Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2013, Article ID 157043, 8 pages
http://dx.doi.org/10.1155/2013/157043
Research Article
Positive Solutions for Discrete Boundary Value Problems to
One-Dimensional 𝑝-Laplacian with Delay
Linjun Wang and Xumei Chen
Faculty of Science, Jiangsu University, Zhenjiang 212013, China
Correspondence should be addressed to Xumei Chen; chenxumei@ujs.edu.cn
Received 23 April 2013; Accepted 20 June 2013
Academic Editor: Junjie Wei
Copyright © 2013 L. Wang and X. Chen. This is an open access article distributed under the Creative Commons Attribution
License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly
cited.
We study the existence of positive solutions for discrete boundary value problems to one-dimensional 𝑝-Laplacian with delay. The
proof is based on the Guo-Krasnoselskii fixed-point theorem in cones. Two numerical examples are also provided to illustrate the
theoretical results.
𝑅
Let 𝑠 = − ∫𝑟 2 𝑡−(𝑑−1)/(𝑝−1) 𝑑𝑡, V(𝑠) = 𝑢(𝑟(𝑠)), and 𝜌 =
1. Introduction
The 𝑝-Laplacian differential equations have been vastly applied in many fields such as non-Newtonian mechanics,
economics, ecology, neural networks, and nonlinear flow
laws, [1–5].
One of the important examples was described in [6]. Let
𝑥 = (𝑥1 , 𝑥2 ) be the two Cartesian coordinates in the plane of
the glacier occupying the Lipschitzian domain Ω, we denote
by 𝑢(𝑥) the horizontal velocity component of the ice at the
point 𝑥. After a rescaling of the physical velocity of the ice, 𝑢
satisfies the following equation:
− div (𝜓 (|∇𝑢|) ∇𝑢) = 𝑒,
in Ω,
(1)
where 𝑒 is a hydrostatic pressure force acting on the glacier
and 𝜓 is a function resulting from a constitutive law for the
ice. The typical case of (1) is the following equation:
div (𝜙𝑝 (∇𝑢)) + 𝑏 (|𝑥|) 𝑔 (𝑢) = 0,
𝑥∈
𝐵 (0, 𝑅2 )
,
𝐵 (0, 𝑅1 )
(2)
where 𝜙𝑝 (𝑠) = |𝑠|𝑝−2 𝑠 with 𝑝 > 1 and 𝐵(0, 𝑅𝑖 ) ⊂ R𝑑 are the
open balls centred about the origin with radius 𝑅𝑖 , respectively. People are interested to consider the positive radial
solutions of (2), and then (2) can be reduced to the following
form [7]:
󸀠
𝑟1−𝑑 (𝑟𝑑−1 𝜙𝑝 (𝑢󸀠 )) + 𝑏 (𝑟) 𝑔 (𝑢 (𝑟)) = 0,
𝑅1 < 𝑟 < 𝑅2 . (3)
𝑅
− ∫𝑅 2 𝑡−(𝑑−1)/(𝑝−1) 𝑑𝑡. Then (3) is transformed to
1
󸀠
(𝜙𝑝 (V󸀠 (𝑠))) + 𝑟𝑝(𝑑−1)/(𝑝−1) (𝑠) 𝑏 (𝑟 (𝑠)) 𝑔 (V (𝑠)) = 0,
(4)
𝜌 < 𝑠 < 0.
With variable change 𝑡 = 1−𝑠/𝜌 and 𝑦(𝑡) = V(𝑠), (4) reads
󸀠
(𝜙𝑝 (𝑦󸀠 (𝑡))) + 𝑐 (𝑡) 𝑔 (𝑦 (𝑡)) = 0,
0 < 𝑡 < 1,
(5)
where 𝑐(𝑡) = (−𝜌)𝑝 𝑟(𝑝(𝑑−1))/(𝑝−1) [𝜌(1 − 𝑡)]𝑏[𝑟(𝜌(1 − 𝑡))].
Equation (5) is a typical type of one-dimensional 𝑝-Laplacian
equation.
In the real world, some processes are more reasonably
described as 𝑝-Laplacian differential equations with delay
[2, 8, 9]. The reason is that the differential of the unknown
solutions depends not only on the values of the unknown
solutions at the current time but also on the values prior to
that. Such equations, to a certain extent, reflect much more
exactly the physical reality than the equations without delay.
In reality, (5) is applied together with some boundary
value conditions, see for example, [1, 4, 8–11]. By using
the Guo-Krasnoselskii fixed-point theorem, Jin and Yin [9]
2
Journal of Applied Mathematics
proved the existence of one positive solutions for the following boundary value problem of one-dimensional 𝑝-Laplacian
with delay
󸀠
(𝜙𝑝 (𝑢󸀠 (𝑡))) + 𝜆𝑔 (𝑡, 𝑢 (𝑡 − 𝜏)) = 0,
𝑢 (1) = 0,
𝑢 (𝑡) = 0,
0 < 𝑡 < 1, 𝜏 > 0,
(6)
−𝜏 ≤ 𝑡 ≤ 0.
Based on a fixed-point approach, Bai and Xu [8] obtained
the existence of positive solutions for problem (6) with 𝑔(𝑡, 𝑢)
having some singularities.
The main motivation of our work is twofold. The one is to
prove the existence of positive solutions for discrete boundary
value problems with delay. It is of interest to note here that the
existence of single and multiple positive solutions for discrete
boundary value problems to one-dimensional 𝑝-Laplacian
have been studied in great detail in the literature [12–18] and
the references therein. However, there are few papers dealing
with the existence of positive solutions for discrete boundary
value problems to one-dimensional 𝑝-Laplacian with delay.
The other motivation is coming from the numerical solutions of problem (6). In order to get the numerical solutions
of problem (6), we can apply the standard Euler method to
discretize problem (6) and approximate its solutions numerically. An immediate and natural question arises if the corresponding difference equation together with boundary conditions has positive solutions.
Motivated by above, our purpose in this paper is to
show the existence of positive solutions for the discretization
equations of problem (6). Namely, we will prove the existence
of positive solutions of the following problem:
Δ (𝜙𝑝 (Δ𝑢 (𝑛 − 1))) + 𝜆𝑎 (𝑛) 𝑓 (𝑛, 𝑢 (𝑛 − 𝑛0 )) = 0,
𝑛 ∈ {1, 2, . . . , 𝑁} ,
𝑢 (𝑁 + 1) = 0,
𝑢 (𝑛) = 0,
𝑛 ∈ {−𝑛0 , −𝑛0 + 1, . . . , 0} ,
(7)
where Δ𝑢(𝑛) = 𝑢(𝑛 + 1) − 𝑢(𝑛) and 𝑛0 is a positive integer.
This paper is organized as follows. In Section 2, we introduce some basic definitions and then we state the GuoKrasnoselskii fixed-point theorem. In Section 3, we write
a representation for a solution to problem (7) in terms of
the fixed point of an appropriate operator. Then we prove
that problem (7) has a positive solution with 𝜆 belonging to
an open interval by employing the fixed-point theorem. In
Section 4, two numerical examples are presented to illustrate
the theoretical results.
2. Preliminaries
In this section, we introduce some basic definitions and recall
the Guo-Krasnoselskii fixed-point theorem which plays a
fundamental role in our subsequent analysis.
For convenience, we will let Z(𝑎, 𝑏) denote {𝑎, 𝑎 + 1, . . . ,
𝑏}, where 𝑎 < 𝑏 ∈ Z.
A sequence (𝑢(𝑛))𝑁+1
𝑛=−𝑛0 is said to be a positive solution of
(7), if it satisfies (7) with 𝑢(𝑛) > 0, 𝑛 ∈ Z(1, 𝑁 + 1).
We collect some properties of the function 𝜙𝑝 (⋅).
Lemma 1. 𝜙𝑝 (𝑠) is increasing in 𝑠 and (𝜙𝑝 )−1 = 𝜙𝑞 , with 1/𝑝 +
1/𝑞 = 1; 𝜙𝑝 (𝑠1 ⋅ 𝑠2 ) = 𝜙𝑝 (𝑠1 ) ⋅ 𝜙𝑝 (𝑠2 ).
Definition 2. Let 𝐸 be a real Banach space. A nonempty closed
convex set 𝐾 ⊂ 𝐸 is called a cone if
(i) 𝑥 ∈ 𝐾 and 𝜆 ≥ 0 imply 𝜆𝑥 ∈ 𝐾,
(ii) 𝑥 ∈ 𝐾 and −𝑥 ∈ 𝑃 imply 𝑥 = 𝜃.
Theorem 3 (see [19]). Let 𝐸 be a Banach space and 𝐾 ⊂ 𝐸
a cone. Assume Ω1 , Ω2 are bounded open subsets of 𝐸 with 𝜃 ∈
Ω1 , Ω1 ⊂ Ω2 , and let 𝑇 : 𝐾 ⋂(Ω2 \ Ω1 ) → 𝐾 be a completely
continuous operator such that one of the following holds:
(i) ‖𝑇𝑢‖ ≤ ‖𝑢‖, 𝑢 ∈ 𝐾 ∩ 𝜕Ω1 , ‖𝑇𝑢‖ ≥ ‖𝑢‖, 𝑢 ∈
𝐾 ∩ 𝜕Ω2 ;
(ii) ‖𝑇𝑢‖ ≥ ‖𝑢‖, 𝑢 ∈ 𝐾 ∩ 𝜕Ω1 , ‖𝑇𝑢‖ ≤ ‖𝑢‖, 𝑢 ∈ 𝐾 ∩ 𝜕Ω2 .
Then 𝑇 has a fixed point in 𝐾 ⋂(Ω2 \ Ω1 ).
3. Existence of Positive Solutions
In this section, we study the existence of positive solutions for
the following equation with delay:
Δ (𝜙𝑝 (Δ𝑢 (𝑛 − 1))) + 𝜆𝑎 (𝑛) 𝑓 (𝑛, 𝑢 (𝑛 − 𝑛0 )) = 0,
𝑛 ∈ Z (1, 𝑁) ,
(8)
subject to the boundary condition
𝑢 (𝑁 + 1) = 0,
𝑢 (𝑛) = 0,
𝑛 ∈ Z (−𝑛0 , 0) .
(9)
We assume that
(A1) 𝑓 : Z(1, 𝑁 + 1) × R → R+ is a continuous function
(R+ denotes the set of nonnegative reals);
(A2) 𝑎(⋅) is a positive function defined on Z(1, 𝑁 + 1),
∑𝑁+1
𝑖=1 𝑎(𝑖) < ∞.
Remark 4. Recall that a map 𝑓 : Z(1, 𝑁 + 1) × R → R+ is
continuous if it is continuous as a map of the topological space
Z(1, 𝑁 + 1) × R into the topological space R+ . Throughout
this paper, the topology on Z(1, 𝑁 + 1) will be the discrete
topology.
Rewrite (8) as
Δ (𝜙𝑝 (Δ𝑢 (𝑖 − 1))) = −𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) .
(10)
Summing up (10) with respect to 𝑖 from 𝑚 to 𝑁 leads to
𝑚−1
𝜙𝑝 (Δ𝑢 (𝑚 − 1)) − 𝜙𝑝 (Δ𝑢 (0)) = − ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) .
𝑖=1
(11)
Noticing Lemma 1, we have
𝑚−1
Δ𝑢 (𝑚 − 1) = 𝜙𝑞 (𝜙𝑝 (Δ𝑢 (0)) − ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 ))) .
𝑖=1
(12)
Journal of Applied Mathematics
3
Summing up (12) with respect to 𝑚 from 1 to 𝑛 and
noticing the boundary condition (9) lead to
𝑛
𝑚−1
𝑚=1
𝑖=1
𝑢 (𝑛) = ∑ 𝜙𝑞 (𝜙𝑝 (Δ𝑢 (0)) − ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 ))) ,
Then by substituting (19) in (21), we obtain
𝑁+1
𝑚−1
𝑚=𝑛+1
𝑖=𝑛1 +1
𝑢 (𝑛) = ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) − 𝜔) ,
𝑛 > 𝑛1 − 1.
(13)
where Δ𝑢(0) satisfies
Since (19) is the solution of (14), we have
𝑁+1
𝑚−1
𝐹 (𝑥) = ∑ 𝜙𝑞 (𝜙𝑝 (𝑥) − ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 ))) = 0.
𝑚=1
𝑖=1
𝑁+1
𝑛1
𝑚=1
𝑖=1
𝐹 (𝑥) = ∑ 𝜙𝑞 (𝜙𝑝 (𝜙𝑞 (∑𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) + 𝜔))
(14)
𝑚−1
Obviously, 𝐹(𝑥) is continuous and strictly increasing. We
have
− ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) ) = 0.
𝑖=1
(23)
𝑁+1
𝐹 (𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )))) > 0. (15)
𝐹 (0) < 0,
(22)
𝑖=1
Namely,
Thus, there exists a unique solution to (14)
𝑁+1
𝑛1
𝑚=1
𝑖=1
∑ 𝜙𝑞 ((∑𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) + 𝜔)
𝑁+1
𝜉𝑢 ∈ (0, 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )))) .
(16)
𝑖=1
𝑚−1
(24)
− ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) ) = 0.
Let
𝑖=1
𝑛
𝑔 (𝑛) = 𝜙𝑞 (∑𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 ))) .
(17)
𝑖=1
Since 𝑔(𝑛) is increasing and 𝑔(0) = 0, we have 𝜉𝑢 ∈
(𝑔(0), 𝑔(𝑁 + 1)). There exists 𝑛1 ∈ Z(0, 𝑁) that satisfied
Thus,
𝑛1
𝑛1
𝑚=1
𝑖=𝑚
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) + 𝜔)
𝑛1
𝜙𝑞 (∑𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )))
𝑁+1
𝑚−1
𝑚=𝑛1 +1
𝑖=𝑛1 +1
(25)
= ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) − 𝜔) .
𝑖=1
(18)
𝑛1 +1
≤ 𝜉𝑢 < 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 ))) .
𝑖=1
Therefore, if 𝑢(𝑛) is a positive solution of the boundary
value problems (8) and (9), then it can be expressed by
𝑢 (𝑛)
Hence,
𝑛1
𝜉𝑢 = 𝜙𝑞 (∑𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) + 𝜔) ,
(19)
𝑖=1
where 𝜔 ∈ [0, 𝜆𝑎(𝑛1 + 1)𝑓(𝑛1 + 1, 𝑢(𝑛1 + 1 − 𝑛0 ))).
By (13) and (19), together with the definition of the positive solution, it is easy to get
𝑛
𝑛1
𝑚=1
𝑖=𝑚
𝑢 (𝑛) = ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) + 𝜔) ,
𝑛 < 𝑛1 + 1.
(20)
Similarly, summing up (12) with respect to 𝑚 from 𝑛 + 1
to 𝑁 + 1 and noticing (9), we have
𝑁+1
𝑢 (𝑛) = − ∑ 𝜙𝑞 (𝜙𝑝 (Δ𝑢 (0))
𝑚=𝑛+1
𝑚−1
− ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 ))) ,
𝑖=1
𝑛 > 𝑛1 − 1.
(21)
𝑛
𝑛1
{
{
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) + 𝜔) ,
{
{
{
𝑚=1
𝑖=𝑚
{
{
{
{
{
{
𝑛 ∈ Z (1, 𝑛1 ) ,
{
{
{
{
{
𝑁+1
𝑚−1
{
= { ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) − 𝜔) ,
{
{
𝑚=𝑛+1
𝑖=𝑛1 +1
{
{
{
{
{
{
𝑛 ∈ Z (𝑛1 + 1, 𝑁) ,
{
{
{
{
{
{
{
{ 0,
𝑛 ∈ Z (−𝑛0 , 0) , 𝑛 = 𝑁 + 1,
{
(26)
where 𝜔 ∈ [0, 𝜆𝑎(𝑛1 + 1)𝑓(𝑛1 + 1, 𝑢(𝑛1 + 1 − 𝑛0 ))).
Let 𝐶(Z(−𝑛0 , 𝑁 + 1), R) denote the class of maps 𝑢 continuous on Z(−𝑛0 , 𝑁 + 1) (discrete topology). We introduce a
function space 𝐸, namely,
𝐸 = {𝑢 ∈ 𝐶 (Z (−𝑛0 , 𝑁 + 1) , R) , 𝑢 (𝑁 + 1)
= 0; 𝑢 (𝑛) = 0, 𝑛 ∈ Z (−𝑛0 , 0)}
(27)
4
Journal of Applied Mathematics
with norm ‖ ⋅ ‖ given by ‖𝑢‖ = max𝑛∈Z(−𝑛0 ,𝑁+1) |𝑢(𝑛)|. Then
(𝐸, ‖ ⋅ ‖) is a Banach space. Define a cone 𝐾 ⊂ 𝐸 as follows:
𝐾 = {𝑢 ∈ 𝐸 : 𝑢 (𝑛) ≥ 𝜇 (𝑛) ‖𝑢‖ , 𝑛 ∈ Z (0, 𝑁 + 1)} ,
For convenience, we introduce some notations:
(28)
𝑓∞ = lim sup max
𝑢 → ∞ 𝑛∈Z(1,𝑁+1)
(29)
𝑛1
{
{
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) + 𝜔) ,
{
{
{
𝑚=1
𝑖=𝑚
{
{
{
{
{
{
𝑛 ∈ Z (1, 𝑛1 ) ,
{
{
{
{
{
𝑚−1
{ 𝑁+1
= { ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) − 𝜔) ,
{
{
𝑚=𝑛+1
𝑖=𝑛1 +1
{
{
{
{
{
{
𝑛 ∈ Z (𝑛1 + 1, 𝑁) ,
{
{
{
{
{
{
{
{ 0,
𝑛 ∈ Z (−𝑛0 , 0) , 𝑛 = 𝑁 + 1,
{
(30)
where 𝜔 ∈ [0, 𝜆𝑎(𝑛1 + 1)𝑓(𝑛1 + 1, 𝑢(𝑛1 + 1 − 𝑛0 ))).
Lemma 5. Suppose that (𝐻1) and (𝐻2) hold. Then the operator 𝑇 : 𝐾 → 𝐾 is completely continuous.
𝑓 (𝑛, 𝑢)
.
𝑢𝑝−1
Theorem 6. Suppose that (A1) and (A2) hold. Assume that
the delay time is appropriately small, say 𝑛0 is a positive integer
not bigger than [𝑁/2]. If 𝑓∞ > 0, 𝑓0 < ∞, then there exists at
least one positive solution to the boundary value problems (8)
and (9) for 𝜆 ∈ (max{𝑎1 , 𝑎2 }, 𝑏), where
𝑎1 =
1
𝑓∞ ⋅
(∑[𝑁/2]
𝑚=2
𝜙𝑞 (∑[𝑁/2]
𝑖=𝑚
𝑝−1
𝑎 (𝑖 + 𝑛0 ) (𝜇 (𝑖))
))
𝑝−1
,
(33)
𝑎2
=
1
𝑓∞ ⋅
𝑁+1−𝑛0
(∑𝑚=2+[𝑁/2]
𝜙𝑞 (∑𝑚−1
𝑖=1+[𝑁/2]
𝑝−1
𝑎 (𝑖 + 𝑛0 ) ((𝜇 (𝑖))
𝑝−1
))
,
(34)
𝑏=
1
𝑓0
𝑝−1
⋅ ((𝑁 + 1) 𝜙𝑞 (∑𝑁+1
𝑖=1 𝑎 (𝑖)))
;
(35)
while if 𝑓0 > 0, 𝑓∞ < ∞, then for any 𝜆 ∈ (max{𝑐1 , 𝑐2 }, 𝑑),
where
1
,
𝑐1 =
𝑝−1 𝑝−1
[𝑁/2]
[𝑁/2]
𝑓0 ⋅ (∑𝑚=2 𝜙𝑞 (∑𝑖=𝑚 𝑎 (𝑖 + 𝑛0 ) (𝜇 (𝑖)) ))
𝑐2
Proof. For any 𝑢 ∈ 𝐾, it is easily to check
=
Δ (𝜙𝑝 (Δ (𝑇𝑢) (𝑛 − 1)))
− 𝜙𝑝 ((𝑇𝑢) (𝑛) − (𝑇𝑢) (𝑛 − 1))
(32)
If 𝑢 is a positive solution of problems (8) and (9), then it
is a fixed point of 𝑇. So the existence of positive solution is
transformed into the existence of fixed point. The following
theorem is the main result of this paper.
(𝑇𝑢) (𝑛)
= 𝜙𝑝 ((𝑇𝑢) (𝑛 + 1) − (𝑇𝑢) (𝑛))
𝑓 (𝑛, 𝑢)
,
𝑢𝑝−1
𝑛∈Z(1,𝑁+1)
𝑓 (𝑛, 𝑢)
𝑓0 = lim inf
min
,
+
𝑢 → 0 𝑛∈Z(1,𝑁+1) 𝑢𝑝−1
Moreover, for any 𝑢 ∈ 𝐾, we have 𝑢(𝑛) ≥ (1/(𝑁 + 1))‖𝑢‖,
𝑛 ∈ Z(1, 𝑁).
For any 𝑢 ∈ 𝐾, define an operator 𝑇 on 𝐾 by
𝑛
𝑓0 = lim sup max
𝑢 → 0+
𝑁+1−𝑛 𝑛
, }
𝑁+1 𝑁
𝑁
𝑛
{
,
𝑛 ∈ Z (1, [ ]) ,
{
{𝑁
2
={
{𝑁 + 1 − 𝑛
𝑁
{
, 𝑛 ∈ Z ([ ] + 1, 𝑁) .
{ 𝑁+1
2
𝑓 (𝑛, 𝑢)
,
𝑢𝑝−1
𝑢 → ∞ 𝑛∈Z(1,𝑁+1)
where 𝜇(𝑛) = min{(𝑁 + 1 − 𝑛)/(𝑁 + 1), 𝑛/𝑁}.
From the definition of 𝜇(𝑛), we have
𝜇 (𝑛) = min {
min
𝑓∞ = lim inf
(31)
= −𝜆𝑎 (𝑛) 𝑓 (𝑛, 𝑢 (𝑛 − 𝑛0 )) ≤ 0.
According to Lemma 3.2 in [1], we have (𝑇𝑢)(𝑛) ≥
𝜇(𝑛)‖𝑇𝑢‖. Thus, 𝑇(𝐾) ⊂ 𝐾. Moreover, 𝜙𝑝 and 𝑓 are both continuous, and 𝐸 is a finite space. This implies that the operator
𝑇 is a completely continuous. This completes the proof.
1
𝑓0 ⋅
𝑑=
𝑁+1−𝑛0
(∑𝑚=2+[𝑁/2]
𝜙𝑞 (∑𝑚−1
𝑖=1+[𝑁/2]
𝑝−1
𝑎 (𝑖 + 𝑛0 ) (𝜇 (𝑖))
1
𝑓∞
⋅ ((𝑁 + 1) 𝜙𝑞 (∑𝑁+1
𝑖=1 𝑎 (𝑖)))
𝑝−1
𝑝−1
))
,
,
(36)
problems (8) and (9) admit at least one positive solution 𝑢 ∈ 𝐾.
Proof. From Lemma 5, we see that 𝑇 : 𝐾 → 𝐾 is a completely continuous operator. Since 𝜆 ∈ (max{𝑎1 , 𝑎2 }, 𝑏), there
exists an 𝜖 > 0 such that
max {𝑎1𝜖 , 𝑎2𝜖 } ≤ 𝜆 ≤ 𝑏𝜖 ,
(37)
Journal of Applied Mathematics
5
where
1/(𝑝−1)
≤ max {(𝜆 (𝑓0 + 𝜖))
𝑎1𝜖
=
1
(𝑓∞ − 𝜖) ⋅
(∑[𝑁/2]
𝑚=2
𝜙𝑞 (∑[𝑁/2]
𝑖=𝑚
𝑝−1
𝑎 (𝑖 + 𝑛0 ) (𝜇 (𝑖))
))
𝑝−1
,
𝑎2𝜖
=
(𝜆 (𝑓0 + 𝜖))
≤ (𝜆 (𝑓0 + 𝜖))
1
(𝑓∞ − 𝜖) ⋅
𝜙𝑞 (∑𝑚−1
𝑖=1+[𝑁/2]
𝑎 (𝑖 + 𝑛0 ) (𝜇 (𝑖))
𝑝−1
1
𝑏𝜖 =
(𝑓0
𝑝−1
+ 𝜖) ⋅ ((𝑁 + 1) 𝜙𝑞 (∑𝑁+1
𝑖=1 𝑎 (𝑖)))
))
𝑝−1
(38)
Let 𝜖 be fixed. By 𝑓0 < ∞, there exists an 𝑀1 > 0 such
that for 0 ≤ 𝑢 ≤ 𝑀1 ,
0
𝑓 (𝑖, 𝑢) ≤ (𝑓 + 𝜖) 𝑢
𝑝−1
,
𝑖 ∈ Z (1, 𝑁 + 1) .
(39)
𝑝−1
,
𝑖 ∈ Z (1, 𝑁+1) .
(40)
In view of Lemma 1 and the definition of operator 𝑇, we
further have
‖𝑇𝑢‖ = max {(𝑇𝑢) (𝑛1 ) , (𝑇𝑢) (𝑛1 + 1)} .
(41)
𝑚=1
𝑖=𝑚
𝑁+1
𝑚−1
𝑚=𝑛1 +2
𝑖=𝑛1 +1
‖𝑢‖ ∑ 𝜙𝑞 ( ∑ 𝑎 (𝑖))}
‖𝑢‖ (𝑁 + 1) 𝜙𝑞 ( ∑ 𝑎 (𝑖)) ≤ ‖𝑢‖ .
𝑖=1
(42)
̃2 >
Next, by 𝑓∞ > 0, it is easy to see that there exists an 𝑀
̃
0 such that for all 𝑢 > 𝑀2 ,
𝑓 (𝑖, 𝑢) ≥ (𝑓∞ − 𝜖) 𝑢𝑝−1 ,
𝑖 ∈ Z (1, 𝑁 + 1) .
(43)
̃2 }. Let Ω2 = {𝑢 ∈ 𝐸, ‖𝑢‖ <
Take 𝑀2 = max{2𝑀1 , (𝑁+1)𝑀
𝑀2 }. For any 𝑢 ∈ 𝐾 ∩ 𝜕Ω2 , we have
Let Ω1 = {𝑢 ∈ 𝐸 : ‖𝑢‖ < 𝑀1 }. Then for any 𝑢 ∈ 𝐾 ∩ 𝜕Ω1 ,
we have
𝑓 (𝑖, 𝑢 (𝑖−𝑛0 )) ≤ (𝑓0 +𝜖) (𝑢 (𝑖−𝑛0 ))
1/(𝑝−1)
,
.
𝑛1 +1
𝑁+1
1/(𝑝−1)
𝑁+1−𝑛0
(∑𝑚=2+[𝑁/2]
𝑛1
‖𝑢‖ ∑ 𝜙𝑞 ( ∑ 𝑎 (𝑖)) ,
𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) ≥ (𝑓∞ − 𝜖) (𝑢 (𝑖 − 𝑛0 ))
𝑝−1
,
(44)
∀𝑖 ∈ Z (𝑛0 + 2, 𝑁 + 1) .
Thus, if 𝑛1 < 𝑛0 + [𝑁/2], we have
‖𝑇𝑢‖
= max {(𝑇𝑢) (𝑛1 ) , (𝑇𝑢) (𝑛1 + 1)} ≥ (𝑇𝑢) (𝑛1 + 1)
𝑁+1
𝑚−1
𝑚=𝑛1 +2
𝑖=𝑛1 +1
= ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) − 𝜔)
Thus,
‖𝑇𝑢‖
= max {(𝑇𝑢) (𝑛1 ) , (𝑇𝑢) (𝑛1 + 1)}
𝑛1
𝑛1
𝑚=1
𝑖=𝑚
𝑛1
𝑖=𝑛1 +1
≥
≥ ‖𝑢‖
𝑖=𝑛1 +1
𝑛1 +1
𝑚=1
𝑖=𝑚
𝑝−1
𝑚−1
0
𝑖=𝑛1 +1
𝑚−1
∑
𝑖=[𝑁/2]+1
𝑁+1−𝑛0
∑
𝜙𝑞 (
)}
×
𝑁+1−𝑛0
∑
𝑚=[𝑁/2]+2
𝜙𝑞 (
)
𝜆𝑎 (𝑖 + 𝑛0 ) (𝑓∞ − 𝜖) (𝑢 (𝑖))𝑝−1 )
𝑚−1
𝑝−1
∑
𝑖=[𝑁/2]+1
1/(𝑝−1)
≥ (𝜆 (𝑓∞ − 𝜖))
),
𝑝−1
𝜆𝑎 (𝑖) (𝑓∞ − 𝜖) (𝑢 (𝑖 − 𝑛0 ))
𝜙𝑞 (
𝑚=[𝑁/2]+2
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) (𝑓 + 𝜖) (𝑢 (𝑖 − 𝑛0 ))
𝑚=𝑛1 +2
∑
𝑚−1
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )))}
𝑁+1
𝑁+1−𝑛0
𝑚=[𝑁/2]+2
≤ max { ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) (𝑓0 + 𝜖) (𝑢 (𝑖 − 𝑛0 ))
𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )))
𝑝−1
∑
𝑖=𝑛0 +[𝑁/2]+1
𝑖=𝑚
𝑛1
𝑚−1
×(
𝑛1 +1
𝑚=𝑛1 +2
∑
𝑖=𝑛0 +[𝑁/2]+1
𝜙𝑞
∑
𝑚−1
𝑁+1
𝑚−1
𝑚=𝑛0 +[𝑁/2]+2
≤ max { ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 ))) ,
𝑚=1
𝑁+1
≥
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) − 𝜔)}
𝑚=𝑛1 +2
𝜙𝑞 (
∑
𝑚=𝑛0 +[𝑁/2]+2
= max { ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) + 𝜔) ,
𝑁+1
𝑁+1
≥
𝜆𝑎 (𝑖 + 𝑛0 ) (𝑓∞ − 𝜖) (𝜇 (𝑖))
)
‖𝑢‖
𝑚−1
∑
𝑖=[𝑁/2]+1
𝑝−1
𝑎 (𝑖 + 𝑛0 ) (𝜇 (𝑖))
) ≥ ‖𝑢‖ ;
(45)
6
Journal of Applied Mathematics
while if 𝑛1 ≥ 𝑛0 + [𝑁/2], then we have
If 𝑛1 < 𝑛0 + [𝑁/2], we have
‖𝑇𝑢‖
‖𝑇𝑢‖ = max {(𝑇𝑢) (𝑛1 ) , (𝑇𝑢) (𝑛1 + 1)}
= max {(𝑇𝑢) (𝑛1 ) , (𝑇𝑢) (𝑛1 + 1)}
≥ (𝑇𝑢) (𝑛1 )
𝑛1
𝑛1
𝑚=1
𝑖=𝑚
≥ (𝑇𝑢) (𝑛1 + 1)
= ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) + 𝜔)
≥
𝑛0 +[𝑁/2]
𝑛0 +[𝑁/2]
𝑚=𝑛0 +2
𝑖=𝑚
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )))
[𝑁/2]
[𝑁/2]
𝑚=2
𝑖=𝑚
(46)
≥ (𝜆 (𝑓∞ − 𝜖))
[𝑁/2]
‖𝑢‖
𝑁+1
∑
𝑁+1
∑
𝑚=𝑛0 +[𝑁/2]+2
[𝑁/2]
𝑝−1
) ≥ ‖𝑢‖ .
𝑖=𝑚
𝑚−1
𝜙𝑞 (
𝑚=[𝑁/2]+2
𝑚−1
𝜙𝑞 (
𝜆𝑎 (𝑖 + 𝑛0 ) (𝑓0 − 𝜖) (𝜇 (𝑖))
∑
𝑝−1
)
𝑖=[𝑁/2]+1
1/(𝑝−1)
∑
)
𝑖=𝑛0 +[𝑁/2]+1
∑
𝑁+1−𝑛0
𝑝−1
𝜆𝑎 (𝑖) (𝑓0 − 𝜖) (𝑢 (𝑖 − 𝑛0 ))
∑
≥ (𝜆 (𝑓0 − 𝜖))
×
𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )))
∑
𝑖=𝑛0 +[𝑁/2]+1
𝑚=[𝑁/2]+2
(47)
𝑚−1
𝜙𝑞 (
𝑁+1−𝑛0
≥ ‖𝑢‖
max {𝑐1𝜖 , 𝑐2𝜖 } ≤ 𝜆 ≤ 𝑑𝜖 ,
𝜙𝑞 (
‖𝑢‖
𝑚−1
∑
𝑖=[𝑁/2]+1
𝑎 (𝑖 + 𝑛0 ) (𝜇 (𝑖))
𝑝−1
) ≥ ‖𝑢‖ .
(51)
Conversely, if 𝑛1 ≥ 𝑛0 + [𝑁/2], then we have
‖𝑇𝑢‖
where
= max {(𝑇𝑢) (𝑛1 ) , (𝑇𝑢) (𝑛1 + 1)}
1
(𝑓0 − 𝜖) ⋅
(∑[𝑁/2]
𝑚=2
𝜙𝑞 (∑[𝑁/2]
𝑖=𝑚
𝑝−1
𝑎 (𝑖 + 𝑛0 ) (𝜇 (𝑖))
𝑝−1
))
,
≥ (𝑇𝑢) (𝑛1 )
𝑛1
𝑛1
𝑚=1
𝑖=𝑚
= ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) + 𝜔)
𝑐2𝜖
=
𝑖=𝑛1 +1
𝑚=𝑛0 +[𝑁/2]+2
Combining the two estimates above, we deduce ‖𝑇𝑢‖ ≥
‖𝑢‖, when 𝑢 ∈ 𝐾∩𝜕Ω1 . Therefore, by Theorem 3, 𝑇 has a fixed
point 𝑢 ∈ 𝐾 ∩ (Ω2 \ Ω1 ). Namely, 𝑢(𝑡) is a positive solution of
problems (8) and (9).
The proof of the second half is parallel to above. There
exists an 𝜖 > 0 such that
𝑐1𝜖 =
𝑚=𝑛1 +2
≥
× ∑ 𝜙𝑞 ( ∑ 𝑎(𝑖 + 𝑛0 ) (𝜇 (𝑖))
𝑚=2
𝑚−1
≥
≥ ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖 + 𝑛0 ) (𝑓∞ − 𝜖) (𝑢 (𝑖))𝑝−1 )
1/(𝑝−1)
𝑁+1
= ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) − 𝜔)
1
,
𝑝−1 𝑝−1
𝑁+1−𝑛0
𝑚−1
(𝑓0 − 𝜖) ⋅ (∑𝑚=2+[𝑁/2]
𝜙𝑞 (∑𝑖=1+[𝑁/2] 𝑎 (𝑖 + 𝑛0 ) (𝜇 (𝑖)) ))
𝑑𝜖 =
1
(𝑓∞ + 𝜖) ⋅ ((𝑁 +
1) 𝜙𝑞 (∑𝑁+1
𝑖=1
𝑝−1
𝑎 (𝑖)))
.
(48)
Let 𝜖 be fixed. By 𝑓0 > 0, then it is not difficult to choose
an 𝑀1 > 0 such that 0 ≤ 𝑢 ≤ 𝑀1 ,
𝑓 (𝑖, 𝑢) ≥ (𝑓0 − 𝜖) 𝑢𝑝−1 ,
𝑖 ∈ Z (1, 𝑁 + 1) .
(49)
Let Ω1 = {𝑢 ∈ 𝐸, ‖𝑢‖ < 𝑀1 }. Then for 𝑢 ∈ 𝐾 ∩ 𝜕Ω1 , it can
be obtained that
𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) ≥ (𝑓0 − 𝜖) (𝑢 (𝑖 − 𝑛0 ))
𝑝−1
,
≥
𝑖 ∈ Z (1, 𝑁 + 1) .
(50)
𝑛0 +[𝑁/2]
𝑛0 +[𝑁/2]
𝑚=𝑛0 +2
𝑖=𝑚
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )))
[𝑁/2]
[𝑁/2]
𝑚=2
𝑖=𝑚
≥ ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖 + 𝑛0 ) (𝑓0 − 𝜖) (𝑢 (𝑖))𝑝−1 )
1/(𝑝−1)
≥ (𝜆 (𝑓0 − 𝜖))
[𝑁/2]
[𝑁/2]
𝑚=2
𝑖=𝑚
𝑝−1
‖𝑢‖ ∑ 𝜙𝑞 ( ∑ 𝑎 (𝑖 + 𝑛0 ) (𝜇 (𝑖))
)
≥ ‖𝑢‖ .
(52)
̃2
𝑀
̃2 > 0 such that for 𝑢 >
Next, by 𝑓∞ < ∞, there exists 𝑀
𝑓 (𝑖, 𝑢) ≤ (𝑓∞ + 𝜖) 𝑢𝑝−1 ,
𝑖 ∈ Z (1, 𝑁 + 1) .
(53)
Journal of Applied Mathematics
7
If 𝑓 is bounded, then there exists an 𝑁1 > 0 such that
𝑓(𝑖, 𝑢) ≤ 𝑁1 for 𝑖 ∈ Z(1, 𝑁 + 1), 0 ≤ 𝑢 < ∞. Take 𝑀2 =
max{2𝑀1 , (𝜆𝑁1 )1/(𝑝−1) (𝑁 + 1) 𝜙𝑞 (∑𝑁+1
𝑖=1 𝑎(𝑖))} and Ω2 = {𝑢 ∈
𝐸, ‖𝑢‖ < 𝑀2 }. Then for any 𝑢 ∈ 𝐾 ∩ 𝜕Ω2 , we have
‖𝑇𝑢‖
𝑛1 +1
𝑚=1
𝑖=𝑚
𝑁+1
𝑚−1
𝑚=𝑛1 +2
𝑖=𝑛1 +1
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖0 , 𝑀2 ))}
1/(𝑝−1)
≤ max {(𝜆 (𝑓∞ + 𝜖))
= max {(𝑇𝑢) (𝑛1 ) , (𝑇𝑢) (𝑛1 + 1)}
𝑛1
𝑛1
≤ max { ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖0 , 𝑀2 )) ,
𝑛1
𝑛1 +1
𝑚=1
𝑖=𝑚
𝑀2 ∑ 𝜙𝑞 ( ∑ 𝑎 (𝑖)) ,
𝑛1
= max { ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) + 𝜔) ,
𝑚=1
1/(𝑝−1)
(𝜆 (𝑓∞ + 𝜖))
𝑖=𝑚
𝑁+1
𝑁+1
𝑚−1
𝑚=𝑛1 +2
𝑖=𝑛1 +1
𝑀2 ∑ 𝜙𝑞 ( ∑ 𝑎 (𝑖))}
𝑚−1
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) − 𝜔)}
𝑚=𝑛1 +2
𝑖=𝑛1 +1
1/(𝑝−1)
≤ (𝑁 + 1) (𝜆 (𝑓∞ + 𝜖))
𝑁+1
𝑀2 𝜙𝑞 ( ∑ 𝑎 (𝑖)) ≤ 𝑀2 = ‖𝑢‖ .
𝑖=1
𝑛1
𝑛1 +1
(55)
𝑚=1
𝑖=𝑚
Therefore, by Theorem 3, 𝑇 has a fixed point 𝑢 ∈ 𝐾∩(Ω2 \
Ω1 ). Namely, 𝑢 is a positive solution of problema (8) and (9).
This completes the proof.
≤ max { ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 ))) ,
𝑁+1
𝑚−1
𝑚=𝑛1 +2
𝑖=𝑛1 +1
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )))}
𝑛1
𝑛1 +1
𝑚=1
𝑖=𝑚
4. Numerical Illustration
≤ max { ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑁1 ) ,
𝑁+1
𝑚−1
𝑚=𝑛1 +2
𝑖=𝑛1 +1
In this section, we present two numerical experiments to
illustrate our results.
Example 7. Consider the following boundary value problem:
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑁1 )}
1/(𝑝−1)
≤ (𝑁 + 1) (𝜆𝑁1 )
𝑢 (𝑛 + 1) − 2𝑢 (𝑛) + 𝑢 (𝑛 − 1) + 𝑎 (𝑛) 𝑓 (𝑛, 𝑢 (𝑛 − 5)) = 0,
𝑁+1
𝑛 ∈ Z (1, 99) ,
𝜙𝑞 ( ∑ 𝑎 (𝑖)) ≤ 𝑀2 = ‖𝑢‖ .
𝑖=1
(54)
While if 𝑓 is unbounded, it is not difficult to see that there
̃2 }, 𝑖0 ∈ Z(1, 𝑁) such that 0 ≤ 𝑢 ≤
exists an 𝑀2 > max{2𝑀1 , 𝑀
𝑀2 , 𝑓(𝑖, 𝑢) ≤ 𝑓(𝑖0 , 𝑀2 ). Take Ω2 = {𝑢 ∈ 𝐸 : ‖𝑢‖ < 𝑀2 }, then
for 𝑢 ∈ 𝐾 ∩ 𝜕Ω2 , we get
𝑢 (100) = 0,
𝑢 (𝑛) = 0,
𝑛 ∈ Z (−5, 0) ,
(56)
where 𝑎(𝑛) = 𝑛/100, 𝑓(𝑛, 𝑢(𝑛 − 5)) = (1/100)2 (𝑢(𝑛 − 5))2 .
It is easy to see that 𝑓∞ = +∞, 𝑓0 = 0. By Theorem 6,
problem (56) has a positive solution. In Figure 1, we show the
numerical solution to problem (56).
Example 8. Consider the following boundary value problem
‖𝑇𝑢‖
= max {(𝑇𝑢) (𝑛1 ) , (𝑇𝑢) (𝑛1 + 1)}
𝑛1
𝑛1
𝑚=1
𝑖=𝑚
= max { ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) + 𝜔) ,
𝑁+1
𝑚−1
𝑚=𝑛1 +2
𝑖=𝑛1 +1
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )) − 𝜔)}
𝑛1
𝑛1 +1
𝑚=1
𝑖=𝑚
≤ max { ∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 ))) ,
𝑁+1
𝑚−1
𝑚=𝑛1 +2
𝑖=𝑛1 +1
∑ 𝜙𝑞 ( ∑ 𝜆𝑎 (𝑖) 𝑓 (𝑖, 𝑢 (𝑖 − 𝑛0 )))}
Δ𝜙4 (Δ𝑢 (𝑛 − 1)) + 𝑎 (𝑛) 𝑓 (𝑛, 𝑢 (𝑛 − 3)) = 0,
𝑢 (100) = 0,
𝑢 (𝑛) = 0,
𝑛 ∈ Z (1, 99) ,
𝑛 ∈ Z (−3, 0) ,
(57)
where 𝑎(𝑛) ≡ 1, 𝑓(𝑛, 𝑢(𝑛 − 3)) = (1/100)4 (𝑢(𝑛 − 3))2 . We
check that 𝑓∞ = 0, 𝑓0 = +∞. Then applying Theorem 6,
problem (57) has a positive solution. Numerically, we obtain
the solution of problem (57) which is shown in Figure 1.
Acknowledgments
This work is supported by Tian Yuan Fund of China
(11226308), NNSF of China (11071102), Natural Science
Fund for Colleges and Universities in Jiangsu Province
(11KJD110001), the Research Fund (10JDG124) for Highlevel
8
Journal of Applied Mathematics
25
[5]
20
15
[6]
u
10
[7]
5
0
−20
0
20
40
n
60
80
100
[8]
(a) The numerical solution to problem (56)
[9]
0.018
0.016
0.014
[10]
0.012
u
0.01
[11]
0.008
0.006
[12]
0.004
0.002
[13]
0
−20
0
20
40
n
60
80
100
(b) The numerical solution to problem (57)
[14]
Figure 1: The figures of the numerical solutions.
[15]
Group of Jiangsu University, and Postdoctoral Science Foundation (2011M500874, 1002030C).
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