Construction of upper and lower solutions for singular discrete initial and boundary value problems via inequality theory

CONSTRUCTION OF UPPER AND LOWER SOLUTIONS FOR

SINGULAR DISCRETE INITIAL AND BOUNDARY VALUE

PROBLEMS VIA INEQUALITY THEORY

HAISHEN L ¨U AND DONAL O’REGAN

Received 25 May 2004

We present new existence results for singular discrete initial and boundary value prob-lems. In particular our nonlinearity may be singular in its dependent variable and is al-lowed to change sign.

1. Introduction

An upper- and lower-solution theory is presented for the singular discrete boundary value problem

−∆ϕp∆u(k−1)=q(k)fk,u(k), k∈N= {1,...,T},

u(0)=u(T+ 1)=0, (1.1)

and the singular discrete initial value problem

∆u(k−1)=q(k)fk,u(k), k∈N= {1,...,T},

u(0)=0, (1.2)

whereϕp(s)=|s|p−2_{s, p >1,∆u(k−1)=u(k)−u(k−1),T∈{1, 2,...},N}+_{={0, 1,...,T},}

andu:N+_{→R.Throughout this paper, we will assumef :N×(0,∞)→Ris continuous.}

As a result, our nonlinearity f(k,u) may be singular atu=0 and may change sign. Remark 1.1. Recall a map f :N×(0,∞)→Ris continuous if it is continuous as a map of the topological spaceN×(0,∞) into the topological spaceR. Throughout this paper, the topolopy onNwill be the discrete topology.

We will letC(N+_{,R) denote the class of mapucontinuous onN}+_{(discrete topology)}

with normu =maxk∈N+u(k). By a solution to (1.1) (resp., (1.2)) we mean au∈

C(N+_{,R) such thatusatisfies (1.1) (resp., (1.2)) fori∈N andusatisfies the boundary}

(resp., initial) condition.

It is interesting to note here that the existence of solutions to singular initial and boundary value problems in the continuous case have been studied in great detail in

the literature (see [2,4,5,6,7,9,10,11] and the references therein). However, only a few papers have discussed the discrete singular case (see [1,3,8] and the references therein).

In [7], the following result has been proved.

Theorem1.2. Letn0∈ {1, 2,...}be fixed and suppose the following conditions are satisfied: f :N×(0,∞)−→Ris continuous, (1.3)

q∈CN, (0,∞), (1.4)

there exists a functionα∈CN+,Rwith

α(0)=α(T+ 1)=0, α >0onNsuch that

q(k)fk,α(k)≥ −∆ϕpα(k−1) fork∈N,

(1.5)

there exists a functionβ∈C(N+,R)with

β(k)≥α(k), β(k)≥_n1

0 fork∈N +_with q(k)fk,β(k)≤ −∆ϕpβ(k−1) fork∈N.

(1.6)

Then (1.1) has a solutionu∈C(N+_{,R)withu(k)≥α(k)fork∈N}+_.

In [1], the following result has been proved.

Theorem1.3. Letn0∈ {1, 2,...}be fixed and suppose the following conditions are satisfied: f :N×(0,∞)−→Ris continuous, (1.7)

q∈CN, (0,∞), (1.8)

there exists a functionα∈CN+_,Rwith α(0)=0, α >0onNsuch that

q(k)fk,α(k)≥∆α(k−1) fork∈N,

(1.9)

there exists a functionβ∈CN+_,Rwith β(k)≥α(k), β(k)> 1

n0 fork∈N +_with q(k)fk,β(k)≤∆β(k−1) fork∈N.

(1.10)

Then (1.2) has a solutionu∈C(N+_{,R)withu(k)≥α(k)fork∈N}+_.

Also some results from the literature, which will be needed inSection 2are presented.

Lemma1.4 [8]. Letu∈C(N+_{,R)satisfyu(k)≥0fork∈N}+_{. Ifu∈C(N}+_,R)satisfies

−∆2_{u(k−1)=u(k), k∈N= {1, 2,...,T},}

then

In Theorems1.2and1.3the construction of a lower solutionαand an upper solution

βis critical. We present an easily verifiable condition inSection 2.

2. Main results

We begin with a result for boundary value problems.

Theorem2.1. Let n0∈ {1, 2,...}be fixed and suppose (1.3), (1.4) hold. Also assume the following conditions are satisfied:

there exists a constantc0>0such that

q(k)f(k,u)≥c0 fork∈N, 0< u≤ 1 n0,

(2.1)

Proof. First we construct the lower solutionαin (1.5). Letα(k)=cv(k), k∈N+_{, where} v∈C(N+_{, [0,∞)) is the solution of}

−∆ϕp∆v(k−1)=1, k∈N,

v(0)=v(T+ 1)=0, (2.5)

0< c <min

c1/(p−1)

0 ,

1

n0v

. (2.6)

Since−∆(ϕp(∆v(k−1)))>0 implies∆2_{v(k−1)<0 fork∈N, it follows fromLemma 1.4}

thatv(k)≥µ(k)vfork∈N+_{. Thus,}

0< α(k)≤ 1

n0

fork∈N, (2.7)

−∆ϕp∆α(k−1)=cp−1_≤c

0 fork∈N,

α(0)=α(T+ 1)=0. (2.8)

As a result (1.5) holds, since

q(k)fk,α(k)≥c0≥ −∆

ϕp∆α(k−1) fork∈N. (2.9) Next we discuss the boundary value problem

−∆ϕp∆u(k−1)=q(k)h(M), k∈N,

u(0)=u(T+ 1)=_n1

0.

(2.10)

It follows from [8] that (2.10) has a solutionu∈C(N+_{,R). Letv(k)=u(k)−1/n} 0 for k∈N+_{. Then∆(ϕp(∆u(k−1)))= −∆(ϕp(∆v(k−1)))≤0 fork∈N, andv(0)=v(T+}

1)=0.Lemma 1.5guarantees thatv(k)≥0 and sou(k)≥1/n0fork∈N+. Next we prove u(k)≤Mfork∈N+_{. Now since∆(ϕp(∆u(k−1)))≤0 onNimplies∆}2_{u(k−1)≤0 on} N, then there existsk0∈Nwith∆u(k)≥0 on [0,k0)= {0, 1,...,k0−1}and∆u(k)≤0 on

[k0,T+ 1)= {k0,k0+ 1,...,T}, andu(k0)= u. Supposeu(k0)> M.

Also notice that fork∈N, we have

−∆ϕp∆u(k−1)=q(k)h(M). (2.11) We sum (2.11) fromj+ 1 (0≤j < k0) tok0to obtain

ϕp∆u(j)=ϕp∆uk0

+h(M)

k0

k=j+1

q(k). (2.12)

Now since∆u(k0)≤0, we have

ϕp∆u(j)≤h(M)

k0

k=j+1

that is,

Now (2.15) and (2.18) imply

Example 2.2. Consider the boundary value problem

To see this, we will applyTheorem 2.1with

q(k)=1, f(k,u)=_uk_α+uβ−A. (2.24)

Next we present a result for initial value problems.

Theorem2.3. Let n0∈ {1, 2,...}be fixed and suppose (1.2), (1.3) hold. Also assume the following conditions are satisfied:

there exists a constantc0>0such that

q(k)f(k,u)≥c0 fork∈N, 0< u≤_n1 0,

(2.28)

Proof. First we construct the lower solutionαin (1.9). Let

Next we discuss the initial value problem

∆u(k−1)=q(k)f∗k,u(k), k∈N,

Then (2.34) is equivalent to

u(k)=

Suppose (2.37) is not true. Then there exists aτ∈Nsuch that

u(τ)< 1

n0, u(τ−1)≥

1

sinceu(0)=1/n0. Thus we have, from (2.28)

∆u(τ−1)=q(τ)f∗τ,u(τ)=q(τ)f

τ, 1

n0

>0, (2.39)

so

u(τ)− 1

n0 > u(τ−1)−

1

n0≥0, (2.40)

a contradiction. Thus (2.37) is satisfied. Next we show

u(k)≤M fork∈N+_{. (2.41)}

Suppose (2.41) is false. Then sinceu(0)=1/n0, there existsτ∈Nsuch that

u(τ)> M, u(k)≤M fork∈ {0, 1,...,τ−1}. (2.42)

Thus, we have

∆u(τ−1)=u(τ)−u(τ−1)≤q(τ)h(M), ∆u(τ−2)=u(τ−1)−u(τ−2)≤q(τ−1)h(M),

.. .

∆u(0)=u(1)−u(0)≤q(1)h(M).

(2.43)

Adding both sides of the above formula gives

u(τ)−u(0)≤h(M)

τ

k=1

q(k)≤h(M)

T

k=1

q(k), (2.44)

that is,

M− 1

n0 ≤h(M) T

k=1

q(k). (2.45)

This contradicts (2.30). Thus, we have (2.20). Letβ(k)≡u(k) fork∈N+_{. By (2.7) and}

(2.37), we haveα(k)≤β(k) fork∈N+_{. Then}

β∈CN+_,Rwith β(k)≥α(k), β(k)> 1

n0

fork∈N+_with q(k)fk,β(k)=∆β(k−1) fork∈N.

NowTheorem 1.3guarantees that (1.2) has a solutionu∈C(N+_{,R) withu(k)≥α(k)>0}

fork∈N.

Example 2.4. Consider the initial value problem

∆u(k−1)=ku(k)−α+u(k)β−A, k∈N,

u(0)=0 (2.47)

withα >0, 0≤β <1, andA >0. Now (2.47) has a solutionu∈C(N+_{,R) withu(k)>0}

fork∈N.

To see this we will applyTheorem 2.3with (2.24). Letn0>(2A)1/αandc0=A.Then for k∈Nand 0< u≤1/n0, (2.25) holds and so (2.28) is satisfied. Leth(u)=uβ+nα0T+A.

Then (2.29) is immediate. Also since 0≤β <1, we see

there existM > 1

n0

such thatM− 1

n0> T

Mβ_+nα 0T+A

, (2.48)

so (2.30) holds. Theorem 2.3guarantees that (2.47) has a solution u∈C(N+_{,R) with} u(k)>0 fork∈N.

Acknowledgment

The research is supported by National Natural Science Foundation (NNSF) of China Grant 10301033.

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Haishen L¨u: Department of Applied Mathematics, Hohai University, Nanjing 210098, China

E-mail address:[email protected]

Donal O’Regan: Department of Mathematics, National University of Ireland, Galway, Ireland