Periodic and Neumann problems for discrete p(⋅) -Laplacian

J. Math. Anal. Appl. 399 (2013) 75–87

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Periodic and Neumann problems for discrete p(·)-Laplacian Cristian Bereanu a,∗ , Petru Jebelean b , Călin Şerban b a

Institute of Mathematics ‘‘Simion Stoilow’’, Romanian Academy 21, Calea Griviţei, RO-010702-Bucharest, Sector 1, Romania

b

Department of Mathematics, West University of Timişoara, 4, Blvd. V. Pârvan RO-300223-Timişoara, Romania

article

info

Article history: Received 23 April 2012 Available online 4 October 2012 Submitted by Juan J. Nieto Keywords: Discrete p(·)-Laplacian operator Critical point Palais–Smale condition Lower and upper solutions

abstract Using critical point theory, we obtain the existence of solutions for some periodic boundary value problems involving the discrete p(·)-Laplacian. These extend and improve known results for similar problems with discrete p-Laplacian. Similar results for Neumann problems are also provided. As applications we prove upper and lower solutions theorems for both considered cases. © 2012 Elsevier Inc. All rights reserved.

1. Introduction Let T be a positive integer, [a, b] be the discrete interval {a, a + 1, . . . , b} for a, b ∈ N and a < b, p : [0, T ] → (1, ∞) and hp(k) : R → R be the homeomorphism defined by hp(k) (x) = |x|p(k)−2 x, x ∈ R (k ∈ [0, T ]). In this paper we deal with the existence of solutions for the periodic problem



−∆p(k−1) x(k − 1) = f (k, x(k)), (∀) k ∈ [1, T ], x(0) − x(T + 1) = 0 = ∆x(0) − ∆x(T ),

(PP )

and for the Neumann problem



−∆p(k−1) x(k − 1) = f (k, x(k)), ∆x(0) = 0 = ∆x(T ),

(∀) k ∈ [1, T ],

(PN )

where ∆x(k) = x(k + 1) − x(k) is the forward difference operator, ∆p(·) stands for the discrete p(·)-Laplacian operator, i.e.,

∆p(k−1) x(k − 1) := ∆(hp(k−1) (∆x(k − 1)))

= hp(k) (∆x(k)) − hp(k−1) (∆x(k − 1))

(1.1)

and f : [1, T ] × R → R is a continuous function. Throughout the paper, we assume that the variable exponent p satisfies p(0) = p(T )

(1.2)

when we refer to the periodic problem (PP ). In the last years the study of discrete boundary value problems by means of variational techniques has captured a special attention (see e.g., [1,4,6,11,12,19–22] and the references therein). The research concerning the discrete anisotropic

∗

Corresponding author. E-mail addresses: [email protected] (C. Bereanu), [email protected] (P. Jebelean), [email protected] (C. Şerban).

0022-247X/$ – see front matter © 2012 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2012.09.047

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C. Bereanu et al. / J. Math. Anal. Appl. 399 (2013) 75–87

boundary value problems has just been started; most of the papers in this direction deals with homogeneous Dirichlet boundary conditions. In this view, we refer the reader to the recent works of Galewski and Gła¸b [7,8], Koné and Ouaro [13], Mashiyev et al. [15], Mihăilescu et al. [17]. Also, existence results for discrete p(·)-Laplacian equations subjected to a general potential type boundary condition were recently obtained by Bereanu et al. in [3]. Recently, Guiro et al. [9] and Koné and Ouaro [14] proved existence and uniqueness of solutions for a family of discrete anisotropic Neumann and mixed boundary value problems, respectively. We note that, in the Neumann case, the results and the setting in [9] are of different type from those in this paper. We use the critical point theory to establish existence results for problems (PP ) and (PN ). The solutions which we obtain appear either as minimizers or as saddle points of the corresponding Euler–Lagrange functional. The case when the potential of the nonlinear term f has convexity properties is particularly discussed. As applications we prove upper and lower solutions theorems for both of problems (PP ) and (PN ). It is worth to point out that the results proved in [11] for p = constant are immediate consequences of the corresponding ones obtained in this paper for the periodic problem (PP ). The rest of the paper is organized as follows. The functional framework and the variational setting are described in Section 2. The main existence results for problem (PP ) without convexity assumptions are presented in Section 3. In Section 4, we obtain existence results for problem (PP ) with convex potential. The solvability of the periodic problem (PP ) is considered in the presence of well ordered lower and upper solutions in Section 5. Section 6 is devoted to the Neumann problem; we discuss here the manner in which the results from the periodic case are adapted for problem (PN ). 2. Preliminaries To establish the existence results we shall use the classical critical point theory [16,18]. For the periodic problem (PP ) we introduce the spaces XP := {x : [0, T + 1] → R | x(0) = x(T + 1)} and

 T   x˜ (k) = 0 , := x˜ ∈ XP  

X P ,0

k=1

while in the case of Neumann problem (PN ), we shall use XN := {x : [0, T + 1] → R} and

 T   x˜ (k) = 0 . := x˜ ∈ XN  

X N ,0

k=1

Further, for the convenience in notations we generically denote by X one of the spaces XP or XN and X0 will stand for XP ,0 and XN ,0 , respectively.

For x ∈ X we set x := (1/T ) k=1 x(k) ∈ R and x˜ := x − x ∈ X0 . This enables us to split X = R ⊕ X0 . The space X0 will be endowed with the Luxemburg norm

T



∥˜x∥p(·)

T +1   = inf ν > 0 k=1

    ∆x˜ (k − 1) p(k−1)   ≤1 ,  p(k − 1)  ν 1

(∀) x˜ ∈ X0

and X will be endowed with the norm

∥x∥X = |x| + ∥˜x∥p(·) ,

(∀) x = x + x˜ ∈ R ⊕ X0 .

Since X0 and X are finite dimensional, any norm on X0 (resp. X ) is equivalent with ∥ · ∥p(·) (resp. ∥ · ∥X ). Also, we shall invoke the norm

∥x∥∞ = max |x(k)| (x ∈ X ). k∈[0,T +1]

From now on, we shall employ the usual notations: p− = min p(k), k∈[0,T ]

p+ = max p(k). k∈[0,T ]

Now, let ϕX : X → R be defined by

ϕX (x) =

T +1 

1

p(k − 1) k=1

|∆x(k − 1)|p(k−1) ,

(∀) x ∈ X .

(2.1)

C. Bereanu et al. / J. Math. Anal. Appl. 399 (2013) 75–87

77

Standard arguments show that ϕX ∈ C 1 (X , R) and

⟨ϕX′ (x), y⟩ =

T +1 

hp(k−1) (∆x(k − 1))∆y(k − 1),

(∀) x, y ∈ X .

(2.2)

k=1

Also, it is easy to check that, for all x˜ ∈ X0 , one has +

−

if ∥˜x∥p(·) < 1;

−

+

if ∥˜x∥p(·) > 1.

∥˜x∥pp(·) ≤ ϕX (˜x) ≤ ∥˜x∥pp(·) , ∥˜x∥pp(·) ≤ ϕX (˜x) ≤ ∥˜x∥pp(·) ,

(2.3)

Setting

 R

[1,T ]

:= {ℓ : [1, T ] → R},

L :=

 T  ℓ(k) = 0 , 

[1,T ] 

ℓ∈R

k=1

for given ℓ ∈ R[1,T ] , we define EX : X → R by EX (x) = ϕX (x) −

T 

ℓ(k)x(k),

(∀) x ∈ X ,

(2.4)

k=1

with ϕX given by (2.1). It is clear that EX ∈ C 1 (X , R) and on account of (2.2), it holds T +1 

⟨EX′ (x), y⟩ =

hp(k−1) (∆x(k − 1))∆y(k − 1) −

k=1

T 

ℓ(k)y(k),

(∀) x, y ∈ X .

(2.5)

k=1

Now, let us consider the following simple periodic problem



−∆p(k−1) x(k − 1) = ℓ(k), (∀) k ∈ [1, T ], x(0) − x(T + 1) = 0 = ∆x(0) − ∆x(T ).

(2.6)

The search of solutions for problem (2.6) is reduced to finding critical points of EXP by the following. Proposition 2.1. Assume that hypothesis (1.2) holds true. A function x ∈ XP is a solution of problem (2.6) if and only if it is a critical point of EXP . Proof. Using (1.1), (2.5) and the summation by parts formula (see e.g., [11,19]), a straightforward computation shows that

⟨EX′ P (x), y⟩ = hp(T ) (∆x(T ))y(T + 1) − hp(0) (∆x(0))y(0) −

T    ∆p(k−1) x(k − 1) + ℓ(k) y(k),

(2.7)

k=1

for all x, y ∈ XP . From (1.2) and (2.7), if x is a solution of problem (2.6), then it is obvious that it is a critical point of EXP . To prove the converse implication, let x ∈ XP be such that ⟨EX′ P (x), y⟩ = 0 for all y ∈ XP . Again, by (2.7) one obtains that T    ∆p(k−1) x(k − 1) + ℓ(k) y(k) = 0, k=1

for all y ∈ XP with y(0) = 0 = y(T + 1). This yields

− ∆p(k−1) x(k − 1) = ℓ(k),

(∀) k ∈ [1, T ].

(2.8)

We infer hp(T ) (∆x(T ))y(T + 1) = hp(0) (∆x(0))y(0),

(∀) y ∈ XP .

On account of (1.2), taking y ∈ XP with y(0) = 1 = y(T + 1), we get

∆x(T ) = ∆x(0).

(2.9)

From (2.8) and (2.9), we conclude that x is a solution of problem (2.6). Next, denoting by F : [1, T ] × R → R the primitive of f , i.e., F (k, t ) =

t



f (k, τ )dτ , 0

(∀) k ∈ [1, T ], t ∈ R,



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C. Bereanu et al. / J. Math. Anal. Appl. 399 (2013) 75–87

we define

FX (x) =

T 

F (k, x(k)),

(∀) x ∈ X .

k=1

It is a simple matter to check that FX ∈ C 1 (X , R) and

⟨FX′ (x), y⟩ =

T 

f (k, x(k))y(k),

(∀) x, y ∈ X .

(2.10)

k=1

The energy functional corresponding to problem (PP ) (resp. (PN )) is

ΦX (x) =

T +1 

1

k=1

p(k − 1)

|∆x(k − 1)|p(k−1) −

T 

F (k, x(k)),

(∀) x ∈ X ,

k=1

with X = XP (resp. X = XN ). Clearly,

ΦX (x) = ϕX (x) − FX (x), therefore ΦX ∈ C 1 (X , R) and by (2.2) and (2.10) one has

⟨ΦX′ (x), y⟩ =

T +1 

hp(k−1) (∆x(k − 1))∆y(k − 1) −

k=1

T 

f (k, x(k))y(k),

(∀) x, y ∈ X .

(2.11)

k=1

Proposition 2.2. Assume that (1.2) holds true. A function x ∈ XP is a solution of problem (PP ) if and only if it is a critical point of ΦXP . Proof. By virtue of (2.11), Proposition 2.1 applies with ℓ(k) = f (k, x(k)), for all k ∈ [1, T ].



Using the Neumann problem



−∆p(k−1) x(k − 1) = ℓ(k), ∆x(0) = 0 = ∆x(T )

(∀) k ∈ [1, T ],

(2.12)

instead of (2.6) and exactly the same strategy as above, but with EXN (resp. ΦXN ) instead of EXP (resp. ΦXP ), we obtain the following. Proposition 2.3. A function x ∈ XN is a solution of problem (2.12) if and only if it is a critical point of EXN . Proposition 2.4. A function x ∈ XN is a solution of problem (PN ) if and only if it is a critical point of ΦXN . Remark 2.5. Assumption (1.2) is not required in the case of Neumann boundary conditions. 3. Problem (PP ) without convexity assumptions We begin by a simple result concerning the solvability of problem (2.6). From (2.5) we see that if (2.6) is solvable then necessarily one has ℓ ∈ L. Actually, this is also a sufficient condition, as shown in the next proposition. Proposition 3.1. If ℓ ∈ L and (1.2) holds true, then problem (2.6) has a unique solution x˜ 0 ∈ XP ,0 . Moreover, x˜ 0 is the unique minimizer of EXP on XP ,0 . Proof. First, we prove that EXP is coercive on (XP ,0 , ∥ · ∥p(·) ). For this, let x˜ ∈ XP ,0 be with ∥˜x∥p(·) > 1. Using (2.3) we estimate EXP as follows p−

EXP (˜x) ≥ ∥˜x∥p(·) −

T 

−

|ℓ(k)˜x(k)| ≥ ∥˜x∥pp(·) − ∥ℓ∥∞

k =1

T +1 

|˜x(k)|

k=0

and, since any norm on XP ,0 is equivalent with ∥ · ∥p(·) , there is a constant C > 0 so that p−

EXP (˜x) ≥ ∥˜x∥p(·) − C ∥ℓ∥∞ ∥˜x∥p(·) → +∞,

as ∥˜x∥p(·) → ∞.

This ensures that there is some x˜ 0 ∈ XP ,0 with EXP (˜x0 ) = inf EXP (˜x) | x˜ ∈ XP ,0 . From (2.5) we get



T +1  k=1

hp(k−1) (∆x˜ 0 (k − 1))∆y˜ (k − 1) =

T  k =1

ℓ(k)˜y(k),

(∀) y˜ ∈ XP ,0 .



(3.1)

C. Bereanu et al. / J. Math. Anal. Appl. 399 (2013) 75–87

Let y ∈ XP be arbitrarily chosen. Taking y˜ = y − (1/T ) T +1 

hp(k−1) (∆x˜ 0 (k − 1))∆y(k − 1) =

k=1

T 

T

ℓ(k)y(k),

k =1

79

y(k) in (3.1) and because ℓ ∈ L, one obtains

(∀) y ∈ XP ,

k=1

which by Proposition 2.1 means that x˜ 0 is a solution of problem (2.6). The uniqueness of x˜ 0 as a solution of problem (2.6) in XP ,0 and as a minimizer of EXP on XP ,0 follows easily by (3.1) and the strict monotonicity of hp(·) .  Next, for given ℓ ∈ L, we consider the slightly more general problem:



−∆p(k−1) x(k − 1) = f (k, x(k)) + ℓ(k), (∀) k ∈ [1, T ], x(0) − x(T + 1) = 0 = ∆x(0) − ∆x(T ).

(3.2)

Instead of ΦXP , the corresponding energy functional to problem (3.2) will be ΨXP : XP → R,

ΨXP (x) = EXP (˜x) −

T 

F (k, x(k)),

(∀) x = x + x˜ ∈ R ⊕ XP ,0 ,

(3.3)

k=1

with EXP given by (2.4). In the following result we consider periodic nonlinearities of pendulum type. The corresponding classical result is due to Hamel [10]. Theorem 3.2. Assume that (1.2) holds true, f : [1, T ] × R → R is continuous and there exists ω > 0 such that F (k, t + ω) = F (k, t ), for all k ∈ [1, T ] and t ∈ R. Then, for each ℓ ∈ L, problem (3.2) has at least one solution which minimizes ΨXP on XP . Proof. Since F is continuous and periodic with respect to the second variable, there exists M > 0 such that

|F (k, t )| ≤ M ,

(∀) k ∈ [1, T ], t ∈ R.

Let x˜ 0 ∈ XP ,0 be the solution of problem (2.6) (see Proposition 3.1). We infer

ΨXP (x) ≥ EXP (˜x0 ) − TM ,

(∀) x ∈ XP

and hence, ΨXP is bounded from below. To conclude the proof it remains to show that ΨXP admits a bounded minimizing sequence. Let {xn } ⊂ XP be a minimizing sequence for ΨXP . Next, we use the same argument as in the proof of Theorem 3.5 in [11]. So, for each n ∈ N there exists mn ∈ Z such that xn + mn ω ∈ [0, ω]. Setting yn = xn + mn ω we have y˜ n = yn − (1/T ) T ˜ n and yn = yn − y˜ n = xn + mn ω ∈ [0, ω]. By the periodicity of F (k, ·) we obtain that ΨXP (xn ) = ΨXP (yn ). k=1 yn (k) = x Hence, {yn } is also a minimizing sequence for ΨXP . Using that {ΨXP (yn )} is bounded, we can find constants M1 , C1 > 0 such that p−

M1 ≥ ΨXP (yn ) ≥ ∥˜yn ∥p(·) − C1 ∥ℓ∥∞ ∥˜yn ∥p(·) − TM , provided that ∥˜yn ∥p(·) > 1 (see (2.3)). Thus, {∥˜yn ∥p(·) } is bounded and then, since ∥yn ∥XP = |yn | + ∥˜yn ∥p(·) and yn ∈ [0, ω], we infer that {yn } is bounded in XP and the proof is complete.  Theorem 3.3. Assume hypothesis (1.2), ℓ ∈ L, f : [1, T ] × R → R is continuous and let x˜ 0 be the unique solution of problem (2.6) in XP ,0 . If there exists a constant M ≥ 0 such that F (k, t ) ≤ M ,

(∀) t ∈ R and k ∈ [1, T ]

(3.4)

and T  k=1

lim sup F (k, t ) < |t |→∞

T 

F (k, x˜ 0 (k)),

k=1

then problem (3.2) has at least one solution which minimizes ΨXP on XP . Proof. By Proposition 3.1, (3.3) and (3.4) we get

ΨXP (x) ≥ EXP (˜x0 ) − TM ,

(∀) x ∈ XP

and hence, ΨXP is bounded from below. It remains to show that ΨXP admits a bounded minimizing sequence. Setting c := inf{ΨXP (x) | x ∈ XP } (∈ R),

(3.5)

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C. Bereanu et al. / J. Math. Anal. Appl. 399 (2013) 75–87

let {xn } ⊂ XP be a minimizing sequence for ΨXP , i.e.,

ΨXP (xn ) → c ,

as n → ∞.

(3.6)

By virtue of (3.3) and (3.5) one has c ≤ ΨXP (˜x0 ) = EXP (˜x0 ) −

T 

F (k, x˜ 0 (k))

k=1

< EXP (˜x0 ) −

T  k=1

lim sup F (k, t ).

(3.7)

|t |→∞

There are constants M1 , C1 > 0 such that M1 ≥ ΨXP (xn ) = EXP (˜xn ) −

T 

F (k, xn (k))

k=1 −

≥ ∥˜xn ∥pp(·) − C1 ∥ℓ∥∞ ∥˜xn ∥p(·) − TM ,

if ∥˜xn ∥p(·) > 1 (see (2.3)),

which implies that {˜xn } is bounded in (XP ,0 , ∥ · ∥p(·) ) and hence, we can find M2 > 0 so that

|˜xn (k)| ≤ M2 (∀) k ∈ [0, T + 1], (∀) n ∈ N.

(3.8)

Suppose, by contradiction, that {xn } is not bounded in XP . We may assume that it holds ∥xn ∥XP → ∞. As {∥˜xn ∥p(·) } is bounded, it follows

|xn | = ∥xn ∥XP − ∥˜xn ∥p(·) → +∞,

as n → ∞,

and, from (3.8), we have

|xn (k)| ≥ |xn | − |˜xn (k)| → +∞,

as n → ∞,

(3.9)

for all k ∈ [0, T + 1]. Then, as lim sup F (k, t ) ≥ lim sup F (k, xn (k)), n→∞

|t |→∞

passing to ‘‘lim supn→∞ ’’ in the inequality

ΨXP (xn ) +

T 

F (k, xn (k)) ≥ EXP (˜x0 )

k=1

and using (3.6) and (3.9), one obtains T 

c+

k=1

lim sup F (k, t ) ≥ EXP (˜x0 ), |t |→∞

which contradicts (3.7). Consequently, {xn } ⊂ XP is bounded and the proof is complete.



Remark 3.4. (i) The proof of Theorem 3.3 is inspired from [11]. However, we note that Theorem 3.8 in [11] contains a little gap which is remediable as done here. Accordingly, only the statement of Proposition 4.1 in [11] slightly changes (see Proposition 5.2 below). (ii) From [3, Theorem 3.1] we have that if (1.2) holds true, f : [1, T ] × R → R is continuous and lim sup |t |→∞

F (k, t )

|t |p(k)

< 0,

(∀) k ∈ [1, T ],

then problem (PP ) has at least one solution which minimizes ΦXP on XP . It is easy to see that this also can be inferred from Theorem 3.3 with ℓ = 0. The following Ahmad–Lazer–Paul type result (see [2]) treats the case when the nonlinearity f is bounded. Theorem 3.5. Assume that (1.2) holds true and f : [1, T ] × R → R is continuous and bounded. If either T  k=1

F (k, t ) → −∞,

as |t | → ∞

(3.10)

C. Bereanu et al. / J. Math. Anal. Appl. 399 (2013) 75–87

81

or T 

F (k, t ) → +∞,

as |t | → ∞,

(3.11)

k=1

then problem (PP ) has at least one solution x0 ∈ XP . Moreover, if (3.10) holds true then x0 minimizes ΦXP on XP . Proof. If (3.10) holds true we prove that ΦXP is coercive on XP . Then, by the direct method in the calculus of variations it is bounded from below and attains its infimum at some x0 ∈ XP , which is a critical point of ΦXP and from Proposition 2.2, it is a solution of problem (PP ). In the second case we apply the Saddle Point Theorem (see e.g., [18, Theorem 4.6], [16, Theorem 4.7]). Since f is bounded, there exists M > 0 such that

|f (k, t )| ≤ M ,

(∀) k ∈ [1, T ], t ∈ R.

(3.12)

For x ∈ XP we write x = x + x˜ , where x = (1/T )

ΦXP (x) = ϕXP (˜x) −

T 

F (k, x) −

k=1

= ϕXP (˜x) −

T 

−

T 

k=1

x(k). Then, using (2.3) and (3.12), we estimate ΦXP as follows

(F (k, x(k)) − F (k, x))

k=1

F (k, x) −

k=1

≥ ∥˜x∥pp(·) −

T 

T

T   k=1

F (k, x) − M

1

f (k, x + sx˜ (k))˜x(k)ds

0

T +1 

|˜x(k)|,

for ∥˜x∥p(·) > 1.

(3.13)

k=0

k=1

Taking into account the equivalence of the norms on XP ,0 , one obtains p−

ΦXP (x) ≥ ∥˜x∥p(·) − C1 ∥˜x∥p(·) −

T 

F (k, x),

(∀) x ∈ XP , ∥˜x∥p(·) > 1,

(3.14)

k=1

with C1 a positive constant. If (3.10) holds true then, using (3.14), a standard reasoning shows that

ΦXP (x) → +∞,

as ∥x∥XP = |x| + ∥˜x∥p(·) → ∞,

hence ΦXP is coercive on XP and the conclusion follows. If (3.11) holds, then ΦXP is neither bounded from below, nor from above. Indeed, if x = c ∈ R is a constant function then

ΦXP (c ) = −

T 

F (k, c ) → −∞,

as |c | → ∞,

(3.15)

k=1

while, if x˜ ∈ XP ,0 , then from (3.14), we have

ΦXP (˜x) → +∞,

as ∥˜x∥p(·) → ∞.

(3.16)

From (3.15) and (3.16), there exists some R > 0 such that sup ΦXP < inf ΦXP , XP ,0

SR

where SR = {x ∈ R | |x| = R}. It remains to show that ΦXP satisfies the Palais–Smale condition, which will conclude the proof. Let {xn } ⊂ XP be a sequence for which {ΦXP (xn )} is bounded and

ΦX′ P (xn ) → 0,

as n → ∞.

Since XP is finite dimensional, we have to prove that {xn } is bounded. From (3.17) there is n0 ∈ N such that

|⟨ΦX′ P (xn ), z ⟩| ≤ ∥z ∥XP , for all n ≥ n0 and z ∈ XP . Thus, we infer that

  T +1 T    p(k−1)    ∆x˜ n (k − 1) |⟨ΦXP (xn ), x˜ n ⟩| =  − f (k, xn (k))˜xn (k) ≤ ∥˜xn ∥p(·) .  k=1  k=1 ′

(3.17)

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This, together with (2.3), (3.12) and the equivalence of the norms on XP ,0 , implies p−

p− ∥˜xn ∥p(·) − ∥˜xn ∥p(·) ≤ MC2 ∥˜xn ∥p(·) ,

(∀) n ≥ n0 , if ∥˜xn ∥p(·) > 1,

with C2 > 0 a constant and hence we get that {∥˜xn ∥p(·) } is bounded. On the other hand, using (2.3), (3.13) and the fact that {ΦXP (xn )} is bounded, we can find C3 ∈ R so that p+ xn p(·)

∥˜ ∥

−

T 

F (k, xn ) −

T  

k=1

k=1

1

f (k, xn + sx˜ n (k))˜xn (k)ds ≥ C3 ,

(∀) n ∈ N

0

and then, by the boundedness of the sequence {∥˜xn ∥p(·) } and condition (3.12), we have that T 

F (k, xn ) ≤ C4 ,

(∀) n ∈ N.

k=1

Hence, by (3.11) we obtain that {|xn |} is bounded. Consequently, {xn } is bounded in XP and the proof is complete.



Example 3.6 (See [11, Example 3.6] for p = constant). The problem

 −∆p(k−1) x(k − 1) = sin(a sign x(k) − x(k)) − sin(a sign x(k)) + ℓ(k), x(0) − x(T + 1) = 0 = ∆x(0) − ∆x(T ) is solvable for any a ∈ R and ℓ ∈ L. Indeed, for a ∈ R \ {mπ | m ∈ Z} we have F (k, x) = −|x| sin a + cos(a − |x|) − cos a + ℓ(k)x and the conclusion follows from Theorem 3.5. Also, if a ∈ {mπ | m ∈ Z}, then Theorem 3.2 applies with F (k, x) = cos(a − |x|) − cos a. 4. Problem (PP ) with convex potential Under the assumption (3.10), the boundedness condition on the nonlinearity f in Theorem 3.5 can be removed when F is concave with respect to the second variable. Notice that in this case the potential ΦXP is convex. Theorem 4.1. Assume (1.2) and let f : [1, T ] × R → R be a continuous function such that F (k, ·) is concave for all k ∈ [1, T ]. If (3.10) holds true then problem (PP ) has at least one solution which minimizes ΦXP on XP . Proof. By assumption (3.10), the function t →

T 

F (k, t )

k=1

has a maximum at some point x ∈ R for which T 

f (k, x) = 0.

(4.1)

k=1

We show that ΦXP admits a bounded minimizing sequence, which will conclude the proof. So, let {xn } ⊂ XP be a minimizing sequence for ΦXP . By the convexity of −F (k, ·), we know that

− F (k, y) ≥ −F (k, z ) − f (k, z )(y − z ),

(∀) y, z ∈ R.

From (4.1) and (4.2) it follows that

ΦXP (xn ) ≥ ϕXP (xn ) −

T 

F (k, x) −

k=1

= ϕXP (xn ) −

T  k=1

T 

f (k, x)(xn (k) − x)

k=1

F (k, x) −

T  k=1

f (k, x)˜xn (k),

(4.2)

C. Bereanu et al. / J. Math. Anal. Appl. 399 (2013) 75–87

where xn = xn + x˜ n , xn = (1/T )

ΦXP (xn ) ≥ ϕXP (xn ) −

T 

83

xn (k). By the equivalence of the norms on XP ,0 and (2.3), we obtain

T

k =1

F (k, x) −

T 

k=1

|f (k, x)||˜xn (k)|

k=1

−

≥ ∥˜xn ∥pp(·) − C1 ∥˜xn ∥p(·) − C2 ,

for ∥˜xn ∥p(·) > 1,

where C1 , C2 > 0 are constants. As {ΦXP (xn )} is bounded, we can find some M > 0 such that p−

M ≥ ΦXP (xn ) ≥ ∥˜xn ∥p(·) − C1 ∥˜xn ∥p(·) − C2 ,

for ∥˜xn ∥p(·) > 1,

showing that {∥˜xn ∥p(·) } is bounded. Next, by convexity,

  1 1 −F (k, xn /2) = −F k, (1/2)(xn (k) − x˜ n (k)) ≤ − F (k, xn (k)) − F (k, −˜xn (k)), 2

2

for all k ∈ [1, T ] and n ∈ N. This, together with (2.3), implies T 

p−

M ≥ ΦXP (xn ) ≥ ∥˜xn ∥p(·) − 2

F (k, xn /2) +

k=1

T 

F (k, −˜xn (k)),

for ∥˜xn ∥p(·) > 1.

k=1

Thus, by the boundedness of the sequences {∥˜xn ∥p(·) } and {∥˜xn ∥∞ }, we get M ≥ ΦXP (xn ) ≥ −2

T 

F (k, xn /2) − C3 ,

k=1

with C3 > 0 and hence, T 

F (k, xn /2) ≥ −

M + C3

k=1

2

,

(∀) n ∈ N.

Therefore, k=1 F (k, xn /2) is bounded from below and from (3.10) one obtains that {xn } is bounded. Since ∥xn ∥XP = |xn | + ∥˜xn ∥p(·) , we infer that {xn } is bounded in XP . 

T

Theorem 4.2. Let f : [1, T ] × R → R be continuous and assume that (1.2) holds true. If F (k, ·) is strictly concave, for all k ∈ [1, T ], then the following conditions are equivalent: (i) Problem (PP ) is solvable; (ii) There exists t ∈ R such that (iii)

T

k=1

T

k=1

f (k, t ) = 0;

F (k, t ) → −∞ when |t | → ∞.

Proof. Assume (i) and let x ∈ XP be a solution of problem (PP ). Summing from 1 to T in the equation, using (1.2) and the boundary conditions, one obtains T 

f (k, x(k)) = 0.

(4.3)

k=1

Let x = x + x˜ , where x = (1/T )

G(t ) = −

T 

F (k, t ),

T

k =1

x(k) and define the strictly convex functions G and G˜ on R by

(∀) t ∈ R,

k =1

respectively,

G˜ (t ) = −

T 

F (k, x˜ (k) + t ),

(∀) t ∈ R.

k=1

From (4.3), we have that G˜ ′ (x) = 0, which by [16, Proposition 1.5], implies

G˜ (t ) → +∞,

as |t | → ∞.

(4.4)

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C. Bereanu et al. / J. Math. Anal. Appl. 399 (2013) 75–87

Also, by the strict convexity of −F (k, ·), one obtains

G˜ (t ) < −

T 1

2 k=1

F (k, 2t ) −

T 1

2 k=1

F (k, 2x˜ (k)) =

1 2

G(2t ) + C (˜x)

and by (4.4) it follows that G(t ) → +∞ as |t | → ∞. Hence, from [16, Proposition 1.5] we get that there exists t ∈ R such T that k=1 f (k, t ) = 0, i.e. (ii). Again, from [16, Proposition 1.5] it results that (ii) implies (iii). By Theorem 4.1 we have that (iii) implies (i), which concludes the proof.  Remark 4.3. Theorems 4.1 and 4.2 are discrete p(·)-Laplacian variants of Theorems 1.8 and 1.9 in [16]. 5. Lower and upper solutions for problem (PP ) In this section we use Theorem 3.3 to derive the solvability of problem (PP ) by the method of lower and upper solutions. Definition 5.1. A function α ∈ XP (resp. β ∈ XP ) is called a lower solution (resp. upper solution) for problem (PP ) if

∆α(0) ≥ ∆α(T )

(resp. ∆β(0) ≤ ∆β(T ))

and the inequality

− ∆p(k−1) α(k − 1) ≤ f (k, α(k)),



resp. − ∆p(k−1) β(k − 1) ≥ f (k, β(k))



(5.1)

holds for all k ∈ [1, T ]. Such a lower or upper solution will be called strict if the inequalities in (5.1) are strict. We need the following simple consequence of Theorem 3.3. Proposition 5.2. Let g : [1, T ] × R → R be continuous and bounded. If (1.2) holds true and a ∈ R[1,T ] is such that a(k) > 0 (k ∈ [1, T ]), then problem

 −∆p(k−1) x(k − 1) + a(k)x(k) = g (k, x(k)),

(∀) k ∈ [1, T ],

x(0) − x(T + 1) = 0 = ∆x(0) − ∆x(T ) has at least one solution. Proof. Theorem 3.3 applies with ℓ = 0 and f (k, t ) = g (k, t ) − a(k)t, for all k ∈ [1, T ] and t ∈ R.



Theorem 5.3. If (1.2) holds and problem (PP ) has a lower solution α and an upper solution β such that α ≤ β , then (PP ) has a solution x with α ≤ x ≤ β . Moreover, if α and β are strict, then α < x < β . Proof. Let γ : [1, T ] × R → R be the continuous function defined by

  β(k), γ (k, t ) = t ,   α(k),

t > β(k),

α(k) ≤ t ≤ β(k),

(5.2)

t < α(k)

and consider the modified problem

 −∆p(k−1) x(k − 1) + x(k) = f (k, γ (k, x(k))) + γ (k, x(k)), x(0) − x(T + 1) = 0 = ∆x(0) − ∆x(T ).

(∀) k ∈ [1, T ],

(5.3)

Using Proposition 5.2 with a ≡ 1 and g (k, t ) = f (k, γ (k, t )) + γ (k, t ), for all k ∈ [1, T ] and t ∈ R, problem (5.3) has at least one solution x ∈ XP . Then, following exactly the same arguments as in Step I from the proof of [5, Theorem 1], we get that if x is a solution of (5.3), then α ≤ x ≤ β (respectively α < x < β , if α and β are strict), so x in fact solves problem (PP ).  Example 5.4. Let f1 : R → R be a continuous function. Assume that (1.2) holds and that there exist α, β ∈ R such that α ≤ β and f1 (α) ≥ 0 ≥ f1 (β), then the following problem  −∆p(k−1) x(k − 1) = f1 (x(k)), (∀) k ∈ [1, T ], x(0) − x(T + 1) = 0 = ∆x(0) − ∆x(T ) has at least one solution x with α ≤ x(k) ≤ β , for all k ∈ [0, T + 1].

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85

6. The Neumann problem (PN ) We reformulate for problem (PN ), the existence results that we have proved in the periodic case. We point out that these are obtained as the corresponding ones for problem (PP ) by no longer than ‘mutatis mutandis’ arguments. Theorem 6.1. Let f : [1, T ] × R → R be continuous and assume that there exists ω > 0 such that F (k, t + ω) = F (k, t ), for all k ∈ [1, T ] and x ∈ R. Then, for each ℓ ∈ L, the problem



−∆p(k−1) x(k − 1) = f (k, x(k)) + ℓ(k), ∆x(0) = 0 = ∆x(T )

(∀) k ∈ [1, T ],

(6.1)

has at least one solution which minimizes on XN the corresponding Euler–Lagrange functional ΨXN : XN → R, defined by

ΨXN (x) =

T +1 

1

k=1

p(k − 1)

|∆x˜ (k − 1)|p(k−1) −

for all x ∈ XN , x = x + x˜ , where x = (1/T )

T 

ℓ(k)˜x(k) −

k=1

T

k=1

T 

F (k, x(k)),

k=1

x(k).

Theorem 6.2. Let f : [1, T ] × R → R be continuous, ℓ ∈ L and let x˜ 0 ∈ XN ,0 be the unique solution of problem (2.12). Assume that there exists M ≥ 0 such that (3.4) and (3.5) hold true. Then problem (6.1) has at least one solution which minimizes ΨXN on XN . Theorem 6.3. Assume that f : [1, T ] × R → R is continuous and bounded. If either (3.10) or (3.11) is satisfied, then problem (PN ) has at least one solution x0 ∈ XN . Moreover, if (3.10) holds true then x0 minimizes ΦXN on XN . Theorem 6.4. Let f : [1, T ] × R → R be a continuous function such that F (k, ·) is concave, for all k ∈ [1, T ]. If (3.10) holds true, then problem (PN ) has at least one solution which minimizes ΦXN on XN . Theorem 6.5. Let f : [1, T ] × R → R be a continuous function such that F (k, ·) is strictly concave, for all k ∈ [1, T ]. Then, the following conditions are equivalent: (i) Problem (PN ) is solvable; T (ii) There exists t ∈ R such that k=1 f (k, t ) = 0; T (iii) F ( k , t ) → −∞ when | t | → ∞. k=1 Analogously, for the Neumann problem the concepts of a lower and an upper solution are defined by the following. Definition 6.6. A function α ∈ XN is called a lower solution for problem (PN ) if it satisfies

− ∆p(k−1) α(k − 1) ≤ f (k, α(k)),

(∀) k ∈ [1, T ],

(6.2)

together with

∆α(T ) ≤ 0 ≤ ∆α(0). In the same way, an upper solution is a function β ∈ XN satisfying the reversed inequalities, i.e.,

− ∆p(k−1) β(k − 1) ≥ f (k, β(k)),

(∀) k ∈ [1, T ],

(6.3)

respectively

∆β(0) ≤ 0 ≤ ∆β(T ). Also, a lower or an upper solution will be called strict if the inequalities in (6.2), respectively (6.3) are strict. The existence of a solution for problem (PN ) lying between the lower and the upper solution can be treated similarly to the periodic problem (PP ). Hence, Proposition 5.2 will become the following proposition. Proposition 6.7. Let g : [1, T ] × R → R be continuous and bounded. If a ∈ R[1,T ] is such that a(k) > 0 (k ∈ [1, T ]), then the problem



−∆p(k−1) x(k − 1) + a(k)x(k) = g (k, x(k)), ∆x(0) = 0 = ∆x(T )

has at least one solution.

(∀) k ∈ [1, T ],

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C. Bereanu et al. / J. Math. Anal. Appl. 399 (2013) 75–87

Theorem 6.8. If problem (PN ) has a lower solution α and an upper solution β such that α ≤ β , then (PN ) has a solution x with α ≤ x ≤ β . Moreover, if α and β are strict, then α < x < β . Proof. We consider the following problem



−∆p(k−1) x(k − 1) + x(k) = f (k, γ (k, x(k))) + γ (k, x(k)), ∆x(0) = 0 = ∆x(T ),

(∀) k ∈ [1, T ],

(6.4)

with γ defined in (5.2). By Proposition 6.7, problem (6.4) has at least one solution x ∈ XN . It remains to show that α ≤ x ≤ β and hence x will solve (PN ). Suppose by contradiction that there is some k0 ∈ [0, T + 1] such that α(k0 ) − x(k0 ) > 0 so that α(m) − x(m) = maxk∈[0,T +1] (α(k) − x(k)) > 0. We show that m can be chosen in [1, T ]. Assuming that α(k) − x(k) < α(0) − x(0) for all k ∈ [1, T + 1], we obtain 0 > α(1) − x(1) − (α(0) − x(0)) = ∆α(0) − ∆x(0) = ∆α(0) ≥ 0, a contradiction. It follows that max (α(k) − x(k)) =

k∈[0,T +1]

max (α(k) − x(k)).

k∈[1,T +1]

Next, if α(k) − x(k) < α(T + 1) − x(T + 1) for all k ∈ [1, T ], then 0 > α(T ) − x(T ) − (α(T + 1) − x(T + 1)) = −∆α(T ) + ∆x(T ) = −∆α(T ) ≥ 0, a contradiction again. Therefore, it holds max (α(k) − x(k)) = max (α(k) − x(k)).

k∈[1,T +1]

k∈[1,T ]

Then, from

α(m) − x(m) = max (α(k) − x(k)), k∈[1,T ]

m can be chosen in [1, T ]. This implies

∆α(m) ≤ ∆x(m) and ∆α(m − 1) ≥ ∆x(m − 1). Taking into account (1.1) and that hp(·) is an increasing homeomorphism, we infer that

∆p(m−1) α(m − 1) ≤ ∆p(m−1) x(m − 1) and by virtue of (6.2), one obtains

∆p(m−1) α(m − 1) ≤ = = <

∆p(m−1) x(m − 1) −f (m, γ (m, x(m))) − γ (m, x(m)) + x(m) −f (m, α(m)) + (x(m) − α(m)) −f (m, α(m)) ≤ ∆p(m−1) α(m − 1),

a contradiction. Similarly, it can be shown that x ≤ β . If α, β are strict, the fact that α < x < β follows as above, in a standard way.  Acknowledgments The research of C. Şerban was supported by the strategic grant POSDRU/CPP107/DMI1.5/S/78421, Project ID 78421 (2010), co-financed by the European Social Fund – Investing in People, within the Sectoral Operational Programme Human Resources Development 2007–2013. Also, the support for C. Bereanu and P. Jebelean from the grant TE-PN-II-RU-TE-2011-3-0157 (CNCS-Romania) is gratefully acknowledged. References [1] A. Agarwal, K. Perera, D. O’Regan, Multiple positive solutions of singular discrete p-Laplacian problems via variational methods, Adv. Difference Equ. 2005 (2005) 93–99. [2] S. Ahmad, A.C. Lazer, J.L. Paul, Elementary critical point theory and perturbations of elliptic boundary value problems at resonance, Indiana Math. J. 25 (1976) 933–944. [3] C. Bereanu, P. Jebelean, C. Şerban, Ground state and mountain pass solutions for discrete p(·)-Laplacian, Bound. Value Probl. 2012 (104) (2012). [4] C. Bereanu, J. Mawhin, Existence and multiplicity results for periodic solutions of nonlinear difference equations, J. Difference Equ. Appl. 12 (2006) 677–695. [5] C. Bereanu, H.B. Thompson, Periodic solutions of nonlinear diference equations with discrete φ -Laplacian, J. Math. Anal. Appl. 330 (2007) 1002–1015. [6] A. Cabada, A. Iannizzotto, S. Tersian, Multiple solutions for discrete boundary value problems, J. Math. Anal. Appl. 356 (2009) 418–428. [7] M. Galewski, S. Gła¸b, On the discrete boundary value problem for anisotropic equation, J. Math. Anal. Appl. 386 (2012) 956–965.

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