positive solutions of singular boundary value problems

28 3 Singular Boundary Value Problems for Ordinary Differential Equations 333.1 Second Order.. In this dissertation we will be concerned with the existence of positive solutions of singu

Positive Solutions of Singular Boundary Value Problems

Curtis J KunkelAdvisor: Johnny Henderson, Ph.D

In this dissertation, we focus on singular boundary value problems with mixedboundary conditions We study a variety of types, to all of which we seek a positivesolution We begin by considering the discrete (or difference equation) case, fromwhich we proceed to look at the continuous (or ordinary differential equation) case

In all cases, we make use of a lower and upper solutions method and the Brouwerfixed point theorem in conjunction with perturbation methods to approximate regularproblems

byCurtis J Kunkel

A DissertationApproved by the Department of Mathematics

Robert Piziak, Ph.D., Chairperson

Submitted to the Graduate Faculty ofBaylor University in Partial Fulfillment of the

Requirements for the Degree

ofDoctor of Philosophy

Approved by the Dissertation Committee

Johnny Henderson, Ph.D., Chairperson

J Larry Lyon, Ph.D., Dean

Page bearing signatures is kept on file in the Graduate School.

3247565 2007

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ACKNOWLEDGMENTS v

2 Singular Boundary Value Problems for Difference Equations 42.1 Third Order 42.1.1 Preliminaries 42.1.2 Lower and Upper Solutions Method for Regular Problems 62.1.3 Existence Result 132.2 Higher Order 172.2.1 Preliminaries 172.2.2 Lower and Upper Solutions Method for Regular Problems 182.2.3 Existence Result 28

3 Singular Boundary Value Problems for Ordinary Differential Equations 333.1 Second Order 333.1.1 Preliminaries 333.1.2 Lower and Upper Solutions Method for Regular Problems 343.1.3 Existence Result 383.2 Third Order 423.2.1 Preliminaries 423.2.2 Lower and Upper Solutions Method for Regular Problems 433.2.3 Existence Result 483.3 Higher Order 533.3.1 Preliminaries 533.3.2 Lower and Upper Solutions Method for Regular Problems 54

iii

BIBLIOGRAPHY 64

iv

I would like to thank Baylor University for giving me the opportunity to work

on my dissertation and earn my doctorate The professors at Baylor University willever have my thanks in preparing me to do the research required for this dissertationand for preparing me to enter the field of mathematics as their colleague Specialthanks go to Dr Johnny Henderson for advising me throughout this process I ameternally grateful for the time he spent reading and re-reading this document

I would like to thank my family and friends for keeping me motivated throughoutthis process and for making my transition to Texas all the easier I would like to thank

my parents for always believing in me and helping me become the person I am today.Finally, I would like to thank my wife Barbara for supporting me in all that I

do, the least of which is this dissertation

Thanks!

v

In this dissertation we will be concerned with the existence of positive solutions

of singular boundary value problems We will begin in Chapter 2 by looking forsolutions u of singular discrete boundary value problems of the form,

(−1)n∆nu(t − (n − 1)) + f (t, u(t), , ∆n−1u(t − (n − 1))) = 0, t ∈ [n − 1, T + 1], (1.1)

satisfying boundary conditions,

∆n−1u(0) = ∆n−2u(T + 2) = · · · = u(T + n) = 0, (1.2)

where [n − 1, T + 1] = {n − 1, n, , T, T + 1}, T ∈ N, n ≥ 2 We prove existence ofpositive solutions of (1.1), (1.2) by means of a lower and upper solutions method, theBrouwer fixed point theorem, and by perturbation methods to approximate regularproblems We will first look at the case where n = 3 in order to get a better under-standing of the methods involved in the higher order case Following the n = 3 case,

we will give the generalization

Discrete boundary value problems (sometimes referred to as boundary valueproblems for difference equations) have been studied extensively in the recent past.Most notable is the work done by Agarwal and his collaborators For a few of theirworks, see [1], [6] and [8] Other mathematicians have also studied discrete boundaryvalue problems The interested reader should refer to [3], [16], [17], [21], [22], [24],[27], [28], [29], [34], [35], [36] and [37] The class of singular discrete boundary valueproblems has been studied in [27], [28] and [32]

As is stated above, we employ the use of a lower and upper solutions method

to prove the existence of positive solutions The lower and upper solutions methodshave been employed by many others, and a brief list of works include [2], [9], [12], [13],

1

[38] and [40] If one were to look to study positive solutions, one may look towards[7], [10], [19], [20] and [41] as a starting point It was also stated that we will use aperturbation method to approximate our regular problem Again, others have usedthis technique, and a brief summary of their works can be found in [1], [31] and [33].Primary motivation for the results of this dissertation is the recent paper onthe existence of singular boundary value problems by Rach ˙unkov´a and Rach ˙unek [39].They studied a second order boundary value problem for the discrete p-Laplacian,

ϕp(x) = |x|p−2x, p > 1 In particular, Rach ˙unkov´a and Rach ˙unek dealt with thediscrete boundary value problem,

∆ (ϕp(∆u(t − 1))) + f (t, u(t), ∆u(t − 1)) = 0, t ∈ [1, T + 1],

∆u(0) = u(T + 2) = 0

(1.3)

We begin our study by focussing on p = 2 for (1.3), or in particular, for (1.1), (1.2),when n = 2 We do not include this case in the dissertation because it would effectivelyamount to the removal of the p-Laplacian in their paper The discrete p-Laplacian inreference to boundary value problems is an active field, and some other work in thearea can be found in [11], [26] and of course in [39]

Moving along to Chapter 3, we make a transition and consider singular ary value problems for ordinary differential equations of the form,

difficulties arose for higher order problems which were not present in the second ordercase In particular, we were forced to use a slightly different approach to the lowerand upper solution method for the higher order results In view of this, we includethe third order results as a means to more fully understand the methods of proofinvolved in the generalization We then conclude the chapter with the higher ordergeneralization that will in fact work for all values of n ≥ 2, however the initial methodfor n = 2 will be a slightly stronger version than this generalization.

Boundary value problems for ordinary differential equations have been studiedextensively in the recent past To see just a few of these recent works, the interestedreader should refer to [1], [5], [6], [14], [15] and [25] The class of singular boundaryvalue problems also has been studied a great deal; see [5], [4], [23], [25] and [32]

As was stated, we again employ the use of a lower and upper solutions method

to prove the existence of positive solutions This method has been employed by manyothers, and a brief list of works includes [14], [15] and [18] For works somewhatrelated to those of this dissertation in the study of positive solutions, one may looktowards [4], [23] and [30] Again, we will use a perturbation method to approximateour regular problem We mention that others have also used this technique in thedifferential equations setting, and a few such works can be found in [1], [4] and [30]

Singular Boundary Value Problems for Difference Equations

This chapter is devoted to the study of solutions of singular boundary valueproblems for difference equations of the form,

(−1)n∆nu(t − (n − 1)) + f (t, u(t), , ∆n−1u(t − (n − 1))) = 0, t ∈ [n − 1, T + 1], (2.1)satisfying boundary conditions,

∆n−1u(0) = ∆n−2u(T + 2) = · · · = u(T + n) = 0, (2.2)where [n − 1, T + 1] = {n − 1, n, , T, T + 1}, T ∈ N, n ≥ 2

In order to better understand this type of difference equation, we must firstdefine a few terms which will be used through the duration of this chapter We begin

by defining the discrete interval and the standard forward difference operator.Definition 2.1 Let a < b be integers Define the discrete interval [a, b] := {a, a +

in detail existence results for the singular third order nonlinear difference equation,

−∆3u(t − 2) + f (t, u(t), ∆u(t − 1), ∆2u(t − 2)) = 0, t ∈ [2, T + 1], (2.3)with mixed boundary conditions,

∆2u(0) = ∆u(T + 2) = u(T + 3) = 0 (2.4)Our goal is to prove the existence of a positive solution of the problem (2.3),(2.4) To this end, we must define what we refer to as a positive solution

Definition 2.3 By a solution u of problem (2.3), (2.4) we mean a function u : [0, T +3] → R such that u satisfies the difference equation (2.3) on [2, T +1] and the boundaryconditions (2.4) If u(t) > 0 for t ∈ [2, T + 1], we say u is a positive solution of theproblem (2.3), (2.4)

Now that we have defined what it is to be a solution, we can look more closely atour difference equation (2.3) and begin to place some of the underlying assumptions

on our function f Before we can do this, however, we need a few more definitions.Definition 2.4 Let D ⊆ R3 We say that f is continuous on [2, T + 1] × D, if

f (·, x0, x1, x2) is defined on [2, T + 1] for each (x0, x1, x2) ∈ D and if f (t, ·, ·, ·) iscontinuous on D for each t ∈ [2, T + 1]

Definition 2.5 If D = R3, problem (2.3), (2.4) is called regular If D ( R3 and f hassingularities on the boundary of D, ∂D, then problem (2.3), (2.4) is singular

We will assume throughout this section that the following assumptions hold:A: D = (0, ∞) × R2

B: f is continuous on [2, T + 1] × D

C: f (t, x0, x1, x2) has a singularity at x0 = 0, i.e lim sup

x 0 →0 +

|f (t, x0, x1, x2)| = ∞for each t ∈ [2, T + 1] and for some (x1, x2) ∈ R2

2.1.2 Lower and Upper Solutions Method for Regular Problems

In this section, we will look at a regular problem that is similar to (2.3), (2.4)

To this end, let h be continuous on [2, T + 1] × R3 and let us consider the differenceequation,

−∆3u(t − 2) + h(t, u(t), ∆u(t − 1), ∆2u(t − 2)) = 0, t ∈ [2, T + 1] (2.5)

We establish a lower and upper solutions method for this regular problem (2.5), (2.4).However, before we can establish this, we must first define what it means to be a lowersolution and an upper solution of (2.5), (2.4)

Definition 2.6 α : [0, T + 3] → R is called a lower solution of (2.5), (2.4) if,

−∆3α(t − 2) + h(t, α(t), ∆α(t − 1), ∆2α(t − 2)) ≥ 0, t ∈ [2, T + 1], (2.6)

satisfying boundary conditions,

∆2α(0) ≤ 0,

∆α(T + 2) ≥ 0,α(T + 3) ≤ 0

(2.9)

We now have the necessary definitions to introduce our lower and upper tions method result

solu-Theorem 2.1 (Lower and Upper Solutions Method) Let α and β be lower and uppersolutions of (2.5), (2.4), respectively, and α ≤ β on [2, T + 1] Let h(t, x0, x1, x2) becontinuous on [2, T + 1] × R3 and nonincreasing in its x2 variable Then (2.5), (2.4)has a solution u(t) satisfying,

for y < ∆β(t − 1),where

|eh(t, x, y, z)| ≤ M, t ∈ [2, T + 1], (x, y, z) ∈ R3

We now study the auxiliary equation,

−∆3u(t − 2) + eh(t, u(t), ∆u(t − 1), ∆2u(t − 2)) = 0, t ∈ [2, T + 1], (2.11)

satisfying boundary conditions (2.4) Our immediate goal is to prove the existence of

a solution to the boundary value problem (2.11), (2.4)

Step 2 We now lay the foundation to use the Brouwer fixed point theorem To thisend, define

E = {u : [0, T + 3] → R : ∆2u(0) = ∆u(T + 2) = u(T + 3) = 0},

and also define

kuk = max{|u(t)| : t ∈ [0, T + 3]}

E is a Banach space Further, we define an operator T : E → E by,

(T u)(t) =

T +2X

j 2 =t+1

T +2X

j 1 =j 2

j 1Xi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2)) (2.12)

Clearly, T is a continuous operator Let

r >

T +2X

j 2 =1

T +2X

j 1 =j 2(j1− 1)M,

u ∈ E and thus, satisfies (2.4) Further,

∆u(t − 1) = u(t) − u(t − 1)

= T u(t) − T u(t − 1)

=

T +2X

j 2 =t+1

T +2X

j 1 =j 2

j 1Xi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

−

T +2X

j 2 =t

T +2X

j 1 =j 2

j 1Xi=2

eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

= −

T +2X

j 1 =t

j 1Xi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

And applying this equality, we see that,

∆2u(t − 2) = ∆u(t − 1) − ∆u(t − 2)

T +2X

j 1 =t

j 1Xi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

!

T +2X

j 1 =t−1

j 1Xi=2

eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

!

=

t−1Xi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

In a similar fashion, we apply this equality to yield,

∆3u(t − 2) = ∆2u(t − 1) − ∆2u(t − 2)

=

tXi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

−

t−1Xi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

= eh(t, u(t), ∆u(t − 1), ∆2u(t − 2))

In particular,

−∆3u(t − 2) + eh(t, u(t), ∆u(t − 1), ∆2u(t − 2)) = 0,

and (2.11) is satisfied

Now assume u solves (2.11), (2.4) Then u ∈ E and from (2.4), we see that,

∆3u(0) = ∆2u(1) − ∆2u(0)

= ∆2u(1)

= eh(2, u(2), ∆u(1), ∆2u(0))

Thus, ∆2u(1) = eh(2, u(2), ∆u(1), ∆2u(0)) Also,

∆3u(1) = ∆2u(2) − ∆2u(1)

= ∆2u(2) − eh(2, u(2), ∆u(1), ∆2u(0))

= eh(3, u(3), ∆u(2), ∆2u(1))

Thus, ∆2u(2) =

3Pi=2

eh(i, u(i), ∆u(i − 1), ∆2u(i − 2)) Continuing inductively, we clude

con-∆2u(t − 1) =

tXi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2)) (2.13)Note: We adopt the convention that, if the upper summation index is smaller thanthe lower summation index, then the sum is zero

From (2.13) and (2.4), we see that

∆2u(T + 1) = ∆u(T + 2) − ∆u(T + 1)

= −∆u(T + 1)

=

T +2Xi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

Thus, ∆u(T + 1) = −

T +2Pi=2

eh(i, u(i), ∆u(i − 1), ∆2u(i − 2)) Also,

∆2u(T ) = ∆u(T + 1) − ∆u(T )

= −

T +2Xi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2)) − ∆u(T )

= −

T +1Xi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

It follows that,

∆u(T ) = −

T +2X

j 1 =T +1

j 1Xi=2

eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

Continuing inductively, we conclude

∆u(t) = −

T +2X

j 1 =t+1

j 1Xi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2)) (2.14)

From (2.14) and (2.4), we see that

∆u(T + 2) = u(T + 3) − u(T + 2)

= −u(T + 2)

= 0

In particular, u(T + 2) = 0 Also,

∆u(T + 1) = u(T + 2) − u(T + 1)

= −u(T + 1)

= −

T +2X

j 1 =T +2

j 1Xi=2

eh(i, u(i), ∆u(i − 1), ∆2u(i − 2)),and so,

u(T + 1) =

T +2X

j 1 =T +2

j 1Xi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

Also,

∆u(T ) = u(T + 1) − u(T )

=

T +2X

j 1 =T +2

j 1Xi=2

eh(i, u(i), ∆u(i − 1), ∆2u(i − 2)) − u(T )

= −

T +2X

j 1 =T +1

j 1Xi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

So we have,

u(T ) =

T +2X

j =T +1

T +2X

j =j

j 1Xi=2

eh(i, u(i), ∆u(i − 1), ∆2u(i − 2))

Continuing inductively, we conclude

u(t) =

T +2X

j 2 =t+1

T +2X

j 1 =j 2

j 1Xi=2eh(i, u(i), ∆u(i − 1), ∆2u(i − 2)) (2.15)

It is clear from (2.15) that u(t) = T u(t) Thus our claim holds

Step 4 We now show that solutions u(t) of (2.11), (2.4) satisfy (2.10)

Consider the case of obtaining u(t) ≤ β(t) We show in fact that ∆β(t) ≤ ∆u(t)from which we then argue u(t) ≤ β(t) Let v(t) = β(t) − u(t) Then ∆v(t) = ∆β(t) −

∆u(t) For the sake of establishing a contradiction, assume that max{∆v(t) : t ∈[0, T + 2]} := ∆v(l) > 0

Conditions (2.4) and (2.9) imply that l ∈ [1, T + 1] Thus, ∆v(l + 1) ≤ ∆v(l)and ∆v(l − 1) ≤ ∆v(l) Consequently, ∆2v(l) ≤ 0 and ∆2v(l − 1) ≥ 0 This in turnimplies that ∆3v(l − 1) ≤ 0 Therefore,

On the other hand, since h is nonincreasing in its fourth variable, from (2.5),

∆3β(l − 1) − ∆3u(l − 1) = −eh(l + 1, u(l + 1), ∆u(l), ∆2u(l − 1)) + ∆3β(l − 1)

At this we note from the definition of eh that u(t) is a solution of (2.5), (2.4) Hence,

T +3Xl=t

∆β(l) ≤

T +3Xl=t

∆u(l)

This, in turn, yields

β(T + 3) − β(t) ≤ u(T + 3) − u(t),β(T + 3) − β(t) ≤ −u(t),

u(t) ≤ β(t) − β(T + 3),u(t) ≤ β(t)

A similar argument shows that α(t) ≤ u(t)

Thus, the conclusion of the theorem holds and our proof is complete

Proof We again proceed through a sequence of steps

Step 1 For k ∈ N, t ∈ [2, T + 1], and x0, x1, x2 ∈ R, define

Then fkis continuous on [2, T +1]×R3and nonincreasing for t ∈ [2, T +1], x0 ∈ [−c, c].Assumption (F) implies that there exists k0, such that, for all k ≥ k0,

−∆3u(t − 2) + fk(t, u(t), ∆u(t − 1), ∆2u(t − 2)) = 0, t ∈ [2, T + 1] (2.17k)

We now have a sequence of regular problems {(2.17k), (2.4)}∞k=k0 Let α(t) = 0 andβ(t) = c Then, for each k ≥ k0, α and β are lower and upper solutions for (2.17k),(2.4) and α(t) ≤ β(t) on [0, T + 3] Thus, by Theorem 2.1, there exists uk(t) a solution

of (2.17k), (2.4) satisfying 0 ≤ uk(t) ≤ c, t ∈ [0, T + 3], k ≥ k0 Consequently,

|∆uk(t)| ≤ c, t ∈ [0, T + 2] (2.18)Step 2 Let k ∈ N, k ≥ k0 Since uk(t) solves (2.17k), we get from our work inTheorem 2.1,

∆uk(t) = −

T +2X

j 1 =t+1

j 1Xi=2

fk(i, uk(i), ∆uk(i − 1), ∆2uk(i − 2)) (2.19)

By assumption (F), there exists ε1 ∈0,k1

0

such that if k ≥ ε1

j 1 =2

j 1Xi=2

fk(i, uk(i), ∆uk(i − 1), ∆2uk(i − 2))

= −

T +2X

j 1 =3

j 1Xi=2

fk(i, uk(i), ∆uk(i − 1), ∆2uk(i − 2))

−fk(2, uk(2), ∆uk(1), ∆2uk(0))

< −c −

T +2X

j 1 =3

j 1Xi=2

fk(i, uk(i), ∆uk(i − 1), ∆2uk(i − 2))

< −c

But this contradicts (2.18) Hence uk(1) ≥ ε1, for all k ≥ ε1

1.Denote m2 = max{|fk(2, x0, x1, x2)| : x0 ∈ [ε1, c], x1 ∈ [−c, c]} By assumption(F), there exists ε2 ∈ (0, ε1] such that, if k ≥ ε1

j 1 =3

j 1Xi=2

fk(i, uk(i), ∆uk(i − 1), ∆2uk(i − 2))

= −

T +2X

j 1 =3

j 1Xi=3

fk(i, uk(i), ∆uk(i − 1), ∆2uk(i − 2))

−T · fk(2, uk(2), ∆uk(1), ∆2uk(0))

= −

T +2X

j 1 =4

j 1Xi=3

fk(i, uk(i), ∆uk(i − 1), ∆2uk(i − 2))

−fk(3, uk(3), ∆uk(2), ∆2uk(1))

−T · fk(2, uk(2), ∆uk(1), ∆2uk(0))

≤ −

T +2X

j 1 =4

j 1Xi=3

fk(i, uk(i), ∆uk(i − 1), ∆2uk(i − 2))

−fk(3, uk(3), ∆uk(2), ∆2uk(1)) + T · m2

< −

T +2X

j 1 =4

j 1Xi=3

fk(i, uk(i), ∆uk(i − 1), ∆2uk(i − 2))

−(c + T · m2) + T · m2

< −c

But this contradicts (2.18) Hence uk(2) ≥ ε2, for all k ≥ ε1

2.Continuing similarly for t = 3, 4, , T, we get 0 < εT < · · · < ε2 < ε1 suchthat uk(t) ≥ εT, for t ∈ [1, T ]

For 2 ≤ i ≤ T, denote mi = max{|fk(i, x0, x1, x2)| : x0 ∈ [εi, c], x1 ∈ [−c, c]} Byassumption (F), there exists εT +1 ∈ (0, εT] such that, if k ≥ ε1

T +1 and uk(T +1) < εT +1,then

fk(T + 1, x0, x1, x2) > c +

TXi=2

mi, x0 ∈ (0, εT], x1 ∈ [−c, c]

∆uk(T + 1) = −

T +2X

j 1 =T +2

j 1Xi=2

fk(i, uk(i), ∆uk(i − 1), ∆2uk(i − 2))

= −

T +2Xi=2

fk(i, uk(i), ∆uk(i − 1), ∆2uk(i − 2))

= −

T +1Xi=2

fk(i, uk(i), ∆uk(i − 1), ∆2uk(i − 2))

−fk(T + 2, uk(T + 2), ∆uk(T + 1), ∆2uk(T ))

< −

T +1Xi=2

mi− (c −

T +1Xi=2

mi)

< −c

But this contradicts (2.18) Hence uk(T + 1) ≥ εT +1, for all k ≥ ε1

T +1 Therefore, byletting ε = εT +1, we get

j 1 =t+1

j 1Xi=2

f (i, ukn(i), ∆ukn(i − 1), ∆2ukn(i − 2)),

and so letting n → ∞ and from the continuity of f, we get that

∆u(t) = −

T +2X

j 1 =t+1

j 1Xi=2

f (i, u(i), ∆u(i − 1), ∆2u(i − 2))

Consequently,

∆2u(t − 1) =

tXi=2

f (i, u(i), ∆u(i − 1), ∆2u(i − 2))

Thus,

∆3u(t − 2) = f (t, u(t), ∆u(t − 1), ∆2u(t − 2))

Therefore, u satisfies (2.3) and by (2.21), our theorem holds

We have therefore proved the existence of a positive solution of the third orderboundary value problem, and as such, have exhibited the methods to be used in thehigher order case, which is to come in the next section of this chapter We first,however consider an example to illustrate the third order result.

∆2u(0) = ∆u(T + 2) = u(T + 3) = 0

yield a positive solution, u(t)

2.2 Higher Order

2.2.1 Preliminaries

We now consider a generalization of the results given in the previous sectionconcerning the singular discrete boundary value problem (2.1), (2.2) when n = 3 Re-ferring back to the definitions of the discrete interval, Definition 2.1, and the standardforward difference operator, Definition 2.2, we now consider the higher order singulardifference equation,

(−1)n∆nu(t−(n−1))+f (t, u(t), , ∆n−1u(t−(n−1))) = 0, t ∈ [n−1, T +1], (2.22)

with mixed boundary conditions,

∆n−1u(0) = ∆n−2u(T + 2) = ∆n−3u(T + 3) = · · · = u(T + n) = 0 (2.23)

Again, ∆ denotes the forward difference operator with step size 1, i.e ∆u(t) =u(t + 1) − u(t) and for n > 1, ∆nu(t) = ∆(∆n−1u(t)) Our goal is to prove theexistence of a positive solution of problem (2.22), (2.23) when n ≥ 2.

Definition 2.8 By a solution u of problem (2.22), (2.23) we mean u : [0, T + n] → Rsuch that u satisfies the difference equation (2.22) on [n − 1, T + 1] and the boundaryconditions (2.23) If u(t) > 0 for t ∈ [n − 1, T + 1], we say u is a positive solution ofthe problem (2.22), (2.23)

Definition 2.9 Let D ⊆ Rn such that f : [n − 1, T + 1] × D is continuous If D = Rn,problem (2.22), (2.23) is called regular If D ( Rn and f has singularities on ∂D,then problem (2.22), (2.23) is singular

We will assume throughout this section that the following hold:

2.2.2 Lower and Upper Solutions Method for Regular Problems

Let h be continuous on [n − 1, T + 1] × Rn and let us consider the regular differenceequation,

(−1)n∆nu(t−(n−1))+h(t, u(t), , ∆n−1u(t−(n−1))) = 0, t ∈ [n−1, T +1] (2.24)

We establish a lower and upper solutions method for the regular problem (2.24),(2.23)

Definition 2.10 α : [0, T + n] → R is called a lower solution of (2.24), (2.23) if,(−1)n∆nα(t−(n−1))+h(t, α(t), , ∆n−1α(t−(n−1))) ≥ 0, t ∈ [n−1, T +1], (2.25)

satisfying boundary conditions

(−1)n−1∆n−1α(0) ≤ 0,(−1)n−1∆n−2α(T + 2)) ≥ 0,

.α(T + n) ≤ 0

.β(T + n) ≥ 0

Step 1 For t ∈ [n − 1, T + 1], and (x0, , xn−1) ∈ Rn, define

h(t, x0, , xn−2− σ(t − (n − 2), xn−1)),

for (−1)n−1∆n−2β(t − (n − 2)) ≤ (−1)n−1xn−2

and(−1)n−1xn−2≤ (−1)n−1∆n−2α(t − (n − 2))

h(t, β(t), , ∆n−2β(t − (n − 2)),

∆n−2β(t − (n − 2)) − σ(t − (n − 2), xn−1))

−(−1)n−1 (−1) n−1 (∆ n−2 β(t−(n−2))−x n−2 )

(−1) n−1 (∆ n−2 β(t−(n−2))−x n−2 )+1,for (−1)n−1xn−2< (−1)n−1∆n−2β(t − (n − 2))

where the function σ(t − (n − 2), z) is defined by

∆n−2β(t − (n − 1)), (−1)n−1z < (−1)n−1∆n−2β(t − (n − 1)),This function, eh, is a bounded modification of the function h by means of thegiven lower and upper solutions In addition, it is clear that eh is continuous Thus,e

h is continuous on [n − 1, T + 1] × Rn and there exists M > 0 so that,

|eh(t, x0, , xn−1)| ≤ M, t ∈ [n − 1, T + 1], (x0, , xn−1) ∈ Rn

We now study the auxiliary equation,

(−1)n−1∆nu(t − (n − 1)) + eh(t, u(t), , ∆n−1u(t − (n − 1))) = 0, t ∈ [n − 1, T + 1],

(2.30)satisfying boundary conditions (2.23) Our immediate goal is to prove the existence

of a solution of (2.30), (2.23)

Step 2 We lay the foundation to use the Brouwer fixed point theorem To this end,define

E = {u : [0, T + 3] → R : ∆n−1u(0) = ∆n−2u(T + 2) = · · · = u(T + n) = 0}

and also define

kuk = max{|u(t)| : t ∈ [0, T + n]}

E is a Banach space Further, we define an operator T : E → E by,

(T u)(t) = −

T +n−1X

j n−1 =t+n−2

· · ·

T +n−1X

j 1 =j 2

j 1Xi=n−1eh(i, u(i), , ∆n−1u(i − (n − 1))) (2.31)

T is a continuous operator Moreover, from the bounds placed on eh in Step 1 andfrom (2.31), along with B(r) = {u ∈ E : kuk < r}, we have that T (B(r)) ⊂ B(r).More precisely, if

r >

T +n−1X

j n−1 =1

· · ·

T +n−1Xs=j 2(s − (n − 2))M,and if u ∈ E, with kuk < r, then for each t ∈ [n − 1, T + 1], we have that

|T u(t)| ≤

−

T +n−1X

j n−1 =t+n−2

· · ·

T +n−1X

j 1 =j 2

j 1Xi=n−1eh(i, u(i), , ∆n−1u(i − (n − 1)))

... u(t).

At this, we note from the definition of eh that u(t) is a solution of (2.24), (2.23).Thus, the conclusion of the theorem holds and our proof is complete

Theorem 2.4 Assume conditions... class="page_container" data-page="34">

Step We now show that solutions u(t) of (2.30), (2.23) satisfy,

α(t) ≤ u(t) ≤ β(t), t ∈ [0, T + 2]

Consider the case of obtaining u(t) ≤ β(t) We show in fact that... solution u satisfying,

0 < u(t) ≤ c, t ∈ [0, T + 1]

Proof Again for the proof, we proceed through a sequence of steps

Step For k ∈ N, t ∈ [n − 1, T + 1], and (x0,