Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 165382, 28 pages
http://dx.doi.org/10.1155/2013/165382
Research Article
Existence and Iterative Algorithms of Positive Solutions for
a Higher Order Nonlinear Neutral Delay Differential Equation
Zeqing Liu,1 Ling Guan,1 Sunhong Lee,2 and Shin Min Kang2
1
2
Department of Mathematics, Liaoning Normal University, Dalian, Liaoning 116029, China
Department of Mathematics and RINS, Gyeongsang National University, Jinju 660-701, Republic of Korea
Correspondence should be addressed to Sunhong Lee; sunhong@gnu.ac.kr and Shin Min Kang, smkang@gnu.ac.kr
Received 23 October 2012; Accepted 10 December 2012
Academic Editor: Jifeng Chu
Copyright © 2013 Zeqing Liu et al. 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.
This paper is concerned with the higher order nonlinear neutral delay differential equation [𝑎(𝑡)(𝑥(𝑡) + 𝑏(𝑡)𝑥(𝑡 − 𝜏))(𝑚) ](𝑛−𝑚) +
[ℎ(𝑡, 𝑥(ℎ1 (𝑡)), . . . , 𝑥(ℎ𝑙 (𝑡)))](𝑖) + 𝑓(𝑡, 𝑥(𝑓1 (𝑡)), . . . , 𝑥(𝑓𝑙 (𝑡))) = 𝑔(𝑡), for all 𝑡 ≥ 𝑡0 . Using the Banach fixed point theorem, we establish
the existence results of uncountably many positive solutions for the equation, construct Mann iterative sequences for approximating
these positive solutions, and discuss error estimates between the approximate solutions and the positive solutions. Nine examples
are included to dwell upon the importance and advantages of our results.
1. Introduction and Preliminaries
In recent years, the existence problems of nonoscillatory solutions for neutral delay differential equations of first, second,
third, and higher order have been studied intensively by using
fixed point theorems; see, for example, [1–12] and the references therein.
Using the Banach, Schauder, and Krasnoselskii fixed
point theorems, Zhang et al. [9] and Liu et al. [7] considered
the existence of nonoscillatory solutions for the following first
order neutral delay differential equations:
󸀠
[𝑥 (𝑡) + 𝑐𝑥 (𝑡 − 𝜏)]󸀠󸀠 + 𝑄1 (𝑡) 𝑥 (𝑡 − 𝜎1 )
− 𝑄2 (𝑡) 𝑥 (𝑡 − 𝜎2 ) = 0,
[𝑟 (𝑡) (𝑥 (𝑡) + 𝑃 (𝑡) 𝑥 (𝑡 − 𝜏))󸀠 ]
𝑚
∀𝑡 ≥ 𝑡0 ,
[𝑥 (𝑡) + 𝑐 (𝑡) 𝑥 (𝑡 − 𝜏)]󸀠
+ ℎ (𝑡) 𝑓 (𝑥 (𝑡 − 𝜎1 ) , 𝑥 (𝑡 − 𝜎2 ) , . . . , 𝑥 (𝑡 − 𝜎𝑘 )) = 𝑔 (𝑡) ,
∀𝑡 ≥ 𝑡0 ,
(1)
where 𝑃 ∈ 𝐶([𝑡0 , +∞), R \ {±1}) and 𝑐 ∈ 𝐶([𝑡0 , +∞), R).
Making use of the Banach and Krasnoselskii fixed point
theorems, Kulenović and Hadžiomerspahić [2] and Zhou [10]
∀𝑡 ≥ 𝑡0 ,
󸀠
+ ∑𝑄𝑖 (𝑡) 𝑓𝑖 (𝑥 (𝑡 − 𝜎𝑖 )) = 0,
𝑖=1
[𝑥 (𝑡) + 𝑃 (𝑡) 𝑥 (𝑡 − 𝜏)] + 𝑄1 (𝑡) 𝑥 (𝑡 − 𝜏1 )
− 𝑄2 (𝑡) 𝑥 (𝑡 − 𝜏2 ) = 0,
studied the existence of a nonoscillatory solution for the
following second order neutral differential equations:
(2)
∀𝑡 ≥ 𝑡0 ,
where 𝑐 ∈ R \ {±1} and 𝑃 ∈ 𝐶([𝑡0 , ∞), R). Zhou and
Zhang [11], Zhou et al. [12], and Liu et al. [4], respectively,
investigated the existence of nonoscillatory solutions for the
following higher order neutral delay differential equations:
[𝑥 (𝑡) + 𝑐𝑥 (𝑡 − 𝜏)](𝑛)
+ (−1)𝑛+1 [𝑃 (𝑡) 𝑥 (𝑡 − 𝜎) − 𝑄 (𝑡) 𝑥 (𝑡 − 𝛿)] = 0,
∀𝑡 ≥ 𝑡0 ,
2
Abstract and Applied Analysis
[𝑥 (𝑡) + 𝑃 (𝑡) 𝑥 (𝑡 − 𝜏)](𝑛)
𝑚
+ ∑𝑄𝑖 (𝑡) 𝑓𝑖 (𝑥 (𝑡 − 𝜎𝑖 )) = 𝑔 (𝑡) ,
CB([𝛾, +∞), R) stands for the Banach space of all continuous and bounded functions in [𝛾, +∞) with norm ‖𝑥‖ =
sup𝑡≥𝛾 |𝑥(𝑡)|, and for any 𝑀 > 𝑁 > 0
∀𝑡 ≥ 𝑡0 ,
𝑖=1
[𝑥 (𝑡) + 𝑎𝑥 (𝑡 − 𝜏)](𝑛)
+ (−1)
= 𝑔 (𝑡) ,
𝑛+1
Ω1 (𝑁, 𝑀) = {𝑥 ∈ CB ([𝛾, +∞) , R) :
𝑓 (𝑡, 𝑥 (𝑡 − 𝜎1 ) , 𝑥 (𝑡 − 𝜎2 ) , . . . , 𝑥 (𝑡 − 𝜎𝑘 ))
∀𝑡 ≥ 𝑡0 ,
(3)
𝑁 ≤ 𝑥 (𝑡) ≤ 𝑀, ∀𝑡 ≥ 𝛾} ,
Ω2 (𝑁, 𝑀) = {𝑥 ∈ CB ([𝛾, +∞) , R) :
where 𝑐 ∈ R \ {±1}, 𝑃 ∈ 𝐶([𝑡0 , ∞), R) and 𝑎 ∈ R \ {−1}.
Candan [1] proved the existence of a bounded nonoscillatory
solution for the higher order nonlinear neutral differential
equation:
[𝑟 (𝑡) (𝑥 (𝑡) + 𝑃 (𝑡) 𝑥 (𝑡 − 𝜏))(𝑛−1) ]
≤ 𝑥 (𝑡) ≤
≤−
∀𝑡 ≥ 𝑡0 ,
(4)
(𝑛−𝑚)
= +[ℎ (𝑡, 𝑥 (ℎ1 (𝑡)) , . . . , 𝑥 (ℎ𝑙 (𝑡)))]
(𝑖)
= +𝑓 (𝑡, 𝑥 (𝑓1 (𝑡)) , . . . , 𝑥 (𝑓𝑙 (𝑡))) = 𝑔 (𝑡) ,
∀𝑡 ≥ 𝑡0 ,
(5)
where 𝑚, 𝑛 ∈ N and 𝑖 ∈ N0 with 𝑖 ≤ 𝑛 − 𝑚 − 1, 𝜏 > 0,
𝑎 ∈ 𝐶([𝑡0 , +∞), R \ {0}), 𝑏, 𝑔, 𝑓𝑗 , ℎ𝑗 ∈ 𝐶([𝑡0 , +∞), R), ℎ ∈
𝐶𝑖 ([𝑡0 , +∞) × R𝑙 , R) and 𝑓 ∈ 𝐶([𝑡0 , +∞) × R𝑙 , R) with
lim ℎ𝑗 (𝑡) = lim 𝑓𝑗 (𝑡) = +∞,
𝑡 → +∞
𝑗 ∈ {1, 2, . . . , 𝑙} .
(6)
It is clear that (5) includes (1)–(4) as special cases. Utilizing the Banach fixed point theorem, we prove several
existence results of uncountably many positive solutions for
(5), construct a few Mann iterative schemes, and discuss
error estimates between the sequences generated by the Mann
iterative schemes and the positive solutions. Nine examples
are given to show that the results presented in this paper
extend substantially the existing ones in [1, 2, 4, 5, 8, 9, 11].
Throughout this paper, we assume that R = (−∞, +∞),
R+ = [0, +∞), N denotes the set of all positive integers, N0 =
N ∪ {0},
1
,
(𝑚 − 1)! (𝑛 − 𝑚 − 𝑗 − 1)!
𝛾 = min {𝑡0 − 𝜏, inf ℎ𝑗 (𝑡) , inf 𝑓𝑗 (𝑡) : 𝑗 ∈ {1, 2, . . . , 𝑙}} ,
𝑡≥𝑡0
𝑡≥𝑡0
(7)
𝑀
, ∀𝑡 ∈ [𝛾, 𝑇)} .
𝑏 (𝑇 + 𝜏)
(8)
It is easy to check that Ω1 (𝑁, 𝑀), Ω2 (𝑁, 𝑀) and Ω3 (𝑁, 𝑀)
are closed subsets of CB([𝛾, +∞), R).
By a solution of (5), we mean a function 𝑥 ∈ 𝐶([𝛾, +∞),
R) for some 𝑇 > 1+|𝑡0 |+𝜏+|𝛾|, such that 𝑎(𝑡)(𝑥(𝑡)+𝑏(𝑡)𝑥(𝑡−
𝜏))(𝑚) are 𝑛 − 𝑚 times continuously differentiable in [𝑇, +∞)
and such that (5) is satisfied for 𝑡 ≥ 𝑇.
Lemma 1. Let 𝜏 > 0, 𝑐 ≥ 0, 𝐹 ∈ 𝐶([𝑐, +∞)3 , R+ ) and 𝐺 ∈
𝐶([𝑐, +∞)2 , R+ ). Then
+∞
(a) ∫𝑐
+∞
∫𝑟
+∞
∫𝑢
+∞
+∞
∑∞
𝑗=0 ∫𝑐+𝑗𝜏 ∫𝑟
+∞
(b) ∫𝑐
+∞
∫𝑢
+∞
+∞
(c) if ∫𝑐
+∞
∞
+∞
∫𝑟
+∞
∑∫
≤
𝑟𝐹(𝑠, 𝑢, 𝑟)𝑑𝑠 𝑑𝑢 𝑑𝑟 < + ∞ ⇔
+∞
∫𝑢
𝐹(𝑠, 𝑢, 𝑟)𝑑𝑠 𝑑𝑢 𝑑𝑟 < +∞;
𝑢𝐺(𝑠, 𝑢)𝑑𝑠 𝑑𝑢 < + ∞ ⇔
∑∞
𝑗=0 ∫𝑐+𝑗𝜏 ∫𝑢
𝑗=1 𝑡+𝑗𝜏
𝑗 ∈ {0, 𝑖} ,
𝑁
≤ 𝑥 (𝑡)
𝑏 (𝑡 + 𝜏)
𝑀
𝑁
, ∀𝑡 ≥ 𝑇; −
𝑏 (𝑡 + 𝜏)
𝑏 (𝑇 + 𝜏)
≤ 𝑥 (𝑡) ≤ −
where 𝑃 ∈ 𝐶([𝑡0 , ∞), R \ {±1}).
Motivated by the results in [1–12], in this paper we
consider the following higher order nonlinear neutral delay
differential equation:
𝐻𝑗 =
𝑀
, ∀𝑡 ∈ [𝛾, 𝑇)} ,
𝑏 (𝑇 + 𝜏)
Ω3 (𝑁, 𝑀) = {𝑥 ∈ CB ([𝛾, +∞) , R) : −
−𝑄2 (𝑡) 𝑔2 (𝑥 (𝑡 − 𝜎2 )) − 𝑓 (𝑡)] = 0,
𝑡 → +∞
𝑀
𝑁
, ∀𝑡 ≥ 𝑇;
𝑏 (𝑡 + 𝜏)
𝑏 (𝑇 + 𝜏)
≤
󸀠
+ (−1)𝑛 [𝑄1 (𝑡) 𝑔1 (𝑥 (𝑡 − 𝜎1 ))
[𝑎 (𝑡) (𝑥 (𝑡) + 𝑏 (𝑡) 𝑥 (𝑡 − 𝜏))(𝑚) ]
𝑁
≤ 𝑥 (𝑡)
𝑏 (𝑡 + 𝜏)
∫
𝐺(𝑠, 𝑢)𝑑𝑠 𝑑𝑢 < +∞;
+∞
∫𝑢
+∞
𝑟
∫
𝑟𝐹(𝑠, 𝑢, 𝑟)𝑑𝑠 𝑑𝑢 𝑑𝑟 < +∞, then
+∞
𝑢
𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
1 +∞ +∞ +∞
∫ ∫ ∫ 𝑟𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
𝜏 𝑡+𝜏 𝑟
𝑢
< +∞,
∀𝑡 ≥ 𝑐;
(9)
Abstract and Applied Analysis
+∞
+∞
(d) if ∫𝑐
∫𝑢
∞
∑∫
3
+∞
𝑢𝐺(𝑠, 𝑢)𝑑𝑠 𝑑𝑢 < +∞, then
+∞
𝑗=1 𝑡+𝑗𝜏
≤
∫
+∞
𝑢
𝐺 (𝑠, 𝑢) 𝑑𝑠 𝑑𝑢
∞
∑∫
+∞
+∞
1
∫ ∫
𝜏 𝑡+𝜏 𝑢
< +∞,
+∞
+∞
Assume that ∫𝑐 ∫𝑟 ∫𝑢 𝑟𝐹(𝑠, 𝑢, 𝑟)𝑑𝑠 𝑑𝑢 𝑑𝑟 < +∞. As
in the proof of (a), we infer that
𝑗=1 𝑡+𝑗𝜏
(10)
𝑢𝐺 (𝑠, 𝑢) 𝑑𝑠 𝑑𝑢
+∞
+∞
∫
𝑟
=∫
+∞
𝑢
+∞
𝑡+𝜏
∀𝑡 ≥ 𝑐.
∫
∫
+∞
𝑟
+∞
𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
+∞
∫
𝑢
+∞
[
𝑟−𝑡
] 𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
𝜏
+∞
Proof. Let [𝑡] denote the largest integral number not exceeding 𝑡 ∈ R. Note that
≤
1
∫ ∫
𝜏 𝑡+𝜏 𝑟
[(𝑟 − 𝑐) /𝜏] + 1 1
lim
= ,
𝑟 → +∞
𝑟
𝜏
𝑟−𝑐
𝑐 + 𝑛𝜏 ≤ 𝑟 < 𝑐 + (𝑛 + 1) 𝜏 ⇐⇒ 𝑛 ≤
< 𝑛 + 1,
𝜏
≤
1 +∞ +∞ +∞
∫ ∫ ∫ 𝑟𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
𝜏 𝑐
𝑟
𝑢
(11)
∀𝑛 ∈ N0 .
(12)
Clearly (12) means that
∞
∑∫
+∞
∫
+∞
𝑗=0 𝑐+𝑗𝜏 𝑟
=∫
+∞
∫
𝑢
+∞
+∞
∫
𝑐
𝑟
+∞
+∫
𝑐+𝜏
𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
∫
+∞
𝑢
∫
+∞
∫
𝑟
+∞
+∫
∫
+∫
∫
+∞
∫
𝑐+𝜏
𝑐
+∞
∫
+∞
𝑐+𝜏
+ 3∫
𝑐+3𝜏
𝑐+2𝜏
+ 4∫
𝑐+4𝜏
𝑐+3𝜏
∞
= ∑∫
∫
𝑟
𝑐+2𝜏
+∞
∫
+∞
𝑟
∫
+∞
∞
+∞
+∞
∫
𝑟
+∞
∫
𝑢
∫
+∞
𝑟
𝑐+(𝑛+1)𝜏
𝑛=0 𝑐+𝑛𝜏
∫
+∞
𝑢
+∞
𝑟
∫
∫
+∞
𝑢
𝑟−𝑐
] + 1)
𝜏
+∞
𝑐
+∞
∫
𝑟
∫
+∞
𝑢
([
𝑟−𝑐
] + 1) 𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟.
𝜏
(13)
Thus (a) follows from (11) and (13).
𝑀−𝑁
,
2𝑀
|𝑏 (𝑡)| ≤ 𝑏0 eventually; (15)
(16)
∀ (𝑡, 𝑢1 , . . . , 𝑢𝑙 , 𝑢1 , . . . , 𝑢𝑙 ) ∈ [𝑡0 , +∞) × [𝑁, 𝑀]2𝑙 ;
󵄨󵄨
󵄨
󵄨󵄨𝑓 (𝑡, 𝑢1 , . . . , 𝑢𝑙 )󵄨󵄨󵄨 ≤ 𝑄 (𝑡) ,
󵄨󵄨
󵄨
󵄨󵄨ℎ (𝑡, 𝑢1 , . . . , 𝑢𝑙 )󵄨󵄨󵄨 ≤ 𝑊 (𝑡) ,
∀ (𝑡, 𝑢1 , . . . , 𝑢𝑙 ) ∈ [𝑡0 , +∞) × [𝑁, 𝑀]𝑙 ;
(17)
(𝑛 + 1) 𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
([
𝑏0 <
󵄨󵄨
󵄨
󵄨󵄨𝑓 (𝑡, 𝑢1 , . . . , 𝑢𝑙 ) − 𝑓 (𝑡, 𝑢1 , . . . , 𝑢𝑙 )󵄨󵄨󵄨
󵄨
󵄨
≤ 𝑃 (𝑡) max {󵄨󵄨󵄨󵄨𝑢𝑗 − 𝑢𝑗 󵄨󵄨󵄨󵄨 : 1 ≤ 𝑗 ≤ 𝑙} ,
󵄨󵄨
󵄨
󵄨󵄨ℎ (𝑡, 𝑢1 , . . . , 𝑢𝑙 ) − ℎ (𝑡, 𝑢1 , . . . , 𝑢𝑙 )󵄨󵄨󵄨
󵄨
󵄨
≤ 𝑅 (𝑡) max {󵄨󵄨󵄨󵄨𝑢𝑗 − 𝑢𝑗 󵄨󵄨󵄨󵄨 : 1 ≤ 𝑗 ≤ 𝑙} ,
𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
× 𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
=∫
Theorem 2. Assume that there exist three constants 𝑀, 𝑁, and
𝑏0 and four functions 𝑃, 𝑄, 𝑅, 𝑊 ∈ 𝐶([𝑡0 , +∞), R+ ) satisfying
𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟 + ⋅ ⋅ ⋅
∀𝑡 ≥ 𝑐,
Now we show the existence of uncountably many positive
solutions for (5) and discuss the convergence of the Mann
iterative sequences to these positive solutions.
𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟 + ⋅ ⋅ ⋅
∫
(14)
2. Existence of Uncountably Many Positive
Solutions and Mann Iterative Schemes
0 < 𝑁 < 𝑀,
𝑢
+∞
𝑐+(𝑛+1)𝜏
𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
𝑢
𝑟
∫
𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
𝑢
𝑟𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
that is, (c) holds.
Similar to the proofs of (a) and (c), we conclude that (b)
and (d) hold. This completes the proof.
𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
𝑢
𝑛=0 𝑐+𝑛𝜏
= ∑∫
+∞
𝑢
∫
+ 2∫
+∞
𝑢
𝑐+3𝜏 𝑟
=∫
+∞
𝑢
𝑐+2𝜏 𝑟
+∞
𝐹 (𝑠, 𝑢, 𝑟) 𝑑𝑠 𝑑𝑢 𝑑𝑟
< +∞,
∫
∫
+∞
𝑡0
∫
+∞
𝑢
|𝑢|𝑚−1 𝑛−𝑚−1
󵄨
󵄨
max {𝑃(𝑠),𝑄(𝑠),󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨}
[|𝑠|
|𝑎(𝑢)|
+|𝑠|𝑛−𝑚−𝑖−1 max {𝑅 (𝑠),𝑊 (𝑠)}] 𝑑𝑠 𝑑𝑢 < +∞.
(18)
Then
(a) for any 𝐿 ∈ (𝑏0 𝑀 + 𝑁, (1 − 𝑏0 )𝑀), there exist 𝜃 ∈ (0, 1)
and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| such that for each 𝑥0 ∈ Ω1 (𝑁, 𝑀),
4
Abstract and Applied Analysis
the Mann iterative sequence {𝑥𝑘 }𝑘∈N0 generated by the following
scheme
𝑥𝑘+1 (𝑡)
(1 − 𝛼𝑘 ) 𝑥𝑘 (𝑡)
{
{
{
{
{
{
{
{
+𝛼𝑘 {𝐿 − 𝑏 (𝑡) 𝑥𝑘 (𝑡 − 𝜏) + (−1)𝑛 𝐻0
{
{
{
{
{
{
{
{
+∞ +∞
{
{
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡)𝑚−1
{
{
×∫ ∫
{
{
𝑎 (𝑢)
{
𝑡
𝑢
{
{
{
{
{
× [𝑔 (𝑠)−𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)) ,
{
{
{
{
{
{
. . . , 𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
{
{
{
{
𝑛−𝑖−1
{
+(−1)
𝐻
{
𝑖
{
{
{
{
+∞ +∞
{
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑡)𝑚−1
{
{
{
×∫ ∫
{
{
𝑎 (𝑢)
𝑡
𝑢
{
{
{
{
{
×ℎ (𝑥𝑘 (ℎ1 (𝑠)) , . . . ,
{
{
{
{
{
{
{
𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} ,
{
{
{
{
{
{
{
𝑡 ≥ 𝑇, 𝑘 ∈ N0 ,
{
= {(1 − 𝛼 ) 𝑥 (𝑇)
𝑘
𝑘
{
{
{
{
{
{
{
{
+𝛼𝑘 {𝐿 − 𝑏 (𝑇) 𝑥𝑘 (𝑇 − 𝜏)
{
{
{
{
{
{
{
+(−1)𝑛 𝐻0
{
{
{
{
{
+∞ +∞
{
{
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑇)𝑚−1
{
{
×
∫
∫
{
{
𝑎 (𝑢)
{
𝑇
𝑢
{
{
{
{
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)) ,
{
{
{
{
{
{
. . . , 𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
{
{
{
𝑛−𝑖−1
{
{
𝐻𝑖
+(−1)
{
{
{
{
{
+∞
+∞
{
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑇)𝑚−1
{
{
×
∫
∫
{
{
{
𝑎 (𝑢)
𝑇
𝑢
{
{
{
{
{
(ℎ
×ℎ
(𝑠,
𝑥
{
𝑘
1 (𝑠)) , . . . ,
{
{
{
{
{
{
𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} ,
{
{
{
{
𝑡0 ≤ 𝑡 < 𝑇, 𝑘 ∈ N0
{
(19)
converges to a positive solution 𝑥 ∈ Ω1 (𝑁, 𝑀) of (5) and has
the following error estimate:
󵄩󵄩
󵄩
−(1−𝜃) ∑𝑘𝑝=0 𝛼𝑝 󵄩
󵄩󵄩𝑥0 − 𝑥󵄩󵄩󵄩 ,
󵄩󵄩𝑥𝑘+1 − 𝑥󵄩󵄩󵄩 ≤ 𝑒
󵄩
󵄩
|𝑏 (𝑡)| ≤ 𝑏0 ,
𝜃 = 𝑏0 + ∫
+∞
𝑇
+∞
∫
𝑢
∀𝑘 ∈ N0 ,
(20)
∀𝑡 ≥ 𝑇;
+𝐻𝑖 𝑠
∫
+∞
𝑇
+∞
∫
𝑢
(22)
𝑢𝑚−1
[𝐻 𝑠𝑛−𝑚−1 𝑃 (𝑠)
|𝑎 (𝑢)| 0
𝑛−𝑚−𝑖−1
𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢;
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
(21)
𝑘=0
(b) Equation (5) has uncountably many positive solutions
in Ω1 (𝑁, 𝑀).
(24)
< min {(1 − 𝑏0 ) 𝑀 − 𝐿, 𝐿 − 𝑏0 𝑀 − 𝑁} .
Define a mapping 𝑆𝐿 : Ω1 (𝑁, 𝑀) → CB([𝛾, +∞), R) by
𝐿 − 𝑏 (𝑡) 𝑥 (𝑡 − 𝜏) + (−1)𝑛 𝐻0
{
{
{
+∞ +∞
{
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡)𝑚−1
{
{
{
×
∫
∫
{
{
𝑎 (𝑢)
𝑡
𝑢
{
{
{
{
,...,
×
[𝑔
−
𝑓
(𝑠,
𝑥
(𝑓
(𝑠))
(𝑠)
{
1
{
{
{
{
𝑥
(𝑓
𝑑𝑠
𝑑𝑢
(𝑠)))]
𝑙
{
{
𝑆𝐿 𝑥 (𝑡) = { +(−1)𝑛−𝑖−1 𝐻𝑖
{
{
+∞ +∞
{
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑡)𝑚−1
{
{
{ ×∫ ∫
{
{
𝑎 (𝑢)
𝑡
𝑢
{
{
{
{
×ℎ
(𝑠,
𝑥
(ℎ
,
.
.
.
,
𝑥
(ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢,
(𝑠))
1
{
{
{
{
𝑡
≥
𝑇,
𝑥
∈ Ω1 (𝑁, 𝑀) ,
{
{
{
𝛾 ≤ 𝑡 < 𝑇, 𝑥 ∈ Ω1 (𝑁, 𝑀) .
{𝑆𝐿 𝑥 (𝑇) ,
(25)
It is obvious that 𝑆𝐿 𝑥 is continuous for each 𝑥 ∈ Ω1 (𝑁, 𝑀).
By means of (16), (22), (23), and (25), we deduce that for any
𝑥, 𝑦 ∈ Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
󵄨
󵄨󵄨
󵄨󵄨𝑆𝐿 𝑥 (𝑡) − 𝑆𝐿 𝑦 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
≤ |𝑏 (𝑡)| 󵄨󵄨󵄨𝑥 (𝑡 − 𝜏) − 𝑦 (𝑡 − 𝜏)󵄨󵄨󵄨
+∞
𝑡
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡)𝑚−1
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))
󵄨
−𝑓 (𝑠, 𝑦 (𝑓1 (𝑠)) , . . . , 𝑦 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
∞
(23)
𝑢𝑚−1
󵄨
󵄨
[𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
|𝑎 (𝑢)|
+ 𝐻0 ∫
where {𝛼𝑘 }𝑘∈N0 is an arbitrary sequence in [0, 1] such that
∑ 𝛼𝑘 = +∞;
Proof. Firstly, we prove that (a) holds. Set 𝐿 ∈ (𝑏0 𝑀 + 𝑁, (1 −
𝑏0 )𝑀). From (15) and (18), we know that there exist 𝜃 ∈ (0, 1)
and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| satisfying
+ 𝐻𝑖 ∫
+∞
𝑡
+∞
∫
𝑢
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑡)𝑚−1
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))
󵄨
−ℎ (𝑠, 𝑦 (ℎ1 (𝑠)) , . . . , 𝑦 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
Abstract and Applied Analysis
5
󵄩
󵄩 󵄩
󵄩
≤ 𝑏0 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩
+∞
×∫
∫
𝑇
+∞
𝑢
≥ 𝐿 − 𝑏0 𝑀 − ∫
+∞
𝑇
𝑢𝑚−1
[𝐻 𝑠𝑛−𝑚−1 𝑃 (𝑠)
|𝑎 (𝑢)| 0
∫
+∞
𝑢
𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
󵄨
× [𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
󵄩
󵄩
= 𝜃 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
> 𝐿 − 𝑏0 𝑀 − min {(1 − 𝑏0 ) 𝑀 − 𝐿, 𝐿 − 𝑏0 𝑀 − 𝑁}
(26)
≥ 𝑁,
(28)
which yields that
󵄩󵄩
󵄩
󵄩
󵄩
󵄩󵄩𝑆𝐿 𝑥 − 𝑆𝐿 𝑦󵄩󵄩󵄩 ≤ 𝜃 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
∀𝑥, 𝑦 ∈ Ω1 (𝑁, 𝑀) .
(27)
On the basis of (17), (22), (24), and (25), we acquire that for
any 𝑥 ∈ Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
𝑆𝐿 𝑥 (𝑡)
≤ 𝐿 + |𝑏 (𝑡)| 𝑥 (𝑡 − 𝜏)
+∞
+ 𝐻0 ∫
𝑡
+ 𝐻𝑖 ∫
∫
𝑢
+∞
𝑡
∫
𝑠𝑛−𝑚−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨 󵄨
󵄨
× [ 󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . ,
󵄨
𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢
+∞
+∞
𝑢
𝑠𝑛−𝑚−𝑖−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨󵄨
󵄨
× 󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
≤ 𝐿 + 𝑏0 𝑀
+∫
+∞
∫
𝑇
𝑢
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
< 𝐿 + 𝑏0 𝑀 + min {(1 − 𝑏0 ) 𝑀 − 𝐿, 𝐿 − 𝑏0 𝑀 − 𝑁}
≤ 𝑀,
𝑆𝐿 𝑥 (𝑡)
≥ 𝐿 − |𝑏 (𝑡)| 𝑥 (𝑡 − 𝜏)
− 𝐻0 ∫
𝑡
− 𝐻𝑖 ∫
+∞
𝑡
∫
+∞
𝑢
∫
+∞
𝑢
×∫
+∞
𝑡
𝑢
|𝑎 (𝑢)|
󵄨 󵄨
󵄨
× [ 󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . ,
󵄨
𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡)𝑚−1
𝑎 (𝑢)
𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
+ (−1)𝑛−𝑖−1 𝐻𝑖
+∞
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑡)𝑚−1
𝑎 (𝑢)
󵄨󵄨
󵄨
×ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . , 𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} −𝑥 (𝑡) 󵄨󵄨󵄨󵄨
󵄨󵄨
󵄨
󵄨
󵄨
󵄨
≤ (1 − 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨 + 𝛼𝑘 󵄨󵄨󵄨𝑆𝐿 𝑥𝑘 (𝑡) − 𝑆𝐿 𝑥 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
󵄨
󵄨
≤ (1 − 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨 + 𝛼𝑘 𝜃 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
= (1 − (1 − 𝜃) 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄩
󵄩
≤ 𝑒−(1−𝜃)𝛼𝑘 󵄩󵄩󵄩𝑥𝑘 − 𝑥󵄩󵄩󵄩
󵄩
󵄩
≤ 𝑒−(1−𝜃) ∑𝑝=0 𝛼𝑝 󵄩󵄩󵄩𝑥0 − 𝑥󵄩󵄩󵄩 ,
𝑘
𝑛−𝑚−1 𝑚−1
∫
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)) , . . . ,
𝑡
󵄨
󵄨
× [𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
+∞
+ 𝛼𝑘 {𝐿 − 𝑏 (𝑡) 𝑥𝑘 (𝑡 − 𝜏) + (−1)𝑛 𝐻0
×∫
𝑢𝑚−1
|𝑎 (𝑢)|
+∞
which guarantee that 𝑆𝐿 (Ω1 (𝑁, 𝑀)) ⊆ Ω1 (𝑁, 𝑀). Consequently, (27) gives that 𝑆𝐿 is a contraction mapping in
Ω1 (𝑁, 𝑀) and it has a unique fixed point 𝑥 ∈ Ω1 (𝑁, 𝑀). It is
easy to see that 𝑥 ∈ Ω1 (𝑁, 𝑀) is a positive solution of (5).
It follows from (19), (25), and (27) that
󵄨󵄨󵄨𝑥𝑘+1 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
󵄨󵄨
󵄨
= 󵄨󵄨󵄨󵄨 (1 − 𝛼𝑘 ) 𝑥𝑘 (𝑡)
󵄨󵄨
∀𝑘 ∈ N0 , 𝑡 ≥ 𝑇,
(29)
𝑠
𝑛−𝑚−𝑖−1 𝑚−1
𝑠
𝑢
|𝑎 (𝑢)|
󵄨
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
which yields that
󵄩󵄩
󵄩
−(1−𝜃) ∑𝑘𝑝=0 𝛼𝑝 󵄩
󵄩󵄩𝑥0 − 𝑥󵄩󵄩󵄩 ,
󵄩󵄩𝑥𝑘+1 − 𝑥󵄩󵄩󵄩 ≤ 𝑒
󵄩
󵄩
∀𝑘 ∈ N0 .
(30)
That is, (20) holds. Thus (20) and (21) ensure that
lim𝑘 → ∞ 𝑥𝑘 = 𝑥.
Secondly, we show that (b) holds. Let 𝐿 1 , 𝐿 2 ∈ (𝑏0 𝑀 +
𝑁, (1 − 𝑏0 )𝑀) with 𝐿 1 ≠ 𝐿 2 . In light of (15) and (18), we know
that for each 𝑝 ∈ {1, 2}, there exist 𝜃𝑝 ∈ (0, 1), 𝑇𝑝 and 𝑇∗
6
Abstract and Applied Analysis
󵄨
󵄩
󵄩
󵄨
≥ 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − 𝑏0 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩
with 𝑇𝑝 > 1 + |𝑡0 | + 𝜏 + |𝛾| and 𝑇∗ > max{𝑇1 , 𝑇2 } satisfying
(22)–(24) and
󵄩
󵄩
− 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩 ∫
+∞
𝑇∗
+∞
∫
𝑇∗
𝑛−𝑚−1
× [𝐻0 𝑠
𝑢𝑚−1
[𝐻 𝑠𝑛−𝑚−1 𝑃 (𝑠) + 𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢
∫
|𝑎 (𝑢)| 0
𝑢
󵄨
󵄨
< 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 ,
(31)
+∞
+∞
∫
𝑢
𝑢𝑚−1
|𝑎 (𝑢)|
𝑃 (𝑠) + 𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢
󵄨
󵄨
󵄨 󵄩
󵄩
󵄨
≥ 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − (𝑏0 + 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨) 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩 ,
(33)
which implies that
where 𝜃 and 𝑇 are replaced by 𝜃𝑝 and 𝑇𝑝 , respectively. Let
the mapping 𝑆𝐿 𝑝 be defined by (25) with 𝐿 and 𝑇 replaced
by 𝐿 𝑝 and 𝑇𝑝 , respectively. As in the proof of (a), we deduce
easily that the mapping 𝑆𝐿 𝑝 possesses a unique fixed point
𝑧𝑝 ∈ Ω1 (𝑁, 𝑀), that is, 𝑧𝑝 is a positive solution of (5) in
Ω1 (𝑁, 𝑀). In order to prove (b), we need only to show that
𝑧1 ≠ 𝑧2 . In fact, (25) means that for each 𝑡 ≥ 𝑇∗ and 𝑝 ∈ {1, 2}
󵄨󵄨
󵄨
󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨
󵄩󵄩
󵄩
󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩 ≥
󵄨 > 0,
󵄨󵄨
1 + 𝑏0 + 󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨
that is, 𝑧1 ≠ 𝑧2 . This completes the proof.
Theorem 3. Assume that there exist three constants 𝑀, 𝑁, and
𝑏0 and four functions 𝑃, 𝑄, 𝑅, 𝑊 ∈ 𝐶([𝑡0 , +∞), R+ ) satisfying
(16)–(18) and
𝑧𝑝 (𝑡) = 𝐿 𝑝 − 𝑏 (𝑡) 𝑧𝑝 (𝑡 − 𝜏) + (−1)𝑛 𝐻0
×∫
+∞
𝑡
+∞
∫
𝑢
0 < 𝑁 < 𝑀,
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡)𝑚−1
𝑎 (𝑢)
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑧𝑝 (𝑓1 (𝑠)) , . . . , 𝑧𝑝 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
+ (−1)𝑛−𝑖−1 𝐻𝑖 ∫
+∞
𝑡
∫
+∞
𝑛−𝑚−𝑖−1
(𝑠 − 𝑢)
𝑢
(𝑢 − 𝑡)
𝑎 (𝑢)
𝑚−1
× ℎ (𝑠, 𝑧𝑝 (ℎ1 (𝑠)) , . . . , 𝑧𝑝 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢.
(32)
It follows from (16), (22), (31), and (32) that for each 𝑡 ≥ 𝑇∗
󵄨
󵄨󵄨
󵄨󵄨𝑧1 (𝑡) − 𝑧2 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
󵄨
󵄨
≥ 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − |𝑏 (𝑡)| 󵄨󵄨󵄨𝑧1 (𝑡 − 𝜏) − 𝑧2 (𝑡 − 𝜏)󵄨󵄨󵄨
− 𝐻0 ∫
+∞
𝑡
+∞
∫
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡)𝑚−1
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨𝑓 (𝑠, 𝑧1 (𝑓1 (𝑠)) , . . . , 𝑧1 (𝑓𝑙 (𝑠)))
+∞
− 𝐻𝑖 ∫
𝑡
∫
+∞
𝑢
𝑛−𝑚−𝑖−1
(𝑠 − 𝑢)
(𝑢 − 𝑡)
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑧1 (ℎ1 (𝑠)) , . . . , 𝑧1 (ℎ𝑙 (𝑠)))
𝑚−1
󵄨
−ℎ (𝑠, 𝑧2 (ℎ1 (𝑠)) , . . . , 𝑧2 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
𝑏0 <
𝑀−𝑁
,
𝑀
0 ≤ 𝑏 (𝑡) ≤ 𝑏0 eventually.
(35)
Then
(a) for any 𝐿 ∈ (𝑏0 𝑀 + 𝑁, 𝑀), there exist 𝜃 ∈ (0, 1) and
𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| such that for each 𝑥0 ∈ Ω1 (𝑁, 𝑀),
the Mann iterative sequence {𝑥𝑘 }𝑘∈N0 generated by (19)
converges to a positive solution 𝑥 ∈ Ω1 (𝑁, 𝑀) of (5)
and has the error estimate (20), where {𝛼𝑘 }𝑘∈N0 is an
arbitrary sequence in [0, 1] satisfying (21);
(b) Equation (5) has uncountably many positive solutions
in Ω1 (𝑁, 𝑀).
Proof. Let 𝐿 ∈ (𝑏0 𝑀 + 𝑁, 𝑀). Equations (18) and (36) ensure
that there exist 𝜃 ∈ (0, 1) and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| satisfying
(23),
0 ≤ 𝑏 (𝑡) ≤ 𝑏0 ,
+∞
∫
󵄨
−𝑓 (𝑠, 𝑧2 (𝑓1 (𝑠)) , . . . , 𝑧2 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
(34)
𝑇
∫
+∞
𝑢
∀𝑡 ≥ 𝑇;
(36)
𝑢𝑚−1
󵄨
󵄨
[𝐻 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
|𝑎 (𝑢)| 0
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
(37)
< min {𝑀 − 𝐿, 𝐿 − 𝑏0 𝑀 − 𝑁} .
Define a mapping 𝑆𝐿 : Ω1 (𝑁, 𝑀) → CB([𝛾, +∞), R) by
(25). Obviously, 𝑆𝐿 𝑥 is continuous for every 𝑥 ∈ Ω1 (𝑁, 𝑀).
Abstract and Applied Analysis
7
+∞
Using (16), (23), (25), and (36), we conclude that for any
𝑥, 𝑦 ∈ Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
− 𝐻0 ∫
𝑡
󵄨
󵄨󵄨
󵄨󵄨𝑆𝐿 𝑥 (𝑡) − 𝑆𝐿 𝑦 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
≤ 𝑏 (𝑡) 󵄨󵄨󵄨𝑥 (𝑡 − 𝜏) − 𝑦 (𝑡 − 𝜏)󵄨󵄨󵄨
+ 𝐻0 ∫
+∞
∫
𝑡
+∞
𝑛−𝑚−1
(𝑠 − 𝑢)
𝑢
(𝑢 − 𝑡)
|𝑎 (𝑢)|
− 𝐻𝑖 ∫
𝑚−1
+ 𝐻𝑖 ∫
∫
𝑡
𝑛−𝑚−𝑖−1
(𝑠 − 𝑢)
𝑢
(𝑢 − 𝑡)
|𝑎 (𝑢)|
𝑇
∫
𝑢
(39)
𝑚−1
which implies that (27) holds. In light of (17), (25), (36), and
(37), we know that for any 𝑥 ∈ Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
which mean that 𝑆𝐿 (Ω1 (𝑁, 𝑀)) ⊆ Ω1 (𝑁, 𝑀). Equation (27)
guarantees that 𝑆𝐿 is a contraction mapping in Ω1 (𝑁, 𝑀) and
it possesses a unique fixed point 𝑥 ∈ Ω1 (𝑁, 𝑀). As in the
proof of Theorem 2, we infer that 𝑥 ∈ Ω1 (𝑁, 𝑀) is a positive
solution of (5). The rest of the proof is similar to that of
Theorem 2 and is omitted. This completes the proof.
Theorem 4. Assume that there exist three constants 𝑀, 𝑁, and
𝑏0 and four functions 𝑃, 𝑄, 𝑅, 𝑊 ∈ 𝐶([𝑡0 , +∞), R+ ) satisfying
(16)–(18) and
0 < 𝑁 < 𝑀,
𝑆𝐿 𝑥 (𝑡)
𝑡
+ 𝐻𝑖 ∫
+∞
𝑡
≤𝐿+∫
+∞
𝑇
𝑢
∫
∫
+∞
𝑢
𝑠𝑛−𝑚−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨 󵄨
󵄨
× [ 󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . ,
󵄨
𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢
𝑠𝑛−𝑚−𝑖−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
+∞
𝑢
𝑢𝑚−1
󵄨
󵄨
[𝐻 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
|𝑎 (𝑢)| 0
< 𝐿 + min {𝑀 − 𝐿, 𝐿 − 𝑏0 𝑀 − 𝑁}
𝑆𝐿 𝑥 (𝑡)
≥ 𝐿 − |𝑏 (𝑡)| 𝑥 (𝑡 − 𝜏)
𝑏0 <
𝑀−𝑁
,
𝑀
−𝑏0 ≤ 𝑏 (𝑡) ≤ 0 eventually.
(40)
Then
(a) for any 𝐿 ∈ (𝑁, (1 − 𝑏0 )𝑀), there exist 𝜃 ∈ (0, 1) and
𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| such that for each 𝑥0 ∈ Ω1 (𝑁, 𝑀),
the Mann iterative sequence {𝑥𝑘 }𝑘∈N0 generated by (19)
converges to a positive solution 𝑥 ∈ Ω1 (𝑁, 𝑀) of (5)
and has the error estimate (20), where {𝛼𝑘 }𝑘∈N0 is an
arbitrary sequence in [0, 1] satisfying (21);
(b) Equation (5) has uncountably many positive solutions
in Ω1 (𝑁, 𝑀).
Proof. Set 𝐿 ∈ (𝑁, (1 − 𝑏0 )𝑀). It follows from (18) and (40)
that there exist 𝜃 ∈ (0, 1) and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| satisfying
(23),
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
≤ 𝑀,
𝑢
𝑢𝑚−1
|𝑎 (𝑢)|
≥ 𝑁,
𝑢
[𝐻 𝑠𝑛−𝑚−1 𝑃 (𝑠)
|𝑎 (𝑢)| 0
+∞
+∞
> 𝐿 − 𝑏0 𝑀 − min {𝑀 − 𝐿, 𝐿 − 𝑏0 𝑀 − 𝑁}
(38)
∫
∫
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
󵄩
󵄩
= 𝜃 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
+∞
+∞
󵄨
󵄨
× [𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢
≤ 𝐿 + 𝐻0 ∫
𝑢
𝑇
󵄩
󵄩 󵄩
󵄩
≤ 𝑏0 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩
×∫
𝑠𝑛−𝑚−𝑖−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
+∞
𝑚−1
󵄨
−ℎ (𝑠, 𝑦 (ℎ1 (𝑠)) , . . . , 𝑦 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
+∞
∫
≥ 𝐿 − 𝑏0 𝑀 − ∫
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))
+∞
𝑢
+∞
𝑡
󵄨
−𝑓 (𝑠, 𝑦 (𝑓1 (𝑠)) , . . . , 𝑦 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
+∞
𝑠𝑛−𝑚−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨 󵄨
󵄨󵄨
× [󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . ,
+∞
󵄨
𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢
󵄨
× 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))
+∞
∫
−𝑏0 ≤ 𝑏 (𝑡) ≤ 0,
∫
+∞
𝑇
+∞
∫
𝑢
∀𝑡 ≥ 𝑇;
(41)
𝑢𝑚−1
󵄨
󵄨
[𝐻 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
|𝑎 (𝑢)| 0
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
< min {𝐿 − 𝑁, (1 − 𝑏0 ) 𝑀 − 𝐿} .
(42)
8
Abstract and Applied Analysis
Define a mapping 𝑆𝐿 : Ω1 (𝑁, 𝑀) → CB([𝛾, +∞), R) by
(25). Distinctly, 𝑆𝐿 𝑥 is continuous for each 𝑥 ∈ Ω1 (𝑁, 𝑀).
In terms of (16), (23), (25), and (41), we reason that for any
𝑥, 𝑦 ∈ Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
< 𝐿 + 𝑏0 𝑀 + min {𝐿 − 𝑁, (1 − 𝑏0 ) 𝑀 − 𝐿}
≤ 𝑀,
𝑠𝑛−𝑚−1 𝑢𝑚−1
|𝑎 (𝑢)|
𝑡
𝑢
󵄨󵄨 󵄨󵄨
󵄨󵄨
× [󵄨󵄨𝑔 (𝑠)󵄨󵄨 + 󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . ,
󵄨
𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢
𝑆𝐿 𝑥 (𝑡) ≥ 𝐿 − 𝐻0 ∫
󵄨
󵄨󵄨
󵄨󵄨𝑆𝐿 𝑥 (𝑡) − 𝑆𝐿 𝑦 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
≤ |𝑏 (𝑡)| 󵄨󵄨󵄨𝑥 (𝑡 − 𝜏) − 𝑦 (𝑡 − 𝜏)󵄨󵄨󵄨
+ 𝐻0 ∫
+∞
∫
𝑡
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡)𝑚−1
|𝑎 (𝑢)|
− 𝐻𝑖 ∫
∫
𝑡
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑡)𝑚−1
|𝑎 (𝑢)|
≥𝐿−∫
𝑇
+∞
𝑢
𝑢𝑚−1
[𝐻 𝑠𝑛−𝑚−1 𝑃 (𝑠)
|𝑎 (𝑢)| 0
+𝐻𝑖 𝑠
≥ 𝑁,
(44)
𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢
󵄩
󵄩
= 𝜃 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
(43)
which means that (27) holds. Owing to (17), (25), (41), and
(42), we earn that for any 𝑥 ∈ Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
𝑆𝐿 𝑥 (𝑡) ≤ 𝐿 + |𝑏 (𝑡)| 𝑥 (𝑡 − 𝜏)
+ 𝐻0 ∫
𝑡
+∞
+ 𝐻𝑖 ∫
𝑡
∫
∫
𝑠𝑛−𝑚−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨 󵄨
󵄨󵄨
× [󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . ,
󵄨
𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢
+∞
𝑢
+∞
𝑢
𝑠𝑛−𝑚−𝑖−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . ,
󵄨
𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
+∞
≤ 𝐿 + 𝑏0 𝑀 + ∫
𝑇
∫
+∞
𝑢
𝑢
𝑢𝑚−1
|𝑎 (𝑢)|
> 𝐿 − min {𝐿 − 𝑁, (1 − 𝑏0 ) 𝑀 − 𝐿}
𝑛−𝑚−𝑖−1
+∞
+∞
∫
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
󵄩
󵄩 󵄩
󵄩
≤ 𝑏0 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩
∫
𝑠𝑛−𝑚−𝑖−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
󵄨
× [𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
󵄨
−ℎ (𝑠, 𝑦 (ℎ1 (𝑠)) , . . . , 𝑦 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
+∞
+∞
+∞
∫
𝑢
+∞
𝑇
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))
×∫
∫
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . ,
󵄨
𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
󵄨
−𝑓 (𝑠, 𝑦 (𝑓1 (𝑠)) , . . . , 𝑦 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
+∞
+∞
𝑡
󵄨
× 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))
+ 𝐻𝑖 ∫
+∞
𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
󵄨
× [𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
which yield that 𝑆𝐿 (Ω1 (𝑁, 𝑀)) ⊆ Ω1 (𝑁, 𝑀). Thus (27)
ensures that 𝑆𝐿 is a contraction mapping in Ω1 (𝑁, 𝑀) and
it owns a unique fixed point 𝑥 ∈ Ω1 (𝑁, 𝑀). As in the proof of
Theorem 2, we infer that 𝑥 ∈ Ω1 (𝑁, 𝑀) is a positive solution
of (5). The rest of the proof is parallel to that of Theorem 2,
and hence is elided. This completes the proof.
Theorem 5. Assume that there exist three constants 𝑀, 𝑁, and
𝑏0 and four functions 𝑃, 𝑄, 𝑅, 𝑊 ∈ 𝐶([𝑡0 , +∞), R+ ) satisfying
(18) and
𝑀 > 𝑁 > 0,
𝑏0 >
𝑀
,
𝑀−𝑁
𝑏 (𝑡) ≥ 𝑏0 eventually;
(45)
󵄨󵄨
󵄨
󵄨󵄨𝑓 (𝑡, 𝑢1 , . . . , 𝑢𝑙 ) − 𝑓 (𝑡, 𝑢1 , . . . , 𝑢𝑙 )󵄨󵄨󵄨
󵄨
󵄨
≤ 𝑃 (𝑡) max {󵄨󵄨󵄨󵄨𝑢𝑗 − 𝑢𝑗 󵄨󵄨󵄨󵄨 : 1 ≤ 𝑗 ≤ 𝑙} ,
󵄨󵄨
󵄨
󵄨󵄨ℎ (𝑡, 𝑢1 , . . . , 𝑢𝑙 ) − ℎ (𝑡, 𝑢1 , . . . , 𝑢𝑙 )󵄨󵄨󵄨
󵄨
󵄨
≤ 𝑅 (𝑡) max {󵄨󵄨󵄨󵄨𝑢𝑗 − 𝑢𝑗 󵄨󵄨󵄨󵄨 : 1 ≤ 𝑗 ≤ 𝑙} ,
∀ (𝑡, 𝑢1 , . . . , 𝑢𝑙 , 𝑢1 , . . . , 𝑢𝑙 ) ∈ [𝑡0 , +∞) × [0,
󵄨󵄨
󵄨
󵄨󵄨𝑓 (𝑡, 𝑢1 , . . . , 𝑢𝑙 )󵄨󵄨󵄨 ≤ 𝑄 (𝑡) ,
(46)
𝑀 2𝑙
] ;
𝑏0
󵄨󵄨
󵄨
󵄨󵄨ℎ (𝑡, 𝑢1 , . . . , 𝑢𝑙 )󵄨󵄨󵄨 ≤ 𝑊 (𝑡) ,
∀ (𝑡, 𝑢1 , . . . , 𝑢𝑙 ) ∈ [𝑡0 , +∞) × [0,
𝑀 𝑙
].
𝑏0
(47)
Abstract and Applied Analysis
9
+∞
Then
(a) for any 𝐿 ∈ (𝑁 + 𝑀/𝑏0 , 𝑀), there exist 𝜃 ∈ (0, 1) and
𝑇 > 1+|𝑡0 |+𝜏+|𝛾| such that for each 𝑥0 ∈ Ω2 (𝑁, 𝑀), the Mann
iterative sequence {𝑥𝑘 }𝑘∈N0 generated by the following scheme
∫
𝑇
𝑏 (𝑡) ≥ 𝑏0 ,
∀𝑡 ≥ 𝑇;
(49)
< min {𝑀 − 𝐿, 𝐿 −
× [𝐻0 𝑠
𝑃 (𝑠)
𝑛−𝑚−𝑖−1
+𝐻𝑖 𝑠
𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢;
(50)
𝑀
− 𝑁} .
𝑏0
𝑆𝐿 𝑥 (𝑡)
𝑥 (𝑡 + 𝜏) (−1)𝑛 𝐻0
𝐿
{
−
+
{
{
{
𝑏 (𝑡 + 𝜏) 𝑏 (𝑡 + 𝜏) 𝑏 (𝑡 + 𝜏)
{
{
{
{
+∞ +∞
{
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡 − 𝜏)𝑚−1
{
{
{ ×∫ ∫
{
{
𝑎 (𝑢)
𝑡+𝜏 𝑢
{
{
{
{
, . . . , 𝑥 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
×
[𝑔
−
𝑓
(𝑠,
𝑥
(𝑓
(𝑠))
(𝑠)
{
1
{
{
{
{
𝑛−𝑖−1
𝐻𝑖
= { + (−1)
{
{
𝑏
+
𝜏)
(𝑡
{
{
𝑛−𝑚−𝑖−1
{
+∞ +∞
{
(𝑢 − 𝑡 − 𝜏)𝑚−1
{ × ∫ ∫ (𝑠 − 𝑢)
{
{
{
{
𝑎 (𝑢)
𝑡+𝜏 𝑢
{
{
{
{
×ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢,
{
{
{
{
𝑡 ≥ 𝑇, 𝑥 ∈ Ω2 (𝑁, 𝑀) ,
{
{
{
𝑥
,
𝛾
≤
𝑡 < 𝑇, 𝑥 ∈ Ω2 (𝑁, 𝑀) .
𝑆
(𝑇)
{ 𝐿
(52)
In light of (46), (49), (50), and (52), we conclude that for
𝑥, 𝑦 ∈ Ω2 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
󵄨
󵄨󵄨
󵄨󵄨𝑆𝐿 𝑥 (𝑡) − 𝑆𝐿 𝑦 (𝑡)󵄨󵄨󵄨
≤
𝐻0
1
󵄨
󵄨󵄨
󵄨𝑥 (𝑡 + 𝜏) − 𝑦 (𝑡 + 𝜏)󵄨󵄨󵄨 +
𝑏 (𝑡 + 𝜏) 󵄨
𝑏 (𝑡 + 𝜏)
×∫
+∞
𝑡+𝜏
+∞
∫
𝑢
𝑠𝑛−𝑚−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))
󵄨
−𝑓 (𝑠, 𝑦 (𝑓1 (𝑠)) , . . . , 𝑦 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
+
𝐻𝑖
𝑏 (𝑡 + 𝜏)
×∫
+∞
𝑡+𝜏
+∞
∫
𝑢
𝑠𝑛−𝑚−𝑖−1 𝑢𝑚−1
𝑎 (𝑢)
󵄨󵄨
× 󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))
󵄨
−ℎ (𝑠, 𝑦 (ℎ1 (𝑠)) , . . . , 𝑦 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
≤
1 󵄩󵄩
󵄩 1 󵄩
󵄩
󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 + 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩
𝑏0
𝑏0
+∞
𝑇
𝑛−𝑚−1
(51)
Define a mapping 𝑆𝐿 : Ω2 (𝑁, 𝑀) → CB([𝛾, +∞), R) by
×∫
1
1 +∞ +∞ 𝑢𝑚−1
𝜃=
+ ∫ ∫
𝑏0 𝑏0 𝑇
|𝑎 (𝑢)|
𝑢
𝑢𝑚−1
󵄨
󵄨
[𝐻 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
|𝑎 (𝑢)| 0
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
converges to a positive solution 𝑥 ∈ Ω2 (𝑁, 𝑀) of (5) and has
the error estimate (20), where {𝛼𝑘 }𝑘∈N0 is an arbitrary sequence
in [0, 1] with (21);
(b) Equation (5) has uncountably many positive solutions
in Ω2 (𝑁, 𝑀).
Proof. First of all, we show that (a) holds. Set 𝐿 ∈ (𝑁 +
𝑀/𝑏0 , 𝑀). It follows from (18) and (45) that there exist 𝜃 ∈
(0, 1) and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| such that
+∞
𝑢
𝑥𝑘+1 (𝑡)
𝛼𝑘
{(1 − 𝛼𝑘 ) 𝑥𝑘 (𝑡) +
{
{
𝑏
+ 𝜏)
(𝑡
{
{
{
{
{
{
× {𝐿 − 𝑥𝑘 (𝑡 + 𝜏) + (−1)𝑛 𝐻0
{
{
{
{
{
+∞ +∞
{
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡 − 𝜏)𝑚−1
{
{
×
∫
∫
{
{
{
𝑎 (𝑢)
𝑡+𝜏 𝑢
{
{
{
{
×
[𝑔
−
𝑓
(𝑠, 𝑥𝑘 (𝑓1 (𝑠)) ,
(𝑠)
{
{
{
{
{
. . . , 𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
{
{
{
𝑛−𝑖−1
{
+(−1)
𝐻
{
𝑖
{
{
+∞ +∞
{
{
− 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑡 − 𝜏)𝑚−1
(𝑠
{
{
×
∫
∫
{
{
{
𝑎 (𝑢)
𝑡+𝜏 𝑢
{
{
{
(ℎ
×ℎ
(𝑠,
𝑥
(𝑠)) , . . . ,
{
𝑘
1
{
{
{
{
{
{
𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} ,
{
{
{
{
{
{
{
𝑡 ≥ 𝑇, 𝑘 ∈ N0 ,
={
𝛼𝑘
{
+
)
𝑥
(1
−
𝛼
(𝑇)
{
𝑘
𝑘
{
{
𝑏 (𝑡 + 𝜏)
{
{
{
{
{
{
× {𝐿 − 𝑥𝑘 (𝑇 + 𝜏) + (−1)𝑛 𝐻0
{
{
{
{
{
+∞ +∞
{
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑇 − 𝜏)𝑚−1
{
{
{
×
∫
∫
{
{
𝑎 (𝑢)
𝑇+𝜏 𝑢
{
{
{
{
×
[𝑔
−
𝑓
(𝑠, 𝑥𝑘 (𝑓1 (𝑠)) ,
(𝑠)
{
{
{
{
{
. . . , 𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
{
{
{
𝑛−𝑖−1
{
𝐻
+(−1)
{
𝑖
{
{
{
+∞ +∞
{
− 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑇 − 𝜏)𝑚−1
(𝑠
{
{
×
∫
∫
{
{
{
𝑎 (𝑢)
𝑇+𝜏 𝑢
{
{
{
{
×ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . ,
{
{
{
{
{
{
𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} ,
{
{
{
{
{
𝛾 ≤ 𝑡 < 𝑇, 𝑘 ∈ N0
{
(48)
∫
+∞
∫
𝑢
𝑢𝑚−1
[𝐻 𝑠𝑛−𝑚−1 𝑃 (𝑠)
|𝑎 (𝑢)| 0
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢
󵄩
󵄩
= 𝜃 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
(53)
10
Abstract and Applied Analysis
󵄨 󵄨
󵄨
× [󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . ,
which yields that
󵄩
󵄩
󵄩󵄩
󵄩
󵄩󵄩𝑆𝐿 𝑥 − 𝑆𝐿 𝑦󵄩󵄩󵄩 ≤ 𝜃 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
∀𝑥, 𝑦 ∈ Ω2 (𝑁, 𝑀) .
󵄨
𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢
(54)
+∞
− 𝐻𝑖 ∫
In view of (47), (49), (51), and (52), we obtain that for any
𝑥 ∈ Ω2 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
𝑆𝐿 𝑥 (𝑡) ≤
𝑡+𝜏
∫
𝑢
1
𝑏 (𝑡 + 𝜏)
+ 𝐻0 ∫
+∞
≥
𝑠𝑛−𝑚−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
󵄨
× [󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨
󵄨
+ 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) ,
1
𝑏 (𝑡 + 𝜏)
+∞
∫
𝑡+𝜏
𝑢
× {𝐿 −
𝑀
𝑏 (𝑡 + 𝜏)
+∞
−∫
𝑇
∫
+∞
𝑢
+∞
+ 𝐻𝑖 ∫
𝑡+𝜏
∫
𝑠𝑛−𝑚−𝑖−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . ,
+∞
𝑢
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢}
>
󵄨
𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢}
1
𝑏 (𝑡 + 𝜏)
+∫
𝑇
∫
+∞
𝑢
𝑢𝑚−1
|𝑎 (𝑢)|
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢}
1
𝑏 (𝑡 + 𝜏)
× (𝐿 −
≤
𝑆𝐿 𝑥 (𝑡) ≥
𝑀
𝑀
− 𝑁})
− min {𝑀 − 𝐿, 𝐿 −
𝑏 (𝑡 + 𝜏)
𝑏0
𝑁
,
𝑏 (𝑡 + 𝜏)
(55)
󵄨
󵄨
× [𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
<
1
𝑏 (𝑡 + 𝜏)
× (𝐿 −
≥
𝑁
𝑏 (𝑡 + 𝜏)
+∞
𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
󵄨
× [𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
󵄨
. . . , 𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢
× {𝐿 −
𝑠𝑛−𝑚−𝑖−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨󵄨
× 󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . ,
󵄨
𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢}
× {𝐿 − 𝑥 (𝑡 + 𝜏)
≤
+∞
𝑀
𝑁
+ min {𝑀 − 𝐿, 𝐿 −
− 𝑁})
𝑏 (𝑡 + 𝜏)
𝑏0
𝑀
,
𝑏 (𝑡 + 𝜏)
which imply that 𝑆𝐿 (Ω2 (𝑁, 𝑀)) ⊆ Ω2 (𝑁, 𝑀). It follows from
(50) and (54) that 𝑆𝐿 is a contraction mapping in Ω2 (𝑁, 𝑀)
and it has a unique fixed point 𝑥 ∈ Ω2 (𝑁, 𝑀). It is clear that
𝑥 ∈ Ω2 (𝑁, 𝑀) is a positive solution of (5).
Note that (48), (52), and (54) undertake that
󵄨
󵄨󵄨
󵄨󵄨𝑥𝑘+1 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄨󵄨
𝛼𝑘
󵄨
= 󵄨󵄨󵄨(1 − 𝛼𝑘 ) 𝑥𝑘 (𝑡) +
󵄨󵄨
𝑏 (𝑡 + 𝜏)
× {𝐿 − 𝑥𝑘 (𝑡 + 𝜏) + (−1)𝑛 𝐻0
1
𝑏 (𝑡 + 𝜏)
×∫
+∞
𝑡+𝜏
− 𝐻0 ∫
𝑡+𝜏
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡 − 𝜏)𝑚−1
𝑎 (𝑢)
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)) , . . . ,
× {𝐿 − 𝑥 (𝑡 + 𝜏)
+∞
+∞
∫
+∞
∫
𝑢
𝑠𝑛−𝑚−1 𝑢𝑚−1
|𝑎 (𝑢)|
𝑠𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
+ (−1)𝑛−𝑖−1 𝐻𝑖
Abstract and Applied Analysis
+∞
×∫
𝑡+𝜏
∫
+∞
𝑢
11
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑡 − 𝜏)𝑚−1
𝑎 (𝑢)
Due to (46), (51), (57), and (58), we conclude that for each
𝑡 ≥ 𝑇3
× ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . ,
󵄨󵄨
󵄨
𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} − 𝑥 (𝑡) 󵄨󵄨󵄨󵄨
󵄨󵄨
󵄨
󵄨
󵄨
󵄨
≤ (1 − 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨 + 𝛼𝑘 󵄨󵄨󵄨𝑆𝐿 𝑥𝑘 (𝑡) − 𝑆𝐿 𝑥 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
󵄨
󵄨
≤ (1 − 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨 + 𝛼𝑘 𝜃 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
= (1 − (1 − 𝜃) 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄩
󵄩
≤ 𝑒−(1−𝜃)𝛼𝑘 󵄩󵄩󵄩𝑥𝑘 − 𝑥󵄩󵄩󵄩
𝑘
󵄩
󵄩
≤ 𝑒−(1−𝜃) ∑𝑝=0 𝛼𝑝 󵄩󵄩󵄩𝑥0 − 𝑥󵄩󵄩󵄩 ,
󵄨󵄨
𝑧 (𝑡 + 𝜏) 𝑧2 (𝑡 + 𝜏) 󵄨󵄨󵄨󵄨
󵄨󵄨
−
󵄨
󵄨󵄨𝑧1 (𝑡) − 𝑧2 (𝑡) + 1
󵄨󵄨
𝑏 (𝑡 + 𝜏)
𝑏 (𝑡 + 𝜏) 󵄨󵄨󵄨
≥
1
𝑏 (𝑡 + 𝜏)
󵄨
󵄨
× ( 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨
− 𝐻0 ∫
+∞
𝑇3
𝑢
+∞
𝑧𝑝 (𝑡) =
−
𝑏 (𝑡 + 𝜏)
×∫
+∞
𝑡+𝜏
+∞
∫
𝑢
𝑏 (𝑡 + 𝜏)
+
− 𝐻𝑖 ∫
𝑇3
×∫
𝑡+𝜏
∫
𝑢
+∞
𝑢
𝑠𝑛−𝑚−𝑖−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
−ℎ (𝑠, 𝑧2 (𝑓1 (𝑠)) , . . . , 𝑧2 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢)
≥
1
𝑏 (𝑡 + 𝜏)
󵄨 󵄩
󵄩
󵄨
× ( 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩
𝑛
(−1) 𝐻0
𝑏 (𝑡 + 𝜏)
×∫
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡 − 𝜏)𝑚−1
𝑎 (𝑢)
+∞
𝑇3
∫
+∞
𝑢
𝑢𝑚−1
[𝐻 𝑠𝑛−𝑚−1 𝑃 (𝑠)
|𝑎 (𝑢)| 0
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢)
(−1)𝑛−𝑖−1 𝐻𝑖
+
𝑏 (𝑡 + 𝜏)
+∞
∫
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑧1 (𝑓1 (𝑠)) , . . . , 𝑧1 (𝑓𝑙 (𝑠)))
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑧𝑝 (𝑓1 (𝑠)) , . . . , 𝑧𝑝 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
+∞
𝑠𝑛−𝑚−1 𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
−𝑓 (𝑠, 𝑧2 (𝑓1 (𝑠)) , . . . , 𝑧2 (𝑓𝑙 (𝑠))))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
which indicates that (20) holds. Thus (20) and (21) assure that
lim𝑘 → ∞ 𝑥𝑘 = 𝑥.
Next we prove that (b) holds. Let 𝐿 1 , 𝐿 2 ∈ (𝑁 + 𝑀/𝑏0 , 𝑀)
with 𝐿 1 ≠ 𝐿 2 . As in the proof of (a) we infer that for each
𝑝 ∈ {1, 2} there exist 𝜃𝑝 ∈ (0, 1), 𝑇𝑝 > 1 + |𝑡0 | + 𝜏 + |𝛾| and
𝑆𝐿 𝑝 satisfying (49)–(52), where 𝐿, 𝜃, 𝑇, and 𝑆𝐿 are replaced by
𝐿 𝑝 , 𝜃𝑝 , 𝑇𝑝 , and 𝑆𝐿 𝑝 , respectively, and 𝑆𝐿 𝑝 has a unique fixed
point 𝑧𝑝 ∈ Ω2 (𝑁, 𝑀), which is a positive solution of (5) in
Ω2 (𝑁, 𝑀). It follows that for each 𝑡 ≥ 𝑇𝑝 and 𝑝 ∈ {1, 2}
𝑧𝑝 (𝑡 + 𝜏)
+∞
󵄨
× 󵄨󵄨󵄨𝑓 ((𝑠, 𝑧1 (𝑓1 (𝑠)) , . . . , 𝑧1 (𝑓𝑙 (𝑠)))
∀𝑘 ∈ N0 , 𝑡 ≥ 𝑇,
(56)
𝐿𝑝
∫
>
󵄨
󵄨
1
󵄨 󵄩
󵄩 󵄨󵄨𝐿 − 𝐿 󵄨󵄨
󵄨
(󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩 󵄨 1󵄩󵄩 2 󵄨 󵄩󵄩 )
𝑏 (𝑡 + 𝜏)
1 + 2 󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩
󵄨󵄨
󵄨
󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨
>
2𝑏 (𝑡 + 𝜏)
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑡 − 𝜏)𝑚−1
𝑎 (𝑢)
> 0,
× ℎ (𝑠, 𝑧𝑝 (ℎ1 (𝑠)) , . . . , 𝑧𝑝 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢.
(59)
(57)
On behalf of proving (b), we need only to show that 𝑧1 ≠ 𝑧2 .
Notice that (18) guarantees that there exits 𝑇3 > max{𝑇1 , 𝑇2 }
satisfying
+∞
∫
𝑇3
𝑢𝑚−1
[𝐻0 𝑠𝑛−𝑚−1 𝑃 (𝑠) + 𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢
(𝑢)|
|𝑎
𝑢
󵄨󵄨
󵄨
󵄨𝐿 − 𝐿 󵄨󵄨
< 󵄨 1󵄩󵄩 2 󵄨 󵄩󵄩 .
1 + 2 󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩
(58)
∫
which yields that 𝑧1 ≠ 𝑧2 . This completes the proof.
Theorem 6. Assume that there exist three constants 𝑀, 𝑁, and
𝑏0 and four functions 𝑃, 𝑄, 𝑅, 𝑊 ∈ 𝐶([𝑡0 , +∞), R+ ) satisfying
(18), (46), (47), and
+∞
0 < 𝑁 < 𝑀,
Then
𝑀
< 𝑏0 ,
𝑀−𝑁
𝑏 (𝑡) ≤ −𝑏0 eventually.
(60)
12
Abstract and Applied Analysis
(a) for any 𝐿 ∈ (𝑁, (1 − 1/𝑏0 )𝑀), there exist 𝜃 ∈ (0, 1) and
𝑇 > 1+|𝑡0 |+𝜏+|𝛾| such that for each 𝑥0 ∈ Ω3 (𝑁, 𝑀), the Mann
iterative sequence {𝑥𝑘 }𝑘∈N0 generated by the following scheme
𝑥𝑘+1 (𝑡)
𝛼𝑘
(1 − 𝛼𝑘 ) 𝑥𝑘 (𝑡) +
{
{
𝑏 (𝑡 + 𝜏)
{
{
{
{
{
{
× { − 𝐿 − 𝑥𝑘 (𝑡 + 𝜏) + (−1)𝑛 𝐻0
{
{
{
{
{
{
+∞ +∞
{
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡 − 𝜏)𝑚−1
{
{
×∫ ∫
{
{
{
𝑎 (𝑢)
𝑡+𝜏 𝑢
{
{
{
{
×
[𝑔
−
𝑓
(𝑠,
𝑥
(𝑓
(𝑠)
𝑘
1 (𝑠)) , . . . ,
{
{
{
{
(𝑓
𝑥
𝑑𝑠
𝑑𝑢
(𝑠)))]
{
𝑘
𝑙
{
{
𝑛−𝑖−1
{
{
𝐻𝑖
+(−1)
{
{
{
+∞ +∞
{
{
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑡 − 𝜏)𝑚−1
{
{
×
∫
∫
{
{
𝑎 (𝑢)
{
𝑡+𝜏 𝑢
{
{
{
{
{
×ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . , 𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} ,
{
{
{
{
{
{
{
𝑡 ≥ 𝑇, 𝑘 ∈ N0 ,
={
𝛼𝑘
{
)
𝑥
(1
−
𝛼
+
(𝑇)
{
𝑘
𝑘
{
{
𝑏 (𝑡 + 𝜏)
{
{
{
{
{
{
× { − 𝐿 − 𝑥𝑘 (𝑇 + 𝜏) + (−1)𝑛 𝐻0
{
{
{
{
{
+∞ +∞
{
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑇 − 𝜏)𝑚−1
{
{
{
×
∫
∫
{
{
𝑎 (𝑢)
𝑇+𝜏 𝑢
{
{
{
{
×
[𝑔
−
𝑓
(𝑠,
𝑥
(𝑓1 (𝑠)) , . . . ,
(𝑠)
{
𝑘
{
{
{
{
𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
{
{
{
𝑛−𝑖−1
{
+(−1)
𝐻
{
𝑖
{
{
{
+∞ +∞
{
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑇 − 𝜏)𝑚−1
{
{
×
∫
∫
{
{
{
𝑎 (𝑢)
𝑇+𝜏 𝑢
{
{
{
{
{
{
×ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . , 𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} ,
{
{
{
{
𝛾 ≤ 𝑡 < 𝑇, 𝑘 ∈ N0
{
Define a mapping 𝑆𝐿 : Ω3 (𝑁, 𝑀) → CB([𝛾, +∞), R) by
𝑥 (𝑡 + 𝜏) (−1)𝑛 𝐻0
−𝐿
{
{
−
+
{
{
𝑏 (𝑡 + 𝜏) 𝑏 (𝑡 + 𝜏) 𝑏 (𝑡 + 𝜏)
{
{
{
{
𝑛−𝑚−1
+∞ +∞
{
{
(𝑢 − 𝑡 − 𝜏)𝑚−1
{ × ∫ ∫ (𝑠 − 𝑢)
{
{
{
𝑎 (𝑢)
{
𝑡+𝜏 𝑢
{
{
{
{
{
{ × [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . ,
{
{
{
{
{
𝑥 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
{
{
{
𝑛−𝑖−1
𝐻𝑖
(−1)
𝑆𝐿 𝑥 (𝑡) = {
+
{
{
{
𝑏 (𝑡 + 𝜏)
{
{
{
{
𝑛−𝑚−𝑖−1
+∞ +∞
{
(𝑢 − 𝑡 − 𝜏)𝑚−1
{
{ × ∫ ∫ (𝑠 − 𝑢)
{
{
{
𝑎 (𝑢)
𝑡+𝜏 𝑢
{
{
{
{
{
{
×ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢,
{
{
{
{
{
{
𝑡 ≥ 𝑇, 𝑥 ∈ Ω3 (𝑁, 𝑀) ,
{
{
{
𝑥
,
𝛾
≤
𝑡 < 𝑇, 𝑥 ∈ Ω3 (𝑁, 𝑀) .
𝑆
(𝑇)
{ 𝐿
(61)
By virtue of (47), (62), and (63), we know that for any 𝑥 ∈
Ω3 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
𝑆𝐿 𝑥 (𝑡)
≤
+∞
×∫
𝑡+𝜏
Proof. Put 𝐿 ∈ (𝑁, (1 − 1/𝑏0 )𝑀). It follows from (18) and (60)
that there exist 𝜃 ∈ (0, 1) and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| satisfying
(50) and
∫
+∞
𝑇
∫
+∞
𝑢
𝑛−𝑚−𝑖−1
+𝐻𝑖 𝑠
< min {𝑀 (1 −
𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
1
) − 𝐿, 𝐿 − 𝑁} .
𝑏0
+∞
− 𝐻𝑖 ∫
𝑡+𝜏
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡 − 𝜏)𝑚−1
|𝑎 (𝑢)|
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑡 − 𝜏)𝑚−1
|𝑎 (𝑢)|
󵄨
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢)
≤
1
𝑏 (𝑡 + 𝜏)
𝑀
𝑏 (𝑡 + 𝜏)
+∞
−∫
(62)
∫
󵄨 󵄨
󵄨
× [󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . ,
󵄨
𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢
× (−𝐿 +
∀𝑡 ≥ 𝑇;
𝑢𝑚−1
󵄨
󵄨
[𝐻 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
|𝑎 (𝑢)| 0
1
𝑏 (𝑡 + 𝜏)
× ( − 𝐿 − 𝑥 (𝑡 + 𝜏) − 𝐻0
converges to a positive solution 𝑥 ∈ Ω3 (𝑁, 𝑀) of (5) and has
the error estimate (20), where {𝛼𝑘 }𝑘∈N0 is an arbitrary sequence
in [0, 1] satisfying (21);
(b) Equation (5) has uncountably many positive solutions
in Ω3 (𝑁, 𝑀).
𝑏 (𝑡) ≤ −𝑏0 ,
(63)
𝑇
∫
+∞
𝑢
𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
󵄨
× [𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢)
Abstract and Applied Analysis
<
𝑀
1
(−𝐿 +
𝑏 (𝑡 + 𝜏)
𝑏 (𝑡 + 𝜏)
− min {𝑀 (1 −
≤
13
Theorem 7. Let 𝑚 ≥ 2. Assume that there exist two constants
𝑀, 𝑁 with 𝑀 > 𝑁 > 0 and four functions 𝑃, 𝑄, 𝑅, 𝑊 ∈
𝐶([𝑡0 , +∞), R+ ) satisfying (16)–(18) and
1
) − 𝐿, 𝐿 − 𝑁})
𝑏0
𝑏 (𝑡) = 1
−𝑀
,
𝑏 (𝑡 + 𝜏)
1
𝑏 (𝑡 + 𝜏)
𝑥𝑘+1 (𝑡)
× ( − 𝐿 − 𝑥 (𝑡 + 𝜏) + 𝐻0
×∫
+∞
𝑡+𝜏
∫
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡 − 𝜏)𝑚−1
|𝑎 (𝑢)|
+∞
𝑢
󵄨 󵄨
󵄨
× [󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . ,
󵄨
𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢
+ 𝐻𝑖 ∫
+∞
∫
𝑡+𝜏
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑡 − 𝜏)𝑚−1
|𝑎 (𝑢)|
󵄨
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢)
≥
1
𝑏 (𝑡 + 𝜏)
× (−𝐿 +
+∫
𝑁
𝑏 (𝑡 + 𝜏)
+∞
𝑇
+∞
∫
𝑢
𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
󵄨
× [𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢)
>
𝑁
1
(−𝐿 +
𝑏 (𝑡 + 𝜏)
𝑏 (𝑡 + 𝜏)
+ min {𝑀 (1 −
≥
(65)
Then
(a) for any 𝐿 ∈ (𝑁, 𝑀), there exist 𝜃 ∈ (0, 1) and 𝑇 >
1 + |𝑡0 | + 𝜏 + |𝛾| such that for each 𝑥0 ∈ Ω1 (𝑁, 𝑀), the Mann
iterative sequence {𝑥𝑘 }𝑘∈N0 generated by the following scheme
𝑆𝐿 𝑥 (𝑡)
≥
eventually.
1
) − 𝐿, 𝐿 − 𝑁})
𝑏0
−𝑁
,
𝑏 (𝑡 + 𝜏)
(64)
which imply that 𝑆𝐿 (Ω3 (𝑁, 𝑀)) ⊆ Ω3 (𝑁, 𝑀). The rest of the
proof is identical with the proof of Theorem 5 and hence is
omitted. This completes the proof.
(1 − 𝛼𝑘 ) 𝑥𝑘 (𝑡) + 𝛼𝑘
{
{
{
{
{
{
× {𝐿 + (−1)𝑛 (𝑚 − 1) 𝐻0
{
{
{
{
{
{ ∞ 𝑡+2𝑗𝜏
+∞ +∞
{
{
{
×∑ ∫
∫ ∫ (((𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2 )
{
{
{
𝑢
{
𝑗=1 𝑡+(2𝑗−1)𝜏 𝑟
{
{
{
{
× (𝑎 (𝑢))−1 )
{
{
{
{
{
{
{ × [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)) , . . . ,
{
{
{
𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢 𝑑𝑟
{
{
{
𝑛−𝑖−1
{
+(−1)
− 1) 𝐻𝑖
{
(𝑚
{
{ ∞ 𝑡+2𝑗𝜏
{
+∞ +∞
{
{
{
×∑ ∫
∫ ∫ (((𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2 )
{
{
{
𝑢
{
𝑗=1 𝑡+(2𝑗−1)𝜏 𝑟
{
{
{
{
× (𝑎 (𝑢))−1 )
{
{
{
{
{
×ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . ,
{
{
{
{
{
{
{
𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢 𝑑𝑟} ,
{
{
{
{
{
{
{
𝑡 ≥ 𝑇, 𝑘 ∈ N0 ,
={
{(1 − 𝛼𝑘 ) 𝑥𝑘 (𝑇) + 𝛼𝑘
{
{
{
{
{
{
× {𝐿 + (−1)𝑛 (𝑚 − 1) 𝐻0
{
{
{
{
{
{
∞ 𝑇+2𝑗𝜏
+∞ +∞
{
{
{ ×∑ ∫
{
∫
∫ (((𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2 )
{
{
{
𝑢
𝑗=1 𝑇+(2𝑗−1)𝜏 𝑟
{
{
{
{
{
× (𝑎 (𝑢))−1 )
{
{
{
{
{
× [𝑔(𝑠)−𝑓(𝑠, 𝑥𝑘 (𝑓1 (𝑠)) , . . . ,
{
{
{
{
𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢 𝑑𝑟
{
{
{
{
{
+(−1)𝑛−𝑖−1 (𝑚 − 1) 𝐻𝑖
{
{
{
∞ 𝑇+2𝑗𝜏
+∞ +∞
{
{
{
{
×∑ ∫
∫
∫ (((𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2 )
{
{
{
𝑇+(2𝑗−1)𝜏
𝑟
𝑢
𝑗=1
{
{
{
{
{
× (𝑎 (𝑢))−1 )
{
{
{
{
{ ×ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . ,
{
{
{
{
{
{
𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢 𝑑𝑟} ,
{
{
{
{
{
𝛾 ≤ 𝑡 < 𝑇, 𝑘 ∈ N0
{
(66)
converges to a positive solution 𝑥 ∈ Ω1 (𝑁, 𝑀) of (5) and has
the error estimate (20), where {𝛼𝑘 }𝑘∈N0 is an arbitrary sequence
in [0, 1] with (21);
(b) Equation (5) has uncountably many positive solutions
in Ω1 (𝑁, 𝑀).
14
Abstract and Applied Analysis
Proof. Let 𝐿 ∈ (𝑁, 𝑀). It follows from (18) and (65) that there
exist 𝜃 ∈ (0, 1) and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| satisfying
𝑏 (𝑡) = 1,
𝜃=∫
+∞
𝑇
∫
∀𝑡 ≥ 𝑇;
𝑢𝑚−1
[𝐻 𝑠𝑛−𝑚−1 𝑃 (𝑠)
|𝑎 (𝑢)| 0
+∞
𝑢
𝑛−𝑚−𝑖−1
+𝐻𝑖 𝑠
∫
+∞
𝑇
∫
+∞
𝑢
(67)
𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢;
𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
𝑡
(69)
Define a mapping 𝑆𝐿 : Ω1 (𝑁, 𝑀) → CB([𝛾, +∞), R) by
𝑆𝐿 𝑥 (𝑡)
With a view to (16), (68), and (70), we derive that for any
𝑥, 𝑦 ∈ Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
󵄨
󵄨󵄨
󵄨󵄨𝑆𝐿 𝑥 (𝑡) − 𝑆𝐿 𝑦 (𝑡)󵄨󵄨󵄨
𝑡+2𝑗𝜏
∫
+∞
𝑗=1 𝑡+(2𝑗−1)𝜏 𝑟
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))
󵄨
−𝑓 (𝑠, 𝑦 (𝑓1 (𝑠)) , . . . , 𝑦 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢 𝑑𝑟
+ (𝑚 − 1) 𝐻𝑖
×∑∫
𝑡+2𝑗𝜏
∫
+∞
∫
+∞
∫
𝑟
+∞
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2
𝑅 (𝑠) 𝑑𝑠 𝑑𝑢 𝑑𝑟
|𝑎 (𝑢)|
𝑡
∫
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡)𝑚−1
𝑃 (𝑠) 𝑑𝑠 𝑑𝑢
|𝑎 (𝑢)|
+∞
𝑢
󵄩
󵄩
+ 𝐻𝑖 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩
+∞
×∫
∫
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑡)𝑚−1
𝑅 (𝑠) 𝑑𝑠 𝑑𝑢
|𝑎 (𝑢)|
+∞
𝑢
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩
+∞
×∫
𝑇
∫
𝑢𝑚−1
[𝐻 𝑠𝑛−𝑚−1 𝑃 (𝑠)
|𝑎 (𝑢)| 0
+∞
𝑢
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢
󵄩
󵄩
= 𝜃 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
(71)
which gives (27). By virtue of (17), (69), and (70), we deduce
that for any 𝑥 ∈ Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
𝑆𝐿 𝑥 (𝑡)
≤ 𝐿 + (𝑚 − 1) 𝐻0
+∞
×∫
𝑡
∫
+∞
𝑟
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
|𝑎 (𝑢)|
󵄨 󵄨
󵄨
󵄨
× [󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢 𝑑𝑟
+ (𝑚 − 1) 𝐻𝑖
≤ (𝑚 − 1) 𝐻0
∞
+∞
×∫
𝑡
𝐿 + (−1)𝑛 (𝑚 − 1) 𝐻0
{
{
{
∞ 𝑡+2𝑗𝜏
+∞ +∞
{
{
{
∫ ∫ (((𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2 )
{× ∑ ∫
{
{
𝑢
{
{ 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑟
{
{
{
× (𝑎 (𝑢))−1 )
{
{
{
{
{
{
{× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢 𝑑𝑟
{
{
{
{
{
𝑛−𝑖−1
{
(𝑚 − 1) 𝐻𝑖
{
{+(−1)
= { ∞ 𝑡+2𝑗𝜏
+∞ +∞
{
{
{
×
∑
∫
∫
∫ (((𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2 )
{
{
{ 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑟
𝑢
{
{
{
{
{
× (𝑎 (𝑢))−1 )
{
{
{
{
{
{×ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢 𝑑𝑟,
{
{
{
{
{
{
{
𝑡 ≥ 𝑇, 𝑥 ∈ Ω1 (𝑁, 𝑀) ,
{
{
{
{
𝛾 ≤ 𝑡 < 𝑇, 𝑥 ∈ Ω1 (𝑁, 𝑀) .
{𝑆𝐿 𝑥 (𝑇) ,
(70)
∞
∫
󵄩
󵄩
= 𝐻0 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩
< min {𝑀 − 𝐿, 𝐿 − 𝑁} .
×∑∫
+∞
×∫
𝑢
󵄨
󵄨
[𝐻 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
|𝑎 (𝑢)| 0
+𝐻𝑖 𝑠
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
𝑃 (𝑠) 𝑑𝑠 𝑑𝑢 𝑑𝑟
|𝑎 (𝑢)|
𝑡
𝑟
𝑢
󵄩
󵄩
+ (𝑚 − 1) 𝐻𝑖 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩
+∞
×∫
(68)
𝑚−1
𝑛−𝑚−𝑖−1
󵄩
󵄩
≤ (𝑚 − 1) 𝐻0 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2
|𝑎 (𝑢)|
𝑡
𝑟
𝑢
󵄨󵄨
󵄨
× 󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢 𝑑𝑟
+∞
×∫
∫
𝑗=1 𝑡+(2𝑗−1)𝜏 𝑟
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))
󵄨
−ℎ (𝑠, 𝑦 (ℎ1 (𝑠)) , . . . , 𝑦 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢 𝑑𝑟
+∞
∫
+∞
≤ 𝐿 + (𝑚 − 1) 𝐻0
+∞
×∫
+∞
∫
𝑡
∫
+∞
𝑟
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
|𝑎 (𝑢)|
󵄨
󵄨
× [󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠)] 𝑑𝑠 𝑑𝑢 𝑑𝑟
+ (𝑚 − 1) 𝐻𝑖
+∞
×∫
𝑡
∫
+∞
𝑟
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2
𝑊 (𝑠) 𝑑𝑠 𝑑𝑢 𝑑𝑟
|𝑎 (𝑢)|
Abstract and Applied Analysis
= 𝐿 + 𝐻0 ∫
+∞
𝑡
+ 𝐻𝑖 ∫
+∞
𝑡
+∞
∫
𝑢
∫
+∞
15
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑡)𝑚−1
|𝑎 (𝑢)|
󵄨
󵄨󵄨
× [󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠)] 𝑑𝑠 𝑑𝑢
𝑛−𝑚−𝑖−1
(𝑠 − 𝑢)
(𝑢 − 𝑡)
|𝑎 (𝑢)|
𝑢
{
+ 𝛼𝑘 {𝐿 + (−1)𝑛 (𝑚 − 1) 𝐻0
{
∞
×∑∫
𝑚−1
≤𝐿+∫
𝑇
∫
+∞
𝑢
∞
×∑∫
𝑆𝐿 𝑥 (𝑡)
≥ 𝐿 − (𝑚 − 1) 𝐻0
𝑡
𝑟
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
𝑎 (𝑢)
𝑡+2𝑗𝜏
+∞
∫
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2
𝑎 (𝑢)
󵄨󵄨
}
󵄨󵄨󵄨
×ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . , 𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢 𝑑𝑟} − 𝑥 (𝑡)󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
}
󵄨
󵄨
󵄨
󵄨
≤ (1 − 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨 + 𝛼𝑘 󵄨󵄨󵄨𝑆𝐿 𝑥𝑘 (𝑡) − 𝑆𝐿 𝑥 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
󵄨
󵄨
≤ (1 − 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨 + 𝛼𝑘 𝜃 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
= (1 − (1 − 𝜃) 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄩
󵄩
≤ 𝑒−(1−𝜃)𝛼𝑘 󵄩󵄩󵄩𝑥𝑘 − 𝑥󵄩󵄩󵄩
≤ 𝑀,
+∞
𝑢
𝑗=1 𝑡+(2𝑗−1)𝜏 𝑟
< 𝐿 + min {𝑀 − 𝐿, 𝐿 − 𝑁}
∫
+∞
+ (−1)𝑛−𝑖−1 (𝑚 − 1) 𝐻𝑖
𝑢𝑚−1
|𝑎 (𝑢)|
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
+∞
∫
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)) , . . . , 𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢 𝑑𝑟
󵄨
󵄨
× [𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
×∫
+∞
∫
𝑗=1 𝑡+(2𝑗−1)𝜏 𝑟
× 𝑊 (𝑠) 𝑑𝑠 𝑑𝑢
+∞
𝑡+2𝑗𝜏
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
|𝑎 (𝑢)|
𝑘
󵄩
󵄩
≤ 𝑒−(1−𝜃) ∑𝑝=0 𝛼𝑝 󵄩󵄩󵄩𝑥0 − 𝑥󵄩󵄩󵄩 ,
󵄨 󵄨
󵄨
× [ 󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . ,
∀𝑘 ∈ N0 , 𝑡 ≥ 𝑇,
(73)
󵄨
𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢 𝑑𝑟
− (𝑚 − 1) 𝐻𝑖
+∞
+∞
+∞
𝑛−𝑚−𝑖−1
(𝑢 − 𝑟)
(𝑢)|
|𝑎
𝑡
𝑟
𝑢
󵄨󵄨
󵄨
× 󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢 𝑑𝑟
×∫
≥𝐿−∫
∫
+∞
𝑇
∫
∫
+∞
𝑢
(𝑠 − 𝑢)
𝑚−2
𝑢𝑚−1
|𝑎 (𝑢)|
󵄨
󵄨
× [𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠)) + 𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
> 𝐿 − min {𝑀 − 𝐿, 𝐿 − 𝑁}
𝑧𝑝 (𝑡) = 𝐿 𝑘 + (−1)𝑛 (𝑚 − 1) 𝐻0
≥ 𝑁,
(72)
which mean that 𝑆𝐿 (Ω1 (𝑁, 𝑀)) ⊆ Ω1 (𝑁, 𝑀). Coupled
with (27) and (68), we get that 𝑆𝐿 is a contraction mapping
in Ω1 (𝑁, 𝑀) and it possesses a unique fixed point 𝑥 ∈
Ω1 (𝑁, 𝑀). Clearly, 𝑥 ∈ Ω1 (𝑁, 𝑀) is a positive solution of
(5).
From (27), (66), and (70), we gain that
󵄨
󵄨󵄨
󵄨󵄨𝑥𝑘+1 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨
= 󵄨󵄨󵄨 (1 − 𝛼𝑘 ) 𝑥𝑘 (𝑡)
󵄨󵄨
󵄨󵄨
which yields (20). It follows from (20) and (21) that
lim𝑘 → ∞ 𝑥𝑘 = 𝑥.
Now we prove that (b) holds. Let 𝐿 1 , 𝐿 2 ∈ (𝑁, 𝑀) and
𝐿 1 ≠ 𝐿 2 . As in the proof of (a), we conclude that for each
𝑝 ∈ {1, 2}, there exist 𝜃𝑝 ∈ (0, 1), 𝑇𝑝 > 1 + |𝑡0 | + 𝜏 + |𝛾| and
𝑆𝐿 𝑝 : Ω1 (𝑁, 𝑀) → Ω1 (𝑁, 𝑀) satisfying (69)–(77), where 𝐿,
𝜃, 𝑇, and 𝑆𝐿 are replaced by 𝐿 𝑝 , 𝜃𝑝 , 𝑇𝑝 , and 𝑆𝐿 𝑝 , respectively,
and 𝑆𝐿 𝑝 has a unique fixed point 𝑧𝑝 ∈ Ω1 (𝑁, 𝑀), which is a
positive solution of (5) in Ω1 (𝑁, 𝑀), that is,
∞
×∑∫
𝑡+2𝑗𝜏
∫
+∞
𝑗=1 𝑡+(2𝑗−1)𝜏 𝑟
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
𝑎 (𝑢)
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑧𝑝 (𝑓1 (𝑠)) , . . . , 𝑧𝑝 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢 𝑑𝑟
+ (−1)𝑛−𝑖−1 (𝑚 − 1) 𝐻𝑖
∞
×∑∫
𝑡+2𝑗𝜏
∫
+∞
𝑗=1 𝑡+(2𝑗−1)𝜏 𝑟
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2
𝑎 (𝑢)
× ℎ (𝑠, 𝑧𝑝 (ℎ1 (𝑠)) , . . . , 𝑧𝑝 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢 𝑑𝑟,
∀𝑡 ≥ 𝑇𝑝 , 𝑝 ∈ {1, 2} .
(74)
16
Abstract and Applied Analysis
For purpose of proving (b), we just need to show that 𝑧1 ≠ 𝑧2 .
It follows from (16), (27), (68), and (74) that
󵄨
󵄨󵄨
󵄨󵄨𝑧1 (𝑡) − 𝑧2 (𝑡)󵄨󵄨󵄨
𝑥𝑘+1 (𝑡)
󵄨
󵄨
󵄩
󵄩
≥ 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − (𝑚 − 1) 𝐻0 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩
×∫
+∞
𝑡
+∞
∫
𝑟
∫
+∞
(((𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2 )
𝑢
×(|𝑎 (𝑢)|)−1 ) 𝑃 (𝑠) 𝑑𝑠 𝑑𝑢 𝑑𝑟
󵄩
󵄩
− (𝑚 − 1) 𝐻𝑖 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩
×∫
+∞
𝑡
+∞
∫
𝑟
∫
+∞
(((𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2 )
𝑢
×(|𝑎 (𝑢)|)−1 ) 𝑅 (𝑠) 𝑑𝑠 𝑑𝑢 𝑑𝑟 (75)
󵄨
󵄨 󵄩
󵄩
≥ 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩
×∫
+∞
+∞
∫
max{𝑇1 ,𝑇2 } 𝑢
𝑢𝑚−1
|𝑎 (𝑢)|
× [𝐻0 𝑠𝑛−𝑚−1 𝑃 (𝑠)
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢
󵄨
󵄨
󵄩
󵄩
> 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − max {𝜃1 , 𝜃2 } 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩 ,
∀𝑡 ≥ max {𝑇1 , 𝑇2 } ,
which yields that
󵄩
󵄩󵄩
󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩 ≥
󵄨󵄨
󵄨
󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨
> 0,
1 + max {𝜃1 , 𝜃2 }
(76)
that is, 𝑧1 ≠ 𝑧2 . This completes the proof.
Theorem 8. Let 𝑚 = 1. Assume that there exist two constants
𝑀, 𝑁 with 𝑀 > 𝑁 > 0 and four functions 𝑃, 𝑄, 𝑅, 𝑊 ∈
𝐶([𝑡0 , +∞), R+ ) satisfying (16), (17), (65), and
∫
+∞
𝑡0
+∞
∫
𝑢
1
|𝑎 (𝑢)|
󵄨
󵄨
× [|𝑠|𝑛−2 max {𝑃 (𝑠) , 𝑄 (𝑠) , 󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨}
+|𝑠|
< +∞.
Then
𝑛−𝑖−2
max {𝑅 (𝑠) , 𝑊 (𝑠)}] 𝑑𝑠 𝑑𝑢
(a) for any 𝐿 ∈ (𝑁, 𝑀), there exist 𝜃 ∈ (0, 1) and 𝑇 >
1 + |𝑡0 | + 𝜏 + |𝛾| such that for each 𝑥0 ∈ Ω1 (𝑁, 𝑀), the Mann
iterative sequence {𝑥𝑘 }𝑘∈N0 generated by the following scheme
(77)
(1 − 𝛼𝑘 ) 𝑥𝑘 (𝑡)
{
{
{
{
{
{
{
{
{
{
(−1)𝑛
{
{
{𝐿
+
+𝛼
{
𝑘
{
(𝑛 − 2)!
{
{
{
{
{
{
∞ 𝑡+2𝑗𝜏
+∞
{
{
(𝑠 − 𝑢)𝑛−2
{
{
×
∑
∫
∫
{
{
𝑎 (𝑢)
{
𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
{
{
{
{
{
{
{
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)) , . . . , 𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
{
{
{
{
{
{
{
{
{
(−1)𝑛−𝑖−1
{
{
{
+
{
{
(𝑛 − 𝑖 − 2)!
{
{
{
{
{
{
∞ 𝑡+2𝑗𝜏
+∞
{
(𝑠 − 𝑢)𝑛−𝑖−2
{
{
{
×
∑
∫
∫
{
{
𝑎 (𝑢)
{
𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
{
{
{
{
{
{
{
{
{
× ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . , 𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} ,
{
{
{
{
{
{
{
{
{
𝑡 ≥ 𝑇, 𝑘 ∈ N0 ,
{
{
{
={
{
{
{
(1 − 𝛼𝑘 ) 𝑥𝑘 (𝑇)
{
{
{
{
{
{
{
{
{
{
(−1)𝑛
{
{
{𝐿
+
+𝛼
{
𝑘
{
{
(𝑛 − 2)!
{
{
{
{
{
∞ 𝑇+2𝑗𝜏
{
+∞
{
(𝑠 − 𝑢)𝑛−2
{
{
×∑ ∫
∫
{
{
{
𝑎 (𝑢)
{
𝑗=1 𝑇+(2𝑗−1)𝜏 𝑢
{
{
{
{
{
{
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)) , . . . , 𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
{
{
{
{
{
{
{
{
{
{
(−1)𝑛−𝑖−1
{
{
+
{
{
{
(𝑛 − 𝑖 − 2)!
{
{
{
{
{
∞ 𝑇+2𝑗𝜏
+∞
{
{
(𝑠 − 𝑢)𝑛−𝑖−2
{
{
×
∑
∫
∫
{
{
𝑎 (𝑢)
{
{
𝑗=1 𝑇+(2𝑗−1)𝜏 𝑢
{
{
{
{
{
{
{
{
× ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . , 𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} ,
{
{
{
{
{
{
{
{
𝛾 ≤ 𝑡 < 𝑇, 𝑘 ∈ N0
{
(78)
converges to a positive solution 𝑥 ∈ Ω1 (𝑁, 𝑀) of (5) and has
the error estimate (20), where {𝛼𝑘 }𝑘∈N0 is an arbitrary sequence
in [0, 1] with (21);
(b) Equation (5) has uncountably many positive solutions
in Ω1 (𝑁, 𝑀).
Abstract and Applied Analysis
17
Proof. Let 𝐿 ∈ (𝑁, 𝑀). It follows from (65) and (77) that there
exist 𝜃 ∈ (0, 1) and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| satisfying (67),
𝜃= ∫
+∞
𝑇
∫
+∞
𝑢
≤
+
1
|𝑎 (𝑢)|
×[
+∞ +∞
1
(𝑠 − 𝑢)𝑛−2
󵄩
󵄩󵄩
𝑃 (𝑠) 𝑑𝑠 𝑑𝑢
󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ∫ ∫
(𝑛 − 2)!
|𝑎 (𝑢)|
𝑇
𝑢
1
󵄩
󵄩󵄩
󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩
(𝑛 − 𝑖 − 2)!
×∫
𝑠𝑛−2
𝑠𝑛−𝑖−2
𝑃 (𝑠) +
𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢,
(𝑛 − 2)!
(𝑛 − 𝑖 − 2)!
+∞
𝑇
∫
+∞
𝑢
+∞
󵄩
󵄩
≤ 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ∫
𝑇
(79)
∫
+∞
𝑇
∫
+∞
𝑢
𝑠𝑛−2 󵄨󵄨
1
󵄨
[
(󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
|𝑎 (𝑢)| (𝑛 − 2)! 󵄨
𝑠𝑛−𝑖−2
+
𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
(𝑛 − 𝑖 − 2)!
(𝑠 − 𝑢)𝑛−𝑖−2
𝑅 (𝑠) 𝑑𝑠 𝑑𝑢
|𝑎 (𝑢)|
∫
+∞
𝑢
1
𝑠𝑛−2
𝑃 (𝑠)
[
|𝑎 (𝑢)| (𝑛 − 2)!
+
𝑠𝑛−𝑖−2
𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢
(𝑛 − 𝑖 − 2)!
󵄩
󵄩
= 𝜃 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
(80)
(82)
< min {𝑀 − 𝐿, 𝐿 − 𝑁} .
Define a mapping 𝑆𝐿 : Ω1 (𝑁, 𝑀) → CB([𝛾, +∞), R) by
which gives (27). It follows from (17), (80), and (81) that for
any 𝑥 ∈ Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
𝑆𝐿 𝑥 (𝑡)
𝑆𝐿 𝑥 (𝑡)
+∞
(𝑠 − 𝑢)𝑛−2
(−1)𝑛 ∞ 𝑡+2𝑗𝜏
{
{
𝐿
+
∑
∫
∫
{
{
𝑎 (𝑢)
(𝑛 − 2)! 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
{
{
{
{
{
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
{
{
{
{
{
{
+∞
{
(𝑠 − 𝑢)𝑛−𝑖−2
(−1)𝑛−𝑖−1 ∞ 𝑡+2𝑗𝜏
={
+
∑
∫
∫
{
𝑎 (𝑢)
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
{
{
{
{
{
{
{
×ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢,
{
{
{
{
{
𝑡 ≥ 𝑇, 𝑥 ∈ Ω1 (𝑁, 𝑀) ,
{
{
𝛾 ≤ 𝑡 < 𝑇, 𝑥 ∈ Ω1 (𝑁, 𝑀) .
{𝑆𝐿 𝑥 (𝑇) ,
(81)
By virtue of (16), (79), and (81), we derive that for any 𝑥, 𝑦 ∈
Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
∞ 𝑡+2𝑗𝜏
+∞
1
(𝑠 − 𝑢)𝑛−2
∑∫
∫
(𝑛 − 2)! 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
|𝑎 (𝑢)|
󵄨 󵄨
󵄨
󵄨
× [󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢
+
∞ 𝑡+2𝑗𝜏
+∞
1
(𝑠 − 𝑢)𝑛−𝑖−2
∑∫
∫
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
|𝑎 (𝑢)|
󵄨
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
≤𝐿+∫
+∞
𝑇
×[
∫
+∞
𝑢
1
|𝑎 (𝑢)|
𝑛−2
𝑠𝑛−𝑖−2
𝑠
󵄨
󵄨
(󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠)) +
𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
(𝑛 − 2)!
(𝑛 − 𝑖 − 2)!
< 𝐿 + min {𝑀 − 𝐿, 𝐿 − 𝑁}
≤ 𝑀,
󵄨
󵄨󵄨
󵄨󵄨𝑆𝐿 𝑥 (𝑡) − 𝑆𝐿 𝑦 (𝑡)󵄨󵄨󵄨
𝑆𝐿 𝑥 (𝑡)
∞ 𝑡+2𝑗𝜏
+∞
1
(𝑠 − 𝑢)𝑛−2
≤
∑∫
∫
(𝑛 − 2)! 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
|𝑎 (𝑢)|
≥𝐿−
󵄨
× 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))
󵄨
−𝑓 (𝑠, 𝑦 (𝑓1 (𝑠)) , . . . , 𝑦 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
+
≤𝐿+
∞
𝑡+2𝑗𝜏
+∞
1
∑∫
∫
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
𝑛−𝑖−2
(𝑠 − 𝑢)
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))
󵄨
−ℎ (𝑠, 𝑦 (ℎ1 (𝑠)) , . . . , 𝑦 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
∞ 𝑡+2𝑗𝜏
+∞
1
(𝑠 − 𝑢)𝑛−2
∑∫
∫
(𝑛 − 2)! 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
|𝑎 (𝑢)|
󵄨 󵄨
󵄨
󵄨
× [󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢
−
∞ 𝑡+2𝑗𝜏
+∞
1
(𝑠 − 𝑢)𝑛−𝑖−2
∑∫
∫
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
|𝑎 (𝑢)|
󵄨
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
≥𝐿−∫
+∞
𝑇
∫
+∞
𝑢
1
|𝑎 (𝑢)|
18
Abstract and Applied Analysis
×[
𝑠𝑛−𝑖−2
𝑠𝑛−2 󵄨󵄨
󵄨
(󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠)) +
𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
(𝑛 − 2)!
(𝑛 − 𝑖 − 2)!
> 𝐿 − min {𝑀 − 𝐿, 𝐿 − 𝑁}
and 𝑆𝐿 𝑝 has a unique fixed point 𝑧𝑝 ∈ Ω1 (𝑁, 𝑀), which is a
positive solution of (5) in Ω1 (𝑁, 𝑀), that is,
𝑧𝑝 (𝑡) = 𝐿 𝑝 +
≥ 𝑁,
(83)
(−1)𝑛
(𝑛 − 2)!
∞
×∑∫
𝑡+2𝑗𝜏
+∞
∫
𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
which mean that 𝑆𝐿 (Ω1 (𝑁, 𝑀)) ⊆ Ω1 (𝑁, 𝑀). Combined
with (27) and (79), we know that 𝑆𝐿 is a contraction mapping
in Ω1 (𝑁, 𝑀) and it possesses a unique fixed point 𝑥 ∈
Ω1 (𝑁, 𝑀). Obviously, 𝑥 ∈ Ω1 (𝑁, 𝑀) is a positive solution
of (5).
In light of (27), (78), and (81), we gain that
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑧𝑝 (𝑓1 (𝑠)) , . . . , 𝑧𝑝 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
+
+∞
(𝑠 − 𝑢)𝑛−𝑖−2
(−1)𝑛−𝑖−1 ∞ 𝑡+2𝑗𝜏
∑∫
∫
𝑎 (𝑢)
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
× ℎ (𝑠, 𝑧𝑝 (ℎ1 (𝑠)) , . . . , 𝑧𝑝 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢,
∀𝑡 ≥ 𝑇𝑝 , 𝑝 ∈ {1, 2} .
󵄨
󵄨󵄨
󵄨󵄨𝑥𝑘+1 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
(85)
󵄨󵄨
󵄨󵄨
= 󵄨󵄨󵄨󵄨 (1 − 𝛼𝑘 ) 𝑥𝑘 (𝑡)
󵄨󵄨
󵄨
In order to prove (b), we just need to show that 𝑧1 ≠ 𝑧2 . In
view of (16), (27), (79), and (85), we get that
+∞
{
(−1)𝑛 ∞ 𝑡+2𝑗𝜏
(𝑠 − 𝑢)𝑛−2
+ 𝛼𝑘 {𝐿 +
∑∫
∫
𝑎 (𝑢)
(𝑛 − 2)! 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
{
󵄨
󵄨󵄨
󵄨󵄨𝑧1 (𝑡) − 𝑧2 (𝑡)󵄨󵄨󵄨
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)) , . . . , 𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
+
∞ 𝑡+2𝑗𝜏
+∞
1
(𝑠 − 𝑢)𝑛−2
∑∫
∫
(𝑛 − 2)! 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
|𝑎 (𝑢)|
󵄨
−𝑓 (𝑠, 𝑧1 (𝑓1 (𝑠)) , . . . , 𝑧1 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
󵄨󵄨
󵄨󵄨
}
󵄨
×ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . , 𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} − 𝑥 (𝑡)󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
}
󵄨
󵄨
󵄨
󵄨
≤ (1 − 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨 + 𝛼𝑘 󵄨󵄨󵄨𝑆𝐿 𝑥𝑘 (𝑡) − 𝑆𝐿 𝑥 (𝑡)󵄨󵄨󵄨
−
∞ 𝑡+2𝑗𝜏
+∞
1
(𝑠 − 𝑢)𝑛−𝑖−2
∑∫
∫
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑧1 (ℎ1 (𝑠)) , . . . , 𝑧1 (ℎ𝑙 (𝑠)))
󵄨
−ℎ (𝑠, 𝑧2 (ℎ1 (𝑠)) , . . . , 𝑧2 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
󵄨
󵄨
󵄨
󵄨
≤ (1 − 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨 + 𝛼𝑘 𝜃 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
= (1 − (1 − 𝜃) 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄩
󵄩
≤ 𝑒−(1−𝜃)𝛼𝑘 󵄩󵄩󵄩𝑥𝑘 − 𝑥󵄩󵄩󵄩
𝑘
󵄨
󵄨
≥ 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 −
󵄨
× 󵄨󵄨󵄨𝑓 (𝑠, 𝑧2 (𝑓1 (𝑠)) , . . . , 𝑧2 (𝑓𝑙 (𝑠)))
+∞
(𝑠 − 𝑢)𝑛−𝑖−2
(−1)𝑛−𝑖−1 ∞ 𝑡+2𝑗𝜏
∑∫
∫
𝑎 (𝑢)
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+(2𝑗−1)𝜏 𝑢
󵄩
󵄩
≤ 𝑒−(1−𝜃) ∑𝑝=0 𝛼𝑝 󵄩󵄩󵄩𝑥0 − 𝑥󵄩󵄩󵄩 ,
(𝑠 − 𝑢)𝑛−2
𝑎 (𝑢)
∀𝑘 ∈ N0 , 𝑡 ≥ 𝑇,
(84)
󵄩
󵄩 +∞ +∞ 𝑛−2
𝑠
󵄨
󵄨 󵄩󵄩𝑧 − 𝑧2 󵄩󵄩󵄩
≥ 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − 󵄩 1
𝑃 (𝑠) 𝑑𝑠 𝑑𝑢
∫ ∫
(𝑛 − 2)! 𝑡
(𝑢)|
|𝑎
𝑢
󵄩
󵄩󵄩
󵄩𝑧 − 𝑧2 󵄩󵄩󵄩 +∞ +∞ 𝑠𝑛−𝑖−2
𝑅 (𝑠) 𝑑𝑠 𝑑𝑢
− 󵄩 1
∫ ∫
(𝑛 − 𝑖 − 2)! 𝑡
|𝑎 (𝑢)|
𝑢
󵄨
󵄨 󵄩
󵄩
≥ 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩
×∫
+∞
∫
+∞
max{𝑇1 ,𝑇2 } 𝑢
which yields (20). It follows from (20) and (21) that
lim𝑘 → ∞ 𝑥𝑘 = 𝑥.
Now we prove that (b) holds. Let 𝐿 1 , 𝐿 2 ∈ (𝑁, 𝑀) and
𝐿 1 ≠ 𝐿 2 . As in the proof of (a), we conclude that for each 𝑝 ∈
{1, 2}, there exist 𝜃𝑝 ∈ (0, 1), 𝑇𝑝 > 1 + |𝑡0 | + 𝜏 + |𝛾| and 𝑆𝐿 𝑝 :
Ω1 (𝑁, 𝑀) → Ω1 (𝑁, 𝑀) satisfying (67) and (79)–(81), where
𝐿, 𝜃, 𝑇, and 𝑆𝐿 are replaced by 𝐿 𝑝 , 𝜃𝑝 , 𝑇𝑝 , and 𝑆𝐿 𝑝 , respectively,
1
|𝑎 (𝑢)|
𝑠𝑛−2
𝑠𝑛−𝑖−2
𝑃 (𝑠) +
𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢
(𝑛 − 2)!
(𝑛 − 𝑖 − 2)!
󵄨
󵄨
󵄩
󵄩
> 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − max {𝜃1 , 𝜃2 } 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩 ,
×[
∀𝑡 ≥ max {𝑇1 , 𝑇2 } ,
(86)
Abstract and Applied Analysis
19
which implies that
󵄨󵄨
󵄨
󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨
󵄩󵄩
󵄩
> 0,
󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩 ≥
1 + max {𝜃1 , 𝜃2 }
(87)
that is, 𝑧1 ≠ 𝑧2 . This completes the proof.
Theorem 9. Let 𝑚 ≥ 2. Assume that there exist two constants
𝑀, 𝑁 with 𝑀 > 𝑁 > 0 and four functions 𝑃, 𝑄, 𝑅, 𝑊 ∈
𝐶([𝑡0 , +∞), R+ ) satisfying (16), (17),
∫
+∞
𝑡0
+∞
∫
𝑟
∫
+∞
𝑢
converges to a positive solution 𝑥 ∈ Ω1 (𝑁, 𝑀) of (5) and has
the error estimate (20), where {𝛼𝑘 }𝑘∈N0 is an arbitrary sequence
in [0, 1] with (21);
(b) Equation (5) has uncountably many positive solutions
in Ω1 (𝑁, 𝑀).
Proof. Set 𝐿 ∈ (𝑁, 𝑀). In view of (88) and (89), there exist
𝜃 ∈ (0, 1) and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| such that
𝑏 (𝑡) = −1,
|𝑟| |𝑢|𝑚
|𝑎 (𝑢)|
𝜃=
󵄨
󵄨
× [|𝑠|𝑛−𝑚−1 max {𝑃 (𝑠) , 𝑄 (𝑠) , 󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨}
𝑚 − 1 +∞ +∞ +∞ 𝑟𝑢𝑚−2
∫ ∫ ∫
𝜏
|𝑎 (𝑢)|
𝑇
𝑟
𝑢
𝑛−𝑚−1
× [𝐻0 𝑠
+|𝑠|𝑛−𝑚−𝑖−1 max {𝑅 (𝑠) , 𝑊 (𝑠)}] 𝑑𝑠 𝑑𝑢 𝑑𝑟
< +∞,
(88)
𝑏 (𝑡) = −1 eventually.
(89)
Then
(a) for any 𝐿 ∈ (𝑁, 𝑀), there exist 𝜃 ∈ (0, 1) and 𝑇 >
1 + |𝑡0 | + 𝜏 + |𝛾| such that for each 𝑥0 ∈ Ω1 (𝑁, 𝑀), the Mann
iterative sequence {𝑥𝑘 }𝑘∈N0 generated by the following scheme
𝑥𝑘+1 (𝑡)
(1 − 𝛼𝑘 ) 𝑥𝑘 (𝑡)
{
{
{
{
{
{
{+𝛼𝑘 {𝐿 + (−1)𝑛−1 (𝑚 − 1) 𝐻0
{
{
{
{
{
∞ +∞ +∞ +∞
{
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
{
{
{
×
∑
∫
∫
∫
{
{
𝑎 (𝑢)
𝑡+𝑗𝜏 𝑟
𝑢
{
{
{ 𝑗=1
{
{
{
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)) , . . . , 𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢 𝑑𝑟
{
{
{
{
+(−1)𝑛−𝑖 (𝑚 − 1) 𝐻𝑖
{
{
{
{ ∞ +∞ +∞ +∞ (𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2
{
{
{
×∑ ∫ ∫ ∫
{
{
{
𝑎 (𝑢)
𝑢
{
𝑗=1 𝑡+𝑗𝜏 𝑟
{
{
{
{
{
{
{ ×ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . , 𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢 𝑑𝑟} ,
{
{
{
{
{
{
𝑡 ≥ 𝑇, 𝑘 ∈ N0 ,
{
{
= {(1 − 𝛼𝑘 ) 𝑥𝑘 (𝑇)
{
{
{
{
{
{
+𝛼𝑘 {𝐿 + (−1)𝑛−1 (𝑚 − 1) 𝐻0
{
{
{
{
{
∞ +∞
+∞ +∞
{
{
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
{
{
×
∑
∫
∫
∫
{
{
{
𝑎 (𝑢)
𝑢
{
𝑗=1 𝑇+𝑗𝜏 𝑟
{
{
{
{
{× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)) , . . . , 𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢 𝑑𝑟
{
{
{
{
{
+(−1)𝑛−𝑖 (𝑚 − 1) 𝐻𝑖
{
{
{
{
{
∞ +∞
+∞ +∞
{
{
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2
{
{
×
∑
∫
∫
∫
{
{
𝑎 (𝑢)
{
𝑢
{
𝑗=1 𝑇+𝑗𝜏 𝑟
{
{
{
{
{
{
{ ×ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . , 𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢 𝑑𝑟} ,
{
{
{
{
𝛾 ≤ 𝑡 < 𝑇, 𝑘 ∈ N0
{
(90)
∀𝑡 ≥ 𝑇;
𝑛−𝑚−𝑖−1
𝑃 (𝑠) + 𝐻𝑖 𝑠
(91)
(92)
𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢 𝑑𝑟;
𝑚 − 1 +∞ +∞ +∞ 𝑟𝑢𝑚−2
∫ ∫ ∫
𝜏
|𝑎 (𝑢)|
𝑇
𝑟
𝑢
󵄨
󵄨
× [𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠)) + 𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢 𝑑𝑟
< min {𝑀 − 𝐿, 𝐿 − 𝑁} .
(93)
Define a mapping 𝑆𝐿 : Ω1 (𝑁, 𝑀) → CB([𝛾, +∞), R) by
𝑆𝐿 𝑥 (𝑡)
𝑛−1
𝐿 + (−1) (𝑚 − 1) 𝐻0
{
{
{
{
{
{ ∞ +∞ +∞ +∞ (𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
{
{
{
×∑ ∫ ∫ ∫
{
{
𝑎 (𝑢)
{
𝑢
𝑗=1 𝑡+𝑗𝜏 𝑟
{
{
{
{
{
{
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢 𝑑𝑟
{
{
{
{
{
{
𝑛−𝑖
= {+(−1) (𝑚 − 1) 𝐻𝑖
{
{
{
∞ +∞ +∞ +∞
{
{
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2
{
{
×
∑
∫
∫
∫
{
{
𝑎 (𝑢)
{
𝑢
𝑗=1 𝑡+𝑗𝜏 𝑟
{
{
{
{
{
×ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢 𝑑𝑟,
{
{
{
{
{
{
𝑡 ≥ 𝑇, 𝑥 ∈ Ω1 (𝑁, 𝑀) ,
{
{
𝑆
𝑥
,
𝛾
≤
𝑡 < 𝑇, 𝑥 ∈ Ω1 (𝑁, 𝑀) .
(𝑇)
{ 𝐿
(94)
By virtue of (16), (92), (94), and Lemma 1, we acquire that for
any 𝑥, 𝑦 ∈ Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
󵄨󵄨
󵄨
󵄨󵄨𝑆𝐿 𝑥 (𝑡) − 𝑆𝐿 𝑦 (𝑡)󵄨󵄨󵄨
∞
≤ (𝑚 − 1) 𝐻0 ∑ ∫
+∞
𝑗=1 𝑡+𝑗𝜏
∫
+∞
𝑟
+∞
∫
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))
󵄨
−𝑓 (𝑠, 𝑦 (𝑓1 (𝑠)) , . . . , 𝑦 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢 𝑑𝑟
20
Abstract and Applied Analysis
∞
+ (𝑚 − 1) 𝐻𝑖 ∑ ∫
+∞
∫
𝑗=1 𝑡+𝑗𝜏
+∞
+∞
∫
𝑟
𝑢
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))
󵄨
−ℎ (𝑠, 𝑦 (ℎ1 (𝑠)) , . . . , 𝑦 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢 𝑑𝑟
≤ (𝑚 − 1)
×∫
+∞
𝑇
𝐻0 󵄩󵄩
󵄩
󵄩𝑥 − 𝑦󵄩󵄩󵄩
𝜏 󵄩
+∞
∫
𝑟
∫
+∞
𝑢
󵄨
󵄨󵄨
󵄨󵄨𝑥𝑘+1 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2 𝑟
𝑃 (𝑠) 𝑑𝑠 𝑑𝑢 𝑑𝑟
|𝑎 (𝑢)|
+ (𝑚 − 1)
𝐻𝑖 󵄩󵄩
󵄩
󵄩𝑥 − 𝑦󵄩󵄩󵄩
𝜏 󵄩
+∞
+∞
×∫
𝑇
+∞
∫
𝑟
∫
which means that 𝑆𝐿 (Ω1 (𝑁, 𝑀)) ⊆ Ω1 (𝑁, 𝑀). It follows
from (27) and (92) that 𝑆𝐿 is a contraction mapping and
it has a unique fixed point 𝑥 ∈ Ω1 (𝑁, 𝑀). It is clear that
𝑥 ∈ Ω1 (𝑁, 𝑀) is a positive solution of (5).
On the basis of (27), (90), and (94), we deduce that
𝑛−𝑚−𝑖−1
(𝑠 − 𝑢)
(𝑢 − 𝑟)
|𝑎 (𝑢)|
𝑢
𝑚−2
𝑟
𝑅 (𝑠) 𝑑𝑠 𝑑𝑢 𝑑𝑟
󵄨󵄨
󵄨󵄨
= 󵄨󵄨󵄨󵄨 (1 − 𝛼𝑘 ) 𝑥𝑘 (𝑡)
󵄨󵄨
󵄨
{
+ 𝛼𝑘 {𝐿 + (−1)𝑛−1 (𝑚 − 1) 𝐻0
{
∞
𝑚 − 1 +∞ +∞ +∞ 𝑟𝑢𝑚−2
≤
∫ ∫ ∫
𝜏
|𝑎 (𝑢)|
𝑇
𝑟
𝑢
×∑∫
+∞
𝑗=1 𝑡+𝑗𝜏
× [𝐻0 𝑠𝑛−𝑚−1 𝑃 (𝑠) + 𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢 𝑑𝑟
+∞
∫
𝑟
which yields that (27) holds. From (17), (94), (98), and
Lemma 1, we obtain that for any 𝑥 ∈ Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
󵄨󵄨󵄨𝑆𝐿 𝑥 (𝑡) − 𝐿󵄨󵄨󵄨
󵄨
󵄨
≤ (𝑚 − 1) 𝐻0 ∑ ∫
𝑗=1 𝑡+𝑗𝜏
∫
+∞
𝑟
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
|𝑎 (𝑢)|
󵄨 󵄨
󵄨
󵄨
× [󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢 𝑑𝑟
∞
+ (𝑚 − 1) 𝐻𝑖 ∑ ∫
+∞
𝑗=1 𝑡+𝑗𝜏
+∞
∫
𝑟
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2
|𝑎 (𝑢)|
󵄨
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢 𝑑𝑟
≤ (𝑚 − 1)
+ (−1)𝑛−𝑖 (𝑚 − 1) 𝐻𝑖
∞
×∑∫
+∞
𝑗=1 𝑡+𝑗𝜏
+∞
∫
𝑟
∫
+∞
𝑢
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2
𝑎 (𝑢)
}
× ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . , 𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢 𝑑𝑟}
}
󵄨󵄨
󵄨󵄨
−𝑥 (𝑡) 󵄨󵄨󵄨󵄨
󵄨󵄨
󵄨
󵄨
󵄨
󵄨󵄨
󵄨
≤ (1 − 𝛼𝑘 ) 󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨 + 𝛼𝑘 󵄨󵄨󵄨𝑆𝐿 𝑥𝑘 (𝑡) − 𝑆𝐿 𝑥 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
󵄨
󵄨
≤ (1 − 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨 + 𝛼𝑘 𝜃 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
𝐻0 +∞ +∞ +∞ 𝑟𝑠𝑛−𝑚−1 𝑢𝑚−2
∫ ∫ ∫
𝜏 𝑇
|𝑎 (𝑢)|
𝑟
𝑢
󵄨
󵄨󵄨
× [󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠)] 𝑑𝑠 𝑑𝑢 𝑑𝑟
󵄨
󵄨
= (1 − (1 − 𝜃) 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄩
󵄩
≤ 𝑒−(1−𝜃)𝛼𝑘 󵄩󵄩󵄩𝑥𝑘 − 𝑥󵄩󵄩󵄩
𝐻 +∞ +∞ +∞ 𝑟𝑠𝑛−𝑚−𝑖−1 𝑢𝑚−2
+ (𝑚 − 1) 𝑖 ∫ ∫ ∫
𝜏 𝑇
|𝑎 (𝑢)|
𝑟
𝑢
𝑘
󵄩
󵄩
≤ 𝑒−(1−𝜃) ∑𝑝=0 𝛼𝑝 󵄩󵄩󵄩𝑥0 − 𝑥󵄩󵄩󵄩 ,
× 𝑊 (𝑠) 𝑑𝑠 𝑑𝑢 𝑑𝑟
=
𝑢
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
𝑎 (𝑢)
𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢 𝑑𝑟
(95)
+∞
+∞
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)) , . . . ,
󵄩
󵄩
= 𝜃 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
∞
∫
𝑚 − 1 +∞ +∞ +∞ 𝑟𝑢𝑚−2
∫ ∫ ∫
𝜏
|𝑎 (𝑢)|
𝑇
𝑟
𝑢
󵄨
󵄨
× [𝐻0 𝑠𝑛−𝑚−1 (󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠))
+𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢 𝑑𝑟
< min {𝑀 − 𝐿, 𝐿 − 𝑁} ,
(96)
∀𝑘 ∈ N0 , 𝑡 ≥ 𝑇,
(97)
which signifies that (20) holds. It follows from (20) and (21)
and that lim𝑘 → ∞ 𝑥𝑘 = 𝑥.
Now we show that (b) holds. Let 𝐿 1 , 𝐿 2 ∈ (𝑁, 𝑀) and
𝐿 1 ≠ 𝐿 2 . As in the proof of (a), we conclude that for each
𝑝 ∈ {1, 2}, there exist 𝜃𝑝 ∈ (0, 1), 𝑇𝑝 > 1 + |𝑡0 | + 𝜏 + |𝛾| and
𝑆𝐿 𝑝 : Ω1 (𝑁, 𝑀) → Ω1 (𝑁, 𝑀) satisfying (91)–(94), where 𝐿,
𝜃, 𝑇, and 𝑆𝐿 are replaced by 𝐿 𝑝 , 𝜃𝑝 , 𝑇𝑝 , and 𝑆𝐿 𝑝 , respectively,
Abstract and Applied Analysis
21
and 𝑆𝐿 𝑝 has a unique fixed point 𝑧𝑝 ∈ Ω1 (𝑁, 𝑀), which is a
positive solution of (5) in Ω1 (𝑁, 𝑀), that is,
Theorem 10. Let 𝑚 = 1. Assume that there exist two constants
𝑀, 𝑁 with 𝑀 > 𝑁 > 0 and four functions 𝑃, 𝑄, 𝑅, 𝑊 ∈
𝐶([𝑡0 , +∞), R+ ) satisfying (16), (17), (89), and
𝑧𝑝 (𝑡)
= 𝐿 𝑝 + (−1)𝑛−1 (𝑚 − 1) 𝐻0
∞
+∞
×∑∫
𝑗=1 𝑡+𝑗𝜏
∫
+∞
𝑟
∫
+∞
𝑢
∫
𝑡0
(𝑠 − 𝑢)𝑛−𝑚−1 (𝑢 − 𝑟)𝑚−2
𝑎 (𝑢)
+ (−1)𝑛−𝑖 (𝑚 − 1) 𝐻𝑖
+∞
×∑∫
𝑗=1 𝑡+𝑗𝜏
∫
+∞
𝑟
∫
+∞
𝑢
∫
+∞
𝑢
|𝑢|
󵄨
󵄨
[|𝑠|𝑛−2 max {𝑃 (𝑠) , 𝑄 (𝑠) , 󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨}
|𝑎 (𝑢)|
+|𝑠|𝑛−𝑖−2 max {𝑅 (𝑠) , 𝑊 (𝑠)}] 𝑑𝑠 𝑑𝑢
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑧𝑝 (𝑓1 (𝑠)) , . . . , 𝑧𝑝 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢 𝑑𝑟
∞
+∞
(101)
< +∞.
Then
(a) for any 𝐿 ∈ (𝑁, 𝑀), there exist 𝜃 ∈ (0, 1) and 𝑇 >
1 + |𝑡0 | + 𝜏 + |𝛾| such that for each 𝑥0 ∈ Ω1 (𝑁, 𝑀), the Mann
iterative sequence {𝑥𝑘 }𝑘∈N0 generated by the following scheme
(𝑠 − 𝑢)𝑛−𝑚−𝑖−1 (𝑢 − 𝑟)𝑚−2
𝑎 (𝑢)
× ℎ (𝑠, 𝑧𝑝 (ℎ1 (𝑠)) , . . . , 𝑧𝑝 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢 𝑑𝑟,
∀𝑡 ≥ 𝑇𝑝 , 𝑝 ∈ {1, 2} .
𝑥𝑘+1 (𝑡)
(98)
In order to prove (b), it is sufficient to show that 𝑧1 ≠ 𝑧2 . Note
that (16), (92), (98), and Lemma 1 lead to
󵄨
󵄨󵄨
󵄨󵄨𝑧1 (𝑡) − 𝑧2 (𝑡)󵄨󵄨󵄨
𝐻 󵄩
󵄨
󵄨
󵄩
≥ 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − (𝑚 − 1) 0 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩
𝜏
+∞
×∫
𝑡+𝜏
∫
+∞
𝑟
− (𝑚 − 1)
∫
+∞
𝑢
𝑟𝑠𝑛−𝑚−1 𝑢𝑚−2
𝑃 (𝑠) 𝑑𝑠 𝑑𝑢 𝑑𝑟
|𝑎 (𝑢)|
𝐻𝑖 󵄩󵄩
󵄩
󵄩𝑧 − 𝑧2 󵄩󵄩󵄩
𝜏 󵄩 1
𝑟𝑠𝑛−𝑚−𝑖−1 𝑢𝑚−2
𝑅 (𝑠) 𝑑𝑠 𝑑𝑢 𝑑𝑟
|𝑎 (𝑢)|
𝑡+𝜏 𝑟
𝑢
󵄩
󵄩
󵄨
󵄨 (𝑚 − 1) 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩
≥ 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 −
𝜏
+∞
×∫
+∞
×∫
∫
+∞
∫
∫
+∞
+∞
max{𝑇1 ,𝑇2 } 𝑟
∫
+∞
𝑢
𝑟𝑢𝑚−2
|𝑎 (𝑢)|
× [𝐻0 𝑠𝑛−𝑚−1 𝑃 (𝑠) + 𝐻𝑖 𝑠𝑛−𝑚−𝑖−1 𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢 𝑑𝑟
󵄨
󵄨
󵄩
󵄩
> 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − max {𝜃1 , 𝜃2 } 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩 ,
∀𝑡 ≥ max {𝑇1 , 𝑇2 } ,
(99)
(1 − 𝛼𝑘 ) 𝑥𝑘 (𝑡)
{
{
{
{
{
{
{
{
{
(−1)𝑛−1 ∞ +∞ +∞ (𝑠 − 𝑢)𝑛−2
{
{
+𝛼
𝐿
+
∑∫ ∫
{
𝑘{
{
{
𝑎 (𝑢)
(𝑛 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
{
{
{
{
{
{
{
{
× [𝑔 (𝑠)−𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)), . . . ,𝑥𝑘 (𝑓𝑙 (𝑠)))]𝑑𝑠 𝑑𝑢
{
{
{
{
{
{
{
{
(−1)𝑛−𝑖 ∞ +∞ +∞ (𝑠 − 𝑢)𝑛−𝑖−2
{
{
+
∑∫ ∫
{
{
{
𝑎 (𝑢)
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
{
{
{
{
{
{
}
{
{
{
×ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . , 𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} ,
{
{
{
{
{
}
{
{
{
𝑡 ≥ 𝑇, 𝑘 ∈ N0 ,
={
{
{(1 − 𝛼𝑘 ) 𝑥𝑘 (𝑇)
{
{
{
{
{
{
{
{
{
(−1)𝑛−1 ∞ +∞ +∞ (𝑠 − 𝑢)𝑛−2
{
{
+𝛼
𝐿
+
∑∫
∫
{ 𝑘{
{
𝑎 (𝑢)
{
(𝑛 − 2)! 𝑗=1 𝑇+𝑗𝜏 𝑢
{
{
{
{
{
{
{
{
× [𝑔 (𝑠)−𝑓(𝑠, 𝑥𝑘 (𝑓1 (𝑠)), . . . , 𝑥𝑘 (𝑓𝑙 (𝑠)))]𝑑𝑠 𝑑𝑢
{
{
{
{
{
{
{
(−1)𝑛−𝑖 ∞ +∞ +∞ (𝑠 − 𝑢)𝑛−𝑖−2
{
{
+
∑∫
∫
{
{
{
𝑎 (𝑢)
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑇+𝑗𝜏 𝑢
{
{
{
{
{
{
}
{
{
{
×ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . , 𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} ,
{
{
{
{
{
}
{
𝛾 ≤ 𝑡 < 𝑇, 𝑘 ∈ N0 ,
{
(102)
which means that
󵄨󵄨
󵄨
󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨
󵄩󵄩
󵄩
> 0,
󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩 ≥
1 + max {𝜃1 , 𝜃2 }
that is, 𝑧1 ≠ 𝑧2 . This completes the proof.
(100)
converges to a positive solution 𝑥 ∈ Ω1 (𝑁, 𝑀) of (5) and has
the error estimate (20), where {𝛼𝑘 }𝑘∈N0 is an arbitrary sequence
in [0, 1] with (21);
(b) Equation (5) has uncountably many positive solutions
in Ω1 (𝑁, 𝑀).
22
Abstract and Applied Analysis
󵄩
󵄩󵄩
󵄩𝑥 − 𝑦󵄩󵄩󵄩 +∞ +∞ 𝑢
≤󵄩
∫ ∫
𝜏
|𝑎 (𝑢)|
𝑇
𝑢
Proof. Set 𝐿 ∈ (𝑁, 𝑀). Due to (101), there exist 𝜃 ∈ (0, 1) and
𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| satisfying (91),
1 +∞ +∞ 𝑢
𝜃= ∫ ∫
𝜏 𝑇
|𝑎 (𝑢)|
𝑢
𝑠𝑛−2
𝑠𝑛−𝑖−2
×[
𝑃 (𝑠) +
𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢,
(𝑛 − 2)!
(𝑛 − 𝑖 − 2)!
(103)
(106)
1 +∞ +∞ 𝑢
∫ ∫
𝜏 𝑇
|𝑎 (𝑢)|
𝑢
×[
𝑠𝑛−2
𝑠𝑛−𝑖−2
𝑃 (𝑠) +
𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢
(𝑛 − 2)!
(𝑛 − 𝑖 − 2)!
󵄩
󵄩
= 𝜃 󵄩󵄩󵄩𝑥 − 𝑦󵄩󵄩󵄩 ,
×[
𝑠𝑛−𝑖−2
𝑠𝑛−2 󵄨󵄨
󵄨
(󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠)) +
𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
(𝑛 − 2)!
(𝑛 − 𝑖 − 2)!
< min {𝑀 − 𝐿, 𝐿 − 𝑁} .
(104)
which means that (27) holds. It follows from (17), (104), (105),
and Lemma 1 that for any 𝑥 ∈ Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
󵄨
󵄨󵄨
󵄨󵄨𝑆𝐿 𝑥 (𝑡) − 𝐿󵄨󵄨󵄨
≤
Define a mapping 𝑆𝐿 : Ω1 (𝑁, 𝑀) → CB([𝛾, +∞), R) by
󵄨 󵄨
󵄨
󵄨
× [󵄨󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨] 𝑑𝑠 𝑑𝑢
𝑆𝐿 𝑥 (𝑡)
+
(−1)𝑛−1 ∞ +∞ +∞ (𝑠 − 𝑢)𝑛−2
{
𝐿
+
∑∫ ∫
{
{
{
𝑎 (𝑢)
(𝑛 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
{
{
{
{
{
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
{
{
{
{
{
∞ +∞ +∞
𝑛−𝑖
{
(𝑠 − 𝑢)𝑛−𝑖−2
= {+ (−1)
∑
∫ ∫
{
𝑎 (𝑢)
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
{
{
{
{
{
{×ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢,
{
{
{
{
{
𝑡 ≥ 𝑇, 𝑥 ∈ Ω1 (𝑁, 𝑀) ,
{
{
𝑥
,
𝛾
≤
𝑡 < 𝑇, 𝑥 ∈ Ω1 (𝑁, 𝑀) .
𝑆
(𝑇)
{ 𝐿
≤
≤
+∞ +∞
1
𝑢𝑠𝑛−𝑖−2
𝑊 (𝑠) 𝑑𝑠 𝑑𝑢
∫ ∫
𝜏 (𝑛 − 𝑖 − 2)! 𝑡+𝜏 𝑢 |𝑎 (𝑢)|
1 +∞ +∞ 𝑢
∫ ∫
𝜏 𝑇
|𝑎 (𝑢)|
𝑢
×[
𝑠𝑛−𝑖−2
𝑠𝑛−2 󵄨󵄨
󵄨
(󵄨󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠)) +
𝑊 (𝑠)] 𝑑𝑠 𝑑𝑢
(𝑛 − 2)!
(𝑛 − 𝑖 − 2)!
< min {𝑀 − 𝐿, 𝐿 − 𝑁} ,
󵄨
󵄨󵄨
󵄨󵄨𝑆𝐿 𝑥 (𝑡) − 𝑆𝐿 𝑦 (𝑡)󵄨󵄨󵄨
(107)
+∞
+∞
1
∑∫ ∫
(𝑛 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
𝑛−2
(𝑠 − 𝑢)
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨𝑓 (𝑠, 𝑥 (𝑓1 (𝑠)) , . . . , 𝑥 (𝑓𝑙 (𝑠)))
󵄨
−𝑓 (𝑠, 𝑦 (𝑓1 (𝑠)) , . . . , 𝑦 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
+
+∞ +∞
1
𝑢𝑠𝑛−2 󵄨󵄨
󵄨
(󵄨𝑔 (𝑠)󵄨󵄨󵄨 + 𝑄 (𝑠)) 𝑑𝑠 𝑑𝑢
∫ ∫
𝜏 (𝑛 − 2)! 𝑡+𝜏 𝑢 |𝑎 (𝑢)| 󵄨
+
In view of (16), (103), (105), and Lemma 1, we achieve that for
any 𝑥, 𝑦 ∈ Ω1 (𝑁, 𝑀) and 𝑡 ≥ 𝑇
≤
∞ +∞ +∞
1
(𝑠 − 𝑢)𝑛−𝑖−2
∑∫ ∫
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
|𝑎 (𝑢)|
󵄨
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
(105)
∞
∞ +∞ +∞
1
(𝑠 − 𝑢)𝑛−2
∑∫ ∫
(𝑛 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
|𝑎 (𝑢)|
∞ +∞ +∞
1
(𝑠 − 𝑢)𝑛−𝑖−2
∑∫ ∫
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑥 (ℎ1 (𝑠)) , . . . , 𝑥 (ℎ𝑙 (𝑠)))
󵄨
−ℎ (𝑠, 𝑦 (ℎ1 (𝑠)) , . . . , 𝑦 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
󵄩
󵄩󵄩
󵄩𝑥 − 𝑦󵄩󵄩󵄩 +∞ +∞ 𝑢𝑠𝑛−2
≤ 󵄩
𝑃 (𝑠) 𝑑𝑠 𝑑𝑢
∫ ∫
𝜏 (𝑛 − 2)! 𝑡+𝜏 𝑢 |𝑎 (𝑢)|
󵄩
󵄩󵄩
+∞ +∞
󵄩𝑥 − 𝑦󵄩󵄩󵄩
𝑢𝑠𝑛−𝑖−2
𝑅 (𝑠) 𝑑𝑠 𝑑𝑢
+ 󵄩
∫ ∫
𝜏 (𝑛 − 𝑖 − 2)! 𝑡+𝜏 𝑢 |𝑎 (𝑢)|
which means that 𝑆𝐿 (Ω1 (𝑁, 𝑀)) ⊆ Ω1 (𝑁, 𝑀). Coupled with
(27), we know that 𝑆𝐿 is a contraction mapping and it has
a unique fixed point 𝑥 ∈ Ω1 (𝑁, 𝑀). It follows that 𝑥 ∈
Ω1 (𝑁, 𝑀) is a positive solution of (5).
In view of (27), (102), and (105), we deduce that
󵄨
󵄨󵄨
󵄨󵄨𝑥𝑘+1 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
= 󵄨󵄨󵄨󵄨 (1 − 𝛼𝑘 ) 𝑥𝑘 (𝑡)
󵄨󵄨
󵄨
{
(−1)𝑛−1 ∞ +∞ +∞ (𝑠 − 𝑢)𝑛−2
+𝛼𝑘{𝐿 +
∑∫ ∫
𝑎 (𝑢)
(𝑛 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
{
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑥𝑘 (𝑓1 (𝑠)) , . . . , 𝑥𝑘 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
Abstract and Applied Analysis
+
23
󵄨
× 󵄨󵄨󵄨ℎ (𝑠, 𝑧1 (ℎ1 (𝑠)) , . . . , 𝑧1 (ℎ𝑙 (𝑠)))
(−1)𝑛−𝑖 ∞ +∞ +∞ (𝑠 − 𝑢)𝑛−𝑖−2
∑∫ ∫
𝑎 (𝑢)
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
󵄨
−ℎ (𝑠, 𝑧2 (ℎ1 (𝑠)) , . . . , 𝑧2 (ℎ𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
󵄨󵄨
󵄨󵄨
}
󵄨
×ℎ (𝑠, 𝑥𝑘 (ℎ1 (𝑠)) , . . . , 𝑥𝑘 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢} − 𝑥 (𝑡)󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
}
󵄨
󵄨
󵄨
󵄨
≤ (1 − 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨 + 𝛼𝑘 󵄨󵄨󵄨𝑆𝐿 𝑥𝑘 (𝑡) − 𝑆𝐿 𝑥 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
󵄨
󵄨
≤ (1 − 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨 + 𝛼𝑘 𝜃 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄨
󵄨
= (1 − (1 − 𝜃) 𝛼𝑘 ) 󵄨󵄨󵄨𝑥𝑘 (𝑡) − 𝑥 (𝑡)󵄨󵄨󵄨
󵄩
󵄩
≤ 𝑒−(1−𝜃)𝛼𝑘 󵄩󵄩󵄩𝑥𝑘 − 𝑥󵄩󵄩󵄩
𝑘
󵄩
󵄩
≤ 𝑒−(1−𝜃) ∑𝑝=0 𝛼𝑝 󵄩󵄩󵄩𝑥0 − 𝑥󵄩󵄩󵄩 ,
󵄩
󵄩 +∞ +∞ 𝑛−2
𝑢𝑠
󵄨
󵄨 󵄩󵄩𝑧 − 𝑧2 󵄩󵄩󵄩
𝑃 (𝑠) 𝑑𝑠 𝑑𝑢
≥ 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − 󵄩 1
∫ ∫
𝜏 (𝑛 − 2)! 𝑡+𝜏 𝑢 |𝑎 (𝑢)|
󵄩󵄩
󵄩
+∞ +∞
󵄩𝑧 − 𝑧2 󵄩󵄩󵄩
𝑢𝑠𝑛−𝑖−2
− 󵄩 1
𝑅 (𝑠) 𝑑𝑠 𝑑𝑢
∫ ∫
𝜏 (𝑛 − 𝑖 − 2)! 𝑡+𝜏 𝑢 |𝑎 (𝑢)|
󵄩
󵄩 +∞
+∞
𝑢
󵄨
󵄨 󵄩󵄩𝑧 − 𝑧2 󵄩󵄩󵄩
≥ 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − 󵄩 1
∫
∫
𝜏
|𝑎 (𝑢)|
max{𝑇1 ,𝑇2 } 𝑢
𝑠𝑛−2
𝑠𝑛−𝑖−2
𝑃 (𝑠) +
𝑅 (𝑠)] 𝑑𝑠 𝑑𝑢
(𝑛 − 2)!
(𝑛 − 𝑖 − 2)!
󵄨
󵄩
󵄩
󵄨
> 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 − max {𝜃1 , 𝜃2 } 󵄩󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩 ,
×[
∀𝑘 ∈ N0 , 𝑡 ≥ 𝑇,
(108)
which signifies that (20) holds. It follows from (20) and (21)
that lim𝑘 → ∞ 𝑥𝑘 = 𝑥.
Now we show that (b) holds. Let 𝐿 1 , 𝐿 2 ∈ (𝑁, 𝑀) and
𝐿 1 ≠ 𝐿 2 . As in the proof of (a), we conclude that for each
𝑝 ∈ {1, 2}, there exist 𝜃𝑝 ∈ (0, 1), 𝑇𝑝 > 1 + |𝑡0 | + 𝜏 + |𝛾|
and 𝑆𝐿 𝑝 : Ω1 (𝑁, 𝑀) → Ω1 (𝑁, 𝑀) satisfying (91) and (103)–
(105), where 𝐿, 𝜃, 𝑇, and 𝑆𝐿 are replaced by 𝐿 𝑝 , 𝜃𝑝 , 𝑇𝑝 and 𝑆𝐿 𝑝 ,
respectively, and 𝑆𝐿 𝑝 has a unique fixed point 𝑧𝑝 ∈ Ω1 (𝑁, 𝑀),
which is a positive solution of (5) in Ω1 (𝑁, 𝑀). It follows that
for any 𝑡 ≥ 𝑇𝑝 and 𝑝 ∈ {1, 2}
𝑧𝑝 (𝑡) = 𝐿 𝑝 +
(−1)𝑛−1 ∞ +∞ +∞ (𝑠 − 𝑢)𝑛−2
∑∫ ∫
𝑎 (𝑢)
(𝑛 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
which yields that
󵄨󵄨
󵄨
󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨
󵄩󵄩
󵄩
> 0,
󵄩󵄩𝑧1 − 𝑧2 󵄩󵄩󵄩 ≥
1 + max {𝜃1 , 𝜃2 }
(−1)𝑛−𝑖 ∞ +∞ +∞ (𝑠 − 𝑢)𝑛−𝑖−2
∑∫ ∫
𝑎 (𝑢)
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
× ℎ (𝑠, 𝑧𝑝 (ℎ1 (𝑠)) , . . . , 𝑧𝑝 (ℎ𝑙 (𝑠))) 𝑑𝑠 𝑑𝑢.
(109)
that is, 𝑧1 ≠ 𝑧2 . This completes the proof.
Remark 11. Theorems 2–10 extend, improve, and unifies
Theorems 1–4 in [1], the theorem in [2], Theorems 2.1–2.4
in [4], Theorems 2.1–2.5 in [5, 8], Theorems 1–3 in [9], and
Theorems 1–4 in [11], respectively. The examples below prove
that Theorems 2–10 extend substantially the corresponding
results in [1, 2, 4, 5, 8, 9, 11]. Note that none of the known
results can be applied to these examples.
Example 12. Consider the higher order nonlinear neutral
delay differential equation
In order to prove (b), we just need to show that 𝑧1 ≠ 𝑧2 . Notice
that (16), (103), (109), and Lemma 1 ensure that
[(𝑡
[
𝑚+1
󵄨
󵄨󵄨
󵄨󵄨𝑧1 (𝑡) − 𝑧2 (𝑡)󵄨󵄨󵄨
∞ +∞ +∞
1
(𝑠 − 𝑢)𝑛−2
󵄨
󵄨
≥ 󵄨󵄨󵄨𝐿 1 − 𝐿 2 󵄨󵄨󵄨 −
∑∫ ∫
(𝑛 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
|𝑎 (𝑢)|
󵄨
× 󵄨󵄨󵄨𝑓 (𝑠, 𝑧2 (𝑓1 (𝑠)) , . . . , 𝑧2 (𝑓𝑙 (𝑠)))
󵄨
−𝑓 (𝑠, 𝑧1 (𝑓1 (𝑠)) , . . . , 𝑧1 (𝑓𝑙 (𝑠)))󵄨󵄨󵄨 𝑑𝑠 𝑑𝑢
−
∞
+∞ +∞
1
(𝑠 − 𝑢)𝑛−𝑖−2
∑∫ ∫
(𝑛 − 𝑖 − 2)! 𝑗=1 𝑡+𝑗𝜏 𝑢
|𝑎 (𝑢)|
(111)
3. Remark and Examples
× [𝑔 (𝑠) − 𝑓 (𝑠, 𝑧𝑝 (𝑓1 (𝑠)) , . . . , 𝑧𝑝 (𝑓𝑙 (𝑠)))] 𝑑𝑠 𝑑𝑢
+
(110)
+ 1) (𝑥 (𝑡) +
sin (2𝑡2 ) − cos (𝑡5 − 1)
7 + 2 sin (8𝑡3 + 2𝑡 − 1)
(𝑚) (𝑛−𝑚)
𝑥 (𝑡−𝜏)) ]
]
(𝑖)
+(
+
𝑡2 𝑥 (𝑡 − 3) 𝑥2 (𝑡 − 4)
)
𝑡𝑛−𝑚−𝑖+3 + 𝑡2 + 1
𝑡3 𝑥3 (𝑡2 − 𝑡) − 𝑥4 (𝑡 − 1)
𝑡𝑛−𝑚+4 + 𝑡 + 2
=
𝑡 ln (1 + 𝑡2 ) − cos2 (𝑡2 − 𝑡 + 1)
𝑡2𝑛−𝑚+3 + 1
,
∀𝑡 ≥ 2,
(112)
24
Abstract and Applied Analysis
where 𝜏 > 0 and 𝑖 ≤ 𝑛 − 𝑚 − 1. Let 𝑙 = 2, 𝑡0 = 2, 𝛾 =
min{2 − 𝜏, −2}, 𝑀 = 10, 𝑁 = 1, 𝑏0 = 2/5 and
where 𝜏 > 0 and 𝑖 ≤ 𝑛 − 𝑚 − 1. Let 𝑙 = 2, 𝑡0 = 1, 𝛾 =
min{1 − 𝜏, 0}, 𝑀 = 6, 𝑁 = 1, 𝑏0 = 3/4 and
ℎ1 (𝑡) = 𝑡 − 3,
ℎ2 (𝑡) = 𝑡 − 4,
ℎ1 (𝑡) = 𝑡 ln 𝑡,
𝑓1 (𝑡) = 𝑡2 − 𝑡,
𝑓2 (𝑡) = 𝑡 − 1,
𝑓1 (𝑡) = 𝑡3 + 𝑡2 ,
𝑚+1
𝑎 (𝑡) = 𝑡
+ 1,
𝑏 (𝑡) =
sin (2𝑡2 ) − cos (𝑡5 − 1)
7 + 2 sin (8𝑡3 + 2𝑡 − 1)
𝑎 (𝑡) = 𝑡𝑛 + 1,
,
𝑡2 𝑢V2
ℎ (𝑡, 𝑢, V) = 𝑛−𝑚−𝑖+3 2
,
𝑡
+𝑡 +1
𝑓 (𝑡, 𝑢, V) =
3 3
2
𝑃 (𝑡) =
𝑓 (𝑡, 𝑢, V) =
4
𝑡 𝑢 −V
,
𝑡𝑛−𝑚+4 + 𝑡 + 2
2
𝑔 (𝑡) =
ℎ (𝑡, 𝑢, V) =
2
𝑡 ln (1 + 𝑡 ) − cos (𝑡 − 𝑡 + 1)
𝑡2𝑛−𝑚+3 + 1
3
𝑀 (3𝑡 + 4𝑀)
𝑡𝑛−𝑚+4 + 𝑡 + 2
3
,
(113)
2
𝑄 (𝑡) =
,
𝑃 (𝑡) =
3
𝑀 (𝑡 + 𝑀)
𝑡𝑛−𝑚+4 + 𝑡 + 2
𝑔 (𝑡) =
,
3𝑀2 𝑡2
𝑅 (𝑡) = 𝑛−𝑚−𝑖+3 2
,
𝑡
+𝑡 +1
𝑊 (𝑡) =
𝑅 (𝑡) =
+
=
(2 − 𝑡2 ) arctan 𝑡 + 𝑡𝑢3 V2
𝑡𝑛+𝑚+3 + V2
,
𝑡2𝑛+𝑚+4
,
2
(𝑡𝑛+𝑚+3 + 𝑁2 )
(2 + 𝑡2 ) arctan 𝑡 + 𝑀5 𝑡
𝑡𝑛+𝑚+3 + 𝑁2
𝑡2 + 2𝑀 (𝑡 + 1)
,
𝑡𝑛+𝑚−𝑖+3
𝑊 (𝑡) =
,
,
𝑀𝑡2 + 𝑀2 (𝑡 + 1)
,
𝑡𝑛+𝑚−𝑖+3
It is easy to check that the conditions of Theorem 3 are
satisfied. Therefore (114) has uncountably many positive
solutions in Ω1 (1, 6), and for any 𝐿 ∈ (11/2, 6), there exist
𝜃 ∈ (0, 1) and 𝑇 > 1+|𝑡0 |+𝜏+|𝛾| such that the Mann iterative
sequence {𝑥𝑘 }𝑘∈N0 generated by (19) and (21) converges to
a positive solution 𝑥 ∈ Ω1 (1, 6) of (114) and has the error
estimate (20).
Example 14. Consider the higher order nonlinear neutral
delay differential equation
(𝑚) (𝑛−𝑚)
]
[(𝑡 + 1) (𝑥 (𝑡) −
𝑡2 𝑥 (𝑡 ln 𝑡) − (𝑡 + 1) 𝑥 (𝑡 ln 𝑡) 𝑥 (√2𝑡)
𝑡𝑛+𝑚−𝑖+3
(𝑖)
)
+(
(114)
(2 − 𝑡2 ) arctan 𝑡 + 𝑡𝑥3 (𝑡3 + 𝑡2 ) 𝑥2 (𝑡2 )
+
𝑡𝑛+𝑚+3 + 𝑥2 (𝑡2 )
√1 − 8𝑡3 + 13𝑡5 + 5𝑡6 cos (𝑡3 − 1)
𝑡2𝑛+𝑚+4
𝑡2 𝑢 − (𝑡 + 1) 𝑢V
,
𝑡𝑛+𝑚−𝑖+3
(115)
Example 13. Consider the higher order nonlinear neutral
delay differential equation
+(
3𝑡2
,
4𝑡2 + 3
∀ (𝑡, 𝑢, V) ∈ [𝑡0 , +∞) × [𝑁, 𝑀]2 .
It is easy to verify that the conditions of Theorem 2 are
satisfied. Thus Theorem 2 ensures that (112) has uncountably
many positive solutions in Ω1 (1, 10), and for any 𝐿 ∈ (5, 6),
there exist 𝜃 ∈ (0, 1) and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| such that the
Mann iterative sequence {𝑥𝑘 }𝑘∈N0 generated by (19) and (21)
converges to a positive solution 𝑥 ∈ Ω1 (1, 10) of (112) and has
the error estimate (20).
3𝑡2
𝑥 (𝑡 − 𝜏))
4𝑡2 + 3
𝑏 (𝑡) =
5𝑀4 𝑡𝑛+𝑚+4 + 2𝑀 (2 + 𝑡2 ) arctan 𝑡 + 5𝑀6 𝑡
∀ (𝑡, 𝑢, V) ∈ [𝑡0 , +∞) × [𝑁, 𝑀]2 .
[(𝑡𝑛 + 1) (𝑥 (𝑡) +
𝑓2 (𝑡) = 𝑡2 ,
√1 − 8𝑡3 + 13𝑡5 + 5𝑡6 cos (𝑡3 − 1)
𝑄 (𝑡) =
𝑀3 𝑡2
,
𝑛−𝑚−𝑖+3
𝑡
+ 𝑡2 + 1
ℎ2 (𝑡) = √2𝑡,
,
∀𝑡 ≥ 1,
=
(𝑚)
arctan 𝑡
𝑥 (𝑡 − 𝜏)) ]
2
𝑡2 𝑥 (𝑡3 + 𝑡) − 𝑥2 (𝑡3 − 1)
𝑡2𝑛+𝑚−𝑖+3 + 𝑥2 (𝑡3 + 𝑡)
(𝑛−𝑚)
(𝑖)
)
(116)
𝑡𝑥 (𝑡 − 1) sin (𝑡𝑥 (𝑡 − sin 𝑡))
𝑡2𝑛+𝑚+3 + 𝑡3 + 2
𝑡√𝑡 + 1sin2 (𝑡2 + 2𝑡 + 1)
𝑡2𝑛+𝑚+3 + 1
,
∀𝑡 ≥ 0,
Abstract and Applied Analysis
25
where 𝜏 > 0 and 𝑖 ≤ 𝑛 − 𝑚 − 1. Let 𝑙 = 2, 𝑡0 = 0, 𝛾 =
min{−𝜏, −1}, 𝑀 = 8, 𝑁 = 1/2, 𝑏0 = 7/8 and
ℎ1 (𝑡) = 𝑡3 + 𝑡,
ℎ2 (𝑡) = 𝑡3 − 1,
𝑓1 (𝑡) = 𝑡 − 1,
𝑓 (𝑡, 𝑢, V) =
𝑃 (𝑡) =
ℎ (𝑡, 𝑢, V, 𝑤) =
𝑡𝑢 sin (𝑡V)
,
𝑡2𝑛+𝑚+3 + 𝑡3 + 2
𝑡√𝑡 + 1sin2 (𝑡2 + 2𝑡 + 1)
𝑡2𝑛+𝑚+3 + 1
𝑄 (𝑡) =
𝑔 (𝑡) =
𝑀𝑡
,
𝑡2𝑛+𝑚+3 + 𝑡3 + 2
𝑡2𝑛+𝑚−𝑖+5 + 𝑀2 𝑡2 + 2𝑀𝑡2𝑛+𝑚−𝑖+3 + 4𝑀3
2
(𝑡2𝑛+𝑚−𝑖+3 + 𝑁2 )
𝑀𝑡2 + 𝑀2
𝑊 (𝑡) = 2𝑛+𝑚−𝑖+3
,
𝑡
+ 𝑁2
𝑃 (𝑡) =
2
∀ (𝑡, 𝑢, V) ∈ [𝑡0 , +∞) × [𝑁, 𝑀] .
It is easy to prove that the conditions of Theorem 4 are satisfied. Hence (116) has uncountably many positive solutions in
Ω1 (1/2, 8), and for any 𝐿 ∈ (1/2, 1), there exist 𝜃 ∈ (0, 1) and
𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| such that the Mann iterative sequence
{𝑥𝑘 }𝑘∈N0 generated by (19) and (21) converges to a positive
solution 𝑥 ∈ Ω1 (1/2, 8) of (116) and has the error estimate
(20).
Example 15. Consider the higher order nonlinear neutral
delay differential equation
2
+(
(𝑚) (𝑛−𝑚)
]
√𝑡 + 1𝑥2 (𝑡 − 1) 𝑥 (𝑡2 )
𝑡𝑛+𝑖+1 + 𝑡 ln (1 + |𝑥 (𝑡 − 2)|) + 4
𝑡2 𝑢V
,
𝑡2𝑛+3 + 𝑡 |𝑤| + 3
𝑀𝑡2 (6𝑏0 + 3𝑀2 𝑡 + 2𝑏0 𝑡2𝑛+3 )
𝑏02 (𝑡2𝑛+3 + 3)
2
,
(119)
𝑀2 𝑡2
,
𝑏02 (𝑡2𝑛+3 + 3)
𝑀2 √𝑡 + 1
𝑅 (𝑡) =
𝑏02 (𝑡𝑛+𝑖+1 + 4)
(117)
[(𝑡2𝑖 + 1) (𝑥 (𝑡) + 2𝑡 +1 𝑥 (𝑡 − 𝜏))
2
𝑏 (𝑡) = 2𝑡 +1 ,
𝑡2 cos (2𝑡) + arctan 𝑡3
,
𝑡𝑛+4 + sin2 (1 − 𝑡3 + 𝑡4 ) + 1
𝑄 (𝑡) =
,
𝑓3 (𝑡) = 𝑡 − 3,
√𝑡 + 1𝑢2 V
,
𝑡𝑛+𝑖+1 + 𝑡 ln (1 + |𝑤|) + 4
𝑓 (𝑡, 𝑢, V, 𝑤) =
,
ℎ3 (𝑡) = 𝑡 − 2,
𝑓2 (𝑡) = 𝑡2 − 9,
𝑎 (𝑡) = 𝑡2𝑖 + 1,
𝑡2 𝑢 − V2
,
2𝑛+𝑚−𝑖+3
𝑡
+ 𝑢2
𝑀𝑡2 + 𝑡
,
2𝑛+𝑚+3
𝑡
+ 𝑡3 + 2
𝑅 (𝑡) =
𝑓1 (𝑡) = 𝑡 − 12,
1
𝑏 (𝑡) = − arctan 𝑡,
2
ℎ (𝑡, 𝑢, V) =
ℎ2 (𝑡) = 𝑡2 ,
ℎ1 (𝑡) = 𝑡 − 1,
𝑓2 (𝑡) = 𝑡 − sin 𝑡,
𝑎 (𝑡) = 𝑡 + 1,
𝑔 (𝑡) =
where 𝜏 > 0 and 𝑖 ≤ 𝑛 − 𝑚 − 1. Let 𝑙 = 3, 𝑡0 = 0, 𝛾 =
min{3 − 𝜏, −9}, 𝑀 = 12, 𝑁 = 5, 𝑏0 = 2 and
2
× [3𝑡𝑛+𝑖+1 + 12 +
𝑊 (𝑡) =
𝑀
𝑀
𝑡 + 3𝑡 ln (1 + )] ,
𝑏0
𝑏0
𝑀3 √𝑡 + 1
,
𝑏03 (𝑡𝑛+𝑖+1 + 4)
∀ (𝑡, 𝑢, V, 𝑤) ∈ [𝑡0 , +∞) × [0,
𝑀 3
].
𝑏0
It is easy to verify that the conditions of Theorem 5 are
satisfied. Hence Theorem 5 ensures that (118) has uncountably
many positive solutions in Ω2 (5, 12), and, for any 𝐿 ∈ (11, 12),
there exist 𝜃 ∈ (0, 1) and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| such that the
Mann iterative sequence {𝑥𝑘 }𝑘∈N0 generated by (48) and (21)
converges to a positive solution 𝑥 ∈ Ω2 (5, 12) of (118) and has
the error estimate (20).
Example 16. Consider the higher order nonlinear neutral
delay differential equation
(𝑖)
)
(118)
2
+
=
2
𝑡 𝑥 (𝑡 − 12) 𝑥 (𝑡 − 9)
(𝑚) (𝑛−𝑚)
[(𝑡3𝑛 + 2𝑡𝑛+1 + 1) (𝑥 (𝑡) − (𝑡2 + 2𝑡 + 4) 𝑥 (𝑡 − 𝜏))
𝑡2𝑛+3 + 𝑡 |𝑥 (𝑡 − 3)| + 3
𝑡2 cos (2𝑡) + arctan 𝑡3
,
𝑡𝑛+4 + sin2 (1 − 𝑡3 + 𝑡4 ) + 1
∀𝑡 ≥ 3,
(𝑖)
+(
𝑡𝑥 (𝑡 − 3) 𝑥2 (𝑡 − 4)
)
𝑡𝑛+10 + 𝑥2 (𝑡 − 3)
]
26
Abstract and Applied Analysis
+
=
√𝑡 + 1𝑥3 (𝑡 ln 𝑡) − 𝑡𝑥4 (𝑡2 − 𝑡)
(𝑡 + 1)
𝑛+3
+
𝑡𝑥2
Example 17. Consider the higher order nonlinear neutral
delay differential equation
(𝑡 ln 𝑡)
𝑡 ln (1 + 𝑡2 ) − √𝑡 + 3cos3 (𝑡3 + 1)
𝑡𝑛+5 + 4𝑡3 − 1 − 𝑡sin5 (𝑡2 − 3)
,
∀𝑡 ≥ 1,
(120)
[(𝑡2 ln 𝑡) (𝑥 (𝑡) + 𝑥 (𝑡 − 𝜏))(𝑚) ]
+(
where 𝜏 > 0, and 𝑖 ≤ 𝑛−𝑚−1. Let 𝑙 = 2, 𝑡0 = 1, 𝛾 = min{1−𝜏,
−3}, 𝑀 = 6, 𝑁 = 2, 𝑏0 = 3 and
+
ℎ1 (𝑡) = 𝑡 − 3,
ℎ2 (𝑡) = 𝑡 − 4,
𝑓1 (𝑡) = 𝑡 ln 𝑡,
𝑓2 (𝑡) = 𝑡2 − 𝑡,
𝑎 (𝑡) = 𝑡3𝑛 + 2𝑡𝑛+1 + 1,
ℎ (𝑡, 𝑢, V) =
𝑓 (𝑡, 𝑢, V) =
𝑔 (𝑡) =
=
𝑡𝑢V2
,
𝑡𝑛+10 + 𝑢2
𝑡3𝑛−𝑚−𝑖+2 + 𝑥2 (𝑡 − cos 𝑡)
√𝑡 + 1𝑢3 − 𝑡V4
(𝑡 + 1)𝑛+3 + 𝑡𝑢2
𝑡2𝑛−𝑚+3 + 1
sin13 (𝑡5 − √𝑡 + 1)
𝑡𝑛+7/2 + 1
,
∀𝑡 ≥ 4,
,
𝑡𝑛+5 + 4𝑡3 − 1 − 𝑡sin5 (𝑡2 − 3)
ℎ1 (𝑡) = √𝑡 − 2,
,
𝑓1 (𝑡) = 𝑡 − 4,
ℎ2 (𝑡) = 2𝑡 − 1,
ℎ (𝑡, 𝑢, V, 𝑤) =
𝑀2 √
𝑀3 2
𝑀
𝑛+3
𝑡
𝑡(𝑡
+
1)
+
6
𝑡 ],
𝑡
+
1
+
4
𝑏0
𝑏02
𝑏03
𝑀3
(𝑡 + 1)
𝑛+8
(√𝑡 + 1 +
𝑀
𝑡) ,
𝑏0
𝑀2
3𝑀2
𝑅 (𝑡) = 2 2𝑛+19 (𝑡𝑛+10 + 2 ) ,
𝑏0 𝑡
𝑏0
𝑊 (𝑡) =
𝑄 (𝑡) =
𝑀3
,
𝑏03 𝑡𝑛+9
∀ (𝑡, 𝑢, V) ∈ [𝑡0 , +∞) × [0,
sin13 (𝑡5 − √𝑡 + 1)
𝑡𝑛+7/2
+1
𝑀2 + 𝑀3
,
𝑡2𝑛−𝑚+3 + 1
𝑊 (𝑡) =
𝑀 2
].
𝑏0
𝑏 (𝑡) = 1,
𝑢+V
,
+ 𝑤2
𝑡3𝑛−𝑚−𝑖+2
𝑓 (𝑡, 𝑢, V, 𝑤) =
𝑔 (𝑡) =
ℎ3 (𝑡) = 𝑡 − cos 𝑡,
𝑓3 (𝑡) = 𝑡 − sin (𝑡9 + 1) ,
𝑓2 (𝑡) = √𝑡 − 1,
𝑎 (𝑡) = 𝑡2 ln 𝑡,
𝑏03
(122)
𝑥2 (𝑡 − 4) + 𝑥 (√𝑡 − 1) 𝑥2 (𝑡 − sin (𝑡9 + 1))
𝑀2
𝑏02 (𝑡 + 1)2𝑛+6
𝑄 (𝑡) =
)
where 𝜏 > 0, 𝑚 ≥ 2 and 𝑖 ≤ 𝑛 − 𝑚 − 1. Let 𝑙 = 3, 𝑡0 = 4,
𝛾 = min{4 − 𝜏, 0}, 𝑀 = 100, 𝑁 = 1 and
𝑡 ln (1 + 𝑡2 ) − √𝑡 + 3cos3 (𝑡3 + 1)
× [3(𝑡 + 1)𝑛+7/2 +
(𝑖)
𝑥 (√𝑡 − 2) + 𝑥 (2𝑡 − 1)
𝑏 (𝑡) = −𝑡2 − 2𝑡 − 4,
𝑃 (𝑡)
=
(𝑛−𝑚)
𝑢2 + V𝑤2
,
𝑡2𝑛−𝑚+3 + 1
,
𝑃 (𝑡) =
𝑅 (𝑡) =
2𝑡3𝑛−𝑚−𝑖+2 + 6𝑀2
(𝑡3𝑛−𝑚−𝑖+2 + 𝑁2 )
2𝑀
𝑡3𝑛−𝑚−𝑖+2
2𝑀 + 3𝑀2
,
𝑡2𝑛−𝑚+3 + 1
+ 𝑁2
2
,
,
∀ (𝑡, 𝑢, V, 𝑤) ∈ [𝑡0 , +∞) × [𝑁, 𝑀]3 .
(123)
(121)
It is easy to check that the conditions of Theorem 6 are
satisfied. Thus Theorem 6 ensures that (120) has uncountably
many positive solutions in Ω3 (2, 6), and, for any 𝐿 ∈ (2, 4),
there exist 𝜃 ∈ (0, 1) and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| such that the
Mann iterative sequence {𝑥𝑘 }𝑘∈N0 generated by (61) and (21)
converges to a positive solution 𝑥 ∈ Ω3 (2, 6) of (120) and has
the error estimate (20).
It is easy to check that the conditions of Theorem 7 are
satisfied. Thus (122) has uncountably many positive solutions
in Ω1 (1, 100), and, for any 𝐿 ∈ (1, 100), there exist 𝜃 ∈ (0, 1)
and 𝑇 > 1+|𝑡0 |+𝜏+|𝛾| such that the Mann iterative sequence
{𝑥𝑘 }𝑘∈N0 generated by (66) and (21) converges to a positive
solution 𝑥 ∈ Ω1 (1, 100) of (122) and has the error estimate
(20).
Abstract and Applied Analysis
27
Example 18. Consider the higher order nonlinear neutral
delay differential equation
[
2 + sin (𝑡 + √𝑡)
𝑡2
+(
[𝑡𝑚+1 ln (4 + sin (𝑡2 − √𝑡)) (𝑥 (𝑡) − 𝑥 (𝑡 − 𝜏))(𝑚) ]
(𝑛−1)
(𝑥 (𝑡) + 𝑥 (𝑡 − 𝜏))󸀠 ]
+(
𝑥3 (𝑡 − 2) − 𝑡2 𝑥4 (√𝑡 + 1 − 1)
𝑡5𝑛+3 + 𝑡 + 1
(𝑖)
)
4
√𝑡2
𝑡𝑛+3
+ ln 𝑡
sin (𝑡 −
+ 1)
,
=
ℎ2 (𝑡) = √𝑡 + 1 − 1,
𝑓1 (𝑡) = 𝑡 − 3,
𝑓2 (𝑡) = 𝑡 − (−1)𝑛 ,
𝑡2
ℎ (𝑡, 𝑢, V) =
𝑔 (𝑡) =
𝑃 (𝑡) =
,
𝑏 (𝑡) = 1,
𝑢3 − 𝑡
,
+ V2
+ ln 𝑡
(𝑡3𝑛+4 + 𝑁2 )
𝑡5𝑛+3 + 𝑡 + 1
,
∀𝑡 ≥ 4,
ℎ1 (𝑡) = 𝑡 − √𝑡,
(125)
,
2
𝑊 (𝑡) =
where 𝜏 > 0, 𝑚 ≥ 2 and 𝑖 ≤ 𝑛 − 𝑚 − 1. Let 𝑙 = 1, 𝑡0 = 4,
𝛾 = min{4 − 𝜏, 1}, 𝑀 = 7, 𝑁 = 5 and
𝑃 (𝑡) =
𝑀 (5𝑀3 + 2𝑡 + 2𝑀𝑡3𝑛+4 )
,
𝑡𝑛 + 2𝑡 − cos3 (𝑡2 − 3)
𝑔 (𝑡) =
𝑡3𝑛+4
sin (𝑡4 − √𝑡2 + 1)
𝑀2 (3 + 4𝑀𝑡2 )
𝑡cos5 (𝑡7 − 𝑡4 + 1)
𝑓 (𝑡, 𝑢) =
𝑅 (𝑡) =
,
𝑀3 (1 + 𝑀𝑡2 )
𝑡5𝑛+3 + 𝑡 + 1
𝑓1 (𝑡) = 𝑡 − 3,
ℎ (𝑡, 𝑢) =
𝑡
𝑡𝑛−𝑚+4
+ 𝑢4
sin 𝑡3 (−2𝑡 + √𝑡3 + 1)
𝑡𝑛+𝑚 + |𝑡 − 𝑢|
𝑡cos5 (𝑡7 − 𝑡4 + 1)
𝑡𝑛 + 2𝑡 − cos3 (𝑡2 − 3)
1
,
𝑡2𝑛+2𝑚
4𝑀3
𝑡2𝑛−2𝑚+7
,
𝑄 (𝑡) =
𝑊 (𝑡) =
,
,
(127)
,
1
,
𝑡𝑛+𝑚
1
𝑡𝑛−𝑚+3
,
∀ (𝑡, 𝑢) ∈ [𝑡0 , +∞) × [𝑁, 𝑀]2 .
𝑀3 + 𝑡
,
𝑄 (𝑡) = 3𝑛+4
𝑡
+ 𝑁2
𝑅 (𝑡) =
𝑡𝑛+𝑚 + |𝑡 − 𝑥 (𝑡 − 3)|
𝑏 (𝑡) = −1,
𝑢3 − 𝑡2 V4
,
+𝑡+1
𝑡𝑛+3
sin (𝑡3 − 2𝑡 + √𝑡3 + 1)
𝑎 (𝑡) = 𝑡𝑚+1 ln (4 + sin (𝑡2 − √𝑡)) ,
𝑡5𝑛+3
𝑓 (𝑡, 𝑢, V) =
𝑡𝑛−𝑚+4 + 𝑥4 (𝑡 − √𝑡)
)
(126)
ℎ1 (𝑡) = 𝑡 − 2,
2 + sin (𝑡 + √𝑡)
(𝑛−𝑚)
(𝑖)
𝑡
∀𝑡 ≥ 3,
where 𝜏 > 0, 𝑚 = 1 and 𝑖 ≤ 𝑛 − 2. Let 𝑙 = 2, 𝑡0 = 3, 𝛾 =
min{3 − 𝜏, 0}, 𝑀 = 10, 𝑁 = 9 and
𝑎 (𝑡) =
+
(124)
𝑥3 (𝑡 − 3) − 𝑡
+ 3𝑛+4
𝑡
+ 𝑥2 (𝑡 − (−1)𝑛 )
=
Example 19. Consider the higher order nonlinear neutral
delay differential equation
,
∀ (𝑡, 𝑢, V) ∈ [𝑡0 , +∞) × [𝑁, 𝑀]2 .
It is easy to check that the conditions of Theorem 9 are
satisfied. Thus Theorem 9 ensures that (126) has uncountably
many positive solutions in Ω1 (5, 7), and, for any 𝐿 ∈ (5, 7),
there exist 𝜃 ∈ (0, 1) and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| such that the
Mann iterative sequence {𝑥𝑘 }𝑘∈N0 generated by (90) and (21)
converges to a positive solution 𝑥 ∈ Ω1 (5, 7) of (126) and has
the error estimate (20).
Example 20. Consider the higher order nonlinear neutral
delay differential equation
It is easy to check that the conditions of Theorem 8 are
satisfied. Thus Theorem 8 ensures that (124) has uncountably
many positive solutions in Ω1 (9, 10), and, for any 𝐿 ∈ (9, 10),
there exist 𝜃 ∈ (0, 1) and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| such that the
Mann iterative sequence {𝑥𝑘 }𝑘∈N0 generated by (78) and (21)
converges to a positive solution 𝑥 ∈ Ω1 (9, 10) of (124) and
has the error estimate (20).
(𝑛−1)
1
[ 3 (𝑥 (𝑡) − 𝑥 (𝑡 − 𝜏))󸀠 ]
𝑡
𝑡 − sin (𝑡8 − 4𝑡5 − 1)
+ ( 𝑛+7 󵄨󵄨
󵄨)
𝑡 + 󵄨󵄨𝑥 (𝑡 − 1) − 𝑥3 (𝑡 − 2)󵄨󵄨󵄨
(𝑖)
28
Abstract and Applied Analysis
+
=
ln (1 + 𝑥2 (𝑡 − arctan 𝑡))
𝑡2𝑛+6
+
𝑥2
(𝑡 − 4)
𝑡 ln (𝑡 + cos (𝑡3 − 1))
𝑡𝑛+8 + 1
,
[5]
∀𝑡 ≥ 5,
(128)
where 𝜏 > 0, 𝑚 = 1 and 𝑖 ≤ 𝑛 − 2. Let 𝑙 = 2, 𝑡0 = 5, 𝛾 =
min{5 − 𝜏, 1}, 𝑀 = 4, 𝑁 = 2 and
ℎ1 (𝑡) = 𝑡 − 1,
ℎ2 (𝑡) = 𝑡 − 2,
1
𝑎 (𝑡) = 3 ,
𝑡
𝑓2 (𝑡) = 𝑡 − 4,
ℎ (𝑡, 𝑢, V) =
ln (1 + 𝑢2 )
𝑓 (𝑡, 𝑢, V) =
𝑡2𝑛+6 + V2
𝑃 (𝑡) =
[7]
𝑏 (𝑡) = −1,
𝑡 − sin (𝑡8 − 4𝑡5 − 1)
󵄨 ,
󵄨
𝑡𝑛+7 + 󵄨󵄨󵄨𝑢 − V3 󵄨󵄨󵄨
,
𝑔 (𝑡) =
2𝑀 (2𝑀2 + 𝑡2𝑛+6 )
𝑅 (𝑡) =
𝑓1 (𝑡) = 𝑡 − arctan 𝑡,
𝑡4𝑛+12
2
2 + 6𝑀
,
𝑡2𝑛+13
,
[6]
[8]
𝑡 ln (𝑡 + cos (𝑡3 − 1))
𝑡𝑛+8 + 1
𝑄 (𝑡) =
𝑊 (𝑡) =
,
𝑀2
,
𝑡2𝑛+6
[9]
[10]
2
,
𝑡𝑛+6
[11]
2
∀ (𝑡, 𝑢, V) ∈ [𝑡0 , +∞) × [𝑁, 𝑀] .
(129)
It is easy to check that the conditions of Theorem 10 are
satisfied. Thus Theorem 10 ensures that (128) has uncountably
many positive solutions in Ω1 (2, 4), and, for any 𝐿 ∈ (2, 4),
there exist 𝜃 ∈ (0, 1) and 𝑇 > 1 + |𝑡0 | + 𝜏 + |𝛾| such that the
Mann iterative sequence {𝑥𝑘 }𝑘∈N0 generated by (102) and (21)
converges to a positive solution 𝑥 ∈ Ω1 (2, 4) of (128) and has
the error estimate (20).
Acknowledgments
The authors would like to thank the referees for useful
comments and suggestions. This paper was supported by the
Science Research Foundation of Educational Department of
Liaoning province (L2012380).
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