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


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

tZ(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

u(t)>u(t1)>Mfor all tZ(
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),

(t,ξ)Z(1,T)×R

. It is easy to verify that


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

CR

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

(t,u)Z(1,T)×[M,M]

. 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

iZ(1,T)

. 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







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

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




0


u
(
t
)


f
(
t
,

s
)
d
(

s
)





=



{
t

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

R{0}

for each

tZ(1,T)

;

((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

tZ(1,T)

.

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

tZ(1,T)

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<

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


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(t,ξ)μξ,(t,ξ)Z(1,T)×(
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


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),

(t,ξ)Z(1,T)×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

tZ(1,T)

. 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

tZ(1,T)

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,


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 )).

Rights and permissions

Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

Disclaimer:

This article is autogenerated using RSS feeds and has not been created or edited by OA JF.

Click here for Source link (https://www.springeropen.com/)