PositiveSolutionsforaClassofDiscreteMixedBoundaryValue...

Research ArticlePositive Solutions for a Class of Discrete Mixed Boundary ValueProblems with the (p q)-Laplacian Operator

Cuiping Li1 and Zhan Zhou 12

1School of Mathematics and Information Science Guangzhou University Guangzhou 510006 China2Center for Applied Mathematics Guangzhou University Guangzhou 510006 China

Correspondence should be addressed to Zhan Zhou zzhou0321hotmailcom

Received 31 December 2019 Accepted 15 February 2020 Published 26 March 2020

Academic Editor Beatrice Di Bella

Copyright copy 2020 Cuiping Li and Zhan Zhou)is is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

In this paper we consider the existence of solutions for the discrete mixed boundary value problems involving (p q)-Laplacianoperator By using critical points theory we obtain the existence of at least two positive solutions for the boundary value problemunder appropriate assumptions on the nonlinearity

1 Introduction

In recent years with the development of mechanical engi-neering control system computer science and economicsthe existence of solutions of difference equations hasattracted wide attention (see [1ndash6]) For example applyingRicceri variational principle to obtain the existence ofmultiple solutions [7ndash9] taking the invariant sets ofdescending flow to prove the existence of sign-changingsolutions [10] making the linking theorem to get the ex-istence and multiplicity of periodic solutions [11] and usingcritical point theory to obtain the existence of homoclinicsolutions [12ndash15] and heteroclinic solutions [16]

As we know the fixed-point method and upper andlower solution techniques are important tools to solve theexistence of solutions for boundary value problems (see

[17 18]) But recently it is more common to use criticalpoint theory to study Dirichlet boundary value problems(see [19ndash23]) More result on difference equations by usingcritical point theories can be referred to [24ndash27]

In [28] DrsquoAguı et al established the existence of at leasttwo positive solutions for the following discrete Dirichletboundary value problem

minusΔ ϕp(Δu(k minus 1))1113872 1113873 + q(k)ϕp(u(k)) λf(k u(k)) k isin Z(1 N)

u(0) u(N + 1) 0

⎧⎨

⎩

(1)

where q(k)ge 0 for all k isin 1 2 N Unlike this DrsquoAguı et al in [29] proved that there are at

least two nonzero weak solutions for the following mixedboundary value problem

minus q(x) uprime(k)1113868111386811138681113868

1113868111386811138681113868pminus2

uprime(x)1113872 1113873prime + s(x)|u(k)|pminus2u(x) λf(x u(x)) x isin [a b]

u(a) uprime(b) 0

⎧⎨

⎩ (2)

where pgt 1 q s isin Linfin([a b]) with q0 ess inf [ab]qgt 0 s0

ess inf [ab]sge 0 f [a b] times R⟶ R is an L1-Caratheodoryfunction and λ is a real positive parameter

As a discrete analogy of the abovementioned problemwe consider the existence of positive solutions for the fol-lowing discrete mixed boundary value problem

HindawiDiscrete Dynamics in Nature and SocietyVolume 2020 Article ID 5414783 9 pageshttpsdoiorg10115520205414783

minusΔ ϕp(Δu(k minus 1))1113872 1113873 + s(k)ϕq(u(k)) λf(k u(k)) k isin Z(1 N)

u(0) Δu(N) 0

⎧⎨

⎩ Df

λ1113872 1113873 (3)

where Z(a b) denote the discrete interval a a + 1 b

for any integers a and b with alt b N be a positive integerf(k u) is continuous in u for each k isin Z(1 N) Δu(k)

u(k + 1) minus u(k) is the forward difference operatorϕr R⟶ R is the r-Laplacian given by ϕr(u) |u|rminus 2u

with u isin R 1lt qleplt +infin s(k)ge 0 for all k isin Z(1 N) andλ is a positive parameter

In this paper under suitable assumptions on the non-linearity f we use the theory of two nonzero critical points(see [30]) to ensure that there are at least two nonzerosolutions for problem (D

f

λ ) )e two nonzero critical pointstheorem is an appropriate combination of local minimumtheorem (see [31]) and classical AmbrosettindashRabinowitztheorem (see [32]) An important hypothesis of mountainpass theorem is PalaisndashSmale condition It satisfies the ap-plication of infinite dimensional space by requiring thecondition that the nonlinear term is stronger than p-superlinearity at infinity In order to obtain the existence oftwo nonzero solutions we can assume the classicalAmbrosettindashRabinowitz condition and nonlinear algebraiccondition (see (40) in )eorem 2) hold that is morewidespread than the p-sublinearity at zero Moreover whenwe require that f(k 0)ge 0 for all k isin Z(1 N) we can usestrongmaximum principle to obtain the existence of positivesolutions which has been proved in Lemma 2

Let slowast min s(k) k isin Z(1 N) a special case of ourmain result is stated as follows

Theorem 1 Let f R⟶ R be a continuous function suchthat

limt⟶0+

f(t)

tpminus14(a)

limt⟶+infin

f(t)

tpminus14(b)

then for each λ isin (0 (1 + slowastNpminus1pNp)min 1113864supcgt0(cq

max|ξ|lec 1113938ξ0 f(t)dt) supcgt0(cpmax|ξ|lec 1113938

ξ0 f(t)dt)1113865) the

problemminusΔ ϕp(Δu(k minus 1))1113872 1113873 + s(k)ϕq(u(k)) λf(u(k)) k isin Z(1 N)

u(0) Δu(N) 0

⎧⎨

⎩

(5)

admits at least two positive solutions

)e structure of the article is as follows In Section 2some basic definitions and properties are given In Section 3we give the main results Under suitable hypothesis Lemma1 is used to obtain that the problem (D

f

λ ) possesses at leasttwo positive solutions Finally some examples are given toillustrate our main results

2 Preliminaries

In this section we recall some definitions notations andproperties Consider the N-dimensional Banach space

S u Z(0 N + 1)⟶ R u(0) Δu(N) 0 (6)

and define the norm

u 1113944N+1

k1|Δu(k minus 1)|

p⎛⎝ ⎞⎠

1p

(7)

and uinfin max |u(k)| k isin Z(1 N) is another norm in S

Proposition 1 -e following inequality holds

uinfin lemaxpNpminus1

1 + slowastNpminus11113888 1113889

1qup

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1q

pNpminus1


1pup

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1p⎧⎨

⎩

⎫⎬

⎭ (8)

Proof Let u isin S then there exist klowast isin Z(1 N) such that|u(klowast)| max |u(k)| k isin Z(1 N)

Since

u klowast

( 11138571113868111386811138681113868

1113868111386811138681113868 1113944

klowast

k1Δu(k minus 1)|le 1113944

klowast

k11)

1minus1p1113944

klowast

k1|Δu(k minus 1)|

p)1p leN

1minus1pu⎛⎝⎛⎝

1113868111386811138681113868111386811138681113868111386811138681113868(9)

2 Discrete Dynamics in Nature and Society

then

upinfin leN

pminus1u

p (10)

If ||u||infin gt 1 then

1 + slowastNpminus1( 1113857uq

infinp

leupinfin + Npminus1 1113936

Nk1 s(k)|u(k)|q

p

leNpminus 1

pu

p+ 1113944

N

k1s(k)|u(k)|

q⎛⎝ ⎞⎠leNpminus1

pu

p+

Npminus1

q1113944

N

k1s(k)|u(k)|

q

(11)

that is

uinfin lepNpminus 1


1qup

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1q

(12)

If uinfin le 1 then

1 + slowastNpminus 1( 1113857u

pinfin

ple

upinfin + Npminus 1 1113936

Nk1 s(k)|u(k)|q

p

leNpminus 1

pu

p+

Npminus 1

q1113944

N

k1s(k)|u(k)|

q

(13)

that is

uinfin lepNpminus 1


1p||u||p

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1p

(14)

In summary we have

uinfin lemaxpNpminus 1


1qup

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1q

pNpminus 1


1pup

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1p⎧⎨

⎩

⎫⎬

⎭ (15)

Put

F(k t) ≔ 1113946t

0f(k ξ)dξ forall(k t) isin Z(1 N) times R (16)

and consider the function Jλ S⟶ R for all λgt 0 by

Jλ Φ minus λΨ (17)

where

Φ ≔ Φ1 +Φ2Φ1(u) up

p

Φ2(u) 1113936

Nk1 s(k)|u(k)|q

q

Ψ(u) ≔ 1113944N

k1F(k u(k))

(18)

It is clear that Φ1Φ2Ψ isin C1(SR) and their Gateauxderivatives at the point u isin S are given by

Φ1prime(u)(v) 1113944N+1

k1ϕp(Δu(k minus 1))v(k)

minus 1113944N+1

k1ϕp(Δu(k minus 1))v(k minus 1)

1113944N

k1ϕp(Δu(k minus 1))v(k) minus 1113944

N

k0ϕp(Δu(k))v(k)

minus 1113944N

k1Δϕp(Δu(k minus 1))v(k)

Φ2prime(u)(v) 1113944N

k1s(k)ϕq(u(k))v(k)

Ψprime(u)(v) 1113944N

k1f(k u(k))v(k)

(19)

Discrete Dynamics in Nature and Society 3

for all u v isin S So we have

Jλprime(u)(v) 1113944N

k1minusΔ ϕp(Δu(k minus 1))1113872 1113873 + s(k)ϕq(u(k)) minus λf(k u(k))1113960 1113961v(k) (20)

Hence a critical point u of Jλ is a solution of problem(D

f

λ )Now we recall a definition and a two nonzero critical

points theorem for the readerrsquos convenience

Definition 1 Let X be a real Banach space we say that aGateaux differentiable function Jλ X⟶ R satisfies the(PS)-condition if any sequence un1113864 1113865nisinNsubeX such that

(i) Jλ(un)⟶ c isin R as n⟶ +infin(ii) Jλprime(un)⟶ 0 as n⟶ +infin has a convergent

subsequence

Lemma 1 Let X be a real Banach space and ΦΨ isin C1(SR)

such that infX(Φ) Φ(0) Ψ(0) 0 Assume that there arer isin R and ω isin X with 0ltΦ(ω)lt r such that

supuisinΦminus1(minusinfinr]Ψ(u)

rltΨ(ω)

Φ(ω) (21)

and for each

λ isin Λ Φ(ω)

Ψ(ω)

r

supuisinΦminus1(minusinfinr]Ψ(u)1113888 1113889 (22)

the functional Jλ Φ minus λΨ satisfies the (PS)-condition and itis unbounded from below

)en for each λ isin Λ the functional Jλ admits at least twononzero critical points uλ1 uλ2 such that Jλ(uλ1)lt 0lt Jλ(uλ2)

In order to obtain the positive solution of problem (Df

λ )we establish the following strong maximum principle

Lemma 2 Fix u isin S such that either

u(k)gt 0 or minusΔ ϕp(Δu(k minus 1))1113872 1113873 + s(k)ϕq(u(k))ge 0

(23)

for all k isin Z(1 N) -en either ugt 0 in Z(1 N) or u equiv 0

Proof Let j isin Z(1 N) such that

u(j) min u(k) k isin Z(1 N) (24)

If u(j)gt 0 then it is easy know that ugt 0 in Z(1 N)If u(j)le 0 then by (23) we have

minusΔ ϕp(Δu(j minus 1))1113872 1113873ge minuss(j)ϕq(u(j))ge 0 (25)

that is

ϕp(Δu(j))leϕp(Δu(j minus 1)) (26)

Since ϕp(u) is increasing in u we have

Δu(j)leΔu(j minus 1) (27)

By the definition of u(j) we know that

Δu(j)ge 0

Δu(j minus 1)le 0(28)

By combining (27) with (28) we get u(j + 1) u(j)

u(j minus 1) If j minus 1 0 we have u(j) u(j minus 1) 0 Other-wise j minus 1 isin Z(1 N) replacing j minus 1 by j we knowu(j minus 2) u(j minus 1) Continuing in this way we haveu(j) u(j minus 1) middot middot middot u(0) 0 Similarly we haveu(j) u(j + 1) middot middot middot u(N + 1) )us u(k) u(0) 0and forallk isin Z(1 N)

Now put

F+(k t) 1113946

t

0f k ξ+

( 1113857dξ forall(k t) isin Z(1 N) times R (29)

where ξ+ max ξ 0

Define J+λ Φ1 +Φ2 minus λΨ+ and

Ψ+(u) 1113936Nk1 F+(k u(k)) Standard arguments show that

J+λ isin C1(S R) and the critical points of J+

λ are precisely thesolutions of the following problem

minusΔ ϕp(Δu(k minus 1))1113872 1113873 + s(k)ϕq(u(k)) λf k u+(k)( ) k isin Z(1 N)

u(0) Δu(N) 0

⎧⎨

⎩ Df+

λ1113872 1113873 (30)

Lemma 3 If f(k 0)ge 0 for each k isin Z(1 N) any nonzerocritical point of the functional J+

λ is a positive solution ofproblem (D

f

λ )

Proof Since a critical point of J+λ is a solution of problem

Df+

λ the conclusion follows by the discrete maximumprinciple ([33] Proposition 1)


Next we suppose that f(k 0)ge 0 and f(k x) f(k 0)

for all xle 0 and for all k isin Z(1 N) Put

Linfin ≔ minkisinZ(1N)

liminft⟶+infin

F(k t)

tp

1113957s 1113944

N

k1s(k)

(31)

we have the following result

Lemma 4 If Linfin gt 0 then Jλ satisfies (PS)-condition and it isunbounded from below for all λ isin (2pN + 1113957s minus 2pminus1qLinfin +infin)

Proof Let λgt 2pN + 1113957s minus 2pminus 1qLinfin We consider a sequenceun1113864 1113865nisinNsube S such that Jλ(un)⟶ c isin R and Jλprime(un)⟶ 0 as

n⟶ +infin Let u+n max un 01113864 1113865 and uminus

n max minusun 01113864 1113865 forall n isin N We first prove that uminus

n1113864 1113865 is bounded On one handwe have

Δuminusn (k minus 1)

11138681113868111386811138681113868111386811138681113868p le minus ϕp Δun(k minus 1)( 1113857Δuminus

n (k minus 1)

s(k)∣uminusn (k)∣q minuss(k)∣un(k)∣qminus2

un(k)uminusn (k)

(32)

for all k isin Z(1 N) So

uminusn

p 1113944

N+1

k1Δuminus

n (k minus 1)1113868111386811138681113868

1113868111386811138681113868p

le minus 1113944N+1

k1ϕp Δun(k minus 1)( 1113857u

minusn (k) + 1113944

N1


minusn (k minus 1)

minus 1113944N


minusn(k) + 1113944

N

k1ϕp Δun(k)( 1113857u

minusn(k)

1113944N

k1Δϕp Δun(k minus 1)( 1113857u

minusn (k) minusΦ1prime un( 1113857 u

minusn( 1113857

1113944

N

k1s(k) u

minusn(k)

11138681113868111386811138681113868111386811138681113868q

minus 1113944N

k1s(k)ϕq un(k)( 1113857u


minusn( 1113857

(33)

On the other hand we assume that

f(k u) f(k u) if ugt 0

f(k 0) if ule 01113896 (34)

for each k isin Z(1 N) then

Ψprime un( 1113857 uminusn( 1113857 1113944

N

k1f k un(k)( 1113857u

minusn (k)ge 0 (35)

)erefore

uminusn

p le uminusn

p+ 1113944

N

k1s(k)|u(k)|

q

le minusΦ1prime un( 1113857 uminusn( 1113857 minusΦ2prime un( 1113857 u

minusn( 1113857 + λΨprime un( 1113857 u

minusn( 1113857

minusJλprime un( 1113857 uminusn( 1113857

(36)

for all n isin N which leads to uminusn pminus 1⟶ 0 as n⟶ +infin

So we have uminusn ⟶ 0 as n⟶ +infin It means that there

exists an Mgt 0 such that uminusn leM From (10) we know that

uminusn infin leN1minus 1pM c for all k isin Z(1 N)

Next we suppose that the sequence un1113864 1113865 is unboundedthat is u+

n1113864 1113865 is unboundedAs Linfin gt 0 we know that there exists an l isin R such that

Linfin gt lgt 2pN + 1113957s minus 2pminus 1λq From the definition of Linfin thereis δk gt 0 such that F(k t)gt l|t|p for all tgt δk Furthermoresince F(k t) is a continuous function there exists a constantC(k)ge 0 such that F(k t)ge l|t|p minus C(k) with t isin [minusc δk]

)us F(k t)ge l|t|p minus C(k) for all sge minus c and k isin Z(1 N)

We can obtain that

1113944

N

k1F k un(k)( 1113857ge 1113944

N

k1l un(k)1113868111386811138681113868

1113868111386811138681113868p

minus Cge l un

p

infin minus C (37)

for all k isin Z(1 N) where C 1113936Nk1 C(k) that is

Ψ un( 1113857ge l un

p

infin minus C (38)

Hence for all un such that uninfinge 1 we conclude that


Jλ un( 1113857 1113936

N+1k1 Δun(k minus 1)

11138681113868111386811138681113868111386811138681113868p

p+

1113936Nk1s(k) un(k)

11138681113868111386811138681113868111386811138681113868q

qminus λΨ un( 1113857

le2pminus 1

p1113944

N

k1un(k)

11138681113868111386811138681113868111386811138681113868p

+ 1113944N

k1un(k minus 1)

11138681113868111386811138681113868111386811138681113868p⎛⎝ ⎞⎠

+1q

1113944

N

k1s(k)∣un(k) ∣q minus λΨ un( 1113857

le2pminus 1(2N minus 1)

pun

p

infin +1113957s

qun

q

infin minus λΨ un( 1113857

le2pN + 1113957s minus 2pminus 1

qminus λl1113888 1113889 un

p

infin + λC

(39)

Since 2pN + 1113957s minus 2pminus 1q minus λllt 0 we can getlimn⟶+infin Jλ(un) minusinfin and this is absurd Hence Jλ sat-isfies (PS)-condition

Let un1113864 1113865 be such that uminusn1113864 1113865 is bounded and u+

n1113864 1113865 is un-bounded From the proof above we can see that Jλ is un-bounded from below

3 Main Results

)e main results of this paper are as follows

Theorem 2 Let f Z(1 N) times R⟶ R be a continuousfunction satisfying f(k 0)ge 0 for all k isin Z(1 N) If there aretwo constants c and d with dlt c such that

pNpminus 1


N

k1max|ξ∣ lec

F(k ξ)max1cq

1cp

1113882 1113883

ltmin1113936

Nk1 F(k d)

dppminus1 + dqqminus11113957s

qLinfin2pN + 1113957s minus 2pminus11113896 1113897

(40)

)en for each λ isin Λ1 with

Λ1 ≔ maxdppminus 1 + dqqminus 11113957s

1113936Nk1 F(k d)

2pN + 1113957s minus 2pminus 1

qLinfin1113896 1113897

1 + slowastNpminus 1pNpminus 1( 1113857min cq cp

1113936Nk1 max |ξ∣lecF(k ξ)

⎛⎝ ⎞⎠ (41)

the problem (Df

λ ) admits at least two positive solutions

Proof Put ΦΨ as in (18) It is clear that infX(Φ) Φ(0)

Ψ(0) 0 According to Lemma 3 we know that a nonzerocritical point in S of the functional J+

λ is precisely a positivesolution of problem (D

f

λ ) Next we just need to provecondition (21) of Lemma 1

We observe that Linfin gt 0 from (40) and Λ1 is nonde-generate Fix λ isin Λ1 Lemma 4 ensures that Jλ satisfies

(PS)-condition for all λgt 2pN + 1113957s minus 2pminus 1qLinfin and it isunbounded from below We let u isin Φminus 1(minusinfin r] that is(upp) + (1113936

Nk1 s(k)|u(k)|qq)le r Put

r 1 + slowastN

pminus 1

pNpminus1 min cq c

p1113864 1113865 (42)

If r (1 + slowastNpminus 1pNpminus 1)cq it means that cge 1

According to (8) we obtain

|u(k)|le uinfin lemaxpNpminus 1


1q

r1q

pNpminus 1


1p

r1p⎧⎨

⎩

⎫⎬

⎭ max c cqp

1113966 1113967 c (43)

If r (1 + slowastNpminus 1pNpminus 1)cp we know 0lt clt 1 then



1q

r1q

pNpminus 1


1p

r1p⎧⎨

⎩

⎫⎬

⎭ max cpq

c1113966 1113967 c (44)

To sum up we know that |u(k)|le c for all k isin Z(1 N)

Furthermore we have

Ψ(u) 1113944N

k1F(k u(k))le 1113944

N

k1max|ξ|lec

F(k ξ) (45)

for all u isin S with Φ(u)le r Hence


rle

pNpminus 1


N

k1max|ξ|lec

F(k ξ)max1cq

1cp

1113882 1113883

(46)


Now let ω(k) d for all k isin Z(1 N) and ω(0)

Δω(N) 0 Clearly ω isin S It is easy to account that Φ(ω)

dppminus 1 + dqqminus 11113957s then

Ψ(ω)

Φ(ω)

1113936Nk1 F(k d)

dppminus1 + dqqminus11113957s (47)

Consequently from (46) (47) and assumption (40) wecan obtain

supuisinΦminus 1(minusinfinr]

Ψ(u)

rltΨ(ω)

Φ(ω)

(48)

Moreover because 0lt dlt c and from (40) we obtain

0ltdpp

minus 1+ d

qq

minus 11113957slt

1 + slowastNpminus 1

pNpminus1 min cq c

p1113864 1113865 (49)

that is mean that 0ltΦ(ω)lt r

Hence the problem (Df

λ ) admits at least two positivesolutions by Lemma 1 and Lemma 3 for all λ isin Λ1

Remark 1 If f(k t) is a nonnegative function and there aretwo positive constants c d with dlt c such that

pNpminus 1

1 + slowastNpminus1 max

1113936Nk1 F(k c)

cq1113936

Nk1 F(k c)

cp1113896 1113897

ltmin1113936

Nk1 F(k d)



(50)

then the result of)eorem 2 is also valid for each λ isin Λ2 with


1113936Nk1 F(k d)


qLinfin1113896 1113897


pNpminus1 mincq

1113936Nk1 F(k c)

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889 (51)

)ere are some consequences of )eorem 2 as follows

Corollary 1 Let g R⟶ [0 +infin) be a continuous functionsuch that f(k t) α(k)g(t) where α(k)gt 0 for allk isin Z(1 N) Put A 1113936

Nk1 α(k) G(t) 1113938

t

0 g(ξ)dξ for allt isin R and Llowastinfin min

kisin[1N]α(k) liminf

t⟶+infin(G(t)tp)gt 0

If there exists cgt dgt 0 such that

pNpminus 1

1 + slowastNpminus1 AG(c)max

1cq

1cp

1113882 1113883

ltminAG(d)


qLlowastinfin2pN + 1113957s minus 2pminus11113896 1113897

(52)

then the problem Df

λ has at least two positive solutions foreach λ isin Λ3 with


AG(d)2pN + 1113957s minus 2pminus 1

qLlowastinfin1113896 1113897


pNpminus1min cq cp

AG(c)1113888 1113889 (53)

Proof Consider the function f Z(1 N) times R⟶ R isgiven as

f(k ξ) α(k)g(ξ) forallk isin Z(1 N) ξ isin R (54)

so that

1113944

N

k1max|ξ|lec

F(k ξ) AG(c) 1113944N

k1F(k d) AG(d) (55)

)en the conclusion can be obtained by)eorem 2

Corollary 2 Assume f be a continuous function withf(k 0)ge 0 and

limsupt⟶0+

F(k t)

tp +infin (56)

limt⟶+infin

F(k t)

tp +infin (57)

for all k isin Z(0 N) Put λlowast (1 + slowastNpminus 1pNpminus 1)min supcgt01113864

(cq1113936Nk1max |ξ∣lecF(k ξ)) supcgt0(cp 1113936

Nk1 max |ξ∣lecF(k ξ))

-en for each λ isin (0 λlowast) the problem (Df

λ ) admits at least twopositive solutions

Proof We know that Linfin +infin from (57) Fix λ isin (0 λlowast)and then there exists cgt 0 such that

λlt1 + slowastN

pminus 1

pNpminus1

middot min supcgt0

cq

1113936Nk1 max|ξ|lecF(k ξ)

supcgt0

cp


⎧⎨

⎩

⎫⎬

⎭

(58)

From (56) we can also obtain

limsupt⟶0+

1113936Nk1 F(k t)

tp +infin (59)

and then there exists d isin (0 c) such that (1113936Nk1 F(k d)

dppminus 1 + dqqminus 11113957s)gt (1λ) )erefore )eorem 2 ensures theconclusion


Remark 2 If f(k t) is a nonnegative function for all(k t) isin Z(1 N) times [0 +infin) As long as condition (56) holdsfor at least one k isin Z(1 N) then Corollary 2 ensures thatthe solutions are obtained for each

λ isin 01 + slowastN

pminus 1

pNpminus1 min supcgt0

cq

1113936Nk1 F(k c)

supcgt0

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889

(60)

Remark 3 When f(k t) f(t) for all k isin Z(1 N) )e-orem 1 can be ensured by Corollary 2 Obviously condition(4(a)) implies f(0)ge 0 Specially if f is nonnegative weonly need condition (4(a)) to get the corresponding so-lution for each


pminus 1

pNpmin sup

cgt0

cq

F(c) sup

cgt0

cp

F(c)1113896 11138971113888 1113889 (61)

Example 1 Let p 4 q 2 N 3 s(k) 12 andf(k t)

et

Put

x(c) c2

ec minus 1 (62)

)en

xprime(c) c (2 minus c)ec minus 2( )

ec minus 1( )2 (63)

Let z(c) (2 minus c)ec minus 2 then zprime(c) (1 minus c)ec So z(c)

is increasing in c isin (0 1) and decreasing in c isin (1 +infin)Since z(0) 0 and z(+infin) minusinfin there exists an uniquec1 isin (1 +infin) such that z(c1) 0 )us x(c) in increasing inc isin (0 c1) and decreasing in c isin (c1 +infin) )is means thatsupcgt0 x(c) x(c1) In fact c1 asymp 15936

Similarly put y(c) c4ec minus 1 we can show that thereexists a unique c2 isin (3 +infin) such that supcgt0y(c) y(c2)

In fact c2 asymp 39207

Since

y c2( 1113857 c42

ec2 minus 1gt

c41ec1 minus 1

c21x c1( 1113857gt x c1( 1113857 (64)

then


pNpmin sup

cgt0x(c) sup

cgt0y(c)1113896 1113897

1 + slowastN

pminus 1

pNpx c1( 1113857 asymp 06496

(65)

)erefore for each λ isin (0 06496) the problem

minusΔ ϕ4(Δu(k minus 1))( 1113857 + 12ϕ2(u(k)) λeu(k) k 1 2 3

u(0) Δu(3) 0

⎧⎨

⎩

(66)


Example 2 Let N 3 p 3 and q 2 and f be a functionas follows

f(k t)

0 if tlt 0kt

3radic

if 0le tle 1kt

3radic

+ 225t2 minus 225 if tgt 1

⎧⎪⎪⎨

⎪⎪⎩(67)

From Remark 1 we can choose c 1 and d 002 Easycalculation shows that

pNpminus 1


N

k1F(k c)max

1cq

1cp

1113882 1113883 pNpminus 1


3

k111139461

0

kt

3radic

dt

asymp 13693

1113936Nk1 F(k d)


11139363k1 1113946

002

0

kt

3radic

dt

dppminus1 + dqqminus11113957sasymp 31386

qLinfin2pN + 1113957s minus 2pminus1

166

limt⟶+infin

kt

3radic

+ 225t2 minus 225t2

asymp 34091

(68)

which satisfy condition (50) )us for each λ isin(03186 07303) the problem

minusΔ ϕ3(Δu(k minus 1))( 1113857 + 8ϕ2(u(k)) λf(k u(k)) k 1 2 3

u(0) Δu(3) 01113896

(69)


Data Availability

No data were used to support the study

Conflicts of Interest

)e authors declare that there are no conflicts of interestregarding the publication of this paper

Authorsrsquo Contributions

All authors contributed equally to the writing of this paperAll authors read and approved the final manuscript

Acknowledgments

)is work was supported by the National Natural ScienceFoundation of China (Grant no 11971126) and Program forChangjiang Scholars and Innovative Research Team inUniversity (Grant no IRTminus16R16)

References

[1] R P Agarwal Difference Equations and Inequalities -eoryMethods and Applications Marcel Dekker Inc New YorkNY USA 2000

[2] A Kristaly M Mihailescu V Radulescu and S TersianldquoSpectral estimates for a nonhomogeneous difference


problemrdquo Communications in Contemporary Mathematicsvol 12 no 6 pp 1015ndash1029 2010

[3] L M Li and Z Zhou ldquoInfinitely many positive solutions for acoupled discrete boundary value problemrdquoDiscrete Dynamicsin Nature and Society vol 2019 Article ID 8052497 7 pages2019

[4] J X Ling and Z Zhou ldquoPositive solutions of the discreteDirichlet problem involving the mean curvature operatorrdquoOpen Mathematics vol 17 no 1 pp 1055ndash1064 2019

[5] Q Q Zhang ldquoHomoclinic orbits for discrete Hamiltoniansystems with local super-quadratic conditionsrdquo Communi-cations on Pure amp Applied Analysis vol 18 no 1 pp 425ndash4342019

[6] Z Zhou and J X Ling ldquoInfinitely many positive solutions fora discrete two point nonlinear boundary value problem withϕc-Laplacianrdquo Applied Mathematics Letters vol 91 pp 28ndash34 2019

[7] G Bonanno and P Candito ldquoInfinitely many solutions for aclass of discrete non-linear boundary value problemsrdquo Ap-plicable Analysis vol 88 no 4 pp 605ndash616 2009

[8] G H Lin and Z Zhou ldquoHomoclinic solutions of discreteφ-Laplacian equations with mixed nonlinearitiesrdquo Commu-nications on Pure amp Applied Analysis vol 17 no 5pp 1723ndash1747 2018

[9] J S Yu and B Zheng ldquoModeling Wolbachia infection inmosquito population via discrete dynamical modelrdquo Journalof Difference Equations and Applications vol 25 no 11pp 1549ndash1567 2019

[10] P Mei Z Zhou and G H Lin ldquoPeriodic and subharmonicsolutions for a 2nth-order ϕc-Laplacian difference equationcontaining both advances and retardationsrdquo Discrete ContDyn-S vol 12 pp 2085ndash2095 2019

[11] M K Moghadam L Li and S Tersian ldquoExistence of threesolutions for a discrete anisotropic boundary value problemrdquoBulletin of the Iranian Mathematical Society vol 44 no 4pp 1091ndash1107 2018

[12] L Erbe B G Jia and Q Q Zhang ldquoHomoclinic solutions ofdiscrete nonlinear systems via variational methodrdquo Journal ofApplied Analysis and Computation vol 9 pp 271ndash294 2019

[13] S Heidarkhani and M Imbesi ldquoMultiple solutions for partialdiscrete Dirichlet problems depending on a real parameterrdquoJournal of Difference Equations and Applications vol 21 no 2pp 96ndash110 2015

[14] G H Lin Z Zhou and J S Yu ldquoGround state solutions ofdiscrete asymptotically linear Schrodinge equations withbounded and non-periodic potentialsrdquo Journal of Dynamicsand Differential Equations 2019

[15] M Mihǎilescu V Rǎdulescu and S Tersian ldquoHomoclinicsolutions of difference equations with variable exponentsrdquoTopological Methods in Nonlinear Analysis vol 38 pp 277ndash289 2011

[16] J Kuang and Z Guo ldquoHeteroclinic solutions for a class of p-Laplacian difference equations with a parameterrdquo AppliedMathematics Letters vol 100 Article ID 106034 2020

[17] C Bereanu and J Mawhin ldquoBoundary value problems forsecond-order nonlinear difference equations with discreteφ-Laplacian and singular φrdquo Journal of Difference Equationsand Applications vol 14 no 10-11 pp 1099ndash1118 2008

[18] W G Kelly and A C Peterson Difference Equations AnIntroduction with Applications Academic Press San DiegoCA USA 1991

[19] G Bonanno P Jebelean and C Serban ldquoSuperlinear discreteproblemsrdquo Applied Mathematics Letters vol 52 pp 162ndash1682016

[20] J Henderson and H B )ompson ldquoExistence of multiplesolutions for second-order discrete boundary value prob-lemsrdquo Computers amp Mathematics with Applications vol 43no 10-11 pp 1239ndash1248 2002

[21] Y H Long and J L Chen ldquoExistence of multiple solutions tosecond-order discrete Neumann boundary problemsrdquo Ap-plied Mathematics Letters vol 83 pp 7ndash14 2018

[22] B Ricceri ldquoA general variational principle and some of itsapplicationsrdquo Journal of Computational and Applied Math-ematics vol 133 no 1-2 pp 401ndash410 2000

[23] Z Zhou and D F Ma ldquoMultiplicity results of breathers for thediscrete nonlinear Schrodinger equations with unboundedpotentialsrdquo Science China Mathematics vol 58 no 4pp 781ndash790 2015

[24] R P Agarwal K Perera and D OrsquoRegan ldquoMultiple positivesolutions of singular discrete p-Laplacian problems via var-iational methodsrdquo Advances in Difference Equationsvol 2015 no 2 Article ID 690272 2005

[25] A Nastasi C Vetro and F Vetro ldquoPositive solutions ofdiscrete boundary value problems with the (p q)-Laplacianoperatorrdquo Electronic Journal of Differential Equationsvol 225 pp 1ndash12 2017

[26] Z Zhou and M T Su ldquoBoundary value problems for 2nth-order ϕc-Laplacian difference equations containing bothadvance and retardationrdquo Applied Mathematics Lettersvol 41 pp 7ndash11 2015

[27] Z Zhou J S Yu and Y M Chen ldquoHomoclinic solutions inperiodic diffrence equations with saturable nonlinearityrdquoScience China Mathematics vol 54 no 1 pp 83ndash93 2011

[28] G DrsquoAguı J Mawhin and A Sciammetta ldquoPositive solutionsfor a discrete two point nonlinear boundary value problemwith p-Laplacianrdquo Journal of Mathematical Analysis andApplications vol 447 no 1 pp 383ndash397 2017

[29] G DrsquoAguı A Sciammetta and E Tornatore ldquoTwo non-zerosolutions for Sturm-Liouville equations with mixed boundaryconditionrdquo Nonlinear Anal Real World Appl vol 47pp 324ndash331 2019

[30] G Bonanno and G DrsquoAguı ldquoTwo non-zero solutions forelliptic Dirichlet problemsrdquo Zeitschrift fur Analysis und ihreAnwendungen vol 35 no 4 pp 449ndash464 2016

[31] G Bonanno ldquoA critical point theorem via the Ekeland var-iational principlerdquo Nonlinear Analysis -eory Methods ampApplications vol 75 no 5 pp 2992ndash3007 2012

[32] A Ambrosetti and P H Rabinowitz ldquoDual variationalmethods in critical point theory and applicationsrdquo Journal ofFunctional Analysis vol 14 no 4 pp 349ndash381 1973

[33] G Bonanno P Candito and G DrsquoAguı ldquoVariational methodson finite dimensional Banach space and discrete problemsrdquoAdvanced Nonlinear Studies vol 14 no 4 pp 915ndash939 2014


minusΔ ϕp(Δu(k minus 1))1113872 1113873 + s(k)ϕq(u(k)) λf(k u(k)) k isin Z(1 N)

u(0) Δu(N) 0

⎧⎨

⎩ Df

λ1113872 1113873 (3)

where Z(a b) denote the discrete interval a a + 1 b

for any integers a and b with alt b N be a positive integerf(k u) is continuous in u for each k isin Z(1 N) Δu(k)

u(k + 1) minus u(k) is the forward difference operatorϕr R⟶ R is the r-Laplacian given by ϕr(u) |u|rminus 2u

with u isin R 1lt qleplt +infin s(k)ge 0 for all k isin Z(1 N) andλ is a positive parameter

In this paper under suitable assumptions on the non-linearity f we use the theory of two nonzero critical points(see [30]) to ensure that there are at least two nonzerosolutions for problem (D

f

λ ) )e two nonzero critical pointstheorem is an appropriate combination of local minimumtheorem (see [31]) and classical AmbrosettindashRabinowitztheorem (see [32]) An important hypothesis of mountainpass theorem is PalaisndashSmale condition It satisfies the ap-plication of infinite dimensional space by requiring thecondition that the nonlinear term is stronger than p-superlinearity at infinity In order to obtain the existence oftwo nonzero solutions we can assume the classicalAmbrosettindashRabinowitz condition and nonlinear algebraiccondition (see (40) in )eorem 2) hold that is morewidespread than the p-sublinearity at zero Moreover whenwe require that f(k 0)ge 0 for all k isin Z(1 N) we can usestrongmaximum principle to obtain the existence of positivesolutions which has been proved in Lemma 2

Let slowast min s(k) k isin Z(1 N) a special case of ourmain result is stated as follows

Theorem 1 Let f R⟶ R be a continuous function suchthat

limt⟶0+

f(t)

tpminus14(a)

limt⟶+infin

f(t)

tpminus14(b)

then for each λ isin (0 (1 + slowastNpminus1pNp)min 1113864supcgt0(cq

max|ξ|lec 1113938ξ0 f(t)dt) supcgt0(cpmax|ξ|lec 1113938

ξ0 f(t)dt)1113865) the

problemminusΔ ϕp(Δu(k minus 1))1113872 1113873 + s(k)ϕq(u(k)) λf(u(k)) k isin Z(1 N)

u(0) Δu(N) 0

⎧⎨

⎩

(5)


)e structure of the article is as follows In Section 2some basic definitions and properties are given In Section 3we give the main results Under suitable hypothesis Lemma1 is used to obtain that the problem (D

f

λ ) possesses at leasttwo positive solutions Finally some examples are given toillustrate our main results

2 Preliminaries

In this section we recall some definitions notations andproperties Consider the N-dimensional Banach space

S u Z(0 N + 1)⟶ R u(0) Δu(N) 0 (6)

and define the norm

u 1113944N+1

k1|Δu(k minus 1)|

p⎛⎝ ⎞⎠

1p

(7)

and uinfin max |u(k)| k isin Z(1 N) is another norm in S

Proposition 1 -e following inequality holds

uinfin lemaxpNpminus1


1qup

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1q

pNpminus1


1pup

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1p⎧⎨

⎩

⎫⎬

⎭ (8)

Proof Let u isin S then there exist klowast isin Z(1 N) such that|u(klowast)| max |u(k)| k isin Z(1 N)

Since

u klowast

( 11138571113868111386811138681113868

1113868111386811138681113868 1113944

klowast

k1Δu(k minus 1)|le 1113944

klowast

k11)

1minus1p1113944

klowast

k1|Δu(k minus 1)|

p)1p leN

1minus1pu⎛⎝⎛⎝

1113868111386811138681113868111386811138681113868111386811138681113868(9)


then

upinfin leN

pminus1u

p (10)



infinp


Nk1 s(k)|u(k)|q

p

leNpminus 1

pu

p+ 1113944

N

k1s(k)|u(k)|


pu

p+

Npminus1

q1113944

N

k1s(k)|u(k)|

q

(11)

that is

uinfin lepNpminus 1


1qup

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1q

(12)

If uinfin le 1 then


pinfin

ple


Nk1 s(k)|u(k)|q

p

leNpminus 1

pu

p+

Npminus 1

q1113944

N

k1s(k)|u(k)|

q

(13)

that is

uinfin lepNpminus 1


1p||u||p

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1p

(14)

In summary we have



1qup

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1q

pNpminus 1


1pup

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1p⎧⎨

⎩

⎫⎬

⎭ (15)

Put

F(k t) ≔ 1113946t




where

Φ ≔ Φ1 +Φ2Φ1(u) up

p

Φ2(u) 1113936

Nk1 s(k)|u(k)|q

q

Ψ(u) ≔ 1113944N

k1F(k u(k))

(18)


Φ1prime(u)(v) 1113944N+1


minus 1113944N+1


1113944N


N

k0ϕp(Δu(k))v(k)

minus 1113944N



k1s(k)ϕq(u(k))v(k)


k1f(k u(k))v(k)

(19)






f





subsequence




rltΨ(ω)

Φ(ω) (21)

and for each

λ isin Λ Φ(ω)

Ψ(ω)

r








(23)






that is





Δu(j)ge 0




Now put

F+(k t) 1113946

t

0f k ξ+


where ξ+ max ξ 0






u(0) Δu(N) 0

⎧⎨

⎩ Df+

λ1113872 1113873 (30)



f

λ )


Df+






liminft⟶+infin

F(k t)

tp

1113957s 1113944

N

k1s(k)

(31)









n (k minus 1)


un(k)uminusn (k)

(32)


uminusn

p 1113944

N+1

k1Δuminus

n (k minus 1)1113868111386811138681113868

1113868111386811138681113868p

le minus 1113944N+1


minusn (k) + 1113944

N1


minusn (k minus 1)

minus 1113944N


minusn(k) + 1113944

N

k1ϕp Δun(k)( 1113857u

minusn(k)

1113944N



minusn( 1113857

1113944

N

k1s(k) u

minusn(k)

11138681113868111386811138681113868111386811138681113868q

minus 1113944N

k1s(k)ϕq un(k)( 1113857u


minusn( 1113857

(33)



f(k 0) if ule 01113896 (34)



N

k1f k un(k)( 1113857u

minusn (k)ge 0 (35)

)erefore

uminusn

p le uminusn

p+ 1113944

N

k1s(k)|u(k)|

q



minusn( 1113857


(36)









We can obtain that

1113944

N

k1F k un(k)( 1113857ge 1113944

N

k1l un(k)1113868111386811138681113868

1113868111386811138681113868p

minus Cge l un

p

infin minus C (37)



p

infin minus C (38)



Jλ un( 1113857 1113936


11138681113868111386811138681113868111386811138681113868p

p+

1113936Nk1s(k) un(k)

11138681113868111386811138681113868111386811138681113868q


le2pminus 1

p1113944

N

k1un(k)

11138681113868111386811138681113868111386811138681113868p

+ 1113944N

k1un(k minus 1)

11138681113868111386811138681113868111386811138681113868p⎛⎝ ⎞⎠

+1q

1113944

N



pun

p

infin +1113957s

qun

q



qminus λl1113888 1113889 un

p

infin + λC

(39)




3 Main Results



pNpminus 1


N

k1max|ξ∣ lec

F(k ξ)max1cq

1cp

1113882 1113883

ltmin1113936

Nk1 F(k d)



(40)



1113936Nk1 F(k d)


qLinfin1113896 1113897



⎛⎝ ⎞⎠ (41)

the problem (Df





f





r 1 + slowastN

pminus 1

pNpminus1 min cq c

p1113864 1113865 (42)





1q

r1q

pNpminus 1


1p

r1p⎧⎨

⎩

⎫⎬

⎭ max c cqp

1113966 1113967 c (43)




1q

r1q

pNpminus 1


1p

r1p⎧⎨

⎩

⎫⎬

⎭ max cpq

c1113966 1113967 c (44)


Furthermore we have

Ψ(u) 1113944N

k1F(k u(k))le 1113944

N

k1max|ξ|lec

F(k ξ) (45)



rle

pNpminus 1


N

k1max|ξ|lec

F(k ξ)max1cq

1cp

1113882 1113883

(46)





Ψ(ω)

Φ(ω)

1113936Nk1 F(k d)




Ψ(u)

rltΨ(ω)

Φ(ω)

(48)


0ltdpp

minus 1+ d

qq

minus 11113957slt


pNpminus1 min cq c

p1113864 1113865 (49)





pNpminus 1


1113936Nk1 F(k c)

cq1113936

Nk1 F(k c)

cp1113896 1113897

ltmin1113936

Nk1 F(k d)



(50)



1113936Nk1 F(k d)


qLinfin1113896 1113897


pNpminus1 mincq

1113936Nk1 F(k c)

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889 (51)



Nk1 α(k) G(t) 1113938

t





pNpminus 1


1cq

1cp

1113882 1113883

ltminAG(d)



(52)

then the problem Df






pNpminus1min cq cp

AG(c)1113888 1113889 (53)



so that

1113944

N

k1max|ξ|lec

F(k ξ) AG(c) 1113944N

k1F(k d) AG(d) (55)



limsupt⟶0+

F(k t)

tp +infin (56)

limt⟶+infin

F(k t)

tp +infin (57)







λlt1 + slowastN

pminus 1

pNpminus1

middot min supcgt0

cq


supcgt0

cp


⎧⎨

⎩

⎫⎬

⎭

(58)


limsupt⟶0+

1113936Nk1 F(k t)

tp +infin (59)






pminus 1


cq

1113936Nk1 F(k c)

supcgt0

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889

(60)



pminus 1

pNpmin sup

cgt0

cq

F(c) sup

cgt0

cp

F(c)1113896 11138971113888 1113889 (61)


et

Put

x(c) c2

ec minus 1 (62)

)en


ec minus 1( )2 (63)





Since

y c2( 1113857 c42

ec2 minus 1gt

c41ec1 minus 1

c21x c1( 1113857gt x c1( 1113857 (64)

then


pNpmin sup

cgt0x(c) sup

cgt0y(c)1113896 1113897

1 + slowastN

pminus 1

pNpx c1( 1113857 asymp 06496

(65)



u(0) Δu(3) 0

⎧⎨

⎩

(66)



f(k t)

0 if tlt 0kt

3radic

if 0le tle 1kt

3radic


⎧⎪⎪⎨

⎪⎪⎩(67)


pNpminus 1


N

k1F(k c)max

1cq

1cp

1113882 1113883 pNpminus 1


3

k111139461

0

kt

3radic

dt

asymp 13693

1113936Nk1 F(k d)


11139363k1 1113946

002

0

kt

3radic

dt



166

limt⟶+infin

kt

3radic

+ 225t2 minus 225t2

asymp 34091

(68)



u(0) Δu(3) 01113896

(69)


Data Availability






Acknowledgments


References





































then

upinfin leN

pminus1u

p (10)



infinp


Nk1 s(k)|u(k)|q

p

leNpminus 1

pu

p+ 1113944

N

k1s(k)|u(k)|


pu

p+

Npminus1

q1113944

N

k1s(k)|u(k)|

q

(11)

that is

uinfin lepNpminus 1


1qup

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1q

(12)

If uinfin le 1 then


pinfin

ple


Nk1 s(k)|u(k)|q

p

leNpminus 1

pu

p+

Npminus 1

q1113944

N

k1s(k)|u(k)|

q

(13)

that is

uinfin lepNpminus 1


1p||u||p

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1p

(14)

In summary we have



1qup

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1q

pNpminus 1


1pup

p+

1113936Nk1 s(k)|u(k)|q

q1113888 1113889

1p⎧⎨

⎩

⎫⎬

⎭ (15)

Put

F(k t) ≔ 1113946t




where

Φ ≔ Φ1 +Φ2Φ1(u) up

p

Φ2(u) 1113936

Nk1 s(k)|u(k)|q

q

Ψ(u) ≔ 1113944N

k1F(k u(k))

(18)


Φ1prime(u)(v) 1113944N+1


minus 1113944N+1


1113944N


N

k0ϕp(Δu(k))v(k)

minus 1113944N



k1s(k)ϕq(u(k))v(k)


k1f(k u(k))v(k)

(19)






f





subsequence




rltΨ(ω)

Φ(ω) (21)

and for each

λ isin Λ Φ(ω)

Ψ(ω)

r








(23)






that is





Δu(j)ge 0




Now put

F+(k t) 1113946

t

0f k ξ+


where ξ+ max ξ 0






u(0) Δu(N) 0

⎧⎨

⎩ Df+

λ1113872 1113873 (30)



f

λ )


Df+






liminft⟶+infin

F(k t)

tp

1113957s 1113944

N

k1s(k)

(31)









n (k minus 1)


un(k)uminusn (k)

(32)


uminusn

p 1113944

N+1

k1Δuminus

n (k minus 1)1113868111386811138681113868

1113868111386811138681113868p

le minus 1113944N+1


minusn (k) + 1113944

N1


minusn (k minus 1)

minus 1113944N


minusn(k) + 1113944

N

k1ϕp Δun(k)( 1113857u

minusn(k)

1113944N



minusn( 1113857

1113944

N

k1s(k) u

minusn(k)

11138681113868111386811138681113868111386811138681113868q

minus 1113944N

k1s(k)ϕq un(k)( 1113857u


minusn( 1113857

(33)



f(k 0) if ule 01113896 (34)



N

k1f k un(k)( 1113857u

minusn (k)ge 0 (35)

)erefore

uminusn

p le uminusn

p+ 1113944

N

k1s(k)|u(k)|

q



minusn( 1113857


(36)









We can obtain that

1113944

N

k1F k un(k)( 1113857ge 1113944

N

k1l un(k)1113868111386811138681113868

1113868111386811138681113868p

minus Cge l un

p

infin minus C (37)



p

infin minus C (38)



Jλ un( 1113857 1113936


11138681113868111386811138681113868111386811138681113868p

p+

1113936Nk1s(k) un(k)

11138681113868111386811138681113868111386811138681113868q


le2pminus 1

p1113944

N

k1un(k)

11138681113868111386811138681113868111386811138681113868p

+ 1113944N

k1un(k minus 1)

11138681113868111386811138681113868111386811138681113868p⎛⎝ ⎞⎠

+1q

1113944

N



pun

p

infin +1113957s

qun

q



qminus λl1113888 1113889 un

p

infin + λC

(39)




3 Main Results



pNpminus 1


N

k1max|ξ∣ lec

F(k ξ)max1cq

1cp

1113882 1113883

ltmin1113936

Nk1 F(k d)



(40)



1113936Nk1 F(k d)


qLinfin1113896 1113897



⎛⎝ ⎞⎠ (41)

the problem (Df





f





r 1 + slowastN

pminus 1

pNpminus1 min cq c

p1113864 1113865 (42)





1q

r1q

pNpminus 1


1p

r1p⎧⎨

⎩

⎫⎬

⎭ max c cqp

1113966 1113967 c (43)




1q

r1q

pNpminus 1


1p

r1p⎧⎨

⎩

⎫⎬

⎭ max cpq

c1113966 1113967 c (44)


Furthermore we have

Ψ(u) 1113944N

k1F(k u(k))le 1113944

N

k1max|ξ|lec

F(k ξ) (45)



rle

pNpminus 1


N

k1max|ξ|lec

F(k ξ)max1cq

1cp

1113882 1113883

(46)





Ψ(ω)

Φ(ω)

1113936Nk1 F(k d)




Ψ(u)

rltΨ(ω)

Φ(ω)

(48)


0ltdpp

minus 1+ d

qq

minus 11113957slt


pNpminus1 min cq c

p1113864 1113865 (49)





pNpminus 1


1113936Nk1 F(k c)

cq1113936

Nk1 F(k c)

cp1113896 1113897

ltmin1113936

Nk1 F(k d)



(50)



1113936Nk1 F(k d)


qLinfin1113896 1113897


pNpminus1 mincq

1113936Nk1 F(k c)

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889 (51)



Nk1 α(k) G(t) 1113938

t





pNpminus 1


1cq

1cp

1113882 1113883

ltminAG(d)



(52)

then the problem Df






pNpminus1min cq cp

AG(c)1113888 1113889 (53)



so that

1113944

N

k1max|ξ|lec

F(k ξ) AG(c) 1113944N

k1F(k d) AG(d) (55)



limsupt⟶0+

F(k t)

tp +infin (56)

limt⟶+infin

F(k t)

tp +infin (57)







λlt1 + slowastN

pminus 1

pNpminus1

middot min supcgt0

cq


supcgt0

cp


⎧⎨

⎩

⎫⎬

⎭

(58)


limsupt⟶0+

1113936Nk1 F(k t)

tp +infin (59)






pminus 1


cq

1113936Nk1 F(k c)

supcgt0

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889

(60)



pminus 1

pNpmin sup

cgt0

cq

F(c) sup

cgt0

cp

F(c)1113896 11138971113888 1113889 (61)


et

Put

x(c) c2

ec minus 1 (62)

)en


ec minus 1( )2 (63)





Since

y c2( 1113857 c42

ec2 minus 1gt

c41ec1 minus 1

c21x c1( 1113857gt x c1( 1113857 (64)

then


pNpmin sup

cgt0x(c) sup

cgt0y(c)1113896 1113897

1 + slowastN

pminus 1

pNpx c1( 1113857 asymp 06496

(65)



u(0) Δu(3) 0

⎧⎨

⎩

(66)



f(k t)

0 if tlt 0kt

3radic

if 0le tle 1kt

3radic


⎧⎪⎪⎨

⎪⎪⎩(67)


pNpminus 1


N

k1F(k c)max

1cq

1cp

1113882 1113883 pNpminus 1


3

k111139461

0

kt

3radic

dt

asymp 13693

1113936Nk1 F(k d)


11139363k1 1113946

002

0

kt

3radic

dt



166

limt⟶+infin

kt

3radic

+ 225t2 minus 225t2

asymp 34091

(68)



u(0) Δu(3) 01113896

(69)


Data Availability






Acknowledgments


References









































f





subsequence




rltΨ(ω)

Φ(ω) (21)

and for each

λ isin Λ Φ(ω)

Ψ(ω)

r








(23)






that is





Δu(j)ge 0




Now put

F+(k t) 1113946

t

0f k ξ+


where ξ+ max ξ 0






u(0) Δu(N) 0

⎧⎨

⎩ Df+

λ1113872 1113873 (30)



f

λ )


Df+






liminft⟶+infin

F(k t)

tp

1113957s 1113944

N

k1s(k)

(31)









n (k minus 1)


un(k)uminusn (k)

(32)


uminusn

p 1113944

N+1

k1Δuminus

n (k minus 1)1113868111386811138681113868

1113868111386811138681113868p

le minus 1113944N+1


minusn (k) + 1113944

N1


minusn (k minus 1)

minus 1113944N


minusn(k) + 1113944

N

k1ϕp Δun(k)( 1113857u

minusn(k)

1113944N



minusn( 1113857

1113944

N

k1s(k) u

minusn(k)

11138681113868111386811138681113868111386811138681113868q

minus 1113944N

k1s(k)ϕq un(k)( 1113857u


minusn( 1113857

(33)



f(k 0) if ule 01113896 (34)



N

k1f k un(k)( 1113857u

minusn (k)ge 0 (35)

)erefore

uminusn

p le uminusn

p+ 1113944

N

k1s(k)|u(k)|

q



minusn( 1113857


(36)









We can obtain that

1113944

N

k1F k un(k)( 1113857ge 1113944

N

k1l un(k)1113868111386811138681113868

1113868111386811138681113868p

minus Cge l un

p

infin minus C (37)



p

infin minus C (38)



Jλ un( 1113857 1113936


11138681113868111386811138681113868111386811138681113868p

p+

1113936Nk1s(k) un(k)

11138681113868111386811138681113868111386811138681113868q


le2pminus 1

p1113944

N

k1un(k)

11138681113868111386811138681113868111386811138681113868p

+ 1113944N

k1un(k minus 1)

11138681113868111386811138681113868111386811138681113868p⎛⎝ ⎞⎠

+1q

1113944

N



pun

p

infin +1113957s

qun

q



qminus λl1113888 1113889 un

p

infin + λC

(39)




3 Main Results



pNpminus 1


N

k1max|ξ∣ lec

F(k ξ)max1cq

1cp

1113882 1113883

ltmin1113936

Nk1 F(k d)



(40)



1113936Nk1 F(k d)


qLinfin1113896 1113897



⎛⎝ ⎞⎠ (41)

the problem (Df





f





r 1 + slowastN

pminus 1

pNpminus1 min cq c

p1113864 1113865 (42)





1q

r1q

pNpminus 1


1p

r1p⎧⎨

⎩

⎫⎬

⎭ max c cqp

1113966 1113967 c (43)




1q

r1q

pNpminus 1


1p

r1p⎧⎨

⎩

⎫⎬

⎭ max cpq

c1113966 1113967 c (44)


Furthermore we have

Ψ(u) 1113944N

k1F(k u(k))le 1113944

N

k1max|ξ|lec

F(k ξ) (45)



rle

pNpminus 1


N

k1max|ξ|lec

F(k ξ)max1cq

1cp

1113882 1113883

(46)





Ψ(ω)

Φ(ω)

1113936Nk1 F(k d)




Ψ(u)

rltΨ(ω)

Φ(ω)

(48)


0ltdpp

minus 1+ d

qq

minus 11113957slt


pNpminus1 min cq c

p1113864 1113865 (49)





pNpminus 1


1113936Nk1 F(k c)

cq1113936

Nk1 F(k c)

cp1113896 1113897

ltmin1113936

Nk1 F(k d)



(50)



1113936Nk1 F(k d)


qLinfin1113896 1113897


pNpminus1 mincq

1113936Nk1 F(k c)

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889 (51)



Nk1 α(k) G(t) 1113938

t





pNpminus 1


1cq

1cp

1113882 1113883

ltminAG(d)



(52)

then the problem Df






pNpminus1min cq cp

AG(c)1113888 1113889 (53)



so that

1113944

N

k1max|ξ|lec

F(k ξ) AG(c) 1113944N

k1F(k d) AG(d) (55)



limsupt⟶0+

F(k t)

tp +infin (56)

limt⟶+infin

F(k t)

tp +infin (57)







λlt1 + slowastN

pminus 1

pNpminus1

middot min supcgt0

cq


supcgt0

cp


⎧⎨

⎩

⎫⎬

⎭

(58)


limsupt⟶0+

1113936Nk1 F(k t)

tp +infin (59)






pminus 1


cq

1113936Nk1 F(k c)

supcgt0

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889

(60)



pminus 1

pNpmin sup

cgt0

cq

F(c) sup

cgt0

cp

F(c)1113896 11138971113888 1113889 (61)


et

Put

x(c) c2

ec minus 1 (62)

)en


ec minus 1( )2 (63)





Since

y c2( 1113857 c42

ec2 minus 1gt

c41ec1 minus 1

c21x c1( 1113857gt x c1( 1113857 (64)

then


pNpmin sup

cgt0x(c) sup

cgt0y(c)1113896 1113897

1 + slowastN

pminus 1

pNpx c1( 1113857 asymp 06496

(65)



u(0) Δu(3) 0

⎧⎨

⎩

(66)



f(k t)

0 if tlt 0kt

3radic

if 0le tle 1kt

3radic


⎧⎪⎪⎨

⎪⎪⎩(67)


pNpminus 1


N

k1F(k c)max

1cq

1cp

1113882 1113883 pNpminus 1


3

k111139461

0

kt

3radic

dt

asymp 13693

1113936Nk1 F(k d)


11139363k1 1113946

002

0

kt

3radic

dt



166

limt⟶+infin

kt

3radic

+ 225t2 minus 225t2

asymp 34091

(68)



u(0) Δu(3) 01113896

(69)


Data Availability






Acknowledgments


References








































liminft⟶+infin

F(k t)

tp

1113957s 1113944

N

k1s(k)

(31)









n (k minus 1)


un(k)uminusn (k)

(32)


uminusn

p 1113944

N+1

k1Δuminus

n (k minus 1)1113868111386811138681113868

1113868111386811138681113868p

le minus 1113944N+1


minusn (k) + 1113944

N1


minusn (k minus 1)

minus 1113944N


minusn(k) + 1113944

N

k1ϕp Δun(k)( 1113857u

minusn(k)

1113944N



minusn( 1113857

1113944

N

k1s(k) u

minusn(k)

11138681113868111386811138681113868111386811138681113868q

minus 1113944N

k1s(k)ϕq un(k)( 1113857u


minusn( 1113857

(33)



f(k 0) if ule 01113896 (34)



N

k1f k un(k)( 1113857u

minusn (k)ge 0 (35)

)erefore

uminusn

p le uminusn

p+ 1113944

N

k1s(k)|u(k)|

q



minusn( 1113857


(36)









We can obtain that

1113944

N

k1F k un(k)( 1113857ge 1113944

N

k1l un(k)1113868111386811138681113868

1113868111386811138681113868p

minus Cge l un

p

infin minus C (37)



p

infin minus C (38)



Jλ un( 1113857 1113936


11138681113868111386811138681113868111386811138681113868p

p+

1113936Nk1s(k) un(k)

11138681113868111386811138681113868111386811138681113868q


le2pminus 1

p1113944

N

k1un(k)

11138681113868111386811138681113868111386811138681113868p

+ 1113944N

k1un(k minus 1)

11138681113868111386811138681113868111386811138681113868p⎛⎝ ⎞⎠

+1q

1113944

N



pun

p

infin +1113957s

qun

q



qminus λl1113888 1113889 un

p

infin + λC

(39)




3 Main Results



pNpminus 1


N

k1max|ξ∣ lec

F(k ξ)max1cq

1cp

1113882 1113883

ltmin1113936

Nk1 F(k d)



(40)



1113936Nk1 F(k d)


qLinfin1113896 1113897



⎛⎝ ⎞⎠ (41)

the problem (Df





f





r 1 + slowastN

pminus 1

pNpminus1 min cq c

p1113864 1113865 (42)





1q

r1q

pNpminus 1


1p

r1p⎧⎨

⎩

⎫⎬

⎭ max c cqp

1113966 1113967 c (43)




1q

r1q

pNpminus 1


1p

r1p⎧⎨

⎩

⎫⎬

⎭ max cpq

c1113966 1113967 c (44)


Furthermore we have

Ψ(u) 1113944N

k1F(k u(k))le 1113944

N

k1max|ξ|lec

F(k ξ) (45)



rle

pNpminus 1


N

k1max|ξ|lec

F(k ξ)max1cq

1cp

1113882 1113883

(46)





Ψ(ω)

Φ(ω)

1113936Nk1 F(k d)




Ψ(u)

rltΨ(ω)

Φ(ω)

(48)


0ltdpp

minus 1+ d

qq

minus 11113957slt


pNpminus1 min cq c

p1113864 1113865 (49)





pNpminus 1


1113936Nk1 F(k c)

cq1113936

Nk1 F(k c)

cp1113896 1113897

ltmin1113936

Nk1 F(k d)



(50)



1113936Nk1 F(k d)


qLinfin1113896 1113897


pNpminus1 mincq

1113936Nk1 F(k c)

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889 (51)



Nk1 α(k) G(t) 1113938

t





pNpminus 1


1cq

1cp

1113882 1113883

ltminAG(d)



(52)

then the problem Df






pNpminus1min cq cp

AG(c)1113888 1113889 (53)



so that

1113944

N

k1max|ξ|lec

F(k ξ) AG(c) 1113944N

k1F(k d) AG(d) (55)



limsupt⟶0+

F(k t)

tp +infin (56)

limt⟶+infin

F(k t)

tp +infin (57)







λlt1 + slowastN

pminus 1

pNpminus1

middot min supcgt0

cq


supcgt0

cp


⎧⎨

⎩

⎫⎬

⎭

(58)


limsupt⟶0+

1113936Nk1 F(k t)

tp +infin (59)






pminus 1


cq

1113936Nk1 F(k c)

supcgt0

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889

(60)



pminus 1

pNpmin sup

cgt0

cq

F(c) sup

cgt0

cp

F(c)1113896 11138971113888 1113889 (61)


et

Put

x(c) c2

ec minus 1 (62)

)en


ec minus 1( )2 (63)





Since

y c2( 1113857 c42

ec2 minus 1gt

c41ec1 minus 1

c21x c1( 1113857gt x c1( 1113857 (64)

then


pNpmin sup

cgt0x(c) sup

cgt0y(c)1113896 1113897

1 + slowastN

pminus 1

pNpx c1( 1113857 asymp 06496

(65)



u(0) Δu(3) 0

⎧⎨

⎩

(66)



f(k t)

0 if tlt 0kt

3radic

if 0le tle 1kt

3radic


⎧⎪⎪⎨

⎪⎪⎩(67)


pNpminus 1


N

k1F(k c)max

1cq

1cp

1113882 1113883 pNpminus 1


3

k111139461

0

kt

3radic

dt

asymp 13693

1113936Nk1 F(k d)


11139363k1 1113946

002

0

kt

3radic

dt



166

limt⟶+infin

kt

3radic

+ 225t2 minus 225t2

asymp 34091

(68)



u(0) Δu(3) 01113896

(69)


Data Availability






Acknowledgments


References





































Jλ un( 1113857 1113936


11138681113868111386811138681113868111386811138681113868p

p+

1113936Nk1s(k) un(k)

11138681113868111386811138681113868111386811138681113868q


le2pminus 1

p1113944

N

k1un(k)

11138681113868111386811138681113868111386811138681113868p

+ 1113944N

k1un(k minus 1)

11138681113868111386811138681113868111386811138681113868p⎛⎝ ⎞⎠

+1q

1113944

N



pun

p

infin +1113957s

qun

q



qminus λl1113888 1113889 un

p

infin + λC

(39)




3 Main Results



pNpminus 1


N

k1max|ξ∣ lec

F(k ξ)max1cq

1cp

1113882 1113883

ltmin1113936

Nk1 F(k d)



(40)



1113936Nk1 F(k d)


qLinfin1113896 1113897



⎛⎝ ⎞⎠ (41)

the problem (Df





f





r 1 + slowastN

pminus 1

pNpminus1 min cq c

p1113864 1113865 (42)





1q

r1q

pNpminus 1


1p

r1p⎧⎨

⎩

⎫⎬

⎭ max c cqp

1113966 1113967 c (43)




1q

r1q

pNpminus 1


1p

r1p⎧⎨

⎩

⎫⎬

⎭ max cpq

c1113966 1113967 c (44)


Furthermore we have

Ψ(u) 1113944N

k1F(k u(k))le 1113944

N

k1max|ξ|lec

F(k ξ) (45)



rle

pNpminus 1


N

k1max|ξ|lec

F(k ξ)max1cq

1cp

1113882 1113883

(46)





Ψ(ω)

Φ(ω)

1113936Nk1 F(k d)




Ψ(u)

rltΨ(ω)

Φ(ω)

(48)


0ltdpp

minus 1+ d

qq

minus 11113957slt


pNpminus1 min cq c

p1113864 1113865 (49)





pNpminus 1


1113936Nk1 F(k c)

cq1113936

Nk1 F(k c)

cp1113896 1113897

ltmin1113936

Nk1 F(k d)



(50)



1113936Nk1 F(k d)


qLinfin1113896 1113897


pNpminus1 mincq

1113936Nk1 F(k c)

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889 (51)



Nk1 α(k) G(t) 1113938

t





pNpminus 1


1cq

1cp

1113882 1113883

ltminAG(d)



(52)

then the problem Df






pNpminus1min cq cp

AG(c)1113888 1113889 (53)



so that

1113944

N

k1max|ξ|lec

F(k ξ) AG(c) 1113944N

k1F(k d) AG(d) (55)



limsupt⟶0+

F(k t)

tp +infin (56)

limt⟶+infin

F(k t)

tp +infin (57)







λlt1 + slowastN

pminus 1

pNpminus1

middot min supcgt0

cq


supcgt0

cp


⎧⎨

⎩

⎫⎬

⎭

(58)


limsupt⟶0+

1113936Nk1 F(k t)

tp +infin (59)






pminus 1


cq

1113936Nk1 F(k c)

supcgt0

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889

(60)



pminus 1

pNpmin sup

cgt0

cq

F(c) sup

cgt0

cp

F(c)1113896 11138971113888 1113889 (61)


et

Put

x(c) c2

ec minus 1 (62)

)en


ec minus 1( )2 (63)





Since

y c2( 1113857 c42

ec2 minus 1gt

c41ec1 minus 1

c21x c1( 1113857gt x c1( 1113857 (64)

then


pNpmin sup

cgt0x(c) sup

cgt0y(c)1113896 1113897

1 + slowastN

pminus 1

pNpx c1( 1113857 asymp 06496

(65)



u(0) Δu(3) 0

⎧⎨

⎩

(66)



f(k t)

0 if tlt 0kt

3radic

if 0le tle 1kt

3radic


⎧⎪⎪⎨

⎪⎪⎩(67)


pNpminus 1


N

k1F(k c)max

1cq

1cp

1113882 1113883 pNpminus 1


3

k111139461

0

kt

3radic

dt

asymp 13693

1113936Nk1 F(k d)


11139363k1 1113946

002

0

kt

3radic

dt



166

limt⟶+infin

kt

3radic

+ 225t2 minus 225t2

asymp 34091

(68)



u(0) Δu(3) 01113896

(69)


Data Availability






Acknowledgments


References








































Ψ(ω)

Φ(ω)

1113936Nk1 F(k d)




Ψ(u)

rltΨ(ω)

Φ(ω)

(48)


0ltdpp

minus 1+ d

qq

minus 11113957slt


pNpminus1 min cq c

p1113864 1113865 (49)





pNpminus 1


1113936Nk1 F(k c)

cq1113936

Nk1 F(k c)

cp1113896 1113897

ltmin1113936

Nk1 F(k d)



(50)



1113936Nk1 F(k d)


qLinfin1113896 1113897


pNpminus1 mincq

1113936Nk1 F(k c)

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889 (51)



Nk1 α(k) G(t) 1113938

t





pNpminus 1


1cq

1cp

1113882 1113883

ltminAG(d)



(52)

then the problem Df






pNpminus1min cq cp

AG(c)1113888 1113889 (53)



so that

1113944

N

k1max|ξ|lec

F(k ξ) AG(c) 1113944N

k1F(k d) AG(d) (55)



limsupt⟶0+

F(k t)

tp +infin (56)

limt⟶+infin

F(k t)

tp +infin (57)







λlt1 + slowastN

pminus 1

pNpminus1

middot min supcgt0

cq


supcgt0

cp


⎧⎨

⎩

⎫⎬

⎭

(58)


limsupt⟶0+

1113936Nk1 F(k t)

tp +infin (59)






pminus 1


cq

1113936Nk1 F(k c)

supcgt0

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889

(60)



pminus 1

pNpmin sup

cgt0

cq

F(c) sup

cgt0

cp

F(c)1113896 11138971113888 1113889 (61)


et

Put

x(c) c2

ec minus 1 (62)

)en


ec minus 1( )2 (63)





Since

y c2( 1113857 c42

ec2 minus 1gt

c41ec1 minus 1

c21x c1( 1113857gt x c1( 1113857 (64)

then


pNpmin sup

cgt0x(c) sup

cgt0y(c)1113896 1113897

1 + slowastN

pminus 1

pNpx c1( 1113857 asymp 06496

(65)



u(0) Δu(3) 0

⎧⎨

⎩

(66)



f(k t)

0 if tlt 0kt

3radic

if 0le tle 1kt

3radic


⎧⎪⎪⎨

⎪⎪⎩(67)


pNpminus 1


N

k1F(k c)max

1cq

1cp

1113882 1113883 pNpminus 1


3

k111139461

0

kt

3radic

dt

asymp 13693

1113936Nk1 F(k d)


11139363k1 1113946

002

0

kt

3radic

dt



166

limt⟶+infin

kt

3radic

+ 225t2 minus 225t2

asymp 34091

(68)



u(0) Δu(3) 01113896

(69)


Data Availability






Acknowledgments


References







































pminus 1


cq

1113936Nk1 F(k c)

supcgt0

cp

1113936Nk1 F(k c)

1113896 11138971113888 1113889

(60)



pminus 1

pNpmin sup

cgt0

cq

F(c) sup

cgt0

cp

F(c)1113896 11138971113888 1113889 (61)


et

Put

x(c) c2

ec minus 1 (62)

)en


ec minus 1( )2 (63)





Since

y c2( 1113857 c42

ec2 minus 1gt

c41ec1 minus 1

c21x c1( 1113857gt x c1( 1113857 (64)

then


pNpmin sup

cgt0x(c) sup

cgt0y(c)1113896 1113897

1 + slowastN

pminus 1

pNpx c1( 1113857 asymp 06496

(65)



u(0) Δu(3) 0

⎧⎨

⎩

(66)



f(k t)

0 if tlt 0kt

3radic

if 0le tle 1kt

3radic


⎧⎪⎪⎨

⎪⎪⎩(67)


pNpminus 1


N

k1F(k c)max

1cq

1cp

1113882 1113883 pNpminus 1


3

k111139461

0

kt

3radic

dt

asymp 13693

1113936Nk1 F(k d)


11139363k1 1113946

002

0

kt

3radic

dt



166

limt⟶+infin

kt

3radic

+ 225t2 minus 225t2

asymp 34091

(68)



u(0) Δu(3) 01113896

(69)


Data Availability






Acknowledgments


References






































































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