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SOLUTIONS TO A RIGHT-FOCAL BOUNDARYVALUE PROBLEM ON TIME SCALES ILKAY YASLAN KARACA Received 10 October 2005; Revised 19 January 2006; Accepted 30 January 2006 We are concerned with prov

SOLUTIONS TO A RIGHT-FOCAL BOUNDARY

VALUE PROBLEM ON TIME SCALES

ILKAY YASLAN KARACA

Received 10 October 2005; Revised 19 January 2006; Accepted 30 January 2006

We are concerned with proving the existence of one or more than one positive solution

of ann-point right-focal boundary value problem for the nonlinear dynamic equation

(−1)n−1xΔn(t) = λr(t) f (t,x σ(t)) We will also obtain criteria which lead to nonexistence

of positive solutions Here the independent variablet is in a time scale We will use fixed

point theorems for operators on a Banach space

Copyright © 2006 Ilkay Yaslan Karaca This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

Motivated by the work of Anderson [3] on discrete third-order three-point right-focal boundary value problems, in this paper we will study annth-order n-point right-focal

boundary problem on time scales This paper also gives nonexistence and multiplicity results for positive solutions to the time scale boundary value problem

(−1)n−1xΔn(t) = λr(t) f

t,x σ(t)

∀t ∈t1,ρ

t n

x

t1



= xΔ

t2



= ··· = xΔn−1(t n)=0, (1.2) wheren ≥2,t1< t2< ··· < t n−1< t n,λ is a real parameter, and x = x(t) is a desired

solu-tion The arguments are similar to those used in [9,13]

In the third section we obtain multiplicity results for this problem withλ =1 In the fourth section existence, nonexistence, and multiplicity results are given for the eigen-value problem

To understand this so-called dynamic equation (1.1) on a time scale T, we need some

preliminary definitions

Definition 1.1 Let T be a nonempty closed subset ofRand define the forward jump operatorσ(t) at t for t < supT by

σ(t) :=inf

s > t : s ∈T

(1.3)

Hindawi Publishing Corporation

Advances in Di fference Equations

Volume 2006, Article ID 43039, Pages 1 13

DOI 10.1155/ADE/2006/43039

and the backward jump operatorρ(t) at t for t > inf T by

ρ(t) :=sup

s < t : s ∈T

(1.4) for allt ∈T.

We assume throughout that T has the topology that it inherits from the standard

topol-ogy on the real numbersR Ifσ(t) > t, we say t is right scattered, while if ρ(t) < t, we say

t is left scattered If σ(t) = t, we say t is right dense, while if ρ(t) = t, we say t is left dense.

We assumeσ0(t) = t, and for any integer n > 0, we have

σ n(t) := σ

σ n−1(t)

Throughout this paper we make the blanket assumption thata ≤ b are points in T Definition 1.2 Define the interval in T:

[a,b] := {t ∈T such thata ≤ t ≤ b}. (1.6) Other types of intervals are defined similarly

We are concerned with calculus on time scales which is a unified approach to contin-uous and discrete calculus In [4,11], Aulbach and Hilger have initiated the development

of this calculus Since then, efforts have been made in the context of time scales, in estab-lishing that some results for boundary value problems for ordinary differential equations and their discrete analogues are special cases of more general results on time scales; for a wide variety of problems addressed, see many references [1,5,6,8–10,14]

Definition 1.3 Assume x : T → Rand fixt ∈T such thatt < supT, then xΔ(t) is defined

to be the number (provided it exists) with the property that, given any > 0, there is a

neighborhoodU of t such that

x

σ(t)

− x(s)

− xΔ(t)

σ(t) − s< σ(t) − s (1.7)

for alls ∈ U xΔ(t) is called the delta derivative of x at t.

It can be shown that ifx : T → Ris continuous att ∈T,t < supT, and t is right

scat-tered, then

xΔ(t) = x σ(t) − x(t)

Note, if T= Z, whereZis the set of integers, then

xΔ(t) = Δx(t) : = x(t + 1) − x(t). (1.9)

Moreover, if T= R, then

Finally, forn ≥1, define

xΔn(t) :=xΔ(t) Δn−1

(1.11) assumingxΔ0(t) = x(t).

Definition 1.4 If FΔ(t) = f (t), then define the integral of f by

t

Note that in the case T= Rwe have

b

a f (t)Δt =

b

and in the case T= Zwe have

b

a f (t)Δt =

b−1

k=a

wherea,b ∈T witha ≤ b.

2 Preliminaries

As in [2], we introduce the Taylor polynomialsh j: T2→ R,j ∈N0, recursively defined as follows:

h j+1(t,s) =

t

For integersn ≥2 and fori =1, 2, ,n −1, define

u n,i(t,s) ≡ u n,i

t,s : t1,t2, ,t n

witht,s,t j ∈T for 1≤ j ≤ n, as follows:

u n,i(t,s) :=(−1)n+1



















t,t1



h2



t,t1



h n−1



t,t1



c2(s,i) 1 h1



t2,t1



h n−2



t2,t1





t3,t1





t n−1,t1



















where

c j(s,i) := H( j −1− i)h n−j

t j,σ(s)

forj =2, 3, ,n −1 andi =1, 2, ,n −1 Here

H(x) =

⎧

⎨

⎩

0 ifx < 0,

is the usual Heaviside function, andh j(t,s) is as defined in (2.2) In addition, define

v n,i(t,s) := u n,i(t,s) + (−1)n−1h n−1



t,σ(s)

for integersn ≥2 and fori =1, 2, ,n −1

Theorem 2.1 [2] For u n,i(t,s) as in ( 2.4 ) and v n,i(t,s) as in ( 2.7 ),

G n



t,s : t1,t2, ,t n



=

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

s ∈ I1:

⎧

⎨

⎩u n,1

(t,s) if t ≤ σ(s),

v n,1(t,s) if t ≥ σ(s),

s ∈ I2:

⎧

⎨

⎩u n,2

(t,s) if t ≤ σ(s),

v n,2(t,s) if t ≥ σ(s),

.

s ∈ I n−1:

⎧

⎨

⎩u n,n−

1(t,s) if t ≤ σ(s),

v n,n−1(t,s) if t ≥ σ(s),

(2.8)

where I1=[t1,ρ(t2)], andI i =[ρ(t i),ρ(t i+1 )] for i =1, 2, ,n − 1, is Green’s function for the homogeneous problem (−1)n−1xΔn−1(t) = 0 satisfying the boundary conditions ( 1.2 ).

Lemma 2.2 [2] For s ∈[t1,ρ(t2)] andn ≥ 2,

G n

t,s : t1,t2, ,t n⎧⎨

⎩

< 0 if t ∈− ∞,t1



,

> 0 if t ∈t1,σ n−1 

t n

Theorem 2.3 [2] Let u n,i(t,s) and v n,i(t,s) be given as in ( 2.4 ) and ( 2.7 ), respectively Assume for n ≥ 4 that

v n−i, j−i+1

σ n−i

t n

,σ

s j

:t i,t i+1, ,t n−1



for j ∈ {2, 3, ,n −2} and i = j −1,j −2, ,1, and for s j ∈[ρ(t j),ρ(t j+1 )] Then

G n

t,s : t1,t2, ,t n⎧⎨

⎩< 0 if t ∈



− ∞,t1



,

> 0 if t ∈t1,σ n−1 

t n



for s ∈[t1,ρ(t n )] if n is even, or for s ∈[t1,ρ(t n−1)] ifn is odd For odd n ≥ 3, the additional assumption

u n,n−1



σ n−1

t n



+ (−1)n−1h n−1



σ n−1

t n



,t n



yields ( 2.11 ) for s ∈[ρ(t n−1),ρ(t n )] as well.

Throughout this paper, we assume that the time scale T is such thatσ(s) is delta

dif-ferentiable for alls ∈T,t1is right-scattered, and hypotheses ofTheorem 2.3hold Furthermore, we have the following assumptions

(H1)r(s) is a nonnegative continuous function defined on [t1,ρ(t n)] satisfying

0<

t n

t1

fort ∈[σ(t1),σ n−1(t n)]

(H2) f : [t1,ρ(t n)]× R → Ris such that f (t,x) ≥0 forx ∈ R+and continuous with respect tox, whereR +denotes the set of nonnegative real numbers

Let us set

M n:=maxG n(t,s)r(s), m n:=minG n(t,s)r(s) (2.14) fors ∈[t1,ρ(t n)],t ∈[σ(t1),σ n−1(t n)], and

A1n:= max

t∈[σ(t1 ),σ n−1 (t n)]

t n

t1

G n(t,s)r(s)Δs, A2n:= min

t∈[σ(t1 ),σ n−1 (t n)]

t n

t1

G n(t,s)r(s)Δs.

(2.15)

We refer to [7,12] for a discussion of the fixed point index that we use below In particular, we will make frequent use of the following lemma

Lemma 2.4 Let Ꮾ be a Banach space, and let ᏼ ⊂ Ꮾ be a cone in Ꮾ Assume r > 0 and that

Ψ : ᏼr → ᏼ is compact operator such that Ψx = x for x ∈ ∂ᏼ r:= {x ∈ᏼ :x = r} Then the following assertions hold.

(i) If x ≤ Ψx  for all x ∈ ∂ᏼ r , then i(Ψ,ᏼ r,ᏼ)= 0.

(ii) If x ≥ Ψx  for all x ∈ ∂ᏼ r , then i(Ψ,ᏼ r,P) = 1.

Thus, if there existsr1> r2> 0 such that condition (i) holds for x ∈ ∂ᏼ r1and (ii) holds forx ∈ ∂ᏼ r2(or (ii) and (i)), then, from the additivity properties of the index, we know that

i

Ψ,ᏼr1,ᏼ= i

Ψ,ᏼr1\int

ᏼr2



,ᏼ+i

Ψ,ᏼr2,ᏼ. (2.16)

As a consequence ofi(Ψ,ᏼ r1\int(ᏼr2),ᏼ)=0,Ψ has a fixed point (nonzero) whose norm

is betweenr1andr2.

Consider the Banach space of continuous functions on [t1,σ n−1(t n)] with the norm

x =maxx(t),t ∈

σ

t1



,σ n−1 

t n

and coneᏼ in Ꮾ given by

ᏼ=



x ∈ Ꮾ : x(t) ≥0,t ∈σ

t1



,σ n−1

t n



t∈[σ(t1 ),σ n−1 (t n)]x(t) ≥ m n

M n x



. (2.18)

ByTheorem 2.1, solving the BVP (1.1)–(1.2) is equivalent to solving the following integral equation inᏼ:

x(t) = λ

t n

t1

G n(t,s)r(s) f

s,x σ(s)

Δs, t ∈t1,σ n−1 

t n

and consequently, it is equivalent to finding fixed points of the operatorΨn λ:Ꮾ→Ꮾ defined by

Ψn λ x(t) := λ

t n

t1

G n(t,s)r(s) f

s,x σ(s)

Δs, t ∈t1,σ n−1 

t n

First, we prove that for everyλ > 0 given, this operator maps the cone ᏼ in itself Lemma 2.5 Let λ > 0 be given Under the hypotheses (H1) and (H2), the operator Ψ n λ is a compact operator such thatΨn λ(ᏼ)⊂ ᏼ.

Proof ThatΨn λ is a compact operator follows by Arzela-Ascoli’s theorem Next, for all

x ∈ᏼ, by (H1), (H2), and the positivity property of the Green function, we have from (2.20),Ψn λ x(t) ≥0 for allt ∈[σ(t1),σ n−1(t n)] Ifx ∈ᏼ, then

min

t∈[σ(t1 ),σ n−1 (t n)]Ψn λ x(t) ≥ λm n

t n

t1

f

s,x σ(s)

Δs

≥ λ m n

M n

t n

t1



max

t∈[σ(t1 ),σ n−1 (t n)]G n(t,s)r(s)



f

s,x σ(s)

Δs

≥ λ m n

t∈[σ(t1 ),σ n−1 (t n)]

t n

t1

G n(t,s)r(s) f

s,x σ(s)

Δs

= m n

M n

Ψ

n λ x.

(2.21)

3 Noneigenvalue problem

In this section we study the existence of at least two positive solutions to the following BVP:

(−1)n−1xΔn(t) = r(t) f

t,x σ(t)

∀t ∈t1,ρ

t n



,

x

t1



= xΔ

t2



= ··· = xΔn−1

t n

which is problem (1.1)-(1.2) withλ =1 As an application, we also give an example to demonstrate our result

Theorem 3.1 The boundary value problem ( 3.1 ) has at least two positive solutions, x1and

x2, if (H1) and (H2) are satisfied and, in addition, both of the following hold.

(H3) There exists 0 < k < R < +∞ such that

f (t,x) > M n

m2

n



t n − t1

x ∀x ∈[0,k]

[R,+∞],t ∈t1,ρ

t n

(H4) There exists p > 0 such that

f (t,x) < p

M n

t n − t1 ∀ x ∈[0,p], t ∈t1,ρ

t n

where M n and m n are given as ( 2.14 ) Moreover, 0 < x1 < p < x2

Proof Let x ∈ ∂ᏼ k From condition (H3), we have

A1

n  = max

t∈[σ(t1 ),σ n−1 (t n)]

t n

t1

G n(t,s)r(s) f

s,x σ(s)

Δs

> m n M n

m2

n



t n − t1

 t n

t1

x σ(s)Δs

≥ M n

m n

t n − t1

m n

M n x

t n

t1

Δs = x

(3.4)

Ifx ∈ ∂ᏼ R1,R1≥ M n /m n R, we have

min

t∈[σ(t1 ),σ n−1 (t n)]x(t) ≥ m n

M n x = m n

Hencex(s) ≥ R for all s ∈[t1,ρ(t n)] Therefore using condition (H3) again, we arrive at the same conclusion

Now, from (H4), ifx = p,

A1n = max

t∈[σ(t1 ),σ n−1 (t n)]

t n

t1

G n(t,s)r(s) f

s,x σ(s)

Δs

≤

t n

t1

M n f

s,x σ(s)

Δs < x

(3.6)

Since we can choosek > 0 small enough and R1sufficiently large so that k < p < R1, we assure the existence of two solutions:x1∈ᏼp \int(ᏼk) andx2∈ᏼR1\int(ᏼp) 

Example 3.2 We illustrateTheorem 3.1with specific time scale

T=Tc =c m:m ∈ Z∪ {0}, (3.7) wherec > 1 and the following specific parameter values for n =3 Letc =11/10, t1=1,

t2=(11/10)3, andt3=(11/10)4 Bohner and Peterson [6] show that

h j(t,s) =

j−1



ν=0

t − c ν s

ν

for alls,t ∈T Using this formula, we have

G3(t,s)

=

⎧

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎩

s ∈



1,

11

10

 2 

:

⎧

⎪

⎪

⎪

⎪

(t −1)

11

10s −1



ift ≤11

10s,

(t −1)

11

10s −1



+10 21



t −11

10s



t −121

100s



ift ≥11

10s,

s ∈

11

10

 2 ,

11

10

 3 

:

⎧

⎪

⎨

⎪

⎩

331

10s,

331

1000(t −1) +10

21



t −11

10s



t −121

100s



ift ≥11

10s.

(3.9)

Ifr(s) = s, then m3=minG(t,s)r(s) =10−2,M3=maxG(t,s)r(s) =3870659646821·10−13 fort ∈[11/10,(11/10)6],s ∈[1, (11/10)3]

Letk =1/18000, p =1/5, R =2/5, let

f (t,x) =

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

2.104k sin tπ

6 ifx ∈[0,k], L(x)sin tπ

6 ifx ∈[k, p], K(x)sin tπ

2.104k + 1

R x sin

tπ

6 ifx ∈[R,+∞),

(3.10)

where

L(x) =1 + p − x

p − k



2.104k −1

,

K(x) =1 + p − x

p − R



2.104(k + 1) −1

.

(3.11)

Note that f is continuous and nonnegative valued for x ≥0

Fort ∈[1, (11/10)3] andx ∈[0,k]

[R,∞), f (t,x) > (8340,14144, )x Indeed; for

x ∈[0,k], f (t,x) =2.104k sin(tπ/6)≥104k > (8340,14144, )x, for x ∈[R,∞), f (t,x) =

2.104((k + 1)/R)x sin(tπ/6) ≥104((k + 1)/R)x =(900050/36)x > (8340,14144, )x So

(H3) is verified

Forx ∈[0,k], f (t,x) =2.104k sin(tπ/6) < 2.104k < p/0,17963731420896261.

Forx ∈[k, p], f (t,x)=1 +((p − x)/(p − k))[2.104k −1] sin(tπ/6) ≤2.104k sin(tπ/6) <

2.104k < p/0,17963731420896261 Hence it verifies the (H4).

We conclude fromTheorem 3.1that for these parameter values, (3.1) forn =3 has at least two positive solutions,x andx such that 0< x < 1/5 < x

4 Eigenvalue problem

Define the nonnegative extended real numbers f0, f0, f ∞, and f ∞by

f0:=lim inf

x→0 + min

t∈[t1 ,ρ(t n)]

f (t,x)

0:=lim sup

x→0 + max

t∈[t1 ,ρ(t n)]

f (t,x)

f ∞:=lim inf

t∈[t1 ,ρ(t n)]

f (t,x)

∞:=lim sup

t∈[t1 ,ρ(t n)]

f (t,x)

(4.1)

respectively

These numbers can be regarded as generalized super or sublinear conditions on the function f (t,x) at x =0 andx = ∞ Thus, if f0= f0=0 (+∞), then f (t,x) is superlinear

(sublinear) atx =0 and if f ∞ = f ∞ =0 (+∞), then f (t,x) is sublinear (superlinear) at

x =+∞

First, we obtain an existence result forλ belonging to a given interval.

Theorem 4.1 If (H1)-(H2) hold and either

(a)M n /(m n A2n f0)< λ < 1/(A1n f ∞ ), or

(b)M n /(m n A2nf ∞)< λ < 1/(A1nf0)

is satisfied, where M n , m n , A1n, and A2nare given as in ( 2.14 ) and ( 2.15 ), then the eigenvalue problem ( 1.1 )-( 1.2 ) has at least one positive solution.

Proof Assume (a) holds First we consider f0< ∞ Since

M n

there is an > 0 so that

λ

f0− m n

Using the definition of f0, there is anr1> 0, sufficiently small, so that

f0−  < min

t∈[t1 ,ρ(t n)]

f (t,x)

for 0< x ≤ r1.

It follows that f (t,x) > ( f0− )x for 0 < x ≤ r1,t ∈[t1,ρ(t n)]

Assume thatx ∈ ∂ᏼ r1, then

Ψn λ x(t) = λ

t n

t1

G n(t,s)r(s) f

s,x σ(s)

Δs

> λ

f0−  t n

t1

G n(t,s)r(s)x σ(s)Δs

≥ λ

f0− m n

M n xA2n ≥ x

(4.5)

Next, we consider the case f0= ∞ ChooseK > 0 sufficiently large so that

λK m n

for anyt ∈[t1,σ n−1(t n)]

So there existsr1> 0 so that f (t,x) > Kx for 0 < x ≤ r1.

Assume thatx ∈ ∂ᏼ r1, then

Ψn λ x(t) > λK

t n

t1

G n(t,s)r(s)x σ(s)Δs ≥ λK m n

M n xA2n ≥ x (4.7) Finally, we use the assumption

λ < 1

Pick an1> 0 so that

λ

f ∞+1



Using the definition of f ∞, there is anr > r1sufficiently large, so that

max

t∈[t1 ,ρ(t n)]

f (t,x)

x < f

forx ≥ r.

It follows that f (t,x) < ( f ∞+1)x for x ≥ r.

We now show that there is anr2≥ r such that if x ∈ ∂ᏼ r2, thenΨn λ x < x Pickr2≥ rM n /m n > r1 Now assumex ∈ ∂ᏼ r2and consider

Ψn λ x(t) < λ

f ∞+1

 t n

t1

G n(t,s)r(s)x σ(s)Δs ≤ λ( f ∞+1)A1nx ≤ x (4.11)

Therefore, byLemma 2.4,Ψn λ has a fixed point x with r1< x < r2 This shows that condition (a) yields the existence of a positive solution of the eigenvalue problem (1.1

The proof of part (b) is similar.

Our next results give criteria for the existence of one, more than one, or no positive solutions of the eigenvalue problem (1.1)-(1.2) in terms of the superlinear or sublinear behavior of f (t,x) For the next three theorems, in addition to the assumptions (H1) and

(H2) we assume

(H5) f (t,x) > 0 on [t1,ρ(t n)]× R+

Theorem 4.2 If hypotheses (H1), (H2), and (H5) are satisfied, then the following assertions hold.

(a) If f0= ∞ or f ∞ = ∞, then there is a λ0> 0 such that for all 0 < λ ≤ λ0the eigenvalue problem ( 1.1 )-( 1.2 ) has a positive solution.