Open Access

Existence of positive solutions for nonlinear m-point boundary value problems on time scales

Boundary Value Problems20122012:4

DOI: 10.1186/1687-2770-2012-4

Received: 4 May 2011

Accepted: 17 January 2012

Published: 17 January 2012

Abstract

In this article, we study the following m-point boundary value problem on time scales,

( ϕ p ( u Δ ( t ) ) ) + h ( t ) f ( t , u ( t ) ) = 0 , t ( 0 , T ) T , u ( 0 ) - δ u Δ ( 0 ) = i = 1 m - 2 β i u Δ ( ξ i ) , u Δ ( T ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equa_HTML.gif

where T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq1_HTML.gif is a time scale such that 0 , T T , δ , β i > 0 , i = 1 , , m - 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq2_HTML.gif, ϕ p (s) = |s| p-2 s,p > 1,h C ld ((0, T), (0, +∞)), and f C([0,+∞), (0,+∞)), 0 < ξ 1 < ξ 2 < < ξ m - 2 < T T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq3_HTML.gif. By using several well-known fixed point theorems in a cone, the existence of at least one, two, or three positive solutions are obtained. Examples are also given in this article.

AMS Subject Classification: 34B10; 34B18; 39A10.

Keywords

positive solutions cone multi-point boundary value problem time scale

1 Introduction

The study of dynamic equations on time scales goes back to its founder Hilger [1], and is a new area of still theoretical exploration in mathematics. Motivating the subject is the notion that dynamic equations on time scales can build bridges between continuous and discrete mathematics. Further, the study of time scales has led to several important applications, e.g., in the study of insect population models, neural networks, heat transfer, epidemic models, etc. [2].

Multipoint boundary value problems of ordinary differential equations (BVPs for short) arise in a variety of different areas of applied mathematics and physics. For example, the vibrations of a guy wire of a uniform cross section and composed of N parts of different densities can be set up as a multi-point boundary value problem [3]. Many problems in the theory of elastic stability can be handled by the method of multi-point problems [4]. Small size bridges are often designed with two supported points, which leads into a standard two-point boundary value condition and large size bridges are sometimes contrived with multi-point supports, which corresponds to a multi-point boundary value condition [5]. The study of multi-point BVPs for linear second-order ordinary differential equations was initiated by Il'in and Moiseev [6]. Since then many authors have studied more general nonlinear multi-point BVPs, and the multi-point BVP on time scales can be seen as a generalization of that in ordinary differential equations.

Recently, the existence and multiplicity of positive solutions for nonlinear differential equations on time scales have been studied by some authors [711], and there has been some merging of existence of positive solutions to BVPs with p-Laplacian on time scales [1219].

He [20] studied
( ϕ p ( u Δ ( t ) ) ) + a ( t ) f ( t ) = 0 , t ( 0 , T ) T , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ1_HTML.gif
(1.1)
subject to one of the following boundary conditions
u ( 0 ) - B 0 ( u Δ ( η ) ) = 0 , u Δ ( T ) = 0 , u Δ ( 0 ) = 0 , u ( T ) - B 1 ( u Δ ( η ) ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ2_HTML.gif
(1.2)

where η ( 0 , T ) T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq4_HTML.gif. By using a double fixed-point theorem, the authors get the existence of at least two positive solutions to BVP (1.1) and (1.2).

Anderson [21] studied
- u Δ ( t ) = η a ( t ) f ( u ( t ) ) , t ( t 1 , t n ) T , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ3_HTML.gif
(1.3)
subject to one of the following boundary conditions
u ( t 1 ) = i = 2 n - 1 α i u ( t i ) , u Δ ( t n ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ4_HTML.gif
(1.4)
u Δ ( t 1 ) = 0 , u ( t n ) = i = 2 n - 1 α i u ( t i ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ5_HTML.gif
(1.5)

by using a functional-type cone expansion-compression fixed-point theorem, the author gets the existence of at least one positive solution to BVP (1.3), (1.4) and BVP (1.3), (1.5).

However, to the best of the authors' knowledge, up to now, there are few articles concerned with the existence of m-point boundary value problem with p-Laplacian on time scales. So, in this article, we try to fill this gap. Motivated by the article mentioned above, in this article, we consider the following m-point BVP with one-dimensional p-Laplacian,
( ϕ p ( u Δ ( t ) ) ) + h ( t ) f ( t , u ( t ) ) = 0 , t ( 0 , T ) T , u ( 0 ) - δ u Δ ( 0 ) = i = 1 m - 2 β i u Δ ( ξ i ) , u Δ ( T ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ6_HTML.gif
(1.6)

where ϕ p (s) = |s| p-2 s,p > 1,h C ld ((0,T), (0, +∞)), 0 < ξ 1 < ξ 2 < < ξ m - 2 < T T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq5_HTML.gif. δ, β i > 0, i = 1,..., m - 2.

We will assume throughout

(S1) h C ld ((0, T), [0, ∞)) such that 0 T h ( s ) s < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq6_HTML.gif;

(S2) f C([0, ∞), (0, ∞)), f 0 on f 0  on  [ 0 , T ] T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq7_HTML.gif;

(S3) By ϕ q we denote the inverse to ϕ p , where 1 p + 1 q = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq8_HTML.gif;

(S4) By t [a, b] we mean that t [ a , b ] T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq9_HTML.gif, where 0 ≤ abT.

2 Preliminaries

In this section, we will give some background materials on time scales.

Definition 2.1. [7, 22] For t < sup T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq10_HTML.gif and t > inf T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq11_HTML.gif, define the forward jump operator σ and the backward jump operator ρ, respectively,
σ ( t ) = inf { τ T | τ > t } T , ρ ( r ) = sup { τ T | τ < r } T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equb_HTML.gif
for all r , t T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq12_HTML.gif. If σ(t) > t, t is said to be right scattered, and if ρ(r) < r, r is said to be left scattered. If σ(t) = t, t is said to be right dense, and if ρ(r) = r, r is said to be left dense. If T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq1_HTML.gif has a right scattered minimum m, define T κ = T - { m } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq13_HTML.gif; Otherwise set T κ = T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq14_HTML.gif. The backward graininess μ b : T κ 0 + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq15_HTML.gif is defined by
μ b ( t ) = t - ρ ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equc_HTML.gif
If T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq1_HTML.gif has a left scattered maximum M, define T κ = T - { M } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq16_HTML.gif; Otherwise set T κ = T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq17_HTML.gif. The forward graininess μ f : T κ 0 + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq18_HTML.gif is defined by
μ f ( t ) = σ ( t ) - t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equd_HTML.gif
Definition 2.2. [7, 22] For x : T R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq19_HTML.gif and t T κ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq20_HTML.gif, we define the "Δ" derivative of x(t), xΔ(t), to be the number (when it exists), with the property that, for any ε > 0, there is neighborhood U of t such that
[ x ( σ ( t ) ) - x ( s ) ] - x Δ ( t ) [ σ ( t ) - s ] < ε | σ ( t ) - s | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Eque_HTML.gif
for all s U. For x : T R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq19_HTML.gif and t T κ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq21_HTML.gif, we define the "" derivative of x(t),xΔ (t), to be the number(when it exists), with the property that, for any ε > 0, there is a neighborhood V of t such that
[ x ( ρ ( t ) ) - x ( s ) ] - x ( t ) [ ρ ( t ) - s ] < ε | ρ ( t ) - s | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equf_HTML.gif

for all s V.

Definition 2.3. [22] If FΔ (t) = f(t), then we define the "Δ" integral by
a t f ( s ) Δ s = F ( t ) - F ( a ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equg_HTML.gif
If F (t) = f(t), then we define the "" integral by
a t f ( s ) s = F ( t ) - F ( a ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equh_HTML.gif

Lemma 2.1. [23]The following formulas hold:

(i) ( a t f ( t ) Δ s ) Δ = f ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq22_HTML.gif,

(ii) ( a t f ( t ) Δ s ) = f ( ρ ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq23_HTML.gif,

(iii) ( a t f ( t ) s ) Δ = f ( σ ( t ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq24_HTML.gif,

(iv) ( a t f ( t ) s ) = f ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq25_HTML.gif.

Lemma 2.2. [7, Theorem 1.75 in p. 28] If f C rd and t T κ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq20_HTML.gif, then
t σ ( t ) f ( τ ) Δ τ = μ f ( t ) f ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equi_HTML.gif

According to [23, Theorem 1.30 in p. 9], we have the following lemma, which can be proved easily. Here, we omit it.

Lemma 2.3. Let a , b T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq26_HTML.gifand f C ld .

(i) If T = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq27_HTML.gif, then
a b f ( t ) t = a b f ( t ) d t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equj_HTML.gif

where the integral on the right is the usual Riemann integral from calculus.

(ii) If [a, b] consists of only isolated points, then
a b f ( t ) t = t ( a , b ] μ b ( t ) f ( t ) , i f a < b , 0 , i f a = b , - t ( a , b ] μ b ( t ) f ( t ) , i f a > b . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equk_HTML.gif
(iii) If T = h = { h k : k } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq28_HTML.gif, where h > 0, then
a b f ( t ) t = k = a h + 1 b h f ( k h ) h , i f a < b , 0 , i f a = b , - k = b h + 1 a h f ( k h ) h , i f a > b . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equl_HTML.gif
(iv) If T = https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq29_HTML.gif, then
a b f ( t ) t = t = a + 1 b f ( t ) , i f a < b , 0 , i f a = b , - t = b + 1 a f ( t ) , i f a > b . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equm_HTML.gif

In what follows, we list the fixed point theorems that will be used in this article.

Theorem 2.4. [24]Let E be a Banach space and P E be a cone. Suppose Ω1, Ω2 E open and bounded, 0 Ω 1 Ω ¯ 1 Ω 2 Ω ¯ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq30_HTML.gif. Assume A : ( Ω ¯ 2 \ Ω 1 ) P P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq31_HTML.gifis completely continuous. If one of the following conditions holds

(i) Axx, x ∂Ω1P, Axx, x ∂Ω2P;

(ii) Axx, x ∂Ω1P, Axx, x ∂Ω2P.

Then, A has a fixed point in ( Ω ¯ 2 \ Ω 1 ) P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq32_HTML.gif.

Theorem 2.5. [25]Let P be a cone in the real Banach space E. Set
P ( γ , r ) = { u P , γ ( u ) < r } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equn_HTML.gif
If α and γ are increasing, nonnegative continuous functionals on P, let θ be a nonnegative continuous functional on P with θ(0) = 0 such that for some positive constants r, M,
γ ( u ) θ ( u ) α ( u ) a n d u M γ ( u ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equo_HTML.gif
for all u P ( γ , r ) ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq33_HTML.gif. Further, suppose there exists positive numbers a < b < r such that
θ ( λ u ) λ θ ( u ) f o r a l l 0 λ 1 , u P ( θ , b ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equp_HTML.gif

If A : P ( γ , r ) ¯ P https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq34_HTML.gif is completely continuous operator satisfying

(i) γ(Au) > r for all u P(γ, r);

(ii) θ(Au) < b for all u ∂P(θ, r);

(iii) P ( α , b ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq35_HTML.gifand α(Au) > a for all u ∂P(α, a).

Then, A has at least two fixed points u 1 and u 2 such that
a < α ( u 1 ) , w i t h θ ( u 1 ) < b , a n d b < θ ( u 2 ) , w i t h γ ( u 1 ) < r , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equq_HTML.gif

Let a, b, c be constants, P r = {u P : u < r}, P(ψ, b, d) = {u P : aψ(u), ub}.

Theorem 2.6. [26]Let A : P ¯ c P ¯ c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq36_HTML.gifbe a completely continuous map and ψ be a nonnegative continuous concave functional on P such that for u P ¯ c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq37_HTML.gif, there holds ψ(u) ≤ u. Suppose there exist a, b, d with 0 < a < b < dc such that

(i) { u P ( ψ , b , d ) : ψ ( u ) > b } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq38_HTML.gifand ψ(Au) > b for all u P(ψ, b, d);

(ii) Au < a for all u P ¯ a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq39_HTML.gif;

(iii) ψ(Au) > b for all u P(ψ, b, d) with Au > d.

Then, A has at least three fixed points u1,u2, and u3satisfying
u 1 < a , b < α ( u 2 ) , u 3 > a , a n d u 3 < b . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equr_HTML.gif
Let the Banach space E = C ld [0, T] be endowed with the norm u = supt [0,T]u(t), and cone P E is defined as
P = { u E , u ( t ) 0  for  t [ 0 , T ]  and  u Δ ( t ) 0  for  t ( 0 , T ) , u Δ ( T ) = 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equs_HTML.gif
It is obvious that u = u(T) for u P. Define A : PE as
( A u ) ( t ) = 0 t ϕ q s T h ( τ ) f ( τ , u ( τ ) ) τ Δ s + δ ϕ q 0 T h ( s ) f ( s , u ( s ) ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) f ( s , u ( s ) ) s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equt_HTML.gif

for t [0, T].

In what follows, we give the main lemmas which are important for getting the main results.

Lemma 2.7. A : PP is completely continuous.

Proof. First, we try to prove that A : PP.
( A u ) Δ ( t ) = ϕ q t T h ( s ) f ( s , u ( s ) ) s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equu_HTML.gif

Thus, (Au)Δ (T) = 0 and by Lemma 2.1 we have (Au)Δ (t) = -h(t)f(t, u(t)) ≤ 0 for t (0, T). Consequently, A : PP.

By standard argument we can prove that A is completely continuous. For more details, see [27]. The proof is complete.

Lemma 2.8. For u P, there holds u ( t ) t T u f o r t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq40_HTML.giffor t [0,T].

Proof. For u P, we have uΔ (t) ≤ 0, it follows that uΔ (t) is non-increasing. Therefore, for 0 < t < T,
u ( t ) - u ( 0 ) = 0 t u Δ ( s ) Δ s t u Δ ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ7_HTML.gif
(2.1)
and
u ( T ) - u ( t ) = t T u Δ ( s ) Δ s ( T - t ) u Δ ( t ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ8_HTML.gif
(2.2)
thus
u ( T ) - u ( 0 ) T u Δ ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ9_HTML.gif
(2.3)
Combining (2.1) and (2.3) we have
T ( u ( t ) - u ( 0 ) ) T t u Δ ( t ) t ( u ( T ) - u ( 0 ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equv_HTML.gif
as u(0) ≥ 0, it is immediate that
u ( t ) t u ( T ) + ( T - t ) u ( 0 ) T t T u ( T ) = t T u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equw_HTML.gif

The proof is complete.

3 Existence of at least one positive solution

First, we give some notations. Set
Λ = δ + i = 1 m - 2 β i + T ϕ q 0 T h ( s ) s , B = ξ 1 T δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 T ϕ q s T h ( τ ) τ Δ s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equx_HTML.gif

Theorem 3.1. Assume in addition to (S 1) and (S 2), the following conditions are satisfied, there exists 0 < r < ξ 1 ρ T < ρ < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq41_HTML.gif such that

(H1) f ( t , u ) ϕ p ( u Λ ) , f o r t [ 0 , T ] , u [ 0 , r ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq42_HTML.gif;

(H2) f ( t , u ) ϕ p u B , f o r t [ ξ 1 , T ] , u [ ξ 1 ρ T , ρ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq43_HTML.gif.

Then, BVP (1.6) has at least one positive solution.

Proof. Cone P is defined as above. By Lemma 2.7 we know that A : PP is completely continuous. Set Ω r = {u E, u < r}. In view of (H1), for u Ω r P,
A u = ( A u ) ( T ) = δ ϕ q ( 0 T h ( s ) f ( s , u ( s ) ) s ) + i = 1 m 2 β i ϕ q ( ξ i T h ( s ) f ( s , u ( s ) ) s ) + 0 T ϕ q ( s T h ( τ ) f ( τ , u ( τ ) ) τ ) Δ s ( δ + i = 1 m 2 β i + T ) ϕ q ( 0 T ϕ p ( u ( s ) Λ ) h ( s ) s ) u Λ ( δ + i = 1 m 2 β i + T ) ϕ q ( 0 T h ( s ) s ) u , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equy_HTML.gif

which means that for u ∂Ω r P, ||Au|| ≤ ||u||.

On the other hand, for u P, in view of Lemma 2.8, there holds u ( t ) ξ 1 T u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq44_HTML.gif, for t 1, T]. Denote Ω ρ = {u E, u < ρ}. Then for u ∂Ω ρ P, considering (H2), we have
A u = ( A u ) ( T ) = δ ϕ q 0 T h ( s ) f ( s , u ( s ) ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) f ( s , u ( s ) ) s + 0 T ϕ q s T h ( τ ) f ( τ , u ( τ ) ) τ Δ s δ ϕ q 0 T ϕ p u ( s ) B h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T ϕ p u ( s ) B h ( s ) s + 0 T ϕ q s T ϕ p u ( τ ) B h ( τ ) τ Δ s ξ 1 u T B δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 T ϕ q s T h ( τ ) τ Δ s = u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equz_HTML.gif

which implies that for u Ω ρ P, Auu Therefore, the immediate result of Theorem 2.4 is that A has at least one fixed point u ρ r ) ∩ P. Also, it is obvious that the fixed point of A in cone P is equivalent to the positive solution of BVP (1.6), this yields that BVP (1.6) has at least one positive solution u satisfies ruρ. The proof is complete.

Here is an example.

Example 3.2. Let T = 1 , 1 = k = 0 [ 2 k , 2 k + 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq45_HTML.gif. Consider the following four point BVP on time scale 1 , 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq46_HTML.gif.
x Δ ( t ) + f ( t , u ( t ) ) = 0 , t [ 0 , T ] T , x ( 0 ) - 2 x Δ ( 0 ) = x Δ ( 2 ) + x Δ ( 3 ) , x Δ ( 4 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ10_HTML.gif
(3.1)
where
f ( t , u ) = t u 128 , 0 u 100 , 39 t 512 ( u - 100 ) + 25 t 32 , 100 u 500 , t u 16 , u 500 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equaa_HTML.gif
and h(t) = 1, T = 4, ξ1 = 2, ξ2 = 3, δ = 2, β1 = β2 = 1,p = q = 2. In what follows, we try to calculate Λ, B. By Lemmas 2.2 and 2.3, we have
Λ = δ + i = 1 m - 2 β i + T ϕ q 0 T h ( s ) s = ( 2 + 1 + 1 + 4 ) 0 4 s = 8 × 0 1 d s + 2 3 d s + 1 2 s + 3 4 s = 8 × 0 1 d s + 2 3 d s + ν ( 2 ) × 1 + ν ( 4 ) × 1 = 8 × ( 1 + 1 + 1 + 1 ) = 32 . B = ξ 1 T δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 T ϕ q s T h ( τ ) τ Δ s = 2 4 2 0 4 s + 2 4 s + 3 4 s + 0 4 s 4 τ Δ s = 1 2 8 + 2 + 1 + 0 1 s 4 τ Δ s + 1 2 2 4 τ Δ s + 2 3 s 4 τ Δ s + 3 4 s 4 τ Δ s = 1 2 11 + 0 1 s 4 τ Δ s + 1 2 s 4 τ Δ s + 2 3 s 4 τ Δ s + 3 4 s 4 τ Δ s , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equab_HTML.gif
where
0 1 s 4 τ Δ s = 0 1 s 1 τ + 1 4 τ Δ s = 0 1 s 1 d τ d s + 0 1 1 2 τ + 2 3 τ + 3 4 τ Δ s = 0 1 s 1 d τ + 2 3 d τ d s + 0 1 1 2 τ + 3 4 τ Δ s = 1 2 + 1 + 1 + 1 = 7 2 , 2 3 s 4 τ Δ s = 2 3 s 3 τ + 3 4 τ Δ s = 2 3 s 3 d τ d s + 2 3 3 4 τ Δ s = 1 2 + 1 = 3 2 , 1 2 s 4 τ Δ s = σ ( 1 ) × 1 4 τ = 3 , 3 4 s 4 τ Δ s = σ ( 3 ) × 3 4 τ = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equac_HTML.gif

Thus, B = 1 2 11 + 7 2 + 3 2 + 3 + 1 = 10 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq47_HTML.gif. Let r = 100 < 2 4 ρ < ρ = 1000 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq48_HTML.gif. Then, we have

(i) f ( t , u ) 4 u 128 = u 32 = ϕ p ( u Λ ) , f o r t [ 0 , 4 ] , u [ 0 , 100 ] ; https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq49_HTML.gif

(ii) f ( t , u ) 2 u 16 = u 8 > ϕ p ( u B ) , f o r t [ 2 , 4 ] , u [ 500 , 1000 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq50_HTML.gif

Thus, if all the conditions in Theorem 3.1 satisfied, then BVP (3.1) has at least one positive solution lies between 100 and 1000.

4 Existence of at least two positive solutions

In this section, we will apply fixed point Theorem 2.5 to prove the existence of at least two positive solutions to the nonlinear BVP (1.6).

Fix η T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq51_HTML.gif such that
0 < ξ m - 2 η < T , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equad_HTML.gif
and define the increasing, nonnegative, continuous functionals γ, θ,α on P by
γ ( u ) = min t [ ξ 1 , η ] u ( t ) = u ( ξ 1 ) , θ ( u ) = max t [ 0 , ξ m - 2 ] u ( t ) = u ( ξ m - 2 ) , α ( u ) = min t [ η , T ] u ( t ) = u ( η ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equae_HTML.gif
We can see that, for u P, there holds
γ ( u ) θ ( u ) α ( u ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equaf_HTML.gif
In addition, Lemma 2.8 implies that γ ( u ) = u ( ξ 1 ) ξ 1 T | | u | | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq52_HTML.gif which means that
| | u | | T ξ 1 γ ( u ) for u P . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equag_HTML.gif
We also see that
θ ( λ u ) = λ θ ( u ) for λ [ 0 , 1 ] , u P ( θ , b ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equah_HTML.gif
For convenience, we give some notations,
K = δ + i = 1 m - 2 β i + ξ m - 2 ϕ q 0 T h ( s ) s , M = δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 ξ 1 ϕ q s T h ( τ ) τ Δ s , L = δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 η ϕ q s T h ( τ ) τ Δ s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equai_HTML.gif

Theorem 4.1. Assume in addition to (S 1), (S 2) there exist positive constants a < T η a < b < T ξ m - 2 b < c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq53_HTML.gifsuch that the following conditions hold

(H3) f(t, u) > ϕ p (c/M) for t [ξ1,T] u [c,Tc/ξ1];

(H4) f(t, u) < ϕ p (b/K) for t [0,ξm-2], u [b,Tb/ξm-2];

(H5) f(t, u) > ϕ p (a/L) for t [η,T], u [a,Ta/η].

Then BVP (1.6) has at least two positive solutions u1and u2such that
α ( u 1 ) > a , w i t h θ ( u 1 ) < b , a n d b < θ ( u 2 ) , w i t h γ ( u 2 ) < c . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ11_HTML.gif
(4.1)

Proof. From Lemma 2.7 we know that A : P(γ, c) → P is completely continuous. In what follows, we will prove the result step by step.

Step one: To verify (i) of theorem 2.5 holds.

We choose u P(γ,c), then γ ( u ) = min t [ ξ 1 , η ] u ( t ) = u ( ξ 1 ) = c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq54_HTML.gif. This implies that u(t) ≥ c for t [ξ1,T], considering that | | u | | T ξ 1 γ ( u ) = T ξ 1 c https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq55_HTML.gif, we have
c u ( t ) T ξ 1 c for t [ ξ 1 , T ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equaj_HTML.gif
As a consequence of (H3),
f ( t , u ( t ) ) > ϕ p ( c M ) for t [ ξ 1 , T ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equak_HTML.gif
Since Au P, we have
γ ( A u ) = ( A u ) ( ξ 1 ) = δ ϕ q 0 T h ( s ) f ( s , u ( s ) ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) f ( s , u ( s ) ) s + 0 ξ 1 ϕ q s T h ( τ ) f ( τ , u ( τ ) ) τ Δ s > c M δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 ξ 1 ϕ q s T h ( τ ) τ Δ s = c . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equal_HTML.gif

Thus, (i) of Theorem 2.5 is satisfied.

Step two: To verify (ii) of Theorem 2.5 holds.

Let u ∂P(θ,b), then θ ( u ) = max t [ 0 , ξ m - 2 ] = u ( ξ m - 2 ) = b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq56_HTML.gif, this implies that 0 ≤ u(t) ≤ b, t [0,ξm-2] and since u P, we have u = u(T), note that | | u | | T ξ m - 2 u ( ξ m - 2 ) = T ξ m - 2 θ ( u ) = T ξ m - 2 b https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq57_HTML.gif. So,
0 u ( t ) T ξ m - 2 b for t [ 0 , T ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equam_HTML.gif
From (H4) we know that f ( t , u ( t ) ) < ϕ p ( b K ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq58_HTML.gif for t [0, ξm-2] and so
θ ( A u ) = ( A u ) ( ξ m - 2 ) = δ ϕ q 0 T h ( s ) f ( s , u ( s ) ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) f ( s , u ( s ) ) s + 0 ξ m - 2 ϕ q s T h ( τ ) f τ , u ( τ ) τ Δ s < b K δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ T h ( s ) s + 0 ξ m - 2 ϕ q s T h ( τ ) τ Δ s < b K δ + i = 1 m - 2 β i + ξ m - 2 ϕ q 0 T h ( s ) s = b . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equan_HTML.gif

Thus, (ii) of Theorem 2.5 holds.

Step three: To verify (iii) of Theorem 2.5 holds.

Choose u 0 ( t ) = a 2 , t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq59_HTML.gif, obviously, u0(t) P(α, a) and α ( u 0 ) = a 2 < a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq60_HTML.gif, thus P ( α , a ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq61_HTML.gif.

Now, let u ∂P(α, a), then, α(u) = mint[η,T]u(t) = u(η) = a. Recalling that | | u | | T η u ( η ) = T η α ( u ) = T η a https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq62_HTML.gif. Thus, we have
a u ( t ) T η a for t [ η , T ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equao_HTML.gif
From assumption (H5) we know that
f ( t , u ( t ) ) > ϕ p a L for t [ η , T ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equap_HTML.gif
and so
α ( A u ) = ( A u ) ( η ) = δ ϕ q 0 T h ( s ) f ( s , u ( s ) ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) f ( s , u ( s ) ) s + 0 η ϕ q s T h ( τ ) f ( τ , u ( τ ) ) τ Δ s > a L δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 η ϕ q s T h ( τ ) τ Δ s = a . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equaq_HTML.gif

Therefore, all the conditions of Theorem 2.5 are satisfied, thus A has at least two fixed points in P(γ,c), which implies that BVP (1.6) has at least two positive solutions u1,u2 which satisfies (4.1). The proof is complete.

Example 4.2. Let T = { 2 n , n } { 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq63_HTML.gif. Consider the following four point boundary value problem on time scale T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq1_HTML.gif.
( ϕ p ( x Δ ) ) ( t ) + t f ( t , u ( t ) ) = 0 , t [ 0 , 8 ] T , x ( 0 ) - x Δ ( 0 ) = x Δ ( 1 ) + 2 x Δ ( 2 ) , x Δ ( 8 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ12_HTML.gif
(4.2)
where
f ( t , u ) = | sin t | + u 10 5 , 0 u 9 . 3 × 10 6 , | sin t | + 93 , 9 . 3 × 10 6 u 4 × 10 8 , | sin t | + 247 u 6 × 10 8 - 215 3 , u 4 × 10 8 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equar_HTML.gif
and h(t) = t, T = 8, ξ1 = 1, ξ2 = 2, δ = 1, β1 = 1, β2 = 2,p = 3/2, q = 3. In what follows, we try to calculate K, M, L. By Lemmas 2.2 and 2.3, we have
K = δ + i = 1 m - 2 β i + ξ m - 2 ϕ q 0 T h ( s ) s = ( 1 + 1 + 2 + 2 ) ϕ q 0 8 s s = 6 × 0 1 s s + 1 2 s s + 2 4 s s + 4 8 s s 2 = 6 × ( ν ( 1 ) × 1 + ν ( 2 ) × 2 + ν ( 4 ) × 4 + ν ( 8 ) × 8 ) 2 = 6 × ( 1 + 2 + 8 + 32 ) 2 = 6 × 1849 = 11094 . M = δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 ξ i ϕ q s T h ( τ ) τ Δ s = 0 8 s s 2 + 1 8 s s 2 + 2 2 8 s s 2 + 0 1 ϕ q s 8 τ τ Δ s = ( 1 + 2 + 8 + 32 ) 2 + ( 2 + 8 + 32 ) 2 + 2 × ( 8 + 32 ) 2 + 0 8 s s 2 = 2 * ( 1 + 2 + 8 + 32 ) 2 + ( 2 + 8 + 32 ) 2 + 2 × ( 8 + 32 ) 2 = 8662 . L = δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 η ϕ q s T h ( τ ) τ Δ s = 0 8 s s 2 + 1 8 s s 2 + 2 2 8 s s 2 + 0 4 s 8 h ( τ ) τ 2 Δ s = ( 1 + 2 + 8 + 32 ) 2 + ( 2 + 8 + 32 ) 2 + 2 × ( 8 + 32 ) 2 + 0 1 s 8 h ( τ ) τ 2 Δ s + 1 2 s 8 h ( τ ) τ 2 Δ s + 2 4 s 8 h ( τ ) τ 2 Δ s = 6813 + μ ( 0 ) × 0 8 s s 2 + μ ( 1 ) × 0 8 s s 2 + μ ( 2 ) × 0 8 s s 2 = 6813 + ( 1 + 2 + 8 + 32 ) 2 + ( 2 + 8 + 32 ) 2 + 2 × ( 8 + 32 ) 2 = 13626 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equas_HTML.gif

Let a = 106, b = 108, c = 109, then we have

(i) f ( t , u ) 340 > 10 9 8662 1 / 2 = ϕ p c M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq64_HTML.gif, for t [1, 8], u [109, 8 × 109];

(ii) f ( t , u ) 94 < 10 8 11094 1 / 2 = ϕ p b K https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq65_HTML.gif, for t [0, 2], u [108, 4 × 108];

(iii) f ( t , u ) > 9 > 10 6 13326 1 / 2 = ϕ p a L https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq66_HTML.gif, for t [4, 8], u [106, 2 × 106].

Thus, if all the conditions in Theorem 4.1 are satisfied, then BVP (4.2) has at least two positive solutions satisfying (4.1).

5 Existence of at least three positive solutions

Let ψ ( u ) = min t [ ξ 1 , T ] u ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq67_HTML.gif, then 0 < ψ(u) ≤ u. Denote
D = δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 ξ 1 ϕ q s T h ( τ ) τ Δ s , R = δ + i = 1 m - 2 β i + T ϕ q 0 T h ( s ) s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equat_HTML.gif

In this section, we will use fixed point Theorem 2.6 to get the existence of at least three positive solutions.

Theorem 5.1. Assume that there exists positive number d, ν, g satisfying d < ν < min ξ 1 T , D R g < g https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq68_HTML.gif, such that the following conditions hold.

(H6) f(t, u) < ϕ p (d/R), t [0,T],u [0,d];

(H7) f(t, u) > ϕ p (ν/D), t [ξ1, T], u [ν, Tυ/ξ1];

(H8) f(t, u) ≤ ϕ p (g/R), t [0,T],u [0,g],

then BVP (1.6) has at least three positive solutions u1, u2, u3satisfying
| | u 1 | | < d , ψ ( u 2 ) > ν , | | u 3 | | > d , w i t h ψ ( u 3 ) < ν . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ13_HTML.gif
(5.1)

Proof. From Lemma 2.8 we know that A : PP is completely continuous. Now we only need to show that all the conditions in Theorem 2.6 are satisfied.

For u P ̄ g , | | u | | g https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq69_HTML.gif. By (H8), one has
| | A u | | = ( A u ) ( T ) = δ ϕ q ( 0 T h ( s ) f ( s , u ( s ) ) s ) + i = 1 m 2 β i ϕ q ( ξ i T h ( s ) f ( s , u ( s ) ) s ) + 0 T ϕ q ( s T h ( τ ) f ( τ , u ( τ ) ) τ ) Δ s g R ( δ + i = 1 m 2 β i + T ) ϕ q ( 0 T h ( s ) s ) = g . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equau_HTML.gif

Thus, A : P ̄ g P ̄ g https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq70_HTML.gif. Similarly, by (H6), we can prove (ii) of Theorem 2.6 is satisfied.

In what follows, we try to prove that (i) of theorem 2.6 holds. Choose u 1 ( t ) = T ξ 1 ν , t [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq71_HTML.gif, obviously, ψ(u1) > ν, thus { u P ( ψ , ν , T ν / ξ 1 ) : ψ ( u ) > ν } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq72_HTML.gif. For u P(ψ,ν,Tν/ξ1),
ψ ( A u ) = ( A u ) ( ξ 1 ) = δ ϕ q 0 T h ( s ) f ( s , u ( s ) ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) f ( s , u ( s ) ) s + 0 ξ 1 ϕ q s T h ( τ ) f ( τ , u ( τ ) ) τ Δ s > ν D δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 ξ 1 ϕ q s T h ( τ ) τ Δ s = ν . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equav_HTML.gif

It remains to prove (iii) of Theorem 2.6 holds. For u P(ψ, ν, Tυ/ξ1), with Au > /ξ1, in view of Lemma 2.8, there holds ψ ( A u ) - ( A u ) ( ξ 1 ) ξ 1 T | | A u | | > ν https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq73_HTML.gif, which implies that (iii) of Theorem 2.6 holds.

Therefore, all the conditions in Theorem 2.6 are satisfied. Thus, BVP (1.6) has at least three positive solutions satisfying (5.1). The proof is complete.

Example 5.2. Let T = [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq74_HTML.gif. Consider the following four point boundary value problem on time scale T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_IEq1_HTML.gif.
( ϕ p ( x Δ ) ) ( t ) + e t ( t , u ( t ) ) = 0 , t [ 0 , T ] T , x ( 0 ) - 3 x Δ ( 0 ) = 2 x Δ ( 1 / 2 ) + 3 x Δ ( 1 ) , x Δ ( 8 ) = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equ14_HTML.gif
(5.2)
where
f ( t , u ) = t 20 + u 2 840 3 , 0 u 126 , t 20 + 18 . 9 3 , u 126 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equaw_HTML.gif
and h(t) = e t , T = 2, ξ1 = 1/2, ξ2 = 1, δ = 3, β1 = 2, β2 = 3, p = 4, q = 4/3. In what follows, we try to calculate D, R. By Lemmas 2.2 and 2.3, we have
D = δ ϕ q 0 T h ( s ) s + i = 1 m - 2 β i ϕ q ξ i T h ( s ) s + 0 ξ 1 ϕ q s T h ( τ ) τ Δ s = 3 0 2 e s s 1 / 3 + 2 1 / 2 2 e s s 1 / 3 + 3 1 2 e s s 1 / 3 + s 1 / 2 s 2 e τ τ 1 / 3 Δ s = 3 0 1 e s d s + 1 2 e s s 1 / 3 + 2 1 / 2 1 e s d s + 1 2 e s s 1 / 3 + 3 1 2 e s s 1 / 3 + 0 1 / 2 s 1 e τ d τ + 1 2 e τ τ 1 / 2 Δ s = 3 ( e + e 2 - 1 ) 1 / 3 + 2 ( e + e 2 - e 1 / 2 ) 1 / 3 + 3 e 2 / 3 + 3 4 ( e + e 2 - 1 ) 4 / 3 - 3 4 ( e + e 2 - e 1 / 2 ) 4 / 3 17 . 5216 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equax_HTML.gif
R = δ + i = 1 m - 2 β i + T ϕ q 0 T h ( s ) s = ( 3 + 2 + 3 + 2 ) 0 2 e s s 1 / 3 = 10 ( e + e 2 - 1 ) 1 / 3 = 20 . 8832 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-4/MediaObjects/13661_2011_Article_110_Equay_HTML.gif

Let d = 40, ν = 50, g = 400, then we have

(i) f(t, u) < 7.027 = (40/20.8832)3 = ϕ p (d/R), for t [0, 2], u [0, 40];

(ii) f(t, u) > 23.2375 = (50/17.5216)3 = ϕ p (ν/D), for t [1/2, 2], u [50, 200];

(iii) f(t, u) < 7027.305 = (400/20.8832)3 = ϕ p (g/R), for t [0, 2], u [0, 400].

Thus, if all the conditions in Theorem 5.1 are satisfied, then BVP (5.2) has at least three positive solutions satisfying (5.1).

Declarations

Acknowledgements

The authors were very grateful to the anonymous referee whose careful reading of the manuscript and valuable comments enhanced presentation of the manuscript. The study was supported by Pre-research project and Excellent Teachers project of the Fundamental Research Funds for the Central Universities (2011YYL079, 2011YXL047).

Authors’ Affiliations

(1)
School of Science, China University of Geosciences
(2)
Department of Mathematics, Beijing Institute of Technology

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© Zhao et al; licensee Springer. 2012

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