Anti-Periodic Presentation

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Anti-periodic solutions for a higher orderdifference equation with p-Laplacian operator

Kaitlin Rizzo, Jacob Parsley, and NicholasRussell

National Science Foundation REUUniversity of Tennessee at Chattanooga

Chattanooga, TN 37403

Rizzo, Parsley, and Russell Anti-Periodic Solutions

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Overview

...1 ProblemThe Actual ProblemAssumptions

...2 PreliminariesBuilding Our Space

...3 TheoremsTheorem 1Theorem 2

...4 ExtensionsCorollaries

...5 Refer’s


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The Problem

Let us define some notation:

□ Let n,T ∈ N.□ ϕp(s) = |s|p−2s, where s ∈ R and 1 < p < ∞.

□ ∆u(t) = u(t + 1)− u(t) and ∆iu(t) = ∆(∆i−1u(t)) fort ∈ Z and i ∈ N

□ r : Z → R


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The Problem

Define

(Qu)(t) = (−1)n∆n(r(t − n)ϕp(∆nu(t − n))), t ∈ Z.

We are concerned with finding infinitely many anti-periodicsolutions of the higher order difference equation with a p-Laplacianand containing both advance and retardation, i.e.

.. (Qu)(t) = λf (t, u(t+1), u(t), u(t−1))+µg(t, u(t+1), u(t), u(t−1)),(1)

where

□ t ∈ Z; λ, µ > 0;

□ f , g : Z× R3 → R are continuous in the second, third, andfourth variables.


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Anti-Periodic Solutions

Recently, there has been increasing interest in the literature onanti-periodic boundary problems due to their extensiveapplications. For example

interpolation problems

anti-periodic wavelets

the Hill differential operator

neural networks ,

and physics.

One of the early works on anti-periodic solutions can be found in apaper by Okochi, where he used the fixed point theory to obtainthe existence of anti-periodic solutions to a nonlinear parabolicequation in a real Hilbert space.


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Previous Work

Many results in the literature for finding anti-periodic solutionswere obtained by using tools such as upper and lower solutions,monotone iterative techniques, continuation theorem ofcoincidence degree , and critical point theory. However, as far weare aware of, critical point theory has not been used as often asother tools in the literature to study anti-periodic solutions fordifference equations, and the work utilizing variational approachesis even rarer for anti-periodic solutions of a difference equationswith a p-Laplacian operator, advance, and retardation.


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Previous Work

The only work we know of is the paper by Tian and Henderson,where the following 2nth order nonlinear difference equation wasconsidered

∆n(r(t − n)∆nu(t − n)) + f (t, u(t)) = 0, t ∈ Z. (2)

The authors established several existence criteria in [6] foranti-periodic solutions of equation (2) by using a variationalapproach.In this presentation, we apply variational approaches and a criticalpoint theorem that appeared in [3, 4] to study the existence ofinfinitely many anti-periodic solutions of equation.


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Assumptions

(K1) r(t + T ) = r(t) and r(t) > 0 for all t ∈ Z;(K2) there exists a function F : Z× R2 → R such that

F (t, 0, 0) ≡ 0 on Z, F is continuously differentiable in thesecond and third variables, and

f (t, x , y , z) = F ′2(t − 1, y , z) + F ′

3(t, x , y),

where

F ′2(t−1, y , z) =

∂F (t − 1, y , z)

∂yand F ′

3(t, x , y) =∂F (t, x , y)

∂y;

(K3) F ′2 and F ′

3 satisfy

F ′2(t+T ,−t1,−t2) = −F ′

2(t, t1, t2) for all t ∈ Z and t1, t2 ∈ R

and

F ′3(t+T ,−t1,−t2) = −F ′

3(t, t1, t2) for all t ∈ Z and t1, t2 ∈ R.


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Assumptions

(K4) there exists a function G : Z× R2 → R such thatG (t, 0, 0) ≡ 0 on Z, G is continuously differentiable in thesecond and third variables, and

g(t, x , y , z) = G ′2(t − 1, y , z) + G ′

3(t, x , y),

where

G ′2(t−1, y , z) =

∂G (t − 1, y , z)

∂yand G ′

3(t, x , y) =∂G (t, x , y)

∂y;

(K5) G ′2 and G ′

3 satisfy

G ′2(t+T ,−t1,−t2) = −G ′

2(t, t1, t2) for all t ∈ Z and t1, t2 ∈ R

and

G ′3(t+T ,−t1,−t2) = −G ′

3(t, t1, t2) for all t ∈ Z and t1, t2 ∈ R.


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Remarks

(K3) holds when either both F ′2 and F ′

3 are odd periodic functionswith period T or both F ′

2 and F ′3 are even anti-periodic functions

with anti-period T , i.e., (K3) holds if one of the following two setsof equalities holds{

F ′2(t,−t1,−t2) = −F ′

2(t, t1, t2), F ′2(t + T , t1, t2) = F ′

2(t, t1, t2),

F ′3(t,−t1,−t2) = −F ′

3(t, t1, t2), F ′3(t + T , t1, t2) = F ′

3(t, t1, t2),

or{F ′2(t,−t1,−t2) = F ′

2(t, t1, t2), F ′2(t + T , t1, t2) = −F ′

2(t, t1, t2),

F ′3(t,−t1,−t2) = F ′

3(t, t1, t2), F ′3(t + T , t1, t2) = −F ′

3(t, t1, t2).


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Remarks

Similarly, (K5) holds when either both G ′2 and G ′

3 are odd periodicfunctions with period T or both G ′

2 and G ′3 are even anti-periodic

functions with anti-period T , i.e., (K5) holds if one of thefollowing two sets of equalities holds{

G ′2(t,−t1,−t2) = −G ′

2(t, t1, t2), G ′2(t + T , t1, t2) = G ′

2(t, t1, t2),

G ′3(t,−t1,−t2) = −G ′

3(t, t1, t2), G ′3(t + T , t1, t2) = G ′

3(t, t1, t2),

or{G ′2(t,−t1,−t2) = G ′

2(t, t1, t2), G ′2(t + T , t1, t2) = −G ′

2(t, t1, t2),

G ′3(t,−t1,−t2) = G ′

3(t, t1, t2), G ′3(t + T , t1, t2) = −G ′

3(t, t1, t2).


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Remarks

The following observations are obviously true.

(a) .. Condition (K1) implies that there exist r2 ≥ r1 > 0 suchthat r1 ≤ r(t) ≤ r2 on Z.

(b) Conditions (K2) and (K3) imply that

.. f (t + T ,−x ,−y ,−z) = −f (t, x , y , z) (3)

for all t ∈ Z and x , y , z ∈ R.(c) Conditions (K4) and (K5) imply that

.. g(t + T ,−x ,−y ,−z) = −g(t, x , y , z), (4)

for all t ∈ Z and x , y , z ∈ R.


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Building Our Space

Let S be a set of sequences

u = (. . . , u(−t), . . . , u(−1), u(0), u(1), . . . , u(t), . . .) = {u(t)}∞−∞,

i.e., S = {u = u(t) : u(t) ∈ R, t ∈ Z}. For any u, v ∈ S anda, b ∈ R, define au + bv by

au + bv = {au(t) + bv(t)}∞t=−∞.

Then, S is a vector space. Now, we define H as a subspace of Ssuch that

.. H = {u = u(t) ∈ S : u(t + T ) = −u(t), t ∈ Z}, (5)

where T ∈ N, given in the preliminaries.


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Building Our Space

H is isomorphic to RT and H is a finite dimensional Hilbert spaceequipped with the inner product

⟨u, v⟩ =T∑t=1

u(t)v(t) for u, v ∈ H,

which induces to a norm ∥u∥2 defined by

∥u∥2 =

(T∑i=1

|u(t)|2)1/2

for u ∈ H.

For p > 1, we define another norm ∥u∥p on H by

∥u∥p =

(T∑i=1

|u(t)|p)1/p

for u ∈ H.


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Building Our Space

Since any two norms on a finite dimensional normed linear spaceare equivalent, there exists two constants C2 ≥ C1 > 0 such that

.. C1∥u∥2 ≤ ∥u∥p ≤ C2∥u∥2 for u ∈ H. (6)


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Functionals

For any u ∈ H and λ, µ > 0, let the functionals Φ and Ψ bedefined by

.. Φ(u) =1

p

T∑t=1

r(t)|∆nu(t)|p (7)

and

.. Ψ(u) =T∑t=1

F (t, u(t+1), u(t))+µ

λ

T∑t=1

G (t, u(t+1), u(t)). (8)

Lemmas 1–3 below present some properties of the functionals Φand Ψ.


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Lemma 2.1

.Lemma (2.1)..

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Assume that (K1) holds. Then Φ ∈ C 1(H,R) with

.. ⟨Φ′(u), v⟩ = (−1)n

T∑t=1

∆n(r(t − n)ϕp(∆nu(t − n)))v(t) (9)

for all u, v ∈ H.


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Proof of Lemma 2.1

For any x , y ∈ H, by the summation by parts formula, we have

T∑t=1

r(t)ϕp(x(t))∆y(t) = r(t − 1)ϕp(x(t))y(t)∣∣∣T+1

t=1

−T∑t=1

∆(r(t − 1)ϕp(x(t − 1))y(t). (10)

By (K1), we see that r(t − 1)ϕp(x(t))y(t)∣∣∣T+1

t=1= 0. Thus,

..

T∑t=1

r(t)ϕp(x(t))∆y(t) = −T∑t=1

∆(r(t − 1)ϕp(x(t − 1)))y(t). (11)

For any u, v ∈ H, by simple calculations, it follows that

⟨Φ′(u), v⟩ =

T∑t=1

r(t)ϕp(∆nu(t))∆nv(t).


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Proof of Lemma 2.1

Note that ∆nu ∈ H and ∆n−iv ∈ H for i = 1, . . . , n. Then,applying .(11) n times yields that

⟨Φ′(u), v⟩ = −

T∑t=1

∆(r(t − 1)ϕp(∆nu(t − 1)))∆n−1v(t)

= (−1)2T∑t=1

∆2(r(t − 2)ϕp(∆nu(t − 2)))∆n−2v(t)

= . . . . . .

= (−1)nT∑t=1

∆n(r(t − n)ϕp(∆nu(t − n)))v(t).

Hence, .(9) holds. This completes the proof of the lemma.


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Lemma 2.2

.Lemma (2.2)..

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Assume that (K2)–(K5) hold. Then Ψ ∈ C 1(H,R) with

⟨Ψ′(u), v⟩ =T∑t=1

f (t, u(t + 1), u(t), u(t − 1))v(t)

+µ

λ

T∑t=1

g(t, u(t + 1), u(t), u(t − 1))v(t) (12)

for all u, v ∈ H.


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Lemma 2.3

.Lemma (2.3)..

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.. Assume that (K1)–(K5) hold. Then, every critical point u ∈ Hof Φ− λΨ is an anti-periodic solution of equation (1).

Proof: Let u ∈ H be a critical point of Φ− λΨ. Then, u isanti-periodic and

⟨Φ′(u)− λΨ

′(u), v⟩ = 0 for all v ∈ H.


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Proof of Lemma 2.3

Then, by Lemmas 1 and 2, it follows that

(−1)nT∑t=1

∆n(r(t − n)ϕp(∆nu(t − n)))v(t)

− λ

T∑t=1

f (t, u(t + 1), u(t), u(t − 1))v(t)

− µ

T∑t=1

g(t, u(t + 1), u(t), u(t − 1))v(t) = 0,

i.e.,

T∑t=1

[(−1)n∆(r(t − n)ϕp(∆

nu(t − n)))− λf (t, u(t + 1), u(t), u(t − 1))

− µg(t, u(t + 1), u(t), u(t − 1))]v(t) = 0.


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Proof of Lemma 2.3

Thus,

(−1)n∆(r(t − n)ϕp(∆nu(t − n)))− λf (t, u(t + 1), u(t), u(t − 1))

− µg(t, u(t + 1), u(t), u(t − 1)) = 0 for t ∈ [1,T ]Z.

For any t ∈ [1,T ]Z and l ∈ Z, by (K1), parts .(3) and .(4) of ourRemark, and the fact that u ∈ H, we have

(−1)n∆(r(t + lT − n)ϕp(∆nu(t + lT − n)))

−λf (t + lT , u(t + lT + 1), u(t + lT ), u(t + lT − 1))

−µg(t + lT , u(t + lT + 1), u(t + lT ), u(t + lT − 1))

= −(−1)n∆(r(t − n)ϕp(∆nu(t − n))) + λf (t, u(t + 1), u(t), u(t − 1))

+µg(t, u(t + 1), u(t), u(t − 1)) = 0.

Therefore, u satisfies equation (1) for all t ∈ Z. This completesthe proof of the lemma.


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More Remarks

We now define a T × T matrix A by

.. A =

2 −1 0 . . . 0 1−1 2 −1 . . . 0 00 −1 2 . . . 0 0. . . . . . . . . . . . . . . . . .0 0 0 . . . 2 −11 0 0 . . . −1 2

.

As pointed out in [6], A is a positive definite matrix, and so all ofits eigenvalues are positive. Let σi , i = 1, . . . ,T , be theeigenvalues of A satisfying

0 < σ1 ≤ σ2 ≤ . . . ≤ σT .


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Lemma 2.4

The following lemma are take from [6, Lemma 2.3]..Lemma (2.4)..

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.. For u ∈ H, we have

σn1∥u∥22 ≤

T∑t=1

(∆nu(t))2 ≤ σnT∥u∥22.


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Defining Notation

Let the functions F and G be given as in conditions (K2) and(K4), and let the space H be defined by .(5) . For convenience, weintroduce the following notations:

.. α = lim infξ→∞

T∑t=1

max|x|,|y |≤ξ

F (t, x , y)

ξp, β = lim sup

ξ→∞

T−1∑t=1

F (t, ξ, ξ) + F (T ,−ξ, ξ)

ξp,

(13)

.. δ = lim infξ→0+

T∑t=1

max|x|,|y |≤ξ

F (t, x , y)

ξp, η = lim sup

ξ→0+

T−1∑t=1

F (t, ξ, ξ) + F (T ,−ξ, ξ)

ξp.

(14)

From now on, we always assume the following condition holds:

(C) α, β, δ, η ≥ 0.


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Defining Notation

We also use the convention that 1/x = ∞ if x = 0.Let r1 and r2 be given as in the remark, .12 . Then, for u ∈ H,from the previous statement and .(7) , we have

r1p

T∑t=1

|∆nu(t)|p ≤ Φ(u) ≤ r2p

T∑t=1

|∆nu(t)|p.

Thus, by .(6) and Lemma .4 , we see the following:


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Defining Notation

Φ(u) ≥ r1p

T∑t=1

|∆nu(t)|p

=r1p∥∆nu(t)∥pp

≥ r1pCp1 ∥∆

nu(t)∥p2

=r1pCp1

(T∑t=1

|∆nu(t)|2)p/2

≥ r1pCp1 σ

(np)/21 ∥u∥p2

Φ(u) ≤ r2p

T∑t=1

|∆nu(t)|p

=r2p∥∆nu(t)∥pp

≤ r2pCp2 ∥∆

nu(t)∥p2

=r2pCp2

(T∑t=1

|∆nu(t)|2)p/2

≤ r2pCp2 σ

(np)/2T ∥u∥p2


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Defining Notation

Let.. d1 =

r1pCp1 σ

np21 and d2 =

r2pCp2 σ

np2T . (15)

Then, we have

.. d1∥u∥p2 ≤ Φ(u) ≤ d2∥u∥p2 . (16)

The following notation are also needed in the statements of ourresults.

.. λ1 =d2T

p/2

β, λ2 =

d1α, (17)

λ3 =d2T

p/2

η, λ4 =

d1δ.


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Lemma 3.1

The proofs of our theorems are based on following particular caseof the variational principle of Ricceri:Let H be a reflexive real Banach space, and let Φ,Ψ : H → R betwo Gateaux differentiable functionals such that Φ is sequentiallyweakly lower semicontinuous, strongly continuous and coercive,and Ψ is sequentially weakly upper semicontinuous. For everyy > infH Φ, let

φ(y) := infu∈Φ−1(−∞,y)

supv∈Φ−1(−∞,y)Ψ(v)−Ψ(u)

y − Φ(u),

andγ := lim inf

y→∞φ(y), ζ := lim inf

y→(infH Φ)+φ(y).

Then,


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Lemma 3.1

(a) for every y > infH Φ and every λ ∈ (0, 1/φ(y)), the restriction ofthe functional Iλ := Φ− λΨ to Φ−1(−∞, y) admits a globalminimum, which is a critical point (local minimum) of Iλ in H.

(b) If γ < ∞ then, for each λ ∈ (0, 1/γ), the following alternativeholds: either

(b1) Iλ possesses a global minimum, or(b2) there is a sequence {ul} of critical points (local minima) of Iλ

such that liml→∞ Φ(ul) = ∞.

(c) If ζ < ∞ then, for each λ ∈ (0, 1/ζ), the following alternative holds:either

(c1) there is a global minimum of Φ which is a local minimum ofIλ, or

(c2) there is a sequence {ul} of pairwise distinct critical points

(local minima) of Iλ which converges weakly to a global

minimum of Φ.


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Theorem 1

.Theorem (1)..

......

Assume that (K1)–(K5) and the following conditions hold:

(1a) λ1 < λ2;

(2a) G (t, x , y) ≥ 0 on Z× R2;

(3a) G∞ = lim supξ→∞

T∑t=1

max|x|,|y|≤ξ G(t,x ,y)

ξp < ∞.

Then, for each λ ∈ (λ1, λ2) and each µ ∈ [0, µ1), equation (1) hasa sequence of anti-periodic solutions that is unbounded in H, where

.. µ1 =d1 − αλ

G∞. (18)


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Proof of Theorem 1

Before we start, we make the following comments:

condition (C) implies that λi ≥ 0, i = 1, 2, 3, 4;

By assumption (1a), we see that α < d1d2T p/2β;

the interval [0, µ1) is well defined since µ1 > 0 under thecondition that λ < λ2.

For H defined by (5), let the functionals Φ,Ψ : H → R be definedby .(7) and .(8) , respectively. Then, it is clear that Φ and Ψsatisfy all the regularity assumptions in Lemma 3.1.From (16), we have

.. ∥u∥2 ≤(Φ(u)

d1

)1/p

. (19)


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Proof of Theorem 1

From the definition of α in .(13) , there exists a sequence {ξm} ofpositive numbers such that limm→∞ ξm = ∞ and

.. α = limm→∞

T∑t=1

max|x |,|y |≤ξm F (t, x , y)

ξpm. (20)

Let sm = d1ξpm. Then, for any u ∈ H with Φ(u) < sm, from (??),

it follows that

maxt∈[1,T ]Z

|u(t)| < ∥u∥2 ≤(Φ(u)

d1

)1/p

<

(smd1

)1/p

= ξm.

Note that Φ(0) = Ψ(0) = 0 by (K2) and (K4).


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Proof of Theorem 1

Then, we have

φ(sm) = infu∈Φ−1(−∞,sm)

supv∈Φ−1(−∞,sm)Ψ(v)−Ψ(u)

sm − Φ(u)

≤supv∈Φ−1(−∞,sm)Ψ(v)

sm

≤

T∑t=1

max|x |,|y |≤ξm F (t, x , y) +µ

λ

T∑t=1

max|x |,|y |≤ξm G (t, x , y)

sm

=

T∑t=1

max|x |,|y |≤ξm F (t, x , y) +µ

λ

T∑t=1

max|x |,|y |≤ξm G (t, x , y)

d1ξpm

.


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Proof of Theorem 1

Hence, for γ defined in Lemma 3.1, condition (1c) and .(20) implythat

.. γ ≤ limm→∞

φ(sm) ≤1

d1

(α+

µ

λG∞

)< ∞. (21)

We now claim that

.. if λ ∈ (λ1, λ2) and µ ∈ [0, µ1), then λ ∈ (0, 1/γ). (22)

In fact, for λ ∈ (λ1, λ2), we have λ > 0. Then, from .(17) , .(18) ,and .(21) , we see that when G∞ > 0,

γ <1

d1

(α+

µ1

λG∞

)=

1

d1

(α+

d1 − αλ

λ

)=

1

λ

and when G∞ = 0,

γ ≤ α

d1=

1

λ2<

1

λ.

Note that G∞ ≥ 0. Thus, we always have λ < 1/γ. This showsthat (??) holds.


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Proof of Theorem 1

Now, let λ ∈ (λ1, λ2) and µ ∈ [0, µ1) be fixed. Then, by Lemma3.1 (b), we see that one of the following alternatives holds:

(b1) Iλ := Φ− λΨ possesses a global minimum, or

(b2) there is a sequence {um} of critical points of Iλ such thatlim

m→∞∥um∥p = ∞.

In the following, we show that (b1) does not hold, and thus, (b2)must hold.


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Proof of Theorem 1

From the definition of β in .(13) , there exists a sequence {ηm} ofpositive numbers such that limm→∞ ηm = ∞ and

.. β = limm→∞

T−1∑t=1

F (t, ηm, ηm) + F (T ,−ηm, ηm)

ηpm. (23)

For any m ∈ N, we define ωm ∈ H satisfying ωm(t) = ηm for allt ∈ [1,T ]Z. By (16), we have

Φ(ωm) ≤ d2Tp/2ηpm.


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Proof of Theorem 1

By .(8) and (1a.), we see that

Ψ(ωm) =T∑t=1

F (t, ωm(t + 1), ωm(t)) +µ

λ

T∑t=1

G (t, ωm(t + 1), ωm(t))

≥T∑t=1

F (t, ωm(t + 1), ωm(t))

=T−1∑t=1

F (t, ηm, ηm) + F (T ,−ηm, ηm).

Hence,

.. Iλ(ωm) = Φ(ωm)− λΨ(ωm)

≤ d2Tp/2ηpm − λ

(T−1∑t=1


).

(24)


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Proof of Theorem 1

We now consider two cases.Case 1. β < ∞. By λ > λ1 and the definition of λ1 in (17), we

have β − d2T p/2

λ > 0. Choose ϵ > 0 satisfying

.. ϵ ∈

(0, β − d2T

p/2

λ

). (25)

From (23), there exists M1 ∈ N such that

T−1∑t=1

F (t, ηm, ηm) + F (T ,−ηm, ηm) > (β − ϵ)ηpm for all m ≥ M1.

Then, from (24),

Iλ(ωm) <(d2T

p/2 − λ(β − ϵ))ηpm.

Thus, by .(25) and the fact that limm→∞ ηm = ∞, we havelimm→∞ Iλ(ωm) = −∞.


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Proof of Theorem 1

Case 2. β = ∞. Choose

.. L >d2T

p/2

λ. (26)

By .(23) , there exists M2 ∈ N such that

T−1∑t=1

F (t, ηm, ηm) + F (T ,−ηm, ηm) > Lηpm for all m ≥ M2.

Then, from .(24) , we have

Iλ(ωm) < (d2Tp/2 − λL)ηpm.

Hence, by (26) and the fact that limm→∞ ηm = ∞, we obtain thatlimm→∞ Iλ(ωm) = −∞.


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Proof of Theorem 1

Combining the above two cases, we see that the functional Iλ isalways unbounded from below. Thus, (b1) does not hold.Therefore, there exists a sequence {um} of anti-periodic criticalpoints of Iλ such that limm→∞ ∥u∥p = ∞. Finally, an applicationof Lemma .3 completes the proof of the theorem.


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Theorem 2

.Theorem..

......

Assume that (K1)–(K5) and the following conditions hold:

(1b) λ3 < λ4;

(2b) G (t, x , y) ≥ 0 on Z× R2;

(3b) G∞ = lim supξ→∞

T∑t=1

max|x|,|y|≤ξ G(t,x ,y)

ξp < ∞.

Then, for each λ ∈ (λ3, λ4) and each µ ∈ [0, µ2), equation (1) hasa sequence of anti-periodic solutions converging uniformly to zeroin H, where

µ2 =d1 − δλ

G∞.


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Proof of Theorem 2

Using Lemma 3.1 (c) and arguing as in the proof of Theorem 1,Theorem 2 can be proved. For the sake of completeness, we sketcha proof below and make the following remarks:

By assumption (1b), we see that δ < d1d2T p/2 η;

The interval [0, µ2) is well defined since µ2 > 0 under thecondition that λ < λ4.

Let the functionals Φ,Ψ : H → R be defined by .(7) and .(8) ,respectively. Then, as before, Φ and Ψ satisfy all the regularityassumptions in Lemma 3.1. From the definition of δ in .(14) , thereexists a sequence {ξm} of positive numbers such that lim

m→∞ξm = 0

and

δ = limm→∞

T∑t=1

max|x |,|y |≤ξm F (t, x , y)

ξpm.


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Proof of Theorem 2

Note that infH Φ = 0. Then, by the definition of ζ, we haveζ = lim infy→0+ φ(y). Then, as in showing .(21) and .(22) in theproof of Theorem 1, we can prove that ζ < ∞ and that ifλ ∈ (λ3, λ4) and µ ∈ [0, µ2), then λ ∈ (0, 1/ζ).

Then, in view of Lemma 3.1 (c), we see that one of the followingalternatives holds:

(c1) there is a global minimum of Φ which is a local minimum ofIλ = Φ− λΨ, or

(c2) there is a sequence {um} of pairwise distinct critical points(local minima) of Iλ which converges weakly to a globalminimum of Φ.

In what follows, we show that (c1) does not hold, and thus, (c2)must hold.


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Proof of Theorem 2

From the definition of η in .(14) , there exists a sequence {ηm} ofpositive numbers such that limm→∞ ηm = 0 and

η = limm→∞

T−1∑t=1


ηpm.

For any m ∈ N, we again define ωm ∈ H satisfying ωm(t) = ηm forall t ∈ [1,T ]Z.


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Proof of Theorem 2

Then, as in the cases 1 and 2 of the proof of Theorem 1, we canobtain that, for m large enough, if ζ < ∞, then

Iλ(ωm) <(d2T

p/2 − λ(η − ϵ))ηpm,

where

ϵ ∈

(0, η − d2T

p/2

λ

);

if ζ = ∞, then,

Iλ(ωm) < (d2Tp/2 − λL)ηpm,

where L satisfies .(26) . Hence, we always have Iλ(ωm) < 0 for largem.


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Proof of Theorem 2

Thus, sincelim

m→∞Iλ(wm) = Iλ(0) = 0,

we see that 0 is not a local minimum of Iλ. This, together with thefact that 0 is the only global minimum of Φ, shows that (c1) doesnot hold. Therefore, there exists a sequence {um} of critical pointsof Iλ that converges weakly (and thus also strongly) to 0. Finally,in view of Lemma 2.3, the conclusion of the theorem follows.


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Corollaries

The following corollaries are simple extensions of two theorems..Corollary (1)..

......

Assume that (K1)–(K5), (1a), and (2a) with G∞ = 0 hold, α = 0,and β = ∞. Then, for each λ ∈ (0,∞) and each µ ∈ [0,∞),equation (1) has a sequence of anti-periodic solutions that isunbounded in H.

.Corollary (2)..

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Assume that (K1)–(K5), (1b), and (2b) with G∞ = 0 hold, δ = 0,and η = ∞. Then, for each λ ∈ (0,∞) and each µ ∈ [0,∞),equation (1) has a sequence of anti-periodic solutions converginguniformly to zero in H.


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Corollaries

Consider the special case of equation (??) whereµg(t, u(t + 1), u(t), u(t − 1)) ≡ 0, i.e.,

(−1)n∆n(r(t − n)ϕp(∆nu(t − n)))

= λf (t, u(t + 1), u(t), u(t − 1)), t ∈ Z. (27)


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Corollaries

.Corollary (3)..

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Assume that (K1)–(K3) hold, α = 0, and β = ∞. Then, for eachλ ∈ (0,∞), equation (27) has a sequence of anti-periodic solutionsthat is unbounded in H.

.Corollary (4)..

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Assume that (K1)–(K3) hold, δ = 0, and η = ∞. Then, for eachλ ∈ (0,∞), equation (27) has a sequence of anti-periodic solutionsconverging uniformly to zero in H.


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Corollaries

Note that when f (t, u(t + 1), u(t), u(t − 1)) = f (t, u(t)) andg(t, u(t + 1), u(t), u(t − 1)) = g(t, u(t)), equation (1) reduces to

(−1)n∆n(r(t−n)ϕp(∆nu(t−n))) = λf (t, u(t))+µg(t, u(t)), t ∈ Z,

(28)where f , g : Z× R → R are assumed to be continuous functions.Let

F (t, y) =

∫ y

0f (t, s)ds for all (t, y) ∈ Z× R

and

G (t, y) =

∫ y

0g(t, s)ds for all (t, y) ∈ Z× R.


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Corollaries

Define the following constants

α = lim infξ→∞

T∑t=1

max|y |≤ξ

F (t, y)

ξp, β = lim sup

ξ→∞

T∑t=1

F (t, ξ)

ξp,

δ = lim infξ→0+

T∑t=1

max|y |≤ξ

F (t, y)

ξp, η = lim sup

ξ→0+

T∑t=1

F (t, ξ)

ξp.

We make the following assumptions.

(C1) f (t + T ,−y) = −f (t, y) for all t ∈ Z and y ∈ R;(C2) g(t + T ,−y) = −g(t, y) for all t ∈ Z and y ∈ R;(C3) G (t, y) ≥ 0 for all t ∈ Z and y ∈ R;

(C4) G∞ = lim supξ→∞

T∑t=1

max|y|≤ξ G(t,y)

ξp = 0.


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Corollaries

.Corollary (5)..

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Assume that (K1), (C1)–(C4) hold, α = 0, and β = ∞. Then, foreach λ ∈ (0,∞) and each µ ∈ [0,∞), equation (28) has asequence of anti-periodic solutions that is unbounded in H.

.Corollary (6)..

......

Assume that (K1), (C1)–(C4) hold, δ = 0, and η = ∞. Then, foreach λ ∈ (0,∞) and each µ ∈ [0,∞), equation (28) has asequence of anti-periodic solutions converging uniformly to zero inH.


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Corollaries

Consider the following special case of equation (28)

(−1)n∆n(r(t − n)ϕp(∆nu(t − n))) = λf (t, u(t)), t ∈ Z. (29)

.Corollary (7)..

......

Assume that (K1) and (C1) hold, α = 0, and β = ∞. Then, foreach λ ∈ (0,∞), equation (29) has a sequence of anti-periodicsolutions that is unbounded in H.

.Corollary (8)..

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Assume that (K1) and (C1) hold, δ = 0, and η = ∞. Then, foreach λ ∈ (0,∞), equation (29) has a sequence of anti-periodicsolutions converging uniformly to zero in H.


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A Remark

Clearly, if u ∈ H is an anti-periodic solution of equation (1), then uis a 2T–periodic solution of (1). Hence, under the assumptions ofTheorems 1 and 2 and Corollaries 1–8, we can obtain thecorresponding existence criteria for each of those results for theexistence of infinitely many 2T–periodic solutions


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An Example

In this example, we assume that n ∈ N, p = 2, r(t) ≡ 1 on Z, andT = 4. Let F : Z× R2 → R be defined by

.. F (t, t1, t2) = sin2(πt4

)t21 sin(arctan(t

21 + t22 )). (30)

Then, F is continuously differentiable in the second and thirdvariables and satisfies that F (t, 0, 0) ≡ 0 on Z.


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An Example

Moreover, we have

F ′2(t, t1, t2) = 2t1 sin

2(πt4

)(sin(arctan(t21 + t22 ))

+t21 cos(arctan(t

21 + t22 ))

1 + (t21 + t22 )2

)(31)

and

F ′3(t, t1, t2) =

2t21 t2 sin2(πt4

)cos(arctan(t21 + t22 ))

1 + (t21 + t22 )2

(32)

for all t ∈ Z and t1, t2 ∈ R.


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An Example

Now, define two functions f , g : Z× R3 → R by

f (t, x , y , z) = F ′2(t − 1, y , z) + F ′

3(t, x , y) (33)

and

g(t, x , y , z) = 2y

(cos2

(π(t − 1)

4

)+ cos2

(πt4

)).

In view of Remarks and from (30)–(33), we can see that(K2)–(K3) holds. With

G (t, x , y) = cos2(πt4

)(x2 + y2),

from our Remarks, we see that (K4), (K5), (1a), and (2a) whereG∞ = 4, are also satisfied.


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An Example

Since p = 2, we can choose C1 = C2 = 1 in .(6) . Note thatr(t) = 1. Then, (K1) holds, and in .12 , we can take r1 = r2 = 1.Since T = 4, the matrix A defined by .24 becomes

A =

2 −1 0 1−1 2 −1 00 −1 2 −11 0 −1 2

.

By calculating the eigenvalues, we have σ1 = 2−√2 and

σT = 2 +√2. Then, from .(15) , we see that

d1 =(2−

√2)n

2and d2 =

(2 +√2)n

2. (34)


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An Example

Now, for α defined in .(13) , we have

α = lim infξ→∞

4∑t=1

max|x |,|y |≤ξ

F (t, x , y)

ξ2

= lim infξ→∞

4∑t=1

sin2(πt4

)max

|x |,|y |≤ξx2 sin(arctan(x2 + y2))

ξ2

= lim infξ→∞

4∑t=1

sin2(πt4

)ξ2 sin(arctan(2ξ2))

ξ2

= lim infξ→∞

sin(arctan(2ξ2))4∑

t=1

sin2(πt4

)= 0


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. . . . . . . . . . . . .Preliminaries

. . . . . . . . . . . . . . . . .Theorems


An Example

For β defined in (13), we have

β = lim supξ→∞

4∑t=1

max|x|,|y |≤ξ

F (t, x , y)

ξ2

= lim supξ→∞

4∑t=1

sin2(πt4

)max

|x|,|y |≤ξx2 sin(arctan(x2 + y2))

ξ2

= lim supξ→∞

4∑t=1

sin2(πt4

)ξ2 sin(arctan(2ξ2))

ξ2

= lim supξ→∞

sin(arctan(2ξ2))4∑

t=1

sin2(πt4

)=

4∑t=1

sin2(πt4

)= 2


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. . . . . . . . . .Problem

. . . . . . . . . . . . .Preliminaries

. . . . . . . . . . . . . . . . .Theorems


An Example

By .(17) and .(34) , we have that λ1 = (2 +√2)n and λ2 = ∞.

Thus, condition (1a) is satisfied. As well, from .(18) , we have

µ1 =(2−

√2)n

8 . Therefore, by Theorem 1, for each

λ ∈((2 +

√2)n,∞

)and each µ ∈

[0, (2−

√2)n

8

), equation (1) has

a sequence of anti-periodic solutions that is unbounded in H whereT = 4.


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. . . . . . . . . .Problem

. . . . . . . . . . . . .Preliminaries

. . . . . . . . . . . . . . . . .Theorems


References

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Math. Ineqal. 1 (2007), 409–417.Rizzo, Parsley, and Russell Anti-Periodic Solutions

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. . . . . . . . . .Problem

. . . . . . . . . . . . .Preliminaries

. . . . . . . . . . . . . . . . .Theorems


References

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approach, Math. Nachr. 286 (2013), 1537–1547.

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variational approach, J. Difference Equ. Appl. 19 (2013), 1380–1392.

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Thank you!

The research by Jacob Parsley, Kaitlin Rizzo, and Nicholas Russellwas conducted as part of a 2015 Research Experience forUndergraduates at the University of Tennessee at Chattanoogathat was supported by NSF Grant DMS-1261308.Special thank you to Dr. Lingju Kong for being our advisor all theway through the REU. Also, a thank you to Dr. AbhinandanChowdhury for help with the presentation formatting.

Thank you for coming!


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