# Boundary value problems for a second-order difference equation involving the mean curvature operator – Boundary Value Problems

#### ByZhenguo Wang and Qilin Xie

Aug 6, 2022

Now, we state our main results.

### Theorem 3.1

Assume that ((a_{1}))((a_{3})) hold, then there exists a positive constant λ̄, when (lambda >bar{lambda}), and problem (4) admits at least T distinct pairs of nontrivial solutions. Furthermore, there exists a positive constant M such that each solution u satisfies (|u|_{infty}leq M).

### Proof

Using the condition ((a_{2})), there exists a positive real sequence ({d_{n}}) with (lim_{nrightarrow infty}d_{n}=+infty ) such that

$\underset{}{}$
lim

n

+

f(t,
d
n
)<0for all tZ(1,T).

We can find a positive integer (n_{0}) such that (M=d_{n_{0}}>alpha ) and (f(t,M)<0), where α comes from ((a_{2})). First, we consider the following boundary value problem


{

Δ
(

ϕ
c

(
Δ
u
(
t

1
)
)
)
+
q
(
t
)
u
(
t
)
=
λ

f
ˆ

(
t
,
u
(
t
)
)
,

t

Z
(
1
,
T
)
,

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

(12)

where (hat{f}(t,xi )) is a truncation function defined by

begin{aligned} hat{f}(t,xi )= textstylebegin{cases} f(t,M) &text{if } xi >M, \ f(t,xi ) &text{if } vert xi vert leq M, \ f(t,-M) &text{if } xi < -M. end{cases}displaystyle end{aligned}

We show that if u satisfies problem (12), then (|u|_{infty}leq M) and u is a solution of problem (4). Arguing indirectly, there exists a

${}_{}$
k
0
Z(1,T)

such that (|u(k_{0})|>M) and (|u(t)|leq M) for

$t\in \mathbb{Z}\left(1,{}_{}$
k
0
1)

. If (u(k_{0})>M), then (hat{f}(t,u(k_{0}))=f(t,M)<0). We have

begin{aligned} -Delta bigl(phi _{c}bigl(Delta u(k_{0}-1)bigr) bigr)+q(k_{0})u(k_{0})< 0, end{aligned}

or

begin{aligned} frac{u(k_{0})-u(k_{0}+1)}{sqrt{1+ kappa (Delta u(k_{0}))^{2}}}< – frac{u(k_{0})-u(k_{0}-1)}{sqrt{1+ kappa (Delta u(k_{0}-1))^{2}}}-q(k_{0})u(k_{0})< 0, end{aligned}

which implies that (u(k_{0}+1)>u(k_{0})>M). By repeating the above process, we obtain

k
0
+1,T).

Further,

begin{aligned} 0=u(T+1)>u(T)>M, end{aligned}

which is a contradiction. If (u(k_{0})<-M), we can similarly obtain a contradiction. Thus, (|u|_{infty}leq M) holds.

Define the functional Ĵ on E as follows:

begin{aligned} hat{J}(u)=sum^{T}_{t=0} biggl( biggl( frac{sqrt{1+kappa (Delta u(t))^{2}}-1}{kappa} biggr)+ frac{q(t) u^{2}(t)}{2} biggr)-lambda sum^{T}_{t=1}hat{F}bigl(t,u(t)bigr), end{aligned}

(13)

where (hat{F}(t,xi )=int _{0}^{xi}hat{f}(t,s),ds),

$\left(t,\xi \right)\in \mathbb{Z}\left(1,T\right)×\mathbb{R}$

. It is easy to verify that

$\stackrel{}{}$
J
ˆ

C
1
(E,R)

and is even. By using (u(0)=u(T+1)=0), we can compute the Frećhet derivative,

begin{aligned} bigllangle hat{J}'(u),vbigrrangle =sum ^{T}_{t=1} bigl(-Delta bigl(phi _{c} bigl(Delta u(t-1)bigr) bigr)+q(t)u(t)- lambda hat{f}bigl(t,u(t)bigr) bigr)v(t), end{aligned}

for all (u, vin E). It is clear that the critical points of Ĵ are the solutions of problem (12). In what follows, we will prove that Ĵ has at least T distinct pairs of nonzero critical points by Lemma 2.1.

For any sequence ({u_{n}}subset E), if ({hat{J}(u_{n})}) is bounded and (hat{J}'(u_{n})rightarrow 0) as (nrightarrow +infty ), we claim that ({u_{n}}) is bounded. In fact, there exists a positive constant

$C\in \mathbb{R}$

such that (|hat{J}(u_{n})|leq C). Since E is a finite-dimensional real Banach space, there is (|u|_{2}leq |u|leq 2|u|_{2}) for all (uin E) (see [30]). Assume that (|u_{n}|rightarrow +infty ) as (nrightarrow +infty ), then

begin{aligned} C&geq hat{J}(u_{n}) \ &=sum^{T}_{t=0} biggl( biggl( frac{sqrt{1+kappa (Delta u(t))^{2}}-1}{kappa} biggr)+ frac{q(t) u_{n}^{2}(t)}{2} biggr)-lambda sum ^{T}_{t=1} hat{F}bigl(t,u_{n}(t)bigr) \ &geq sum^{T}_{t=1}frac{q(t) u_{n}(t)^{2}}{2}- lambda sum_{ vert u_{n}(t) vert leq M} biglvert hat{F} bigl(t,u_{n}(t)bigr) bigrvert -lambda sum _{ vert u_{n}(t) vert >M} biglvert hat{F}bigl(t,u_{n}(t)bigr) bigrvert \ &geq frac{q_{ast}}{2} Vert u_{n} Vert _{2}^{2}- lambda Dsum_{ vert u_{n}(t) vert leq M} biglvert u_{n}(t) bigrvert -lambda sum^{T}_{t=0} bigglvert int _{0}^{M} hat{f}(t,s),ds biggrvert – lambda sum_{ vert u_{n}(t) vert >M} bigglvert int _{M}^{u_{n}(t)} hat{f}(t,s),ds biggrvert \ &geq frac{q_{ast}}{2} Vert u_{n} Vert _{2}^{2}- lambda Dsum_{ vert u_{n}(t) vert leq M} biglvert u_{n}(t) bigrvert -lambda Dsum_{ vert u_{n}(t) vert >M} biglvert u_{n}(t) bigrvert -2 lambda TDM \ &=frac{q_{ast}}{2} Vert u_{n} Vert _{2}^{2} -lambda Dsum^{T}_{t=1} biglvert u_{n}(t) bigrvert -2 lambda TDM \ &geq frac{q_{ast}}{8} Vert u_{n} Vert ^{2} – lambda D T^{frac{1}{2}} Vert u_{n} Vert -2lambda TDM rightarrow +inftyquad text{as } nrightarrow +infty, end{aligned}

where

${}_{}$
q

=
min

t

Z
(
1
,
T
)

q(t)

and (D=max |f(t,u)|) for

$\left(t,u\right)\in \mathbb{Z}\left(1,T\right)×\left[-M,M\right]$

. This is impossible, since C is a fixed constant. Thus, ({u_{n}}) is bounded in E. This implies that ({u_{n}}) has a convergent subsequence. Then, the functional Ĵ satisfies the P.S. condition.

Moreover, the coerciveness of Ĵ,

begin{aligned} hat{J}(u)geq frac{q_{ast}}{8} Vert u Vert ^{2} -lambda D T^{frac{1}{2}} Vert u Vert -2lambda TDM rightarrow +inftyquad text{as } Vert u Vert rightarrow + infty end{aligned}

implies that Ĵ is bounded from below.

Let ({e_{i}}_{i=1}^{T}) be a base of E and (|e_{i}|=1) for each

$i\in \mathbb{Z}\left(1,T\right)$

. We define

begin{aligned} A(rho )= Biggl{ sum^{T}_{i=1} beta _{i}e_{i}| sum^{T}_{i=1} vert beta _{i} vert ^{2}=rho ^{2} Biggr} , quadrho >0. end{aligned}

Obviously, (theta notin A(rho )), (A(rho )) is closed in E and symmetric with respect to θ. We note that (A(rho )) is homeomorphic to (S^{T-1}) for any (rho >0). For (uin A(rho )), we see that

begin{aligned} Vert u Vert ^{2}&=sum^{T}_{t=0} Bigglvert sum^{T}_{i=1}beta _{i}Delta e_{i}(t) Biggrvert ^{2} leq sum ^{T}_{t=0} Biggl(sum ^{T}_{i=1} vert beta _{i} vert ^{2} sum^{T}_{i=1} biglvert Delta e_{i}(t) bigrvert ^{2} Biggr)\ &=rho ^{2} sum^{T}_{i=1} Vert e_{i} Vert ^{2}leq rho ^{2}(T+1),quad rho >0. end{aligned}

Take (rho =frac{alpha}{T+1}), thus

begin{aligned} Vert u Vert _{infty}leq sum ^{T}_{t=0} biglvert Delta u(t) bigrvert leq (T+1)^{ frac{1}{2}} Vert u Vert leq (T+1)rho < alpha < M. end{aligned}

For (uin A(frac{alpha}{T+1})), we note that (uneq theta ) and (hat{f}(t,u(t))=f(t,u(t))). By ((a_{2})) and ((a_{3})), then

$\begin{array}{}\end{array}$

t
=
1

T

F
ˆ

(
t
,
u
(
t
)
)

=

{
t

Z
(
1
,
T
)
|
u
(
t
)
>
0
}

F
ˆ

(
t
,
u
(
t
)
)

+

{
t

Z
(
1
,
T
)
|
u
(
t
)
<
0
}

F
ˆ

(
t
,
u
(
t
)
)

=

{
t

Z
(
1
,
T
)
|
u
(
t
)
>
0
}

0

u
(
t
)

f
(
t
,
s
)

d
s
+

{
t

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(
1
,
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)
|
u
(
t
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<
0
}

0

u
(
t
)

f
(
t
,

s
)
d
(

s
)

=

{
t

Z
(
1
,
T
)
|
u
(
t
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>
0
}

0

u
(
t
)

f
(
t
,
s
)

d
s
+

{
t

Z
(
1
,
T
)
|
u
(
t
)
<
0
}

0

u
(
t
)

f
(
t
,
s
)

d
s

>
0
.

Let (tau =inf_{uin A(frac{alpha}{T+1}) }sum^{T}_{t=1} hat{F}(t,u(t))) and (bar{lambda}=frac{(2+q^{ast})alpha ^{2}}{2Ttau}). By ((a_{2})), we know (tau >0). If (lambda >bar{lambda}), then

begin{aligned} hat{J}(u) &=sum^{T}_{t=0} biggl( biggl( frac{sqrt{1+kappa (Delta u(t))^{2}}-1}{kappa} biggr)+ frac{q(t) u^{2}(t)}{2} biggr)-lambda sum^{T}_{t=1} hat{F}bigl(t,u(t)bigr) \ &leq sum^{T}_{t=0} biglvert Delta u(t) bigrvert ^{2}+frac{q^{ast}}{2} Vert u Vert _{2}^{2}-lambda sum^{T}_{t=1} hat{F}bigl(t,u(t)bigr) \ &leq frac{2+q^{ast}}{2} Vert u Vert ^{2}-lambda sum ^{T}_{t=1} hat{F}bigl(t,u(t)bigr) \ &leq frac{(2+q^{ast})alpha ^{2}}{2T} -lambda tau \ &< 0, end{aligned}

where

${}^{}$
q

=
max

t

Z
(
1
,
T
)

q(t)

. Since all conditions of Lemma 2.1 hold, problem (4) admits at least T distinct pairs of nontrivial solutions. The proof is completed. □

At the end of this section, we try to prove a pair of constant-sign solutions of the original problem. We first introduce the following assumptions:

((a_{1}’)):

(f(t,cdot )) is a continuous functional on

$\mathbb{R}\mathrm{\setminus }\left\{0\right\}$

for each

$t\in \mathbb{Z}\left(1,T\right)$

;

((a_{2}’)):

$0<{}^{}$
q

f(t,ξ),(t,ξ)Z(1,T)×(0,α)

for some (alpha >0), where

${}^{}$
q

=
max

t

Z
(
1
,
T
)

q(t)

;

((a_{3}’)):

(f(t,xi )=-f(t,-xi )) in (xi neq 0) for any

$t\in \mathbb{Z}\left(1,T\right)$

.

We note that the function (f(t,cdot )) is locally bounded from below for each

$t\in \mathbb{Z}\left(1,T\right)$

in the right-hand side of 0 from ((a_{2}’)) and problem (4) has no trivial solution. When θ is not the solution of the problem, many problems become more complicated [31, 32]. For example, we put (f(t,xi )=frac{1}{sqrt[3]{xi}}), (alpha =1) and (q^{ast}=frac{1}{2}), then

$0<\frac{}{}$

1

2
<
1

ξ
3

,(t,ξ)Z(1,T)×(0,1)

, which satisfies the conditions ((a_{1}’))((a_{3}’)).

Let

begin{aligned} mu _{1}=inf_{uin Ebackslash {theta }} frac{sum^{T}_{t=0}frac{(Delta u(t))^{2}}{sqrt{1+kappa (Delta u(t))^{2}}}}{ Vert u Vert _{2}^{2}}. end{aligned}

We observe that (frac{sum^{T}_{t=0}frac{(Delta u(t))^{2}}{sqrt{1+kappa (Delta u(t))^{2}}}}{|u|_{2}^{2}}) is positive in (Ebackslash {theta }). Thus, (mu _{1}geq 0).

### Theorem 3.2

Assume that ((a_{1}’))((a_{3}’)) hold and

$\underset{}{}$
lim sup

|
ξ
|

f
(
t
,
ξ
)

ξ
<
μ
1
+
q

,tZ(1,T).

(14)

Then, problem (4) has a positive solution and a negative solution for each (lambda in (0,frac{mu _{1}+q_{ast}}{mu})), where (mu in (0,mu _{1}+q_{ast})).

### Proof

We consider the following problem


{

Δ
(

ϕ
c

(
Δ
u
(
t

1
)
)
)
+
q
(
t
)
u
(
t
)
=
λ

q

,

t

Z
(
1
,
T
)
,

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

(15)

Define the variational framework of problem (15)

begin{aligned} J_{q^{ast}}(u)=sum^{T}_{t=0} biggl( biggl( frac{sqrt{1+kappa (Delta u(t))^{2}}-1}{kappa} biggr)+ frac{q(t) u^{2}(t)}{2} biggr)-lambda q^{ast}sum^{T}_{t=1}u(t), end{aligned}

then we have

begin{aligned} J_{q^{ast}}(u)&=sum^{T}_{t=0} biggl( biggl( frac{sqrt{1+kappa (Delta u(t))^{2}}-1}{kappa} biggr)+ frac{q(t) u^{2}(t)}{2} biggr)-lambda q^{ast}sum^{T}_{t=1}u(t) \ &=sum^{T}_{t=0} biggl( frac{(Delta u(t))^{2}}{sqrt{1+kappa (Delta u(t))^{2}}+1}+ frac{q(t) u^{2}(t)}{2} biggr)-lambda q^{ast}sum ^{T}_{t=1}u(t) \ &geq sum^{T}_{t=0} frac{(Delta u(t))^{2}}{2sqrt{1+kappa (Delta u(t))^{2}}}+ frac{q_{ast}}{2} Vert u Vert _{2}^{2}-lambda q^{ast}sqrt{T} Vert u Vert _{2} \ &geq frac{mu _{1}+q_{ast}}{2} Vert u Vert _{2}^{2}-lambda q^{ast}sqrt{T} Vert u Vert _{2} rightarrow +inftyquad text{as } Vert u Vert _{2}rightarrow + infty. end{aligned}

Hence, (J_{q^{ast}}) is coercive on E and has a global minimum point (u_{0}) that is its critical point. Combining with Lemma 2.4, (u_{0}) is a positive solution of problem (15). We take (varepsilon >0) so small that (u_{1}(t)=varepsilon u_{0}(t)<alpha ).

Define a continuous function as follows:

begin{aligned} f_{u_{1}}(t,xi )= textstylebegin{cases} f(t,xi ) &text{if } xi geq u_{1}(t), \ f(t,u_{1}(t)) &text{if } xi < u_{1}(t). end{cases}displaystyle end{aligned}

By (14), there exist a (mu in [0,mu _{1}+q_{ast})) and (M_{1}>u_{1}(t)) such that

$f\left(t,\xi \right)\le \mu \xi ,\phantom{\rule{1em}{0ex}}\left(t,\xi \right)\in \mathbb{Z}\left(1,T\right)×\left({}_{}$
M
1
,).

(16)

Thus,

${}_{}$
f

u
1

(t,ξ)
{

f
(
t
,

u
1

(
t
)
)
+

max

(
t
,
ξ
)

Z
(
1
,
T
)
×
[

u
1

(
t
)
,

M
1

]

f
(
t
,
ξ
)
+
μ
ξ

if
ξ

0
,

q

if
ξ
<
0
,

(17)

and

$\underset{}{}$
lim sup

|
ξ
|

f

u
1

(
t
,
ξ
)

ξ
μ,tZ(1,T).

(18)

Next, we claim that the following problem


{

Δ
(

ϕ
c

(
Δ
u
(
t

1
)
)
)
+
q
(
t
)
u
(
t
)
=
λ

f

u
1

(
t
,
u
(
t
)
)
,

t

Z
(
1
,
T
)
,

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

(19)

admits a positive solution u and (u>u_{1}>0).

We define the following variational framework corresponding to problem (19)

begin{aligned} hat{J}(u)=sum^{T}_{t=0} biggl( biggl( frac{sqrt{1+kappa (Delta u(t))^{2}}-1}{kappa} biggr)+ frac{q(t) u^{2}(t)}{2} biggr)-lambda sum ^{T}_{t=1}F_{u_{1}}bigl(t,u(t)bigr), end{aligned}

where (F_{u_{1}}(t,xi )=int _{0}^{xi}f_{u_{1}}(t,s),ds),

$\left(t,\xi \right)\in \mathbb{Z}\left(1,T\right)×\mathbb{R}$

.

Using (18), there is one positive constant such that

begin{aligned} F_{u_{1}}(t,xi )leq frac{mu}{2} vert xi vert ^{2}+overline{M}. end{aligned}

(20)

Let (eta =frac{mu _{1}+q_{ast}}{mu}). For (eta >lambda >0), we have

begin{aligned} hat{J}(u)&=sum^{T}_{t=0} biggl( biggl( frac{sqrt{1+kappa (Delta u(t))^{2}}-1}{kappa} biggr)+ frac{q(t) u^{2}(t)}{2} biggr)-lambda sum^{T}_{t=1}F_{u_{1}}bigl(t,u(t) bigr) \ &geq sum^{T}_{t=0} frac{(Delta u(t))^{2}}{sqrt{1+kappa (Delta u(t))^{2}}+1}+ frac{q_{ast}}{2} Vert u Vert _{2}^{2}- frac{lambda mu}{2} Vert u Vert _{2}^{2} – lambda T overline{M} \ &geq sum^{T}_{t=0} frac{(Delta u(t))^{2}}{2sqrt{1+kappa (Delta u(t))^{2}}}+ frac{q_{ast}}{2} Vert u Vert _{2}^{2}- frac{lambda mu}{2} Vert u Vert _{2}^{2} – lambda T overline{M} \ &geq frac{mu}{2}(eta -lambda ) Vert u Vert _{2}^{2}- lambda T overline{M} rightarrow +infty quadtext{as } Vert u Vert _{2}rightarrow + infty. end{aligned}

This shows that Ĵ is also coercive. As the functional is coercive and continuous, it has a global minimum point (uin E), which is a critical point. By (17) and Lemma 2.5, u is a positive solution. Moreover, if we can show (u>u_{1}), then u must be one positive solution of problem (4). First, we assume that (uleq u_{1}) for every

$t\in \mathbb{Z}\left(1,T\right)$

. Since

begin{aligned} -Delta bigl(phi _{c}bigl(Delta u(t-1)bigr) bigr)+q(t)u(t)= lambda fbigl(t,u_{1}(t)bigr) geq lambda q^{ast}=-Delta bigl(phi _{c}bigl(Delta u_{0}(t-1)bigr) bigr)+q(t)u_{0}(t), end{aligned}

by Lemma 2.2, we obtain (ugeq u_{0}>u_{1}), contradicting the assumption above. Secondly, we consider that u and (u_{1}) are not ordered vectors. There exist some

${}_{}$
j
0
Z(1,T)

such that (u(j_{0})< u_{1}(j_{0})). Let

$j:=max$
{

j
0

|

j
0

Z
(
1
,
T
)
such that
u
(

j
0

)
<

u
1

(

j
0

)
}
.

From the proof of Lemma 2.2, if (u(j)< u_{1}(j)) holds, we have the following inequality

begin{aligned} lambda fbigl(i,u_{1}(i)bigr)=-Delta bigl(phi _{c}bigl(Delta u(i-1)bigr)bigr)+q(i)u(i)< – Delta bigl(phi _{c}bigl(Delta u_{1}(i-1)bigr)bigr)+q(i)u_{1}(i), end{aligned}

(21)

where (i=1,2,j). Since (q^{ast}leq f(i,u_{1}(i))) from the proof of Lemma 2.2 and ((a_{2}’)), this implies

begin{aligned} lambda q^{ast}=-Delta bigl(phi _{c}bigl( Delta u_{0}(i-1)bigr)bigr)+q(i)u_{0}(i)< – Delta bigl( phi _{c}bigl(Delta u_{1}(i-1)bigr)bigr)+q(i)u_{1}(i). end{aligned}

(22)

We see from (22) that

begin{aligned} 0leq{}& q(i) bigl(u_{0}(i)-u_{1}(i)bigr) \ < {}&Delta bigl(phi _{c}bigl(Delta u_{0}(i-1)bigr) bigr)-Delta bigl(phi _{c}bigl(Delta u_{1}(i-1)bigr) bigr) \ ={}& biggl( frac{varepsilon Delta u_{0}(i-1)}{sqrt{1+ kappa (varepsilon Delta u_{0}(i-1))^{2}}}- frac{Delta u_{0}(i-1)}{sqrt{1+ kappa (Delta u_{0}(i-1))^{2}}} biggr) \ &{} + biggl( frac{Delta u_{0}(i)}{sqrt{1+ kappa (Delta u_{0}(i))^{2}}}- frac{varepsilon Delta u_{0}(i)}{sqrt{1+ kappa (varepsilon Delta u_{0}(i))^{2}}} biggr). end{aligned}

By the strict monotonicity of (phi _{c}), we find that if (Delta u_{0}(i-1)>0), then (Delta u_{0}(i)>0). That is, if (Delta u_{1}(i-1)>0), then (Delta u_{1}(i)>0).

Further, we estimate the inequality (22) from the following three cases. First, if (Delta u_{1}(i-1)leq Delta u_{1}(i)), we have

begin{aligned} lambda q^{ast}< -Delta bigl(phi _{c}bigl( Delta u_{1}(i-1)bigr)bigr)+q(i)u_{1}(i)< varepsilon q(i) u_{0}(i). end{aligned}

(23)

For the second case, if (Delta u_{1}(i-1)>Delta u_{1}(i)>0), then

begin{aligned} begin{aligned} lambda q^{ast}&< -Delta bigl(phi _{c}bigl(Delta u_{1}(i-1)bigr)bigr)+q(i)u_{1}(i) \ &=frac{Delta u_{1}(i-1)}{sqrt{1+ kappa (Delta u_{1}(i-1))^{2}}}- frac{Delta u_{1}(i)}{sqrt{1+ kappa (Delta u_{1}(i))^{2}}}+q(i)u_{1}(i) \ &leq Delta u_{1}(i-1)+q(i)u_{1}(i) \ &leq varepsilon bigl(Delta u_{0}(i-1)+q(i)u_{0}(i) bigr). end{aligned} end{aligned}

(24)

For the last case, if (0>Delta u_{1}(i-1)>Delta u_{1}(i)), then

begin{aligned} begin{aligned} lambda q^{ast}&< -Delta bigl(phi _{c}bigl(Delta u_{1}(i-1)bigr)bigr)+q(i)u_{1}(i) \ &=frac{Delta u_{1}(i-1)}{sqrt{1+ kappa (Delta u_{1}(i-1))^{2}}}- frac{Delta u_{1}(i)}{sqrt{1+ kappa (Delta u_{1}(i))^{2}}}+q(i)u_{1}(i) \ &leq -Delta u_{1}(i)+q(i)u_{1}(i) \ &leq varepsilon bigl(-Delta u_{0}(i)+q(i)u_{0}(i) bigr). end{aligned} end{aligned}

(25)

We note that (Delta u_{1}(i-1)=0) or (Delta u_{1}(i)=0) still satisfies the above cases. Obviously, when ε is taken sufficiently small, (23), (24), and (25) cannot hold. These are contradictions. Thus, (ugeq u_{1}). u is one positive solution of problem (4). Moreover, we see that −u is a negative solution of problem (4) because of ((a_{3}’)). This completes the proof. □

### Example 3.1

Let (lambda =1), we consider the problem


{

Δ
(

ϕ
c

(
Δ
u
(
t

1
)
)
)
+
q
(
t
)
u
(
t
)
=
sin
(
u
(
t
)
)
,

t

Z
(
1
,
T
)
,

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

(26)

we note that the conditions ((a_{1}))((a_{3})) hold from Example 1.1, and λ̄ can be less than 1 by the definition, thus the problem (26) admits at least T distinct pairs of nontrivial solutions by Theorem 3.1.

In fact, in [30], such a problem can be found when (kappa =0) and (q(t)=0) for each

$t\in \mathbb{Z}\left(1,T\right)$

in Corollary 5.1. We see that the conditions of Theorem 3.1 are different from the conditions of Corollary 5.1 of [30], and we find more solutions of problem (26).

### Example 3.2

Let (kappa =0) and (T=3). We consider the problem


{

Δ
2

u
(
t

1
)
+
q
(
t
)
u
(
t
)
=
λ

1

u
(
t
)

3

,

t

Z
(
1
,
3
)
,

u
(
0
)
=
u
(
4
)
=
0
.

(27)

Put (alpha =1), (q_{ast}=frac{1}{4}) and (q^{ast}=frac{1}{2}). The conditions ((a_{1}’))((a_{3}’)) hold from the previous example. We have (mu _{1}=2-sqrt{2}) from [30]. Clearly,

$\underset{}{}$
lim sup

|
ξ
|

f
(
t
,
ξ
)

ξ
=
lim sup

|
ξ
|

1

ξ

4
/
3

=0<

9

4

2
,tZ(1,T).

All conditions of Theorem 3.2 are verified. If we take (mu >0) sufficiently small, then the problem (27) has a positive solution and a negative solution for each (lambda in (0,+infty )).