Open Access

An order-type existence theorem and applications to periodic problems

Boundary Value Problems20132013:37

DOI: 10.1186/1687-2770-2013-37

Received: 8 November 2012

Accepted: 5 February 2013

Published: 21 February 2013

Abstract

Based on the fixed point index and partial order method, one new order-type existence theorem concerning cone expansion and compression is established. As applications, we present sufficient existence conditions for the first- and second-order periodic problems.

MSC:34B15.

Keywords

fixed point index order-type existence theorem cone expansion and compression positive solutions periodic boundary value problems

1 Introduction and preliminaries

Let X, Y be real Banach spaces. Consider a linear mapping L : dom L X Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq1_HTML.gif and a nonlinear operator N : X Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq2_HTML.gif. Here we assume that L is a Fredholm operator of index zero, that is, ImL is closed and dim Ker L = codim Im L < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq3_HTML.gif. Then the solvability of the operator equation
L x = N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equa_HTML.gif

has been studied by many researchers in the literature; see [18] and the references therein. In [1], Cremins established a fixed point index for A-proper semilinear operators defined on cones which includes and improves the results in [5, 8, 9]. Using the fixed point index and the concept of a quasi-normal cone introduced in [10], Cremins established a norm-type existence theorem concerning cone expansion and compression in [11], which generalizes some corresponding results contained in [12].

In this paper, we will use the properties of the fixed point index in [1] and partial order to present a new order-type existence theorem concerning cone expansion and compression which extends the corresponding results in [12]. We recall that a partial order in X induced by a cone K X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq4_HTML.gif is defined by
x y y x K . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equb_HTML.gif

As applications, we study the first- and second-order periodic boundary problems and obtain new existence results. During the last few decades, periodic boundary value problems have been studied by many researchers in the literature; see, for example, [1319] and the references therein. Our new results improve those contained in [13, 18].

Next we recall some notations and results which will be needed in this paper. Let X and Y be Banach spaces, D be a linear subspace of X, { X n } D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq5_HTML.gif and { Y n } Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq6_HTML.gif be the sequences of oriented finite dimensional subspaces such that Q n y y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq7_HTML.gif in Y for every y and dist ( x , X n ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq8_HTML.gif for every x D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq9_HTML.gif, where Q n : Y Y n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq10_HTML.gif and P n : X X n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq11_HTML.gif are sequences of continuous linear projections. The projection scheme Γ = { X n , Y n , P n , Q n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq12_HTML.gif is then said to be admissible for maps from D X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq13_HTML.gif to Y. A map T : D X Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq14_HTML.gif is called approximation-proper (abbreviated A-proper) at a point y Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq15_HTML.gif with respect to an admissible scheme Γ if T n Q n T | D X n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq16_HTML.gif is continuous for each n N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq17_HTML.gif and whenever { x n j : x n j D X n j } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq18_HTML.gif is bounded with T n j x n j y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq19_HTML.gif, then there exists a subsequence { x n j k } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq20_HTML.gif such that x n j k x D https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq21_HTML.gif and T x = y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq22_HTML.gif. T is simply called A-proper if it is A-proper at all points of Y. L : dom L X Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq1_HTML.gif is a Fredholm operator of index zero if ImL is closed and dim  Ker L = codim  Im L < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq23_HTML.gif. As a consequence of this property, X and Y may be expressed as direct sums; X = X 0 X 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq24_HTML.gif, Y = Y 0 Y 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq25_HTML.gif with continuous linear projections P : X Ker L = X 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq26_HTML.gif and Q : Y Y 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq27_HTML.gif. The restriction of L to dom L X 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq28_HTML.gif, denoted L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq29_HTML.gif, is a bijection onto Im L = Y 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq30_HTML.gif with continuous inverse L 1 1 : Y 1 dom L X 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq31_HTML.gif. Since X 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq32_HTML.gif and Y 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq33_HTML.gif have the same finite dimension, there exists a continuous bijection J : Y 0 X 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq34_HTML.gif. Let H = L + J 1 P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq35_HTML.gif, then H : dom L X Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq36_HTML.gif is a linear bijection with bounded inverse. Let K be a cone in a Banach space X. Then K 1 = H ( K dom L ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq37_HTML.gif is a cone in Y. In [20], Petryshyn has shown that an admissible scheme Γ L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq38_HTML.gif can be constructed such that L is A-proper with respect to Γ L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq38_HTML.gif. The following properties of the fixed point index ind K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq39_HTML.gif and two lemmas can be found in [1].

Proposition 1.1 Let Ω X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq40_HTML.gif be open and bounded and Ω K = Ω K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq41_HTML.gif. Assume that Q n K 1 K 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq42_HTML.gif, P + J Q N + L 1 1 ( I Q ) N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq43_HTML.gif maps K to K, and L x N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq44_HTML.gif on Ω K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq45_HTML.gif.

(P1) (Existence property) If ind K ( [ L , N ] , Ω ) { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq46_HTML.gif, then there exists x Ω K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq47_HTML.gif such that L x = N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq48_HTML.gif.

(P2) (Normality) If x 0 Ω K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq49_HTML.gif, then ind K ( [ L , J 1 P + y ˆ 0 ] , Ω ) = { 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq50_HTML.gif, where y ˆ 0 = H x 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq51_HTML.gif and y ˆ 0 ( y ) = y 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq52_HTML.gif for every y H Ω K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq53_HTML.gif.

(P3) (Additivity) If L x N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq44_HTML.gif for x Ω ¯ K ( Ω 1 Ω 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq54_HTML.gif, where Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq55_HTML.gif and Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq56_HTML.gif are disjoint relatively open subsets of Ω K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq57_HTML.gif, then
ind K ( [ L , N ] , Ω ) ind K ( [ L , N ] , Ω 1 ) + ind K ( [ L , N ] , Ω 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equc_HTML.gif

with equality if either of indices on the right is a singleton.

(P4) (Homotopy invariance) If L N ( λ , x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq58_HTML.gif is an A-proper homotopy on Ω K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq57_HTML.gif for λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq59_HTML.gif and ( N ( λ , x ) + J 1 P ) H 1 : K 1 K 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq60_HTML.gif and θ ( L N ( λ , x ) ) ( dom L Ω K ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq61_HTML.gif for λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq59_HTML.gif, then ind K ( [ L , N ( λ , x ) ] , Ω ) = ind K 1 ( T λ , U ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq62_HTML.gif is independent of λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq59_HTML.gif, where T λ = ( N ( λ , x ) + J 1 P ) H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq63_HTML.gif.

Lemma 1.1 If L : dom L Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq64_HTML.gif is Fredholm of index zero, Ω is an open bounded set and Ω K dom L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq65_HTML.gif, θ Ω X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq66_HTML.gif. Let L λ N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq67_HTML.gif be A-proper for λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq59_HTML.gif. Assume that N is bounded and P + J Q N + L 1 1 ( I Q ) N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq43_HTML.gif maps K to K. If L x μ N x ( 1 μ ) J 1 P x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq68_HTML.gif on Ω K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq45_HTML.gif for μ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq69_HTML.gif, then
ind K ( [ L , N ] , Ω ) = { 1 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equd_HTML.gif
Lemma 1.2 If L : dom L Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq64_HTML.gif is Fredholm of index zero, Ω is an open bounded set and Ω K dom L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq65_HTML.gif. Let L λ N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq67_HTML.gif be A-proper for λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq59_HTML.gif. Assume that N is bounded and P + J Q N + L 1 1 ( I Q ) N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq43_HTML.gif maps K to K. If there exists e K 1 { θ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq70_HTML.gif such that
L x N x μ e , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Eque_HTML.gif
for every x Ω K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq71_HTML.gif and all μ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq72_HTML.gif, then
ind K ( [ L , N ] , Ω ) = { 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equf_HTML.gif

2 An abstract result

We will establish an abstract existence theorem concerning cone expansion and compression of order type, which reads as follows.

Theorem 2.1 If L : dom L Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq64_HTML.gif is Fredholm of index zero, let L λ N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq67_HTML.gif be A-proper for λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq73_HTML.gif. Assume that N is bounded and P + J Q N + L 1 1 ( I Q ) N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq43_HTML.gif maps K to K. Suppose further that Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq55_HTML.gif and Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq56_HTML.gif are two bounded open sets in X such that θ Ω 1 Ω ¯ 1 Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq74_HTML.gif, Ω 1 K dom L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq75_HTML.gif and Ω 2 K dom L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq76_HTML.gif. If one of the following two conditions is satisfied:

(C1) ( P + J Q N ) x + L 1 1 ( I Q ) N x x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq77_HTML.gif for all x Ω 1 K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq78_HTML.gif and ( P + J Q N ) x + L 1 1 ( I Q ) N x x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq79_HTML.gif for all x Ω 2 K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq80_HTML.gif;

(C2) ( P + J Q N ) x + L 1 1 ( I Q ) N x x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq79_HTML.gif for all x Ω 1 K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq81_HTML.gif and ( P + J Q N ) x + L 1 1 ( I Q ) N x x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq77_HTML.gif for all x Ω 2 K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq80_HTML.gif.

Then there exists x ( Ω ¯ 2 Ω 1 ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq82_HTML.gif such that L x = N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq48_HTML.gif.

Proof We assume that (C1) is satisfied. First we show that
L x μ N x ( 1 μ ) J 1 P x , for any  x Ω 1 K , μ [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equ1_HTML.gif
(2.1)
In fact, otherwise, there exist x 1 Ω 1 K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq83_HTML.gif and μ 1 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq84_HTML.gif such that
L x 1 = μ 1 N x 1 ( 1 μ 1 ) J 1 P x 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equg_HTML.gif
then we obtain
( L + J 1 P ) x 1 = μ 1 ( N + J 1 P ) x 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equh_HTML.gif
Therefore,
x 1 = μ 1 ( L + J 1 P ) 1 ( N + J 1 P ) x 1 = μ 1 [ ( P + J Q N ) x 1 + L 1 1 ( I Q ) N x 1 ] ( P + J Q N ) x 1 + L 1 1 ( I Q ) N x 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equi_HTML.gif
which contradicts condition (C1). From (2.1) and Lemma 1.1, we have
ind K ( [ L , N ] , Ω 1 ) = { 1 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equ2_HTML.gif
(2.2)
Choosing an arbitrary e K 1 { θ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq70_HTML.gif, next we prove that
L x N x μ e . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equ3_HTML.gif
(2.3)
In fact, otherwise, there exist x 2 Ω 2 K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq85_HTML.gif and μ 2 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq86_HTML.gif such that
L x 2 N x 2 = μ 2 e , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equj_HTML.gif
then we obtain
( L + J 1 P ) x 2 = ( N + J 1 P ) x 2 + μ 2 e 1 ( N + J 1 P ) x 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equk_HTML.gif
in which the partial order is induced by the cone K 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq87_HTML.gif in Y. So,
x 2 ( L + J 1 P ) 1 ( N + J 1 P ) x 2 = ( P + J Q N ) x 2 + L 1 1 ( I Q ) N x 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equl_HTML.gif
which is a contradiction to condition (C1). Hence (2.3) holds, and then by Lemma 1.2, we have
ind K ( [ L , N ] , Ω 2 ) = { 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equ4_HTML.gif
(2.4)
It follows therefore from (2.2), (2.4) and the additivity property (P3) of Proposition 1.1 that
ind K ( [ L , N ] , Ω 2 Ω 1 ) = ind K ( [ L , N ] , Ω 2 ) ind K ( [ L , N ] , Ω 1 ) = { 0 } { 1 } = { 1 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equ5_HTML.gif
(2.5)

Since the index is nonzero, the existence property (P1) of Proposition 1.1 implies that there exists x ( Ω ¯ 2 Ω 1 ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq82_HTML.gif such that L x = N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq48_HTML.gif.

Similarly, when (C2) is satisfied, instead of (2.2), (2.4) and (2.5), we have
ind K ( [ L , N ] , Ω 1 ) = { 0 } , ind K ( [ L , N ] , Ω 2 ) = { 1 } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equm_HTML.gif
and therefore
ind K ( [ L , N ] , Ω 2 Ω 1 ) = { 1 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equn_HTML.gif

Also, we can assert that there exists x ( Ω ¯ 2 Ω 1 ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq88_HTML.gif such that L x = N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq48_HTML.gif. □

3 Applications

3.1 First-order periodic boundary value problems

We consider the following first-order periodic boundary value problem:
{ x ( t ) = f ( t , x ( t ) ) , t ( 0 , 1 ) , x ( 0 ) = x ( 1 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equ6_HTML.gif
(3.1)

where f : [ 0 , 1 ] × [ 0 , + ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq89_HTML.gif is continuous and f ( 0 , x ) = f ( 1 , x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq90_HTML.gif for all x R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq91_HTML.gif.

Consider the Banach spaces X = Y = C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq92_HTML.gif endowed with the norm x = max t [ 0 , 1 ] | x ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq93_HTML.gif. Define the cone K in X by
K = { x X : x ( t ) 0 , t [ 0 , 1 ] } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equo_HTML.gif
Let L be the linear operator from dom L X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq94_HTML.gif to Y with
dom L = { x X : x C [ 0 , 1 ] , x ( 0 ) = x ( 1 ) } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equp_HTML.gif
and
L x ( t ) = x ( t ) , x dom L , t [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equq_HTML.gif
Let us define N : X Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq2_HTML.gif by
N x ( t ) = f ( t , x ( t ) ) , t [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equr_HTML.gif
Then (3.1) is equivalent to the equation
L x = N x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equs_HTML.gif
It is obvious that L is a Fredholm operator of index zero with
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equt_HTML.gif
Next we define the projections P : X X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq95_HTML.gif, Q : Y Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq96_HTML.gif by
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equu_HTML.gif
and the isomorphism J : Im Q Im P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq97_HTML.gif as J y = y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq98_HTML.gif. Note that for y Im L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq99_HTML.gif, the inverse operator
L 1 1 : Im L dom L Ker P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equv_HTML.gif
of
L | dom L Ker P : dom L Ker P Im L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equw_HTML.gif
is given by
( L 1 1 y ) ( t ) = 0 1 K ( t , s ) y ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equx_HTML.gif
where
K ( t , s ) = { s + 1 , 0 s < t 1 , s , 0 t s 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equy_HTML.gif
Set
G ( t , s ) = 1 + K ( t , s ) 0 1 K ( t , s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equz_HTML.gif
We can verify that
G ( t , s ) = { 3 2 ( t s ) , 0 s < t 1 , 1 2 + ( s t ) , 0 t s 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equaa_HTML.gif
and
1 2 G ( t , s ) 3 2 , t , s [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equab_HTML.gif

To state the existence result, we introduce two conditions:

(H1) f ( t , b ) < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq100_HTML.gif for all t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq101_HTML.gif,

(H2) f ( t , x ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq102_HTML.gif for all ( t , x ) [ 0 , 1 ] × [ 0 , a ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq103_HTML.gif.

Theorem 3.1 Assume that there exist two positive numbers 0 < a < b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq104_HTML.gif such that (H1), (H2) and

(H3) f ( t , x ) 2 3 x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq105_HTML.gif for all ( t , x ) [ 0 , 1 ] × [ 0 , b ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq106_HTML.gif

hold. Then (3.1) has at least one positive periodic solution x K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq107_HTML.gif with a x b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq108_HTML.gif.

Proof First, we note that L, as defined, is Fredholm of index zero, L 1 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq109_HTML.gif is compact by the Arzela-Ascoli theorem and thus L λ N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq67_HTML.gif is A-proper for λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq59_HTML.gif by [[20], Lemma 2(a)].

For each x K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq110_HTML.gif, then by condition (H3),
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equac_HTML.gif

Thus ( P + J Q N + L 1 1 ( I Q ) N ) ( K ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq111_HTML.gif.

Let
Ω 1 = { x X : x < a } , Ω 2 = { x X : x < b } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equad_HTML.gif
Clearly, Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq55_HTML.gif and Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq56_HTML.gif are bounded open sets and
θ Ω 1 Ω ¯ 1 Ω 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equae_HTML.gif
We now show that
( P + J Q N ) x + L 1 1 ( I Q ) N x x for any  x Ω 2 K . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equ7_HTML.gif
(3.2)
In fact, if there exists x 3 Ω 2 K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq112_HTML.gif such that
( P + J Q N ) x 3 + L 1 1 ( I Q ) N x 3 x 3 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equaf_HTML.gif
Then
x 3 ( t ) f ( t , x 3 ( t ) ) , t [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equag_HTML.gif
Let t 1 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq113_HTML.gif be such that x 3 ( t 1 ) = b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq114_HTML.gif. Clearly, the function x 3 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq115_HTML.gif attains a maximum on [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq116_HTML.gif at t = t 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq117_HTML.gif. Therefore 2 x 3 ( t 1 ) x 3 ( t 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq118_HTML.gif. As a consequence,
0 = 2 b x 3 ( t 1 ) 2 b f ( t 1 , x 3 ( t 1 ) ) = 2 b f ( t 1 , b ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equah_HTML.gif

which is a contradiction to (H1). Therefore (3.2) holds.

On the other hand, we claim that
( P + J Q N ) x + L 1 1 ( I Q ) N x x for any  x Ω 1 K . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equ8_HTML.gif
(3.3)
In fact, if not, there exists x 4 Ω 1 K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq119_HTML.gif such that
( P + J Q N ) x 4 + L 1 1 ( I Q ) N x 4 x 4 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equai_HTML.gif
For any x 4 Ω 1 K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq119_HTML.gif, we have x 4 = a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq120_HTML.gif, then 0 x 4 ( t ) a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq121_HTML.gif for t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq101_HTML.gif. By condition (H2), we have
x 4 ( t ) ( P + J Q N ) x 4 ( t ) + L 1 1 ( I Q ) N x 4 ( t ) = 0 1 x 4 ( s ) d s + 0 1 G ( t , s ) f ( s , x 4 ( s ) ) d s > 0 1 x 4 ( s ) d s , for any  t [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equaj_HTML.gif

which is a contradiction. As a result, (3.3) is verified.

It follows from (3.2), (3.3) and Theorem 2.1 that there exists x K ( Ω ¯ 2 Ω 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq122_HTML.gif such that L x = N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq123_HTML.gif with a x b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq108_HTML.gif. □

Remark 3.1 In [18], the following condition is required instead of (H2):

(H) there exist a ( 0 , b ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq124_HTML.gif, t 0 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq125_HTML.gif, r ( 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq126_HTML.gif, and continuous functions g : [ 0 , 1 ] [ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq127_HTML.gif, h : ( 0 , a ] [ 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq128_HTML.gif such that f ( t , x ) g ( t ) h ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq129_HTML.gif for all t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq101_HTML.gif and x ( 0 , a ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq130_HTML.gif, h ( x ) / x r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq131_HTML.gif is nonincreasing on ( 0 , a ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq132_HTML.gif with
h ( a ) 2 r 1 0 1 G ( t 0 , s ) g ( s ) d s a . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equak_HTML.gif

Obviously, our condition (H2) is much weaker and less strict compared with (H). Moreover, (H2) is easier to check than (H). So, our result generalizes and improves [[18], Theorem 5].

Remark 3.2 From the proof of Theorem 3.1, we can see that condition (H2) can be replaced by one of the following two relatively weaker conditions:

( H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq133_HTML.gif) f ( t , x ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq134_HTML.gif for all ( t , x ) [ 0 , 1 ] × [ 0 , a ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq135_HTML.gif and f ( t , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq136_HTML.gif is positive for almost everywhere on [ 0 , a ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq137_HTML.gif.

( H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq138_HTML.gif) lim x 0 + min t [ 0 , 1 ] f ( t , x ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq139_HTML.gif.

Remark 3.3 Finally in this section, we note that conditions (H1) and (H2) can be replaced by the following asymptotic conditions:

( H 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq140_HTML.gif) lim x + f ( t , x ) x < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq141_HTML.gif uniformly for t;

( H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq142_HTML.gif) lim x 0 + f ( t , x ) x > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq143_HTML.gif uniformly for t.

Example 3.1 Let the nonlinearity in (3.1) be
f ( t , x ) = c ( t ) x α + μ d ( t ) x β k x , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equal_HTML.gif

where 0 < α < 1 < β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq144_HTML.gif, c ( t ) , d ( t ) C [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq145_HTML.gif are positive 1-periodic functions, k ( 0 , 2 / 3 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq146_HTML.gif and μ > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq147_HTML.gif is a positive parameter. Then (3.1) has at least one positive 1-periodic solution for each 0 < μ < μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq148_HTML.gif, here μ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq149_HTML.gif is some positive constant.

Proof We will apply Theorem 3.1 with f ( t , x ) = c ( t ) x α + μ d ( t ) x β k x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq150_HTML.gif. Since k ( 0 , 2 / 3 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq146_HTML.gif, it is easy to see that (H3) holds. Set
T ( x ) = k x c x α d x β , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equam_HTML.gif
where
c = max t c ( t ) , d = max t d ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equan_HTML.gif
Since 0 < α < 1 < β https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq144_HTML.gif, we have
T ( 0 + ) = , T ( + ) = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equao_HTML.gif
One may easily see that there exists b > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq151_HTML.gif such that
T ( b ) = k b c b α d b β = sup x > 0 T ( x ) > 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equap_HTML.gif
Let
μ = k b c b α d b β . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equaq_HTML.gif
Then, for each μ ( 0 , μ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq152_HTML.gif, we have
f ( t , b ) = c ( t ) b α + μ d ( t ) b β k b < c b α + μ d b β k b = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equar_HTML.gif

which implies that (H1) holds.

On the other hand, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equas_HTML.gif

which implies that ( H 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq142_HTML.gif) holds. Now we have the desired result. □

3.2 Second-order periodic boundary value problems

Let f : [ 0 , 1 ] × [ 0 , + ) R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq89_HTML.gif be continuous and f ( 0 , x ) = f ( 1 , x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq90_HTML.gif for all x R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq91_HTML.gif. We will discuss the existence of positive solutions of the second-order periodic boundary value problem
{ x ( t ) = f ( t , x ) , t ( 0 , 1 ) , x ( 0 ) = x ( 1 ) , x ( 0 ) = x ( 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equ9_HTML.gif
(3.4)

Since some parts of the proof are in the same line as that of Theorem 3.1, we will outline the proof with the emphasis on the difference.

Let X, Y be Banach spaces and the cone K be as in Section 3.1. In this case, we may define
dom L = { x X : x C [ 0 , 1 ] , x ( 0 ) = x ( 1 ) , x ( 0 ) = x ( 1 ) } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equat_HTML.gif
and let the linear operator L : dom L Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq64_HTML.gif be defined by
L x = x , for  x dom L . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equau_HTML.gif
Then L is Fredholm of index zero,
Ker L = { x dom L : x ( t ) constants } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equav_HTML.gif
and
Im L = { y Y : 0 1 y ( s ) d s = 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equaw_HTML.gif
Define N : X Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq153_HTML.gif by
N x ( t ) = f ( t , x ( t ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equax_HTML.gif
Thus it is clear that (3.4) is equivalent to
L x = N x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equay_HTML.gif
We use the same projections P, Q as in Section 3.1 and define the isomorphism J : Im Q Im P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq97_HTML.gif as
J y = β y , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equaz_HTML.gif
where β = 1 6 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq154_HTML.gif. It is easy to verify that the inverse operator L 1 1 : Im L dom L Ker P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq155_HTML.gif of L | dom L Ker P : dom L Ker P Im L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq156_HTML.gif is
( L 1 1 y ) ( t ) = 0 1 Λ ( t , s ) y ( s ) d s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equba_HTML.gif
where
Λ ( t , s ) = { s 2 ( 1 2 t + s ) , 0 s < t 1 , 1 2 ( 1 s ) ( 2 t s ) , 0 t s 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equbb_HTML.gif
Set
H ( t , s ) = 1 6 + Λ ( t , s ) 0 1 Λ ( t , s ) d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equbc_HTML.gif
We can verify that
H ( t , s ) = { 1 4 + s 2 ( 1 2 t + s ) + t 2 2 t 2 , 0 s < t 1 , 1 4 + 1 2 ( 1 s ) ( 2 t s ) + t 2 2 + t 2 , 0 t s 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equbd_HTML.gif
and
1 8 H ( t , s ) 1 4 , t , s [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Eqube_HTML.gif

Theorem 3.2 Assume that there exist two positive numbers 0 < a < b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq104_HTML.gif such that (H1), (H2) and

(H4) f ( t , x ) 4 x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq157_HTML.gif for all ( t , x ) [ 0 , 1 ] × [ 0 , b ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq158_HTML.gif

hold. Then (3.4) has at least one positive periodic solution x K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq107_HTML.gif with a x b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq108_HTML.gif.

Proof It is again easy to show that L λ N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq67_HTML.gif is A-proper for λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq59_HTML.gif by [[20], Lemma 2(a)].

For each x K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq110_HTML.gif, then by condition (H4),
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equbf_HTML.gif

Thus ( P + J Q N + L 1 1 ( I Q ) N ) ( K ) K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq111_HTML.gif.

Let
Ω 3 = { x X : x < a } , Ω 4 = { x X : x < b } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equbg_HTML.gif
Clearly, Ω 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq159_HTML.gif and Ω 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq160_HTML.gif are bounded and open sets and
θ Ω 3 Ω ¯ 3 Ω 4 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equbh_HTML.gif
Next, we show that
( P + J Q N ) x + L 1 1 ( I Q ) N x x , for any  x Ω 4 K . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equ10_HTML.gif
(3.5)
On the contrary, suppose that there exists x 5 Ω 4 K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq161_HTML.gif such that
( P + J Q N ) x 5 + L 1 1 ( I Q ) N x 5 x 5 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equbi_HTML.gif
Then
x 5 ( t ) f ( t , x 5 ( t ) ) , t [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equbj_HTML.gif
Let t 2 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq162_HTML.gif such that x 5 ( t 2 ) = max t [ 0 , 1 ] x 5 ( t ) = b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq163_HTML.gif. Using the boundary conditions, we have t 2 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq164_HTML.gif. In this case, x 5 ( t 2 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq165_HTML.gif, x 5 ( t 2 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq166_HTML.gif. This gives
0 x 5 ( t 2 ) f ( t 2 , x 5 ( t 2 ) ) = f ( t 2 , b ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equbk_HTML.gif

which is a contradiction to condition (H1). Therefore (3.5) holds.

Finally, similar to the proof of (3.3), it follows from condition (H2) that
( P + J Q N ) x + L 1 1 ( I Q ) N x x , for any  x Ω 3 K . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_Equbl_HTML.gif

Consequently all conditions of Theorem 2.1 are satisfied. Therefore, there exists x K ( Ω ¯ 4 Ω 3 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq167_HTML.gif such that L x = N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq123_HTML.gif with x K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq107_HTML.gif and a x b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-37/MediaObjects/13661_2012_Article_291_IEq108_HTML.gif and the assertion follows. □

Declarations

Acknowledgements

JC was supported by the National Natural Science Foundation of China (Grant No. 11171090, No. 11271333 and No. 11271078), the Program for New Century Excellent Talents in University (Grant No. NCET-10-0325), China Postdoctoral Science Foundation funded project (Grant No. 2012T50431). FW was supported by the National Natural Science Foundation of China (Grant No. 10971179) and the Natural Science Foundation of Changzhou University (Grant No. JS201008).

Authors’ Affiliations

(1)
Department of Mathematics, Hohai University
(2)
School of Mathematics and Physics, Changzhou University

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© Chu and Wang; licensee Springer. 2013

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