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Existence of Positive Solutions for Nonlinear m-point Boundary Value Problems on Time Scales Boundary Value Problems 2012, 2012:4 doi:10.1186/1687-2770-2012-4 Junfang Zhao zhao_junfang@1

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Existence of Positive Solutions for Nonlinear m-point Boundary Value Problems

on Time Scales

Boundary Value Problems 2012, 2012:4 doi:10.1186/1687-2770-2012-4

Junfang Zhao (zhao_junfang@163.com)Hairong Lian (lianhr@126.com)Weigao Ge (gew@bit.edu.cn)

ISSN 1687-2770

Article type Research

Submission date 4 May 2011

Acceptance date 17 January 2012

Publication date 17 January 2012

Article URL http://www.boundaryvalueproblems.com/content/2012/1/4

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which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Existence of positive solutions for nonlinear

m-point boundary value problems on time scales

1 School of Mathematics and Physics, China University of Geosciences, Beijing 100083, P.R China

2 Department of Mathematics, Beijing Institute of Technology, Beijing 100081, P.R China

∗ Corresponding author: zhao junfang@163.com

0 < ξ1 < ξ2 < · · · < ξm−2 < T ∈ T By using several well-known fixed point theorems

in a cone, the existence of at least one, two, or three positive solutions are obtained Examplesare also given in this article

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

Keywords: positive solutions; cone; multi-point; boundary value problem; time scale.

The study of dynamic equations on time scales goes back to its founder Hilger [1], and is anew area of still theoretical exploration in mathematics Motivating the subject is the notionthat dynamic equations on time scales can build bridges between continuous and discretemathematics 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, thevibrations of a guy wire of a uniform cross section and composed of N parts of differentdensities can be set up as a multi-point boundary value problem [3] Many problems in thetheory of elastic stability can be handled by the method of multi-point problems [4] Smallsize 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 byIl’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 differentialequations on time scales have been studied by some authors [7–11], and there has been somemerging of existence of positive solutions to BVPs with p-Laplacian on time scales [12–19]

He [20] studied

(φp(u∆(t)))∇+ a(t)f (t) = 0, t ∈ (0, T )T, (1.1)

subject to one of the following boundary conditions

where η ∈ (0, T ) ∩ T 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)

in this article, we consider the following m-point BVP with one-dimensional p-Laplacian,

(S2) f ∈ C([0, ∞), (0, ∞)), f 6≡ 0 on [0, T ]T.

(S3) By φq we denote the inverse to φp, where 1p +1q = 1

(S4) By t ∈ [a, b] we mean that t ∈ [a, b] ∩ T, where 0 ≤ a ≤ b ≤ T

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

Definition 2.1 [7, 22] For t < sup T and t > inf T, define the forward jump operator σand the backward jump operator ρ, respectively,

σ(t) = inf{τ ∈ T|τ > t} ∈ T, ρ(r) = sup{τ ∈ T|τ < r} ∈ Tfor all r, t ∈ T If σ(t) > t, t is said to be right scattered, and if ρ(r) < r, r is said to be leftscattered 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 has a right scattered minimum m, define Tκ = T − {m}; Otherwise set Tκ = T Thebackward graininess µb : Tκ → R+0 is defined by

for all s ∈ U For x : T → R and t ∈ Tκ, we define the “∇” derivative of x(t), x∆(t), to bethe 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|

for all s ∈ V

Definition 2.3 [22] If F∆(t) = f (t), then we define the “∆” integral by

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

(ii) If [a, b] consists of only isolated points, then

(iii) If T = hZ = {hk : k ∈ Z}, where h > 0, then

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 ⊂ Eopen and bounded, 0 ∈ Ω1 ⊂ Ω1 ⊂ Ω2 ⊂ Ω2 Assume A : (Ω2\ Ω1) ∩ P → P is completelycontinuous If one of the following conditions holds

(i) kAxk ≤ kxk, ∀x ∈ ∂Ω1∩ P , kAxk ≥ kxk, ∀x ∈ ∂Ω2∩ P ;

(ii) kAxk ≥ kxk, ∀x ∈ ∂Ω1∩ P , kAxk ≤ kxk, ∀x ∈ ∂Ω2∩ P

Then, A has a fixed point in (Ω2\ Ω1) ∩ P

Theorem 2.5 [25] Let P be a cone in the real Banach space E Set

for all u ∈ P (γ, r) Further, suppose there exists positive numbers a < b < r such that

θ(λu) ≤ λθ(u) for all 0 ≤ λ ≤ 1, u ∈ ∂P (θ, b)

If A : P (γ, r) → P 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) 6= ∅ and α(Au) > a for all u ∈ ∂P (α, a)

Then, A has at least two fixed points u1 and u2 such that

a < α(u1), with θ(u1) < b, and b < θ(u2), with γ(u1) < r,

Let a, b, c be constants, Pr = {u ∈ P : kuk < r}, P (ψ, b, d) = {u ∈ P : a ≤ ψ(u), kuk ≤b}

Theorem 2.6 [26] Let A : Pc→ Pc be a completely continuous map and ψ be a ative continuous concave functional on P such that for ∀u ∈ Pc, there holds ψ(u) ≤ kuk.Suppose there exist a, b, d with 0 < a < b < d ≤ c such that

nonneg-(i) {u ∈ P (ψ, b, d) : ψ(u) > b} 6= ∅ and ψ(Au) > b for all u ∈ P (ψ, b, d);

(ii) kAuk < a for all u ∈ Pa;

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

Then, A has at least three fixed points u1, u2, and u3 satisfying

It is obvious that kuk = u(T ) for u ∈ P Define A : P → E as

Lemma 2.7 A : P → P is completely continuous

Proof First, we try to prove that A : P → P

Combining (2.1) and (2.3) we have

T (u(t) − u(0)) ≥ T tu∆(t) ≥ t(u(T ) − u(0)),

as u(0) ≥ 0, it is immediate that

u(t) ≥ tu(T ) + (T − t)u(0)

3 Existence of at least one positive solution

First, we give some notations Set

Theorem 3.1 Assume in addition to (S1) and (S2), the following conditions are satisfied,there exists 0 < r < ξ1 ρ

T < ρ < ∞ such that

(H1) f (t, u) ≤ φp(Λu), for t ∈ [0, T ], u ∈ [0, r];

(H2) f (t, u) ≥ φp(Bu), for t ∈ [ξ1, T ], u ∈ [ξ1 ρ

T , ρ]

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 : P → P is completelycontinuous Set Ωr= {u ∈ E, kuk < r} In view of (H1), for u ∈ ∂Ωr∩ P,

which means that for u ∈ ∂Ωr∩ P, kAuk ≤ kuk

On the other hand, for u ∈ P, in view of Lemma 2.8, there holds u(t) ≥ ξ1

Tkuk, for

t ∈ [ξ1, T ] Denote Ωρ= {u ∈ E, kuk < ρ} Then for u ∈ ∂Ωρ∩ P, considering (H2), we have

which implies that for u ∈ ∂Ωρ∩P, kAuk ≥ kuk Therefore, the immediate result of Theorem

2.4 is that A has at least one fixed point u ∈ (Ωρ\ Ωr) ∩ 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 r ≤ kuk ≤ ρ The proof is complete

512(u − 100) +

25t

32, 100 ≤ u ≤ 500,tu

16, u ≥ 500,

and h(t) = 1, T = 4, ξ1 = 2, ξ2 = 3, δ = 2, β1 = β2 = 1, p = q = 2 In what follows, we try tocalculate Λ, B By Lemmas 2.2 and 2.3, we have

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

Fix η ∈ T such that

0 < ξm−2 ≤ η < T,

and define the increasing, nonnegative, continuous functionals γ, θ, α on P by

We can see that, for u ∈ P, there holds

γ(u) ≤ θ(u) ≤ α(u)

In addition, Lemma 2.8 implies that γ(u) = u(ξ1) ≥ ξ1

T kuk, which means that

kuk ≤ T

ξ1γ(u) for u ∈ P.

We also see that

θ(λu) = λθ(u) for λ ∈ [0, 1], u ∈ ∂P (θ, b)

For convenience, we give some notations,

Then BVP (1.6) has at least two positive solutions u1 and u2 such that

α(u1) > a, with θ(u1) < b, and b < θ(u2), with γ(u2) < c (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) = mint∈[ξ1,η]u(t) = u(ξ1) = c This implies that

u(t) ≥ c for t ∈ [ξ1, T ], considering that kuk ≤ ξT

h(s)f (s, u(s))∇s

+

h(s)∇s

+

Z ξ 1

0

φq

Z T s

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) = maxt∈[0,ξm−2] = u(ξm−2) = b, this implies that 0 ≤ u(t) ≤

b, t ∈ [0, ξm−2] and since u ∈ P, we have kuk = u(T ), note that kuk ≤ ξT

From (H4) we know that f (t, u(t)) < φp(Kb) for t ∈ [0, ξm−2], and so

Thus, (ii) of Theorem 2.5 holds

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

Choose u0(t) = a2, t ∈ [0, T ], obviously, u0(t) ∈ P (α, a) and α(u0) = a2 < a, thus

for t ∈ [η, T ],

Therefore, all the conditions of Theorem 2.5 are satisfied, thus A has at least two fixedpoints in P (γ, c), which implies that BVP (1.6) has at least two positive solutions u1, u2which satisfies (4.1) The proof is complete.

Example 4.2 Let T = {2n, n ∈ Z} ∪ {0} Consider the following four point boundary valueproblem on time scale T

5 Existence of at least three positive solutions

Let ψ(u) = mint∈[ξ1,T ]u(t), then 0 < ψ(u) ≤ kuk Denote

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

Theorem 5.1 Assume that there exists positive number d, ν, g satisfying d < ν <min{ξ1

T , DR}g < g, such that the following conditions hold

Thus, A : Pg → Pg 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 u1(t) =

It remains to prove (iii) of Theorem 2.6 holds For u ∈ P (ψ, ν, T ν/ξ1), with kAuk >

T ν/ξ1, in view of Lemma 2.8, there holds ψ(Au) = (Au)(ξ1) ≥ ξ1

TkAuk > ν, which impliesthat (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] ∪ N Consider the following four point boundary value problem

we try to calculate D, R By Lemmas 2.2 and 2.3, we have

Z ξ 1

0

φq

Z T s

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