Open Access

Existence and multiplicity of difference ϕ-Laplacian boundary value problems

Advances in Difference Equations20132013:267

https://doi.org/10.1186/1687-1847-2013-267

Received: 5 June 2013

Accepted: 22 August 2013

Published: 8 September 2013

Abstract

Concerned are the difference ϕ-Laplacian boundary value problems. The multiplicity result based on the lower and upper solutions method associated with Brouwer degree is applied to a difference ϕ-Laplacian eigenvalue problem. An existence result of at least three positive solutions is established for the eigenvalue problem with the parameter belonging to an explicit open interval. In addition, an example is given to illustrate the three solutions result.

Keywords

difference ϕ-Laplacian boundary value problemthe lower and upper solutionpositive solutionmultiplicityBrouwer degree

1 Introduction

Recently, Kim [1] studied a one-dimensional differential p-Laplacian boundary value problem with a positive parameter and established an existence result of three positive solutions by the lower and upper solutions method associated with Leray-Schauder degree theory. Kim and Shi [2] studied the global continuum and multiple positive solutions of a p-Laplacian boundary value problem. Motivated by the methods in [1, 2], we consider difference ϕ-Laplacian boundary value problems.

For a , b Z with a < b , let [ a , b ] Z = { a , a + 1 , a + 2 , , b 1 , b } . First, by the upper and lower solutions method associated with Brouwer degree theory, we establish the existence and multiplicity results for the following discrete ϕ-Laplacian boundary value problem:
{ Δ ( ϕ ( Δ u ( k 1 ) ) ) + f ( k , u ( k ) ) = 0 , k [ 1 , T ] Z , u ( 0 ) = u ( T + 1 ) = 0 ,
(1)

where T > 1 is a given positive integer, Δ u ( k ) = u ( k + 1 ) u ( k ) , and

(A1) ϕ : R R is an odd and strictly increasing function;

(A2) f : [ 1 , T ] Z × R R is continuous.

Then, we apply the multiplicity result of (1) to the following eigenvalue problem:
{ Δ ( ϕ ( Δ u ( k 1 ) ) ) + λ p ( k ) g ( u ( k ) ) = 0 , k [ 1 , T ] Z , u ( 0 ) = u ( T + 1 ) = 0 ,
(2)

where λ is a positive parameter. Under some suitable assumptions imposed on g, we establish the existence of three positive solutions of (2) with λ belonging to an explicit open interval.

The function ϕ ( u ) covers two important cases: ϕ ( u ) = u and ϕ ( u ) = | u | p 2 u ( p > 1 ). If ϕ ( u ) = u , then problem (1) is the classical second order difference Dirichlet boundary value problem. For the case that ϕ ( u ) = | u | p 2 u , problem (1) is the well-known discrete p-Laplacian problem. The two cases have been widely studied. To name a few, see [310] and the references therein.

Problem (1) can be viewed as the discrete analogue of the following differential ϕ-Laplacian problem:
{ ( ϕ ( u ) ) + f ( t , u ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 1 ) = 0 ,
(3)

which rises from the study of radial solutions for p-Laplacian equations ( ϕ ( u ) = | u | p 2 u ) on an annular domain (see [11], and references therein). Recently, the differential ϕ-Laplacian problems have been widely studied in many different papers. We refer the readers to [1219] and the references therein.

For discrete ϕ-Laplacian problems, there are fewer study results than for differential ϕ-Laplacian problems. See Cabda [20], Cabada and Espinar [21] and Bondar [22]. To the best of our knowledge, there are no results on the existence and multiplicity of positive solutions for difference ϕ-Laplacian problems.

The remaining part of this paper is organized as follows. In Section 2, we show the lower and upper solutions method and establish the existence and multiplicity of solutions of (1). In Section 3, we establish the existence of three positive solutions of (2). Finally, we give an example to illustrate our main results.

2 The upper and lower solutions method

In this section, we establish the existence and multiplicity results of solutions for problem (1) by lower and upper solutions method associated with Brouwer degree.

Let E = { u : [ 0 , T + 1 ] Z R T + 2 } with the norm u = max t [ 0 , T + 1 ] Z | u ( t ) | .

Definition 2.1 Given u , v , w E , we say that
  1. (1)

    u v if for all k [ 0 , T + 1 ] Z , u ( k ) v ( k ) .

     
  2. (2)

    u [ v , w ] if v u w .

     
  3. (3)

    u v if for all k [ 1 , T ] Z , u ( k ) < v ( k ) and u ( 0 ) v ( 0 ) , u ( T + 1 ) v ( T + 1 ) .

     
Definition 2.2 α E is called a lower solution of problem (1) if
{ Δ ( ϕ ( Δ α ( k 1 ) ) ) + f ( k , α ( k ) ) 0 , k [ 1 , T ] Z , α ( 0 ) 0 , α ( T + 1 ) 0 .

If the first inequality above is strict, then α is called a strict lower solution of (1).

In the same way, we define the upper solution and the strict upper solution of (1) by reversing the inequalities above.

Lemma 2.1 Let (A1) hold. The problem
{ Δ ( ϕ ( Δ u ( k 1 ) ) ) + u ( k ) = 0 , k [ 1 , T ] Z , u ( 0 ) = u ( T + 1 ) = 0
(4)

has the unique solution u 0 .

Proof It is clear that 0 is a trivial solution of (4). Suppose that (4) has a nontrivial solution u. Let | u ( m ) | = max k [ 1 , T ] Z | u ( k ) | = u . If u ( m ) = max k [ 1 , T ] Z u ( k ) , then u ( m ) > 0 and Δ u ( m ) 0 , Δ u ( m 1 ) 0 , which yields a contradiction:
u ( m ) = Δ ( ϕ ( Δ u ( m 1 ) ) ) = ϕ ( Δ u ( m ) ) ϕ ( Δ u ( m 1 ) ) 0 < u ( m ) .

Similarly, if u ( m ) = min k [ 1 , T ] Z u ( k ) , then u ( m ) < 0 and Δ u ( m ) 0 , Δ u ( m 1 ) 0 , which implies that u ( m ) = Δ ( ϕ ( Δ u ( m 1 ) ) ) 0 > u ( m ) , which is a contradiction. The proof is complete. □

Theorem 2.1 Let (A1) and (A2) hold.
  1. (i)

    Assume that there exist α and β, respectively lower and upper solutions of (1) such that α β . Then problem (1) has at least one solution u with α u β .

     
  2. (ii)
    Assume that problem (1) has two pairs of lower and upper solutions ( α 1 , β 1 ) and ( α 2 , β 2 ) with α 2 and β 1 being strict, satisfying that
    α 1 α 2 β 2 , α 1 β 1 β 2 ,
     
and that there exists k 0 [ 0 , T + 1 ] Z such that β 1 ( k 0 ) < α 2 ( k 0 ) . Then problem (1) has at least three solutions u 1 , u 2 , u 3 with
α 1 u 1 β 1 , α 2 u 2 β 2 , u 3 [ α 1 , β 2 ] ( [ α 1 , β 1 ] [ α 2 , β 2 ] ) .

Remark We denote that the result (i) has been proved in [20] by Brouwer fixed point theorem. Here, for the convenience of the proof of (ii), it is proven by Brouwer degree theory. The proof of (ii) is motivated by the idea in [1].

Proof of Theorem 2.1. (i) Define γ : [ 1 , T ] Z × R R by
γ ( k , u ) = { β ( k ) , u > β ( k ) , u , α ( k ) u β ( k ) , α ( k ) , u < α ( k ) .
Consider the modified problem
{ Δ ( ϕ ( Δ u ( k 1 ) ) ) + f ( k , γ ( k , u ( k ) ) ) [ u ( k ) γ ( k , u ( k ) ) ] = 0 , k [ 1 , T ] , u ( 0 ) = u ( T + 1 ) = 0 .
(5)
Clearly, all solutions u of (5) satisfying α u β are solutions of (1). Let u be a solution of (5). By the arguments in [20], we know that α u β . Now, we prove that problem (5) has at least one solution. Let E 0 = { u E : u ( 0 ) = u ( T + 1 ) = 0 } and define operator T ˜ : E 0 R T by
( T ˜ u ) ( k ) = Δ ( ϕ ( Δ u ( k 1 ) ) ) + f ( k , γ ( k , u ( k ) ) ) [ u ( k ) γ ( k , u ( k ) ) ] , k [ 1 , T ] Z .
(6)
Obviously, each solution u of T ˜ u = 0 solves (5). Define homotopic mapping Γ : [ 0 , 1 ] × E 0 R T by
Γ ( λ , u ) ( k ) = ( 1 λ ) [ Δ ( ϕ ( Δ u ( k 1 ) ) ) u ( k ) ] + λ T ˜ u ( k ) = Δ ( ϕ ( Δ u ( k 1 ) ) ) u ( k ) + λ [ f ( k , γ ( k , u ( k ) ) ) + γ ( k , u ( k ) ) ] , k [ 1 , T ] Z .
By the definition of γ and the continuity of f, there exists an R > 0 , such that
max k [ 1 , T ] Z max u R | f ( k , γ ( k , u ) ) + γ ( k , u ) | < R .
Let B R ( 0 ) = { u E 0 : u < R } . We prove that if ( λ , u ) [ 0 , 1 ] × E 0 is a solution of Γ ( λ , u ) = 0 , then u B R ( 0 ) . Let | u ( m ) | = max k [ 1 , T ] Z | u ( k ) | = u . Then there are two cases that u ( m ) = max k [ 1 , T ] Z u ( k ) and u ( m ) = min k [ 1 , T ] Z u ( k ) . For the first case, since u ( m + 1 ) u ( m ) 0 , u ( m ) u ( m 1 ) 0 and ϕ is odd, we have that
u ( m ) λ [ f ( m , γ ( m , u ( m ) ) ) + γ ( m , u ( m ) ) ] = Δ ( ϕ ( Δ u ( m 1 ) ) ) = ϕ ( u ( m + 1 ) u ( m ) ) ϕ ( u ( m ) u ( m 1 ) ) 0 ,
which implies that
u ( m ) λ [ f ( m , γ ( m , u ( m ) ) ) + γ ( m , u ( m ) ) ] < R .
Similarly, for the second case, we have that
u ( m ) λ [ f ( m , γ ( m , u ( m ) ) ) + γ ( m , u ( m ) ) ] > R .
Therefore, u < R , and deg ( Γ ( λ , ) , B R ( 0 ) , 0 ) is well defined. By the homotopy invariance of Brouwer degree, we get that
deg ( T ˜ , B R ( 0 ) , 0 ) = deg ( Γ ( 1 , ) , B R ( 0 ) , 0 ) = deg ( Γ ( 0 , ) , B R ( 0 ) , 0 ) .
By Lemma 2.1, the equation Δ ( ϕ ( Δ u ( k 1 ) ) ) + u ( k ) = 0 has the unique solution u = 0 in E 0 , thus we have
deg ( Γ ( 0 , ) , B R ( 0 ) , 0 ) = 1 .
Therefore, deg ( T ˜ , B R ( 0 ) , 0 ) = 1 , which implies that problem (5) has at least one solution u E 0 .
  1. (ii)
    First, we show that if α and β are strict lower and upper solutions, respectively, such that α β , then deg ( T ˜ , Ω α β , 0 ) = 1 , where Ω α β = { u E 0 , α u β , u < R } . By the arguments above, each solution u of (5) satisfies that α u β . We claim that α u β . In fact, if it is not true, then there exists an m [ 1 , T ] Z such that α ( m ) = u ( m ) . Since Δ u ( m 1 ) Δ α ( m 1 ) , Δ u ( m ) Δ α ( m ) , we have by the monotonicity of ϕ that
    Δ ( ϕ ( Δ u ( m 1 ) ) ) Δ ( ϕ ( Δ α ( m 1 ) ) ) = [ ϕ ( Δ u ( m ) ) ϕ ( Δ α ( m ) ) ] + [ ϕ ( Δ α ( m 1 ) ) ϕ ( Δ u ( m 1 ) ) ] 0 .
     
It yields a contradiction:
Δ ( ϕ ( Δ u ( m 1 ) ) ) = f ( m , γ ( m , u ( m ) ) ) + [ u ( m ) γ ( m , u ( m ) ) ] = f ( m , α ( m ) ) < Δ ( ϕ ( Δ α ( m 1 ) ) ) .
Thus α u . Similarly, one can check that u β . By the excision property of Brouwer degree,
deg ( T ˜ , Ω α β , 0 ) = deg ( T ˜ , B R ( 0 ) , 0 ) = 1 .
Now, consider the following modified problem:
{ Δ ( ϕ ( Δ u ( k 1 ) ) ) + f ( k , γ ( k , u ( k ) ) ) [ u ( k ) γ ( k , u ( k ) ) ] = 0 , k [ 1 , T ] , u ( 0 ) = u ( T + 1 ) = 0 ,
(7)
where γ : [ 1 , T ] Z × R R is defined by
γ ( k , u ) = { β 2 ( k ) , u > β 2 ( k ) , u , α 1 ( k ) u β 2 ( k ) , α 1 ( k ) , u < α 1 ( k ) .
It is easy to see that for sufficiently small ϵ > 0 , ( α 1 ϵ , β 1 ) and ( α 2 , β 2 + ϵ ) are two pairs of strict lower and upper solutions of (7). Similarly to (6), let T ˜ be the operator corresponding to problem (7). For sufficiently large R > 0 , define
Ω α 1 β 1 = { u E 0 , α 1 ϵ u β 1 , u < R } , Ω α 1 β 2 = { u E 0 , α 1 ϵ u β 2 + ϵ , u < R } ,
and
Ω α 2 β 2 = { u E 0 , α 2 u β 2 + ϵ , u < R } .

Then deg ( T ˜ , Ω α 1 β 1 , 0 ) = 1 , deg ( T ˜ , Ω α 1 β 2 , 0 ) = 1 and deg ( T ˜ , Ω α 2 β 2 , 0 ) = 1 . Thus by the additivity property of Brouwer degree, we have deg ( T ˜ , Ω α 1 β 2 ( Ω ¯ α 1 β 1 Ω ¯ α 2 β 2 , 0 ) ) = 1 . Therefore, problem (7) has three solutions u 1 , u 2 and u 3 with u 1 Ω α 1 β 1 , u 2 Ω α 2 β 2 and u 3 Ω α 1 β 2 Ω ¯ α 1 β 1 Ω ¯ α 2 β 2 . By the facts that all solutions of (7) satisfy [ α 1 , β 2 ] and are solution of (1), the proof is complete. □

3 Three positive solutions of eigenvalue problems

Lemma 3.1 Let (A1) hold and u satisfy the following difference inequality:
Δ ( ϕ ( Δ u ( k 1 ) ) ) 0 , k [ 1 , T ] Z
(8)

with u ( 0 ) 0 , u ( T + 1 ) 0 . Then u ( k ) 0 for all k [ 1 , T ] Z , and Δ u ( k 1 ) 0 for k [ 1 , k ] Z , Δ u ( k ) 0 for k [ k , T ] Z , where k [ 0 , 1 + T ] Z satisfies u ( k ) = max k [ 0 , 1 + T ] Z u ( k ) .

Proof Since Δ [ ϕ ( Δ u ( k 1 ) ) ] = ϕ ( Δ u ( k ) ) ϕ ( Δ u ( k 1 ) ) 0 , k [ 1 , T ] Z , we have by the monotonicity of ϕ that Δ u ( k ) Δ u ( k 1 ) , k [ 1 , T ] Z . If k = 0 or T + 1 , the result is clear. Now, we assume that k [ 1 , T ] Z . Since
Δ u ( k 1 ) = u ( k ) u ( k 1 ) 0 , Δ u ( k ) = u ( k + 1 ) u ( k ) 0 ,

we have by the monotonicity of Δ u ( ) that Δ u ( k 1 ) 0 for k [ 1 , k ] Z , Δ u ( k ) 0 for k [ k , T ] Z , which implies that u ( k ) 0 holds for all k [ 1 , T ] Z by the boundary conditions u ( 0 ) 0 , u ( T + 1 ) 0 . □

Remark If inequality (8) is strict, then u ( k ) > 0 for k [ 1 , T ] Z , and there exists k [ 0 , 1 + T ] Z such that u ( k ) = u , and Δ u ( k 1 ) > 0 for k [ 1 , k 1 ] Z , Δ u ( k 1 ) 0 , and Δ u ( k ) < 0 for k [ k , T ] Z .

Consider the following problem:
{ Δ ( ϕ ( Δ u ( k 1 ) ) ) + h ( k ) = 0 , k [ 1 , T ] Z , u ( 0 ) = u ( T + 1 ) = 0 ,
(9)

where h : [ 1 , T ] Z ( 0 , ) .

In the following arguments, we assume that

(B1) ϕ : R R is an odd and strictly increasing homeomorphism.

Lemma 3.2 Let (B1) hold and u solve (9). If h is symmetric on [ 1 , T ] Z , i.e., h ( k ) = h ( T + 1 k ) , k [ 1 , T ] Z , then u ( k ) is symmetric on [ 1 , T ] Z . Moreover,
  1. (i)
    if T + 1 ( T 2 ) is odd, then u = u ( T 2 ) = u ( T 2 + 1 ) , and the solution u of (9) can be expressed as
    u ( k ) = { s = 1 k ϕ 1 ( l = s T 2 h ( l ) ) , k T 2 , s = k T ϕ 1 ( l = T 2 + 1 s h ( l ) ) , k T 2 + 1 .
     
  2. (ii)
    if T + 1 ( T 3 ) is even, then u = u ( T + 1 2 ) , and the solution u of (9) can be expressed as
    u ( k ) = { s = 1 k ϕ 1 ( l = s T + 1 2 1 h ( l ) + 1 2 h ( T + 1 2 ) ) , k T + 1 2 , s = k T ϕ 1 ( l = T + 1 2 + 1 s h ( l ) + 1 2 h ( T + 1 2 ) ) , k T + 1 2 .
     
Proof It is easy to see that
u ( k ) = s = 1 k ϕ 1 ( ϕ ( u ( T ) ) + l = s T h ( l ) ) , k [ 0 , T + 1 ] Z ,
(10)
with
k = 1 T + 1 ϕ 1 ( ϕ ( u ( T ) ) + l = k T h ( l ) ) = 0 .
(11)
Equivalently,
u ( k ) = s = k T ϕ 1 ( ϕ ( u ( 1 ) ) + l = 1 s h ( l ) ) , k [ 0 , T + 1 ] Z ,
(12)
with
k = 0 T ϕ 1 ( ϕ ( u ( 1 ) ) + l = 1 k h ( l ) ) = 0 .
(13)
By (10) or (12), one has
ϕ ( u ( 1 ) ) + ϕ ( u ( T ) ) = l = 1 T h ( l ) .
(14)
The symmetry of h first implies that u ( 1 ) = u ( T ) . In fact, by (11),
0 = k = 1 T + 1 ϕ 1 ( ϕ ( u ( T ) ) + l = k T h ( l ) ) = k = 1 T + 1 ϕ 1 ( ϕ ( u ( T ) ) + l = 1 T + 1 k h ( l ) ) = k = 0 T ϕ 1 ( ϕ ( u ( T ) ) + l = 1 k h ( l ) ) .
Since ϕ 1 is a homeomorphism from R onto itself, the solution C of the equation k = 1 T + 1 ϕ 1 ( C + l = k T h ( l ) ) = 0 is unique. Comparing the equation above with (13), we have ϕ ( u ( 1 ) ) = ϕ ( u ( T ) ) . Thus for k [ 1 , T ] Z ,
u ( k ) = s = k T ϕ 1 ( ϕ ( u ( 1 ) ) + l = 1 s h ( l ) ) = s = k T ϕ 1 ( ϕ ( u ( T ) ) + l = T + 1 s T h ( l ) ) = m = 1 T + 1 k ϕ 1 ( ϕ ( u ( T ) ) + l = m T h ( l ) ) = u ( T + 1 k ) ,
the solution u of (9) is symmetric on [ 1 , T ] Z .
  1. (i)
    Assume that T + 1 ( T 2 ) is odd. Since u ( 1 ) = u ( T ) , by the symmetry of h and (14), we have
    ϕ ( u ( 1 ) ) = ϕ ( u ( T ) ) = l = 1 T 2 h ( l ) = l = T 2 + 1 T h ( l ) .
     
Then for k T 2 ,
u ( k ) = s = 1 k ϕ 1 ( ϕ ( u ( T ) ) + l = s T h ( l ) ) = s = 1 k ϕ 1 ( l = T 2 + 1 T h ( l ) + l = s T h ( l ) ) = s = 1 k ϕ 1 ( l = s T 2 h ( l ) ) ,
and for k T 2 + 1 ,
u ( k ) = s = k T ϕ 1 ( ϕ ( u ( 1 ) ) + l = 1 s h ( l ) ) = s = k T ϕ 1 ( l = 1 T 2 h ( l ) + l = 1 s h ( l ) ) = s = k T ϕ 1 ( l = T 2 + 1 s h ( l ) ) .
Clearly, u = u ( T 2 ) = u ( T 2 + 1 ) .
  1. (ii)
    If T + 1 ( T 3 ) is even, then (14) and the symmetry of h imply that
    ϕ ( u ( 1 ) ) = ϕ ( u ( T ) ) = l = 1 T + 1 2 1 h ( l ) + 1 2 h ( T + 1 2 ) = l = T + 1 2 + 1 T h ( l ) + 1 2 h ( T + 1 2 ) .
     
Thus for k T + 1 2 ,
u ( k ) = s = 1 k ϕ 1 ( ϕ ( u ( T ) ) + l = s T h ( l ) ) = s = 1 k ϕ 1 ( l = T + 1 2 + 1 T h ( l ) 1 2 h ( T + 1 2 ) + l = s T h ( l ) ) = s = 1 k ϕ 1 ( l = s T + 1 2 h ( l ) 1 2 h ( T + 1 2 ) ) = s = 1 k ϕ 1 ( l = s T + 1 2 1 h ( l ) + 1 2 h ( T + 1 2 ) ) ,
and for k T + 1 2 ,
u ( k ) = s = k T ϕ 1 ( ϕ ( u ( 1 ) ) + l = 1 s h ( l ) ) = s = k T ϕ 1 ( l = 1 T + 1 2 1 h ( l ) 1 2 h ( T + 1 2 ) + l = 1 s h ( l ) ) = s = k T ϕ 1 ( l = T + 1 2 s h ( l ) 1 2 h ( T + 1 2 ) ) = s = k T ϕ 1 ( l = T + 1 2 + 1 s h ( l ) + 1 2 h ( T + 1 2 ) ) .

Clearly, u = u ( T + 1 2 ) . The proof is complete. □

Now, we state the existence result of at least three positive solutions of (2). Throughout the following arguments, we suppose that T 4 . Let v be the unique positive solution of the following boundary value problem:
{ Δ ( ϕ ( Δ u ( k 1 ) ) ) + p ( k ) = 0 , k [ 1 , T ] Z , u ( 0 ) = u ( T + 1 ) = 0 ,

and p 0 = min k [ 1 , T ] Z p ( k ) .

We make the following assumptions.

(B2) There exists an increasing homeomorphism ψ : ( 0 , ) ( 0 , ) such that for all μ > 1 ,
ϕ ( μ x ) ϕ ( μ y ) ϕ ( x ) ϕ ( y ) ψ ( μ ) , x , y R : x < y ;

(B3) p : [ 1 , T ] Z ( 0 , ) ;

(B4) g C ( [ 0 , ) , ( 0 , ) ) and lim u g ( u ) ψ ( u v ) = 0 ;

(B5) There exist a, b and M satisfying v < a < b < M such that g is nondecreasing on [ b , M ) and
max { 2 ϕ ( b ) g ( a ) p 0 g ( b ) ψ ( a v ) , T ϕ ( b ) ϕ ( 2 M T + 1 ) } < 1 .

Here g ( u ) = max 0 s u g ( s ) .

We denote that condition (B4) implies that lim u g ( u ) ψ ( u v ) = 0 (see [12], Lemma 2.8). Clearly, g is nondecreasing on [ 0 , ) .

Assumption (B2) is satisfied by two important cases ϕ ( x ) = x and ϕ ( x ) = | x | p 2 x ( p > 1 ). We also provide the following two functions as examples:
ϕ ( x ) = c 1 x 3 + c 2 x , ϕ ( x ) = c 3 x 1 3 + c 4 x 1 5 ,

where c i > 0 ( i = 1 , 2 , 3 , 4 ).

Theorem 3.1 Let (B1)-(B5) hold. Then for λ ( λ 1 , λ 2 ) , problem (2) has at least three positive solutions. Here
λ 1 = 2 ϕ ( b ) p 0 g ( b ) , λ 2 = min { ψ ( a v ) g ( a ) , 2 ϕ ( 2 M T + 1 ) T p 0 g ( b ) } .
Proof Let λ be fixed with λ ( λ 1 , λ 2 ) . Clearly, α 1 0 is a strict lower solution of (2). Let β 1 = a v v . Note that a > v and Δ v ( k ) < Δ v ( k 1 ) , k [ 1 , T ] Z . Then by (B2) and the monotonicity of g , for k [ 1 , T ] Z , we have
Δ ( ϕ ( Δ β 1 ( k 1 ) ) ) = ϕ ( Δ β 1 ( k ) ) ϕ ( Δ β 1 ( k 1 ) ) = ϕ ( a v Δ v ( k ) ) ϕ ( a v Δ v ( k 1 ) ) ψ ( a v ) ( ϕ ( Δ v ( k ) ) ϕ ( Δ v ( k 1 ) ) ) = ψ ( a v ) p ( k ) < λ g ( a ) p ( k ) λ g ( β 1 ( k ) ) p ( k ) λ g ( β 1 ( k ) ) p ( k ) .
Thus β 1 is a strict upper solution of (2). Now, let α 2 solve the following problem:
{ Δ ( ϕ ( Δ u ( k 1 ) ) ) + λ p 0 g ( b ) = 0 , k [ 1 , T ] Z , u ( 0 ) = u ( T + 1 ) = 0 ,
where λ ( λ 1 , λ ) . By the expression (12), we have
α 2 ( 1 ) = s = 1 T ϕ 1 ( ϕ ( α 2 ( 1 ) ) + s λ p 0 g ( b ) ) > ϕ 1 ( ϕ ( α 2 ( 1 ) ) + λ p 0 g ( b ) ) ,
which implies that ϕ ( α 2 ( 1 ) ) > 1 2 λ p 0 g ( b ) > ϕ ( b ) . Thus Lemma 3.2 implies that α 2 ( T ) = α 2 ( 1 ) > b . Consequently, by Lemma 3.1, α 2 ( k ) > b for all k [ 1 , T ] Z . Again by Lemma 3.2, one can see that if T + 1 is odd, then
α 2 = s = 1 T 2 ϕ 1 ( l = s T 2 λ p 0 g ( b ) ) < T + 1 2 ϕ 1 ( T 2 λ p 0 g ( b ) ) < M ,
and that if T + 1 is even, then
α 2 = s = 1 T + 1 2 ϕ 1 ( l = s T + 1 2 1 λ p 0 g ( b ) + 1 2 λ p 0 g ( b ) ) < T + 1 2 ϕ 1 ( T 2 λ p 0 g ( b ) ) < M .
Thus b < α 2 ( k ) < M for k [ 1 , T ] Z . Therefore,
0 = Δ ( ϕ ( Δ α 2 ( k 1 ) ) ) + λ p 0 g ( b ) Δ ( ϕ ( Δ α 2 ( k 1 ) ) ) + λ p 0 g ( α 2 ( k ) ) < Δ ( ϕ ( Δ α 2 ( k 1 ) ) ) + λ p ( k ) g ( α 2 ( k ) ) , k [ 1 , T ] Z ,
which implies that α 2 is a strict lower solution of (2). It is easy to see that
α 2 ( k ) > b > a β 1 ( k ) , k [ 1 , T ] Z .
By lim u g ( u ) ψ ( u v ) = 0 , one can choose a sufficiently large positive number C λ , such that
g ( λ C λ ) ψ ( λ C λ v ) < 1 λ ,
and
λ C λ v > 1 , α 2 β 2 ,
where β 2 = λ C λ v v . Then by (B2) and the monotonicity of g , β 2 is a strict upper solution of (2). In fact, for k [ 1 , T ] Z ,
Δ ( ϕ ( Δ β 2 ( k 1 ) ) ) = ϕ ( Δ β 2 ( k ) ) ϕ ( Δ β 2 ( k 1 ) ) = ϕ ( λ C λ v Δ v ( k ) ) ϕ ( λ C λ v Δ v ( k 1 ) ) ψ ( λ C λ v ) ( ϕ ( Δ v ( k ) ) ϕ ( Δ v ( k 1 ) ) ) = ψ ( λ C λ v ) p ( k ) < λ g ( λ C λ ) p ( k ) λ g ( β 2 ( k ) ) p ( k ) λ g ( β 2 ( k ) ) p ( k ) .

Thus by Theorem 2.1, problem (2) has three positive solutions for λ ( λ 1 , λ 2 ) . □

Remark If g is nondecreasing on [ 0 , ) , then we take M = and λ 2 = ψ ( a v ) g ( a ) .

4 An example

Taking ϕ ( u ) = u 1 3 + u 1 5 , g ( u ) = ( u + 1 ) 1 2 , p ( k ) 1 , T = 4 , consider
{ Δ ( ( Δ u ( k 1 ) ) 1 3 + ( Δ u ( k 1 ) ) 1 5 ) + λ ( u ( k ) + 1 ) 1 2 = 0 , k [ 1 , T ] Z , u ( 0 ) = u ( 5 ) = 0 .
(15)
Let ψ ( u ) = u . It is easy to see that (B1)-(B4) hold. Choose a = 14 , b = 15 , then (B5) is satisfied. In fact, after some simple calculations, we get that v 1.089 and that
λ 1 = 2 ϕ ( b ) p 0 g ( b ) = 1 2 ( 15 1 3 + 15 1 5 ) 2.091 , λ 2 = ψ ( a v ) g ( a ) = 14 15 v 3.319 .

Thus by Theorem 3.1, problem (15) has at least three positive solutions for λ 2.091 < λ < 3.319 .

Declarations

Acknowledgements

The authors are very grateful to the referees for their helpful comments. This research is supported partially by the Research Funds for the Doctoral Program of Higher Education of China (No. 20104410120001, 20114410110002), PCSIRT of China (No. IRT1226) and the Natural Science Fund of China (No. 11171078).

Authors’ Affiliations

(1)
School of Mathematics and Information Science, and Key Laboratory of Mathematics and Interdisciplinary Sciences of Guangdong Higher Education Institutes, Guangzhou University

References

  1. Kim C: The three-solutions theorem for p -Laplacian boundary value problems. Nonlinear Anal. 2012, 75: 924–931. 10.1016/j.na.2011.09.031MathSciNetView ArticleGoogle Scholar
  2. Kim C, Shi J: Global continuum and multiple positive solutions to a p -Laplacian boundary value problem. Electron. J. Differ. Equ. 2012, 106: 1–12.MathSciNetGoogle Scholar
  3. Agarwal RP, O’Regan D: Boundary value problems for discrete equations. Appl. Math. Lett. 1997, 10(4):83–89. 10.1016/S0893-9659(97)00064-5MathSciNetView ArticleGoogle Scholar
  4. Henderson J, Thompson HB: Existence of multiple solutions for second-order discrete boundary value problems. Comput. Math. Appl. 2002, 43: 1239–1248. 10.1016/S0898-1221(02)00095-0MathSciNetView ArticleGoogle Scholar
  5. Rachunkova I, Rachunek L: Solvability of discrete Dirichlet problem via lower and upper functions method. J. Differ. Equ. Appl. 2007, 13(5):423–429. 10.1080/10236190601143302MathSciNetView ArticleGoogle Scholar
  6. Bai D, Xu Y: Nontrivial solutions of boundary value problems of second order difference equations. J. Math. Anal. Appl. 2007, 326(1):297–302. 10.1016/j.jmaa.2006.02.091MathSciNetView ArticleGoogle Scholar
  7. Bereanu C, Mawhin J, Neuve L: Existence and multiplicity results for nonlinear second order difference equations with Dirichlet value conditions. Math. Bohem. 2006, 2: 145–160.Google Scholar
  8. Agarwal RP, Perera K, O’Regan D: Multiple positive solutions of singular discrete p -Laplacian problems via variational methods. Adv. Differ. Equ. 2005, 2: 93–99.MathSciNetGoogle Scholar
  9. Cabada A, Iannizzotto A, Tersianc S: Multiple solutions for discrete boundary value problems. J. Math. Anal. Appl. 2009, 356: 418–428. 10.1016/j.jmaa.2009.02.038MathSciNetView ArticleGoogle Scholar
  10. Jiang D, Pang P, Agrwal RP: Upper and lower solutions method and a superlinear singular discrete boundary value problem. Dyn. Syst. Appl. 2007, 16: 743–754.Google Scholar
  11. Dang H, Oppenheimer S: Existence and uniqueness results for some nonlinear boundary value problems. J. Math. Anal. Appl. 1996, 198: 35–48. 10.1006/jmaa.1996.0066MathSciNetView ArticleGoogle Scholar
  12. Wang H: On the number of positive solutions of nonlinear systems. J. Math. Anal. Appl. 2003, 281: 287–306. 10.1016/S0022-247X(03)00100-8MathSciNetView ArticleGoogle Scholar
  13. Lee E, Lee Y: A multiplicity result for generalized Laplacian system with multiparameters. Nonlinear Anal. 2009, 71: e366-e376. 10.1016/j.na.2008.11.001View ArticleGoogle Scholar
  14. Cabada A, Habets P, Pouso R: Optimal existence conditions for ϕ -Laplacian equations with upper and lower solutions in the reversed order. J. Differ. Equ. 2000, 166: 385–401. 10.1006/jdeq.2000.3803MathSciNetView ArticleGoogle Scholar
  15. Cabada A, Pouso RL:Existence results for the problem ( ϕ ( u ) ) = f ( t , u , u ) with nonlinear boundary conditions. Nonlinear Anal. 1999, 35: 221–231. 10.1016/S0362-546X(98)00009-1MathSciNetView ArticleGoogle Scholar
  16. Arrázola E, Ubilla P: Positive solutions for the 1-dimensional generalized p -Laplacian involving a real parameter. Proyecciones 1998, 17: 189–200.MathSciNetGoogle Scholar
  17. Henderson J, Wang H: An eigenvalue problem for quasilinear systems. Rocky Mt. J. Math. 2007, 37: 215–228. 10.1216/rmjm/1181069327MathSciNetView ArticleGoogle Scholar
  18. Lian W, Wong F: Existence of positive solutions for higher order generalized p -Laplacian BVPs. Appl. Math. Lett. 2000, 13: 35–43.MathSciNetView ArticleGoogle Scholar
  19. Bai D, Chen Y: Three positive solutions for a generalized Laplacian boundary value problem with a parameter. Appl. Math. Comput. 2013, 219: 4782–4788. 10.1016/j.amc.2012.10.100MathSciNetView ArticleGoogle Scholar
  20. Cabada A: Extremal solutions for the difference ϕ -Laplacian problem with nonlinear functional boundary conditions. Comput. Math. Appl. 2001, 42: 593–601. 10.1016/S0898-1221(01)00179-1MathSciNetView ArticleGoogle Scholar
  21. Cabada A, Otero-Espinar V: Existence and comparison results for difference ϕ -Laplacian boundary value problems with lower and upper solutions in reverse order. J. Math. Anal. Appl. 2002, 267: 501–521. 10.1006/jmaa.2001.7783MathSciNetView ArticleGoogle Scholar
  22. Bondar K, Borkar V, Patil S: Existence and uniqueness results for difference ϕ -Laplacian boundary value problems. ITB J. Sci. 2011, 43A(1):51–58.MathSciNetView ArticleGoogle Scholar

Copyright

© Bai and Xu; licensee Springer 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.