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Existence of positive periodic solutions for higher order singular functional difference equations

Jacob D. Johnson

1

, Lingju Kong

B2

, Michael G. Ruddy

3

and Alexander M. Ruys de Perez

4

1Department of Mathematics, University of Tennessee at Knoxville, Knoxville, TN 37916, USA

2Department of Mathematics, University of Tennessee at Chattanooga, Chattanooga, TN 37403, USA

3Department of Mathematics, University of Tennessee at Martin, Martin, TN 38237, USA

4Department of Mathematics,Vanderbilt University, Nashville, TN 37235, USA

Received 14 January 2014, appeared 13 March 2014 Communicated by Paul Eloe

Abstract. We study a higher order singular functional difference equation onZ. Suffi- cient conditions are obtained for the existence of at least one positive periodic solution of the equation. Our proof utilizes the nonlinear alternative of Leray–Schauder.

Keywords: positive periodic solutions, higher order, functional difference equations, singular, nonlinear alternative of Leray–Schauder.

2010 Mathematics Subject Classification:39A23, 39A10.

1 Introduction

Nonlinear difference equations have numerous applications in modeling processes in biology, physics, statistics, and many other areas. For this reason, the existence of positive solutions to these equations is of great interest to many researchers. We refer the reader to [6–12,15–18]

for some recent work on this subject. In this paper, we are concerned with a higher order functional difference equation. To introduce our equation, we let a 6= 1, b 6= 1 be any fixed positive numbers, andm,k,ωbe any fixed positive integers, and for anyu: ZR, define

Lu(n) =u(n+m+k)−au(n+m)−bu(n+k) +abu(n).

Here, we study the existence of positive periodic solutions of the higher order functional dif- ference equation

Lu(n) = f(n,u(n−τ(n))) +r(n), n∈Z, (1.1) where f: Z×(0,∞) → R, τ: ZZ, and r: ZR are ω-periodic on Z, and f(n,x) is continuous inx.

Equation (1.1) withr(n)≡0, i.e., the equation

Lu(n) = f(n,u(n−τ(n))), n∈Z,

BCorresponding author. Email: Lingju-Kong@utc.edu

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has been recently studied by Wang and Chen in [18] using Krasnosel’skii’s fixed point theorem.

When f(n,x)is nonsingular atx = 0, sufficient conditions were found there for the existence of positive periodic solutions. In this paper, we will establish a new existence criterion for equation (1.1). The nonlinear term f(n,x)is allowed to be singular atx =0 in (1.1). The proof will employ a nonlinear alternative of Leray–Schauder. Our approach involves examining a one-parameter family of nonsingular problems constructed from a sequence of nonsingular perturbations of f. For each of these nonsingular problems, we will apply the nonlinear al- ternative of Leray–Schauder to obtain the existence of at least one positive periodic solution.

From this sequence of solutions, we will extract a subsequence that converges to a positive periodic solution of (1.1). This type of technique has been successfully used in obtaining pos- itive solutions for several classes of singular problems, see, for example, [1,2,6,12,13]. Our proofs are partly motivated by these works. Other results on singular problems can be found in [3–5,7,14].

As a simple application of our general existence theorem, we also derive some sufficient conditions for the existence of at least one positive periodic solution of the functional difference equation

Lu=c(n)(u(n−τ(n)))α+µd(n)(u(n−τ(n)))β+r(n), n∈Z, (1.2) whereα ≥ 0 and β≥ 0 are constants,c,d, andr areω-periodic functions onZwithc(n) > 0 andd(n)≥0 onZ, andµ>0 is a parameter.

The remainder of this paper is laid out as follows. In Section2, we present our assumptions and main results. Some preliminary lemmas as well as the proofs are given in Section3.

2 Main results

In this paper, for anyc,d ∈Zwithc ≤d, let[c,d]Zdenote the discrete interval{c, . . . ,d}. For the functionr(n)given in equation (1.1), define a functionγ: ZRby

γ(n) =

ω i=1

ω j=1

G(i,j)r(nij), (2.1)

where

G(i,j) = a

ωbωaibj

(1−aω)(1−bω), i,j∈ [1,ω]Z, (2.2) and

nij = n+ (i−1)k+ (j−1)m. (2.3) Let

δ1 = a

ωbω

|(1−aω)(1−bω)|min{a1,aω} ·min{b1,bω} and

δ2= a

ωbω

|(1−aω)(1−bω)|max{a1,aω} ·max{b1,bω}. Then, (2.2) implies that

( δ1 ≤G(i,j)≤δ2 fori,j∈[1,ω]Z if(a−1)(b−1)>0,

δ1 ≤ −G(i,j)≤δ2 fori,j∈ [1,ω]Z if(a−1)(b−1)<0. (2.4) We make the following assumptions.

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(H1) (k,ω) = (m,ω) =1, where(x,y)denotes the greatest common divisor ofxandy;

(H2) either

(a) (a−1)(b−1)>0,γ(n)≥0 onZ, and there exist continuous, nonnegative functions g(x), h(x), and φ(n) such that g(x) > 0 is nonincreasing on(0,∞), h(x)/g(x) is nondecreasing on(0,∞), and

0≤ f(n,x+γ(n−τ(n)))≤φ(n)(g(x) +h(x)) for(n,x)∈ Z×(0,∞), or

(b) (a−1)(b−1)<0,γ(n)≥0 onZ, and there exist continuous, nonpositive functions g(x), h(x), and nonnegative φ(n) such that g(x) < 0 is nondecreasing on (0,∞), h(x)/g(x)is nondecreasing on(0,∞), and

0≥ f(n,x+γ(n−τ(n)))≥φ(n)(g(x) +h(x)) for(n,x)∈ Z×(0,∞); (H3) for eachq>0, there exists a continuous functionψq(n)such that either

(i) ψq(n)is nonnegative,ψq(n)>0 for somen∈[1,ω]Z, and

f(n,x+γ)≥ψq(n) for(n,x)∈[1,ω]Z×(0,q], if (H2)(a) holds, or

(ii) ψq(n)is nonpositive,ψq(n)<0 for somen∈[1,ω]Z, and

f(n,x+γ)≤ψq(n) for(n,x)∈[1,ω]Z×(0,q], if (H2)(b) holds;

(H4) there existsR>0 such that

R>δ2ωg(δR)

1+ h(R) g(R)

ω

i

=1

φ(i), whereδ=δ12∈(0, 1].

Now, we state our main results.

Theorem 2.1. Assume that (H1)–(H4) hold. Then equation (1.1) has at least one positive periodic solution y(n)satisfying y(n)>γ(n)on[1,ω]Zand0<maxn∈[1,ω]Z|y(n)−γ(n)|< R.

As a consequence of Theorem2.1, we have the following corollary.

Corollary 2.2. Assume that (H1) holds,(a−1)(b−1)>0, andγ(n)≥0onZ. Then, we have (i) ifβ<1, then equation(1.2)has at least one positive periodic solution for eachµ∈(0,); (ii) ifβ≥1, then there existsµ¯ >0such that equation(1.2)has at least one positive periodic solution

for eachµ∈(0, ¯µ).

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3 Proofs of the main results

Throughout this section, letXbe the set of all realω-periodic functions onZ. Then, equipped with the maximum normkuk=maxn∈[1,ω]

Z|u(n)|,Xis a Banach space.

Lemma3.1below can be proved using [18, Lemma 2.1].

Lemma 3.1. Assume (H1) holds. Then for any h∈X, u(n)is a periodic solution of the equation Lu(n) =h(n), n∈Z,

if and only if u(n)is a solution of the summation equation u(n) =

ω i=1

ω j=1

G(i,j)h(nij), where G(i,j)and nijare given by(2.2)and(2.3)respectively.

We refer the reader to [1, Theorem 1.2.3] for the following version of the well known non- linear alternative of Leray–Schauder.

Lemma 3.2. Let K be a convex subset of a normed linear space X, and letΩbe a bounded open subset withp˜ ∈Ω. Then every compact map N: Ω→K has at least one of the following properties:

(i) N has at least one fixed point inΩ;

(ii) there is u∈ ∂Ωandλ∈(0, 1)such that u= (1−λ)p˜+λNu.

For anyh∈X, it is easy to see that

ω i=1

ω j=1

h(nij) =ω

ω i=1

h(i). We will use this identity in the proof of Theorem 2.1.

Now, we are ready to prove our results.

Proof of Theorem2.1. We show the case where (H2)(a) holds. The proof for (H2)(b) is similar.

LetΩ = {u ∈ X : kuk < R}, where R is given in (H4). We first observe that, to prove the theorem, it suffices to show that the equation

Lu(n) = f(n,u(n−τ(n)) +γ(n−τ(n))), n∈Z. (3.1) has a positive periodic solutionu ∈ satisfying 0 < kuk < R. If fact, if this is true, we let y(n) =u(n) +γ(n). Then,y(n)>γ(n)on[1,ω]Z, 0<ky−γk<R, and

Ly(n) =Lu(n) +Lγ(n)

= f(n,u(n−τ(n)) +γ(n−τ(n))) +r(n)

= f(n,y(n−τ(n))) +r(n), n∈Z.

Thus,y(n)is a positive periodic solution of (1.1) with the required properties.

From (H4), there existsk0Nsuch that R> 1

k0 +δ2ωg(δR)

1+ h(R) g(R)

ω

i

=1

φ(i). (3.2)

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LetK0={k0,k0+1, . . .}. For any fixedk ∈K0, consider the family of equations u(n) = 1

k +λ

ω i=1

ω j=1

G(i,j)fk(nij,u(nτij) +γ(nτij)) = 1

k +λTku(n), n∈Z, (3.3) whereλ∈ (0, 1), fk(n,x) = f(n, max{x, 1/k})for(n,x)∈Z×R,nτij =nijτ(nij), and

Tku(n) =

ω i=1

ω j=1

G(i,j)fk(nij,u(nτij) +γ(nτij)). We now prove two claims.

Claim 1. For anyλ∈ (0, 1), any possible solutionu(n)of (3.3) satisfiesu(n)≥δ||u||forn ∈Z, whereδ=δ12∈ (0, 1]is given in (H4).

In fact, from (2.4), we have u(n)≤ 1

k +δ2

ω i=1

ω j=1

f(nij,u(nτij) +γ(nτij)) and

u(n)≥ 1 k +δ1

ω i=1

ω j=1

f(nij,u(nτij) +γ(nτij))

δ 1 k +δ2

ω i=1

ω j=1

f(nij,u(nτij) +γ(nτij))

!

forn∈ Z. Thus,u(n)≥δ||u||onZ, i.e., Claim 1 holds.

Claim 2. For anyλ∈ (0, 1), any possible solution of (3.3) satisfieskuk 6= R.

Suppose the claim is not true and assume thatu(n)is a solution of (3.3) for someλ∈ (0, 1) withkuk= R. SinceλTku(n)≥0 onZ, we haveu(n)≥1/k, which implies thatu(n) +γ(n)≥ u(n)≥1/konZ. Then, (3.3) becomes

u(n) = 1 k +λ

ω i=1

ω j=1

G(i,j)f(nij,u(nτij) +γ(nτij)). (3.4) From (2.4), (H2), and Claim 1, it follows that

R= kuk ≤ 1 k0

+λ

ω i=1

ω j=1

G(i,j)f(nij,u(nτij) +γ(nτij))

1

k0 +δ2ω

ω i=1

f(i,u(i−τ(i)) +γ(i−τ(i)))

1

k0 +δ2ω

ω i=1

φ(i)g(u(i−τ(i)))

1+ h(u(i−τ(i))) g(u(i−τ(i)))

1 k0

+δ2ωg(δkuk)

1+ h(kuk) g(kuk)

ω

i=1

φ(i)

= 1

k0 +δ2ωg(δR)

1+ h(R) g(R)

ω

i

=1

φ(i), which contradicts (3.2). Hence, Claim 2 holds.

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Note that 1/k≤1/k0< Rand (3.3) can be rewritten as u(n) = (1−λ)1

k +λNku(n),

whereNku(n) =Tku(n) +1/k. Clearly,Nk is compact, and as in Claim 1, we have Nku(n)≥δ||Nku|| forn∈Z.

Then,Nk mapsΩintoK, whereK ={u ∈ X : u(n)≥ δkukonZ}. Therefore, by Lemma3.2, Nkhas at least one fixed pointuk. Thus, for anyk∈K0, we have proven that the equation

uk(n) = 1 k +

ω i=1

ω j=1

G(i,j)fk(nij,u(nτij) +γ(nτij)), n∈Z,

has a solutionuk(n)withkukk ≤ R. Sinceuk(n) +γ(n) ≥ uk(n) ≥ 1/k, we see thatuk(n)is actually a solution of the equation

uk(n) = 1 k +

ω i=1

ω j=1

G(i,j)f(nij,u(nτij) +γ(nτij)), n∈Z. (3.5) This, together with (H3), implies that

uk(n)>

ω i=1

ω j=1

G(i,j)f(nij,u(nτij) +γ(nτij))

ωδ1

ω i=1

ψR(i)>0 onZ. (3.6)

Sincekukk ≤ R for all k ∈ K0, we know that the sequence{uk(n)}kK0 has a subsequence, which converges uniformly to a functionu∈ X. For simplicity, we still denote this subsequence by{uk(n)}kK0. If we letk→in (3.5) and (3.6), we obtain

u(n) =

ω i=1

ω j=1

G(i,j)f(nij,u(nτij) +γ(nτij)) (3.7) and

u(n)≥ωδ1

ω i=1

ψR(i)>0

forn ∈ Z. Thus, by Lemma 3.1, u(n) is a positive solution of (3.1). Since ku(n)k ≤ Rand u = lim

kuk, we havekuk ≤ R. By an argument similar to the one used to show Claim 2, we havekuk<R. Hence, 0<kuk<R. This completes the proof of the theorem.

Proof of Corollary2.2. We will apply Theorem2.1. To this end, let f(n,x) =c(n)xα+µd(n)xβ, g(x) = xα, h(x) = µxβ, and φ(n) = max{c(n), d(n)}. Then, (H1), (H2), and (H3) with ψq(n) =qαc(n)hold. LetA=δ2ωωi=1φ(i). Then, 0 < A< and (H4) becomes

µ< δ

αRα+1−A

ARα+β for someR>0.

Hence, equation (1.2) has at least one positive solution for 0<µ<µ¯ :=sup

R>0

δαRα+1−A ARα+β .

Note that ¯µ=ifβ<1 and ¯µ<ifβ≥1. This completes the proof of the corollary.

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Acknowledgments

The research by Jacob D. Johnson, Michael G. Ruddy, and Alexander M. Ruys de Perez was conducted as part of a 2013 Research Experience for Undergraduates at the University of Ten- nessee at Chattanooga that was supported by NSF Grant DMS-1261308.

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