HTML.gif

Define
ζ 2 ( s ) = ζ 1 ( s + s 2 ) + 1 , s > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ15_HTML.gif
(3.6)

Clearly, the function is nondecreasing.

Lemma 3.4 Let u be a solution of (3.2) with 0 ≤ λR and |u'|0ζ2(R) for some R > 0. Then, for any x [0, 1] with |u(x)| ≤ R, we have |u'(x)| ≥ R2.

Proof. Suppose, on the contrary that there exists x0 (0, 1) such that |u(x0)| ≤ R and |u'(x0)| < R2. Then
| u ( x 0 ) | + | u ( x 0 ) | < R + R 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equu_HTML.gif
Combining this with λR < R + R2 and using Lemma 3.3, it concludes that
| u | 0 ζ 1 ( R + R 2 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equv_HTML.gif

However, this is impossible if |u'|0ζ2(R). ■

For fixed R > b1, let us define
χ ( s ) : = ζ 2 ( R ) + ζ 2 ( s ) , s 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ16_HTML.gif
(3.7)
Let us now consider the problem
u + λ u + θ ( | u | 0 χ ( λ ) ) ( g ( u ) + e ( u ) ) = 0 , u X , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ17_HTML.gif
(3.8)

where θ : is a strictly increasing, C-function with θ(s) = 0, s ≤ 1 and θ(s) = 1, s ≥ 2. The nonlinear term in (3.8) is a continuous function of (λ, u) × X and is zero for λ , |u'|0χ(λ), so (3.8) becomes a linear eigenvalue problem in this region, and overall the problem can be regarded as a bifurcation (from u = 0) problem.

The next lemma now follows immediately.

Lemma 3.5 The set of solutions (λ, u) of (3.8) with |u'|0χ(λ) is
{ ( λ , 0 ) : λ } { ( λ k , t ϕ k ) : k 1 , | t | χ ( λ k ) | ϕ k | 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equw_HTML.gif

We also have the following global bifurcation result for (3.8).

Lemma 3.6 For each k ≥ 1 and ν {+, -}, there exists a connected set C k ν × E https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq12_HTML.gif of nontrivial solutions of (3.8) such that C k ν ( λ k , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq13_HTML.gif is closed and connected and:
  1. (i)

    there exists a neighborhood N k of (λ k , 0) in × E such that N k C k ν × Γ k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq14_HTML.gif,

     
  2. (ii)

    C k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq15_HTML.gif meets infinity in × E (that is, there exists a sequence ( λ n , u n ) C k ν , n = 1 , 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq16_HTML.gif, such that |λ n | + |u n | E → ∞).

     
Proof. Since L-1 : YX exists and is bounded, (3.8) can be rewritten in the form
u = λ L - 1 u + θ ( | u | 0 χ ( λ ) ) L - 1 ( g ( u ) + e ( u ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ18_HTML.gif
(3.9)

and since L-1 can be regarded as a compact operator from Y to E, it is clear that finding a solution (λ, u) of (3.8) in × E is equivalent to finding a solution of (3.9) in × E. Now, by the similar method used in the proof of [1, Theorem 4.2]), we may deduce the desired result.

Since e(u)(t) σ 0 in (3.8), nodal properties need not be preserved. However, we will rely on preservation of nodal properties for "large" solutions, encapsulated in the following result.

Lemma 3.7 If (λ, u) is a solution of (3.8) with λ ≥ 0 and |u'|0> χ(λ), then u Γ k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq17_HTML.gif, for some k ≥ 1 and ν {+, -}.

Proof. If u Γ k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq18_HTML.gif for any k ≥ 1 and ν, then one of the following cases must occur:

Case 1. u'(0) = 0;

Case 2. u' (τ) = u″(τ) = 0 for some τ (0, 1].

In the Case 1, u(t) ≡ 0 on [0, 1]. This contradicts the assumption |u'|0> χ(λ) ≥ ζ2(λ). So this case cannot occur.

In the Case 2, we have from (3.8) that
λ u ( τ ) + θ ( | u | 0 χ ( λ ) ) ( g ( u ( τ ) ) + e ( u ) ( τ ) ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ19_HTML.gif
(3.10)
Since |u'|0> χ(λ), we have from the definition of θ that
θ ( | u | 0 χ ( λ ) ) > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ20_HTML.gif
(3.11)
It follows from Lemma 3.4 that |u(τ)| > Rb1. Combining this with (3.11) and (3.3), it concludes that
λ u ( τ ) + θ ( | u | 0 χ ( λ ) ) ( g ( u ( τ ) ) + e ( u ) ( τ ) ) 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ21_HTML.gif
(3.12)

which contradicts (3.10). So, Case 2 cannot occur.

Therefore, u Γ k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq17_HTML.gif for any k ≥ 1 and ν {+, -}. ■

In view of Lemmas 3.5 and 3.7, in the following lemma, we suppose that (λ, u) is an arbitrary nontrivial solution of (3.8) with λ ≥ 0 and u Γ k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq17_HTML.gif, for some k ≥ 1 and ν.

Lemma 3.8. There exists an integer k0 ≥ 1 (depending only on χ(0)) such that for any nontrivial solution u of (3.8) with λ = 0 and χ(0) ≤ |u'|0 ≤ 2χ(0), we have
k  <  k 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ22_HTML.gif
(3.13)
Proof. Let x1, x2 be consecutive zeros of u. Then there exists x3 (x1, x2) such that u'(x3) = 0, and hence, Lemma 3.4, (3.3), and (3.7) yield that |u(x3)| > 1. Since
2 < | u ( x 2 ) - u ( x 3 ) | + | u ( x 3 ) - u ( x 1 ) | (1) = | ( x 2 - x 3 ) u ( τ 1 ) | + | ( x 3 - x 1 ) u ( τ 2 ) | (2) ( x 2 - x 3 ) | u | 0 + ( x 3 - x 1 ) | u | 0 (3) = ( x 2 - x 1 ) | u | 0 (4) (5) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equx_HTML.gif
for some τ1 (x3, x2), τ2 (x1, x3), it follows that
| x 2 - x 1 | > 2 | u | 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ23_HTML.gif
(3.14)
Notice that |u'|0> χ(0) ≥ ζ2(R) implies that u Γ k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq17_HTML.gif for some k ∞ and ν {+, -}, and subsequently, there exist 0 < r1< r2< · · · < rk-1, such that
u ( r j ) = 0 , j = 1 , , k - 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equy_HTML.gif
This together with (3.14) imply that
1 > ( k - 1 ) 2 | u | 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equz_HTML.gif

and accordingly, k < |u'|0/2 + 1 ≤ χ(0) + 1. ■

Now let
V R ( u ) = { x [ 0 , 1 ] : | u ( x ) | R } , W R ( u ) = { x [ 0 , 1 ] : | u ( x ) | < R } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equaa_HTML.gif

Lemma 3.9. Suppose that 0 ≤ λR and |u'|0χ(R). Then W R (u) consists of at least k intervals and at most k + 1 intervals, each of length less than 2/R, and V R (u) consists of at least k intervals and at most k + 1 intervals.

Proof. Lemma 3.4 implies that |u'(x)| ≥ R2 for all x W R (u). For any interval I W R (u), u' does not change sign on I, say,
u ( x ) R 2 , x I . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equab_HTML.gif

We claim that the length of I is less than 2/R.

In fact, for x, y I with x > y, say,
u ( x ) - u ( y ) = y x u ( s ) d s R 2 ( x - y ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equac_HTML.gif
Thus,
x - y R - ( - R ) R 2 = 2 R , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equad_HTML.gif
which implies
| I | 2 R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equae_HTML.gif
The case
u ( x ) - R 2 , x I https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equaf_HTML.gif

can be treated by the similar method. Since u is monotonic in any subinterval containing in W R (u), the desired result is followed. ■

Lemma 3.10. There exists ζ3 with limR→∞ζ3(R) = 0, and η1 ≥ 0 such that, for any Rη1, if either
  1. (a)

    0 ≤ λR and |u'|0 = 2χ(R), or

     
  2. (b)

    λ = R and χ(R) ≤ |u'|0 ≤ 2χ(R),

     

then the length of each interval of V R (u) is less than ζ3(R).

Proof. Define H = H(R) by
H ( R ) 2 : = min { R , min { g ( ξ ) / ξ : | ξ | R } ( C / R + β ) } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equag_HTML.gif

and let ζ3(R) := 2π/H(R). By (1.4), limR→∞H(R) = ∞, so limR→∞ζ3(R) = 0, and we may choose η1b1 sufficiently large that H(R) > 0 for all Rη1.

We firstly show that
| u ( τ ) | > R , for some τ ( 0 , 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ24_HTML.gif
(3.15)
In fact, if |u(x)| ≤ R on [0, 1], then Lemma 3.4 yields that either
u ( x ) R 2 , x [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equah_HTML.gif
or
u ( x ) - R 2 , x [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equai_HTML.gif

However, these contradict the boundary conditions (1.2), since (H1) implies u'(s0) = 0 for some s0 (0, 1). Therefore, (3.15) is valid.

Now, Let us choose x0, x2 such that either
  1. (1)

    u(x 0) = u(x 2) = R and u > R on (x 0, x 2) or

     
  2. (2)

    u(x 0) = R, x 2 = 1 and u > R on (x 0, 1].

     
(the case of intervals on which u < 0 is similar). Let
I = [ x 0 , x 2 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equaj_HTML.gif
By (3.8) and the construction of H(R), if either (a) or (b) holds then
- u ( x ) H ( R ) 2 u ( x ) > 0 , x I , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equak_HTML.gif

and by Lemma 3.4, u'(x0) > 0, and u'(x2) < 0, if x2< 1.

Suppose, now on the contrary that x2 - x0> ζ3(R), that is, l := 2π/(x2 - x0) < H(R). Defining x1 = (x0 + x2)/2 and
v ( x ) = 1 + cos l ( x - x 1 ) , x I , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equal_HTML.gif
we have
v ( x 0 ) = v ( x 2 ) = 0 , v ( x 0 ) = v ( x 2 ) = 0 , (1) v ( x ) = - l 2 ( v ( x ) - 1 ) , x I , (2) (3) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equam_HTML.gif
and hence
0 = x 0 x 2 d d x ( u v - u v ) d x (1) = x 0 x 2 ( u v - u v ) d x (2) x 0 x 2 ( - H 2 u v + l 2 ( v - 1 ) u ) d x (3) = - l 2 x 0 x 2 u d x (4) - l 2 R < 0 , (5) (6) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equan_HTML.gif

and this contradiction shows that x2 - x0ζ3(R), which proves the lemma.

Now, we are in the position to prove Theorem 1.1.

Proof of Theorem 1.1 Now, choose an arbitrary integer kk0 and ν {+, -}, and choose Λ > max{η1, μ k } (Here, we assume Λ > η1, so that Lemma 3.10 could be applied!) such that
( k + 1 ) 2 Λ + ( k + 1 ) ζ 3 ( Λ ) < 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equ25_HTML.gif
(3.16)
Notice that Lemma 3.9 implies that if |u'|0χ(Λ), then the length of each interval of WΛ(u) is less than 2 Λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq19_HTML.gif for 0 ≤ λ ≤ Λ. This together with (3.16) and Lemma 3.10 imply that there exists no solution (λ, u) of (3.8), which satisfies either
  1. (a)

    0 ≤ λ ≤ Λ and |u'|0 = 2χ(Λ) or

     
  2. (b)

    λ = Λ and χ(Λ) ≤ |u'|0 ≤ 2χ(Λ).

     
Now, let us denote
B = { ( λ , u ) : 0 λ Λ , χ ( λ ) | u | 0 2 χ ( Λ ) } , D 1 = { ( λ , u ) : 0 λ Λ , | u | 0 = χ ( λ ) } , D 2 = { ( 0 , u ) : 2 χ ( 0 ) | u | 0 2 χ ( Λ ) } . (4) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equao_HTML.gif
It follows from Lemma 3.5 that C k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq15_HTML.gif "enters" B through the set D1, while from Lemma 3.7, C k ν B × Γ k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq20_HTML.gif. Thus, by Lemma 3.6 and the fact
| u | 0 | u | 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equap_HTML.gif
C k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq15_HTML.gif must "leave" B. (Suppose, on the contrary that C k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq15_HTML.gif does not "leave" B, then
| u | 0 | u | 0 2 χ ( Λ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_Equaq_HTML.gif

which contradicts the fact that C k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq15_HTML.gif joins (μ k , 0) to infinity in × E.) Since C k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq15_HTML.gif is connected, it must intersect ∂B. However, Lemmas 3.8-3.10 (together with (3.16)) show that the only portion of ∂B (other than D1), which C k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq15_HTML.gif can intersect is D2. Thus, there exists a point ( 0 , u k ν ) C k ν D 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq21_HTML.gif, and clearly u k ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-14/MediaObjects/13661_2011_Article_11_IEq22_HTML.gif provides the desired solution of (1.1)-(1.2).

Declarations

Acknowledgements

The authors are grateful to the anonymous referee for his/her valuable suggestions. Supported by the NSFC(No.11061030), the Fundamental Research Funds for the Gansu Universities.

Authors’ Affiliations

(1)
Department of Mathematics, Northwest Normal University
(2)
School of Automation and Electrical Engineering, Lanzhou Jiaotong University

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© Ma et al; licensee Springer. 2011

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