- Email: [email protected]
Monotone iterative technique and positive solutions of Lidstone boundary value problems
Applied Mathematics and Computation 138 (2003) 1–9 www.elsevier.com/locate/amc
Monotone iterative technique and positive solutions of Lidstone boundary value problems Qingliu Yao Department of Applied Mathematics, Fundamental Department, Nanjing Economic University, Nanjing 210003, China
Abstract The purpose of this paper is to investigate the existence and iteration of N symmetric positive solutions for Lidstone boundary value problem: ð1Þn wð2nÞ ðtÞ ¼ f ðt; wðtÞÞ; wð2iÞ ð0Þ ¼ wð2iÞ ð1Þ ¼ 0;
0 6 t 6 1;
0 6 i 6 n 1:
By making use of monotone iterative technique, we are proved that the Lidstone boundary value problem has N symmetric positive solutions under suitable conditions, and given the iterative method for approximating to these solutions, where N is a natural number. Ó 2002 Elsevier Science Inc. All rights reserved. Keywords: Lidstone boundary value problem; Symmetric positive solution; Existence and multiplicity; Monotone iterative technique
1. Introduction The purpose of this paper is to consider the symmetric positive solutions of following Lidstone boundary value problem
E-mail address: [email protected] (Q. Yao). 0096-3003/02/$ - see front matter Ó 2002 Elsevier Science Inc. All rights reserved. PII: S 0 0 9 6 - 3 0 0 3 ( 0 1 ) 0 0 3 1 6 - 2
2
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
ðP Þ
n
ð1Þ wð2nÞ ðtÞ ¼ f ðt; wðtÞÞ; wð2iÞ ð0Þ ¼ wð2iÞ ð1Þ ¼ 0;
0 < t < 1; 0 6 i 6 n 1;
where n P 1. Here, a symmetric positive solution w of ðP Þ will mean a solution w of ðP Þ satisfying w ðtÞ ¼ w ð1 tÞ; t 2 ½0; 1 and
w ðtÞ > 0; t 2 ð0; 1Þ:
The following conditions will be assumed in this paper: (1) f : ½0; 1 ½0; þ1Þ ! ½0; þ1Þ is continuous; (2) f ðt; lÞ is nondecreasing in l for any t 2 ½0; 1; (3) for any l 2 ½0; þ1Þ; f ðt; lÞ ¼ f ð1 t; lÞ;
t 2 ½0; 1:
In recent years Lidstone boundary value problem ðP Þ has attracted considerable attention [1–7]. Very recently, Wong and Agarwal [5] are considered the eigenvalues of problem ðP Þ. They are derived explicit intervals such that for any k in the interval, the existence of a positive solution of ðP Þ is guaranteed. And, Henderson and Thompson [6] are obtained the existence of three symmetric nonnegative solutions for the problem ðP Þ with f ðt; lÞ ¼ f ðlÞ. In this paper, we shall consider the properties of f ðt; lÞ in bounded sets and establish a new existence theorem (Theorem 3.1) by making use of monotone iterative technique (see [8–10]). Then, we shall prove the existence of N symmetric positive solutions, and give the iteration procedures for approximating to these solutions (Theorem 3.3), where N is a natural number. 2. Preliminaries In what follows C½0; 1 will denote the Banach space of continuous functions w: ½0; 1 ! R with the norm kwk ¼ max0 6 t 6 1 jwðtÞtj. A function w 2 C½0; 1 is said to be symmetric, if wðtÞ ¼ wð1 tÞ;
t 2 ½0; 1:
And, function w is said to be concave, if for any t1 , t2 , s 2 ½0; 1, wðst1 þ ð1 sÞt2 Þ P swðt1 Þ þ ð1 sÞwðt2 Þ: We denote C0þ ½0; 1 ¼ fw 2 C½0; 1: wðtÞ P 0; 0 6 t 6 1 K ¼ fw 2
C0þ ½0; 1;
and
wð0Þ ¼ wð1Þ ¼ 0g;
w is symmetric and concaveg:
It is clear that K is a cone of nonnegative functions in C½0; 1. Let qðtÞ ¼ 2 minft; 1 tg;
t 2 ½0; 1;
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
3
then max0 6 t 6 1 qðtÞ ¼ qð1=2Þ ¼ 1, min1=4 6 t 6 3=4 qðtÞ ¼ ð1=2Þ. If w 2 K, then kwk ¼ wð1=2Þ and wðtÞ P kwkqðtÞ;
t 2 ½0; 1:
In particular, max1=4 6 t 6 3=4 wðtÞ P 12 kwk. We denote that Gn ðt; sÞ is the GreenÕs function of homogeneous boundary value problem: wð2nÞ ðtÞ ¼ 0; 0 6 t 6 1; wð2iÞ ð0Þ ¼ wð2iÞ ð1Þ ¼ 0; 1 6 i 6 n 1: By the induction, the GreenÕs function Gn ðt; sÞ can be expressed as (see [5]) Z 1 Gðt; fÞGi1 ðf; sÞ df; 2 6 i 6 n; Gi ðt; sÞ ¼ 0
where G1 ðt; sÞ ¼ Gðt; sÞ ¼
tð1 sÞ; sð1 tÞ;
0 6 t 6 s 6 1; 0 6 s 6 t 6 1:
It is easy to see that Gn ðt; sÞ > 0;
ðt; sÞ 2 ð0; 1Þ ð0; 1Þ:
Lemma 2.1. (see [5]) (1) For any ðt; sÞ 2 ½0; 1 ½0; 1, Gn ðt; sÞ 6
1 sð1 sÞ: 6n1
(2) Let d 2 ð0; 12Þ. Then for any ðt; sÞ 2 ½d; 1 d ½0; 1, Gn ðt; sÞ P hn ðdÞsð1 sÞ P 6n1 hn ðdÞ max Gn ðt; sÞ; 06t61
where hn ðdÞ ¼ dn
4d3 6d2 þ 1 6
n1 :
Define the operator Tn : C½0; 1 ! C½0; 1 as follows Z 1 Gn ðt; sÞf ðs; wðsÞÞ ds; t 2 ½0; 1: ðTn wÞðtÞ ¼ 0
By an immediate calculation, we have
4
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
Lemma 2.2. For any w 2 C½0; 1 and n P 1, ðTn wÞ00 ðtÞ ¼ ðTn1 wÞðtÞ;
t 2 ð0; 1Þ;
where ðT0 wÞðtÞ ¼ f ðt; wðtÞÞ. The foundation of this paper is the following lemma. Lemma 2.3. Tn : K ! K is compactly continuous. Proof Step I. We show Tn : C0þ ½0; 1 ! C0þ ½0; 1 is completely continuous. It is clear that Tn : C0þ ½0; 1 ! C0þ ½0; 1 is continuous. Let R is a bounded set in þ C0 ½0; 1. We denote d ¼ supfkwk: w 2 Rg and Md ¼ maxff ðt; lÞ: ðt; lÞ 2 ½0; 1 ½0; dg: It is easy to prove that, for any t1 ; t2 2 ½0; 1, jGn ðt1 ; sÞ Gn ðt2 ; sÞj 6 2jt1 t2 j: Let w 2 R, then jðTn wÞðt1 Þ ðTn wÞðt2 Þj 6
Z
1
jGn ðt1 ; sÞ Gn ðt2 ; sÞjf ðs; wðsÞÞ ds 0
6 2Md jt1 t2 j: Further, from Lemma 2.1, Z 1 kTn wk ¼ max Gn ðt; sÞf ðs; wðsÞÞ ds 06t61 0 Z 1 1 1 sð1 sÞ ds ¼ n Md : 6 n1 Md max 06t61 0 6 6 According to Arzela–Ascoli theorem (see [8]), the set Tn ðRÞ is compact. Thus, Tn : C0þ ½0; 1 ! C0þ ½0; 1 is completely continuous. Step II. We show Tn : K ! K. Let w 2 K. Since Gn ð0; sÞ ¼ Gn ð1; sÞ ¼ 0, s 2 ½0; 1, we have ðTn wÞð0Þ ¼ ðTn wÞð1Þ ¼ 0. By Lemma 2.2, ðTn wÞ00 ðtÞ ¼ ðTn1 wÞðtÞ 6 0;
t 2 ð0; 1Þ:
We see that Tn w is a nonnegative concave function on ½0; 1. Now, we prove that Tn w is symmetric by the induction. Because h and w are symmetric, let s0 ¼ 1 s, then
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
ðT1 wÞðtÞ ¼
Z
t
sð1 tÞf ðs; wðsÞÞ ds þ 0
¼
Z
5
1
tð1 sÞf ðs; wðsÞÞ ds
t
Z Z
1t
ð1 s0 Þð1 tÞf ð1 s0 ; wð1 s0 ÞÞ ds0
1 0
ts0 f ð1 s0 ; wð1 s0 ÞÞ ds0 1t
¼
Z
1
ð1 s0 Þð1 tÞf ðs0 ; wðs0 ÞÞ ds0 þ 1t
¼
Z
Z
1t
ts0 f ðs0 ; wðs0 ÞÞ ds0 0
1
Gð1 t; s0 Þf ðs0 ; wðs0 ÞÞ ds0 ¼ ðT1 wÞð1 tÞ: 0
Thus T1 w is symmetric. We suppose that Tn1 w is symmetric. If Tn w is not symmetric, then there exists t0 2 ð0; 1Þ such that max jðTn wÞð1 tÞ ðTn wÞðtÞj ¼ jðTn wÞð1 t0 Þ ðTn wÞðt0 Þj > 0:
06t61
Without loss of generality, we may assume max ½ðTn wÞð1 tÞ ðTn wÞðtÞ ¼ ðTn wÞð1 t0 Þ ðTn wÞðt0 Þ > 0:
06t61
Let jðtÞ ¼ ðTn wÞð1 tÞ ðTn wÞðtÞ;
t 2 ½0; 1;
then jðt0 Þ ¼ max0 6 t 6 1 jðtÞ > 0. Since jð0Þ ¼ jð1Þ ¼ 0, it follows that there exists d > 0 such that jðtÞ is a concave function on ½t0 d; t0 þ d, and jðtÞ is not a constant on ½t0 d; t0 þ d. Thus, we assert that there exists t1 2 ðt0 d; t0 þ dÞ such that j00 ðt1 Þ < 0. By the inductive assumption and Lemma 2.2, we have 0 ¼ ½ðTn1 wÞð1 t1 Þ ½ðTn1 wÞðt1 Þ 00
00
¼ ðTn wÞ ð1 t1 Þ ðTn wÞ ðt1 Þ ¼ j00 ðt1 Þ < 0: It is impossible. Thus, Tn : K ! K. The proof is complete.
Noticing that f ðt; lÞ is nondecreasing in l, the proof of following lemma is simple. Lemma 2.4. If w1 ; w2 2 K such that w2 ðtÞ P w1 ðtÞ; t 2 ½0; 1, then ðTn w2 ÞðtÞ P ðTn w1 ÞðtÞ, t 2 ½0; 1.
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Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
Let A¼
Z max
06t61
"
1
1
Gn ðt; sÞ ds
;
B¼
0
Z max
06t61
#1
3=4
Gn ðt; sÞ ds
:
1=4
It is easy to see that 0 < A < B. The constants A, B are not easy to compute explicitly. For convenience, we can replace A by A0 , replace B by B0 , from Lemma 2.1, where 1 Z 1 1 A0 ¼ n1 sð1 sÞ ds ¼ 6n ; 6 0 " Z #1 3=4 1 3n 27n 0 B ¼ hn sð1 sÞ ds ¼ : 4 11n 1=4 Obviously, A0 < A, B < B0 . 3. Main results We denote min f 0 ¼ limt 6 3=4f ðt; lÞ=l;
max f 1 ¼ lim max f ðt; lÞ=l; l!1 0 6 t 6 1
l!0
min f0 ¼ lim
min
l!0 1=4 6 t 6 3=4
f ðt; lÞ=l;
max f1 ¼ lim max f ðt; lÞ=l: l!1 0 6 t 6 1
3.1. Existence of one symmetric positive solution Theorem 3.1. Assume that there exist two positive numbers a > b such that max f ðt; aÞ 6 aA;
06t61
min
1=4 6 t 6 3=4
f ðt; b=2Þ P bB:
Then problem ðP Þ has at least one symmetric positive solution w 2 K such e , i.e., that b 6 kw k 6 a, and w ¼ limk!1 Tnk w
k e ÞðtÞ w ðtÞ ¼ 0; lim max ðTn w k!1 0 6 t 6 1
e ðtÞ ¼ bqðtÞ, t 2 ½0; 1. where w Proof. We denote K½b; a ¼ fw 2 K: b 6 kwk 6 ag. Let w 2 K½b; a, then max0 6 t 6 1 wðtÞ 6 a and min
1=4 6 t 6 3=4
wðtÞ P b
min
1=4 6 t 6 3=4
qðtÞ ¼ b=2:
Since f is nondecreasing, we have f ðt; wðtÞÞ 6 f ðaÞ 6 aA;
t 2 ½0; 1;
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
f ðt; wðtÞÞ P f ðb=2Þ P bB; It follows that kTn wk ¼ max
Z
06t61
7
t 2 ½1=4; 3=4:
1
Gn ðt; sÞf ðs; wðsÞÞ ds 6 aA max
Z
06t61
0
1
Gn ðt; sÞ ds ¼ a
0
and kTn wk P max
Z
06t61
3=4
Gn ðt; sÞf ðs; wðsÞÞ ds P bB max
06t61
1=4
Z
3=4
Gn ðt; sÞ ds ¼ b:
1=4
Thus we assert that Tn : K½b; a ! K½b; a by Lemma 2.3. e 2 K½b; a. Let w e 1 ¼ Tn w e , then w e 1 2 K½b; a. We denote It is clear that w e k; e kþ1 ¼ Tn w w
k ¼ 1; 2; . . .
e k 2 K½b; a, k ¼ 1; 2; . . . From Lemma 2.3, Since Tn ðK½b; aÞ K½b; a, we have w e k g1 Tn is completely continuous. We assert that f w k¼1 has a convergent subse1 e ki gi¼1 and there exists w 2 K½b; a such that w e ki ! w . quence f w e 1 2 K½b; a implies Now, w e 1 kqðtÞ P bqðtÞ ¼ w e ðtÞ; e 1 ðtÞ P k w w
t 2 ½0; 1:
Using Lemma 2.4 and the induction, we have e k ðtÞ; e kþ1 ðtÞ P w w
t 2 ½0; 1; k ¼ 1; 2; . . .
e k ! w , i.e., Tnk w e ! w ðk ! 1Þ. So, Tn w ¼ w . Since Hence, we assert that w kw k P b > 0 and w is a nonnegative, symmetric and concave function on ½0; 1, we assert that w ðtÞ > 0;
t 2 ð0; 1Þ:
It is well-known that the fixed point of operator Tn is the solution of problem ðP Þ. Therefore, w is a positive solution of the problem ðP Þ. Corollary 3.2. Assume that min f 0 > 2B; max f 1 < A ðparticularly, min f0 ¼ þ1, max f1 ¼ 0Þ. Then problem ðP Þ has at least one symmetric positive solution e , i.e., w 2 K, and there exists b > 0 such that w ¼ limk!1 Tnk w
k e ÞðtÞ w ðtÞ ¼ 0; lim max ðTn w k!1 0 6 t 6 1
e ðtÞ ¼ bqðtÞ; t 2 ½0; 1. where w 3.2. Existence of N symmetric positive solutions Theorem 3.3. If there exist 2N positive numbers b1 < a 1 < b 2 < a 2 < < b N < a N such that
8
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
max f ðt; ai Þ 6 ai A;
06t61
min
1=4 6 t 6 3=4
f ðt; bi =2Þ P bi B; i ¼ 1; 2; . . . ; N :
Then problem ðP Þ has N symmetric positive solution wi 2 K, i ¼ 1; 2; . . . ; N , e i Þ ¼ wi , i.e., satisfying bi 6 kwi k 6 ai and limk!1 Tnk ð w
e i ÞðtÞ wi ðtÞ ¼ 0; lim max ðTnk w k!1 0 6 t 6 1
e i ðtÞ ¼ bi qðtÞ, t 2 ½0; 1. where w Proof. The proof is immediate from Theorem 3.1.
Corollary 3.4. Assume that min f 0 > 2B; max f 1 < A ðparticularly, min f0 ¼ þ1, max f1 ¼ 0Þ. If there exist two positive numbers a < b such that max f ðt; aÞ 6 aA;
06t61
min
1=4 6 t 6 3=4
f ðt; b=2Þ P bB:
Then problem ðP Þ has at least two symmetric positive solution w1 ; w2 2 K satisfying 0 < kw1 k 6 a < b 6 kw2 k e i ; i ¼ 1; 2, i.e., and there exists 0 < c < a such that wi ¼ limk!1 Tnk w
k
e i ÞðtÞ wi ðtÞ ¼ 0; lim max ðTn w k!1 0 6 t 6 1
where e 1 ðtÞ ¼ cqðtÞ; w
e 2 ðtÞ ¼ bqðtÞ; t 2 ½0; 1: w
Corollary 3.5. Assume that min f 0 > 2B; max f 1 < A ðparticularly, min f0 ¼ þ1, max f1 ¼ 0Þ. If there exist four positive numbers a1 < b1 < a2 < b2 such that max f ðt; ai Þ 6 ai A;
06t61
min
1=4 6 t 6 3=4
f ðt; bi =2Þ P bi B; i ¼ 1; 2:
Then problem ðP Þ has at least three symmetric positive solution w1 ; w2 ; w3 2 K satisfying 0 < kw1 k 6 a1 < b1 6 kw2 k 6 a2 < b2 6 kw3 k e i , i ¼ 1; 2; 3, i.e., and there exists 0 < c < a1 such that wi ¼ limk!1 Tnk w
k
e i ÞðtÞ wi ðtÞ ¼ 0; lim max ðTn w k!1 0 6 t 6 1
where e 1 ðtÞ ¼ cqðtÞ; w
e 2 ðtÞ ¼ b1 qðtÞ; w
e 3 ðtÞ ¼ b2 qðtÞ; t 2 ½0; 1: w
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
9
References [1] R.P. Agarwal, Boundary Value Problems for Higher Order Differential Equations, World Scientific, Singapore, 1986. [2] R.P. Agarwal, P.J.Y. Wong, On Lidstone polynomials and boundary value problems, Comput. Math. Appl. 17 (1989) 1397–1421. [3] E.H. Twizell, A. Boutayeb, K. Djidjli, Numerical methods for eighth-, tenth-, and twelvethorder eigenvalue problems arising in thermal instability, Adv. Comput. Math. 2 (1994) 407– 436. [4] R.P. Agarwal, Focal Boundary Value Problems for Differential and Difference Equations, Kluwer Academic Publishers, Dordrecht, 1998. [5] P.J.Y. Wong, R.P. Agarwal, Eigenvalues of Lidstone boundary value problems, Appl. Math. Comput. 104 (1999) 15–31. [6] J. Henderson, H.B. Thompson, Multiple symmetric positive solutions for a second order boundary value problem, Proc. Am. Math. Soc. 128 (2000) 2373–2379. [7] Q. Yao, Z. Bai, Existence of positive solutions of BVP for uð4Þ ðtÞ khðtÞf ðuðtÞÞ ¼ 0, Chinese Annal. Math. 20 (A) (1999) 575–578 (in Chinese). [8] G.S. Ladde, V. Lakshmikantham, A.S. Vatsala, Monotone Iterative Techniques for Nonlinear Differential Equations, Pitman Advanced Publishing Program, Boston, MA, 1985. [9] D. Guo, V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, New York, 1988. [10] E. Zeidler, Nonlinear Functional Analysis and its Applications II/B: Nonlinear Monotone Operators, Springer-Verlag, New York, 1992.
Monotone iterative technique and positive solutions of Lidstone boundary value problems Qingliu Yao Department of Applied Mathematics, Fundamental Department, Nanjing Economic University, Nanjing 210003, China
Abstract The purpose of this paper is to investigate the existence and iteration of N symmetric positive solutions for Lidstone boundary value problem: ð1Þn wð2nÞ ðtÞ ¼ f ðt; wðtÞÞ; wð2iÞ ð0Þ ¼ wð2iÞ ð1Þ ¼ 0;
0 6 t 6 1;
0 6 i 6 n 1:
By making use of monotone iterative technique, we are proved that the Lidstone boundary value problem has N symmetric positive solutions under suitable conditions, and given the iterative method for approximating to these solutions, where N is a natural number. Ó 2002 Elsevier Science Inc. All rights reserved. Keywords: Lidstone boundary value problem; Symmetric positive solution; Existence and multiplicity; Monotone iterative technique
1. Introduction The purpose of this paper is to consider the symmetric positive solutions of following Lidstone boundary value problem
E-mail address: [email protected] (Q. Yao). 0096-3003/02/$ - see front matter Ó 2002 Elsevier Science Inc. All rights reserved. PII: S 0 0 9 6 - 3 0 0 3 ( 0 1 ) 0 0 3 1 6 - 2
2
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
ðP Þ
n
ð1Þ wð2nÞ ðtÞ ¼ f ðt; wðtÞÞ; wð2iÞ ð0Þ ¼ wð2iÞ ð1Þ ¼ 0;
0 < t < 1; 0 6 i 6 n 1;
where n P 1. Here, a symmetric positive solution w of ðP Þ will mean a solution w of ðP Þ satisfying w ðtÞ ¼ w ð1 tÞ; t 2 ½0; 1 and
w ðtÞ > 0; t 2 ð0; 1Þ:
The following conditions will be assumed in this paper: (1) f : ½0; 1 ½0; þ1Þ ! ½0; þ1Þ is continuous; (2) f ðt; lÞ is nondecreasing in l for any t 2 ½0; 1; (3) for any l 2 ½0; þ1Þ; f ðt; lÞ ¼ f ð1 t; lÞ;
t 2 ½0; 1:
In recent years Lidstone boundary value problem ðP Þ has attracted considerable attention [1–7]. Very recently, Wong and Agarwal [5] are considered the eigenvalues of problem ðP Þ. They are derived explicit intervals such that for any k in the interval, the existence of a positive solution of ðP Þ is guaranteed. And, Henderson and Thompson [6] are obtained the existence of three symmetric nonnegative solutions for the problem ðP Þ with f ðt; lÞ ¼ f ðlÞ. In this paper, we shall consider the properties of f ðt; lÞ in bounded sets and establish a new existence theorem (Theorem 3.1) by making use of monotone iterative technique (see [8–10]). Then, we shall prove the existence of N symmetric positive solutions, and give the iteration procedures for approximating to these solutions (Theorem 3.3), where N is a natural number. 2. Preliminaries In what follows C½0; 1 will denote the Banach space of continuous functions w: ½0; 1 ! R with the norm kwk ¼ max0 6 t 6 1 jwðtÞtj. A function w 2 C½0; 1 is said to be symmetric, if wðtÞ ¼ wð1 tÞ;
t 2 ½0; 1:
And, function w is said to be concave, if for any t1 , t2 , s 2 ½0; 1, wðst1 þ ð1 sÞt2 Þ P swðt1 Þ þ ð1 sÞwðt2 Þ: We denote C0þ ½0; 1 ¼ fw 2 C½0; 1: wðtÞ P 0; 0 6 t 6 1 K ¼ fw 2
C0þ ½0; 1;
and
wð0Þ ¼ wð1Þ ¼ 0g;
w is symmetric and concaveg:
It is clear that K is a cone of nonnegative functions in C½0; 1. Let qðtÞ ¼ 2 minft; 1 tg;
t 2 ½0; 1;
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
3
then max0 6 t 6 1 qðtÞ ¼ qð1=2Þ ¼ 1, min1=4 6 t 6 3=4 qðtÞ ¼ ð1=2Þ. If w 2 K, then kwk ¼ wð1=2Þ and wðtÞ P kwkqðtÞ;
t 2 ½0; 1:
In particular, max1=4 6 t 6 3=4 wðtÞ P 12 kwk. We denote that Gn ðt; sÞ is the GreenÕs function of homogeneous boundary value problem: wð2nÞ ðtÞ ¼ 0; 0 6 t 6 1; wð2iÞ ð0Þ ¼ wð2iÞ ð1Þ ¼ 0; 1 6 i 6 n 1: By the induction, the GreenÕs function Gn ðt; sÞ can be expressed as (see [5]) Z 1 Gðt; fÞGi1 ðf; sÞ df; 2 6 i 6 n; Gi ðt; sÞ ¼ 0
where G1 ðt; sÞ ¼ Gðt; sÞ ¼
tð1 sÞ; sð1 tÞ;
0 6 t 6 s 6 1; 0 6 s 6 t 6 1:
It is easy to see that Gn ðt; sÞ > 0;
ðt; sÞ 2 ð0; 1Þ ð0; 1Þ:
Lemma 2.1. (see [5]) (1) For any ðt; sÞ 2 ½0; 1 ½0; 1, Gn ðt; sÞ 6
1 sð1 sÞ: 6n1
(2) Let d 2 ð0; 12Þ. Then for any ðt; sÞ 2 ½d; 1 d ½0; 1, Gn ðt; sÞ P hn ðdÞsð1 sÞ P 6n1 hn ðdÞ max Gn ðt; sÞ; 06t61
where hn ðdÞ ¼ dn
4d3 6d2 þ 1 6
n1 :
Define the operator Tn : C½0; 1 ! C½0; 1 as follows Z 1 Gn ðt; sÞf ðs; wðsÞÞ ds; t 2 ½0; 1: ðTn wÞðtÞ ¼ 0
By an immediate calculation, we have
4
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
Lemma 2.2. For any w 2 C½0; 1 and n P 1, ðTn wÞ00 ðtÞ ¼ ðTn1 wÞðtÞ;
t 2 ð0; 1Þ;
where ðT0 wÞðtÞ ¼ f ðt; wðtÞÞ. The foundation of this paper is the following lemma. Lemma 2.3. Tn : K ! K is compactly continuous. Proof Step I. We show Tn : C0þ ½0; 1 ! C0þ ½0; 1 is completely continuous. It is clear that Tn : C0þ ½0; 1 ! C0þ ½0; 1 is continuous. Let R is a bounded set in þ C0 ½0; 1. We denote d ¼ supfkwk: w 2 Rg and Md ¼ maxff ðt; lÞ: ðt; lÞ 2 ½0; 1 ½0; dg: It is easy to prove that, for any t1 ; t2 2 ½0; 1, jGn ðt1 ; sÞ Gn ðt2 ; sÞj 6 2jt1 t2 j: Let w 2 R, then jðTn wÞðt1 Þ ðTn wÞðt2 Þj 6
Z
1
jGn ðt1 ; sÞ Gn ðt2 ; sÞjf ðs; wðsÞÞ ds 0
6 2Md jt1 t2 j: Further, from Lemma 2.1, Z 1 kTn wk ¼ max Gn ðt; sÞf ðs; wðsÞÞ ds 06t61 0 Z 1 1 1 sð1 sÞ ds ¼ n Md : 6 n1 Md max 06t61 0 6 6 According to Arzela–Ascoli theorem (see [8]), the set Tn ðRÞ is compact. Thus, Tn : C0þ ½0; 1 ! C0þ ½0; 1 is completely continuous. Step II. We show Tn : K ! K. Let w 2 K. Since Gn ð0; sÞ ¼ Gn ð1; sÞ ¼ 0, s 2 ½0; 1, we have ðTn wÞð0Þ ¼ ðTn wÞð1Þ ¼ 0. By Lemma 2.2, ðTn wÞ00 ðtÞ ¼ ðTn1 wÞðtÞ 6 0;
t 2 ð0; 1Þ:
We see that Tn w is a nonnegative concave function on ½0; 1. Now, we prove that Tn w is symmetric by the induction. Because h and w are symmetric, let s0 ¼ 1 s, then
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
ðT1 wÞðtÞ ¼
Z
t
sð1 tÞf ðs; wðsÞÞ ds þ 0
¼
Z
5
1
tð1 sÞf ðs; wðsÞÞ ds
t
Z Z
1t
ð1 s0 Þð1 tÞf ð1 s0 ; wð1 s0 ÞÞ ds0
1 0
ts0 f ð1 s0 ; wð1 s0 ÞÞ ds0 1t
¼
Z
1
ð1 s0 Þð1 tÞf ðs0 ; wðs0 ÞÞ ds0 þ 1t
¼
Z
Z
1t
ts0 f ðs0 ; wðs0 ÞÞ ds0 0
1
Gð1 t; s0 Þf ðs0 ; wðs0 ÞÞ ds0 ¼ ðT1 wÞð1 tÞ: 0
Thus T1 w is symmetric. We suppose that Tn1 w is symmetric. If Tn w is not symmetric, then there exists t0 2 ð0; 1Þ such that max jðTn wÞð1 tÞ ðTn wÞðtÞj ¼ jðTn wÞð1 t0 Þ ðTn wÞðt0 Þj > 0:
06t61
Without loss of generality, we may assume max ½ðTn wÞð1 tÞ ðTn wÞðtÞ ¼ ðTn wÞð1 t0 Þ ðTn wÞðt0 Þ > 0:
06t61
Let jðtÞ ¼ ðTn wÞð1 tÞ ðTn wÞðtÞ;
t 2 ½0; 1;
then jðt0 Þ ¼ max0 6 t 6 1 jðtÞ > 0. Since jð0Þ ¼ jð1Þ ¼ 0, it follows that there exists d > 0 such that jðtÞ is a concave function on ½t0 d; t0 þ d, and jðtÞ is not a constant on ½t0 d; t0 þ d. Thus, we assert that there exists t1 2 ðt0 d; t0 þ dÞ such that j00 ðt1 Þ < 0. By the inductive assumption and Lemma 2.2, we have 0 ¼ ½ðTn1 wÞð1 t1 Þ ½ðTn1 wÞðt1 Þ 00
00
¼ ðTn wÞ ð1 t1 Þ ðTn wÞ ðt1 Þ ¼ j00 ðt1 Þ < 0: It is impossible. Thus, Tn : K ! K. The proof is complete.
Noticing that f ðt; lÞ is nondecreasing in l, the proof of following lemma is simple. Lemma 2.4. If w1 ; w2 2 K such that w2 ðtÞ P w1 ðtÞ; t 2 ½0; 1, then ðTn w2 ÞðtÞ P ðTn w1 ÞðtÞ, t 2 ½0; 1.
6
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
Let A¼
Z max
06t61
"
1
1
Gn ðt; sÞ ds
;
B¼
0
Z max
06t61
#1
3=4
Gn ðt; sÞ ds
:
1=4
It is easy to see that 0 < A < B. The constants A, B are not easy to compute explicitly. For convenience, we can replace A by A0 , replace B by B0 , from Lemma 2.1, where 1 Z 1 1 A0 ¼ n1 sð1 sÞ ds ¼ 6n ; 6 0 " Z #1 3=4 1 3n 27n 0 B ¼ hn sð1 sÞ ds ¼ : 4 11n 1=4 Obviously, A0 < A, B < B0 . 3. Main results We denote min f 0 ¼ limt 6 3=4f ðt; lÞ=l;
max f 1 ¼ lim max f ðt; lÞ=l; l!1 0 6 t 6 1
l!0
min f0 ¼ lim
min
l!0 1=4 6 t 6 3=4
f ðt; lÞ=l;
max f1 ¼ lim max f ðt; lÞ=l: l!1 0 6 t 6 1
3.1. Existence of one symmetric positive solution Theorem 3.1. Assume that there exist two positive numbers a > b such that max f ðt; aÞ 6 aA;
06t61
min
1=4 6 t 6 3=4
f ðt; b=2Þ P bB:
Then problem ðP Þ has at least one symmetric positive solution w 2 K such e , i.e., that b 6 kw k 6 a, and w ¼ limk!1 Tnk w
k e ÞðtÞ w ðtÞ ¼ 0; lim max ðTn w k!1 0 6 t 6 1
e ðtÞ ¼ bqðtÞ, t 2 ½0; 1. where w Proof. We denote K½b; a ¼ fw 2 K: b 6 kwk 6 ag. Let w 2 K½b; a, then max0 6 t 6 1 wðtÞ 6 a and min
1=4 6 t 6 3=4
wðtÞ P b
min
1=4 6 t 6 3=4
qðtÞ ¼ b=2:
Since f is nondecreasing, we have f ðt; wðtÞÞ 6 f ðaÞ 6 aA;
t 2 ½0; 1;
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
f ðt; wðtÞÞ P f ðb=2Þ P bB; It follows that kTn wk ¼ max
Z
06t61
7
t 2 ½1=4; 3=4:
1
Gn ðt; sÞf ðs; wðsÞÞ ds 6 aA max
Z
06t61
0
1
Gn ðt; sÞ ds ¼ a
0
and kTn wk P max
Z
06t61
3=4
Gn ðt; sÞf ðs; wðsÞÞ ds P bB max
06t61
1=4
Z
3=4
Gn ðt; sÞ ds ¼ b:
1=4
Thus we assert that Tn : K½b; a ! K½b; a by Lemma 2.3. e 2 K½b; a. Let w e 1 ¼ Tn w e , then w e 1 2 K½b; a. We denote It is clear that w e k; e kþ1 ¼ Tn w w
k ¼ 1; 2; . . .
e k 2 K½b; a, k ¼ 1; 2; . . . From Lemma 2.3, Since Tn ðK½b; aÞ K½b; a, we have w e k g1 Tn is completely continuous. We assert that f w k¼1 has a convergent subse1 e ki gi¼1 and there exists w 2 K½b; a such that w e ki ! w . quence f w e 1 2 K½b; a implies Now, w e 1 kqðtÞ P bqðtÞ ¼ w e ðtÞ; e 1 ðtÞ P k w w
t 2 ½0; 1:
Using Lemma 2.4 and the induction, we have e k ðtÞ; e kþ1 ðtÞ P w w
t 2 ½0; 1; k ¼ 1; 2; . . .
e k ! w , i.e., Tnk w e ! w ðk ! 1Þ. So, Tn w ¼ w . Since Hence, we assert that w kw k P b > 0 and w is a nonnegative, symmetric and concave function on ½0; 1, we assert that w ðtÞ > 0;
t 2 ð0; 1Þ:
It is well-known that the fixed point of operator Tn is the solution of problem ðP Þ. Therefore, w is a positive solution of the problem ðP Þ. Corollary 3.2. Assume that min f 0 > 2B; max f 1 < A ðparticularly, min f0 ¼ þ1, max f1 ¼ 0Þ. Then problem ðP Þ has at least one symmetric positive solution e , i.e., w 2 K, and there exists b > 0 such that w ¼ limk!1 Tnk w
k e ÞðtÞ w ðtÞ ¼ 0; lim max ðTn w k!1 0 6 t 6 1
e ðtÞ ¼ bqðtÞ; t 2 ½0; 1. where w 3.2. Existence of N symmetric positive solutions Theorem 3.3. If there exist 2N positive numbers b1 < a 1 < b 2 < a 2 < < b N < a N such that
8
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
max f ðt; ai Þ 6 ai A;
06t61
min
1=4 6 t 6 3=4
f ðt; bi =2Þ P bi B; i ¼ 1; 2; . . . ; N :
Then problem ðP Þ has N symmetric positive solution wi 2 K, i ¼ 1; 2; . . . ; N , e i Þ ¼ wi , i.e., satisfying bi 6 kwi k 6 ai and limk!1 Tnk ð w
e i ÞðtÞ wi ðtÞ ¼ 0; lim max ðTnk w k!1 0 6 t 6 1
e i ðtÞ ¼ bi qðtÞ, t 2 ½0; 1. where w Proof. The proof is immediate from Theorem 3.1.
Corollary 3.4. Assume that min f 0 > 2B; max f 1 < A ðparticularly, min f0 ¼ þ1, max f1 ¼ 0Þ. If there exist two positive numbers a < b such that max f ðt; aÞ 6 aA;
06t61
min
1=4 6 t 6 3=4
f ðt; b=2Þ P bB:
Then problem ðP Þ has at least two symmetric positive solution w1 ; w2 2 K satisfying 0 < kw1 k 6 a < b 6 kw2 k e i ; i ¼ 1; 2, i.e., and there exists 0 < c < a such that wi ¼ limk!1 Tnk w
k
e i ÞðtÞ wi ðtÞ ¼ 0; lim max ðTn w k!1 0 6 t 6 1
where e 1 ðtÞ ¼ cqðtÞ; w
e 2 ðtÞ ¼ bqðtÞ; t 2 ½0; 1: w
Corollary 3.5. Assume that min f 0 > 2B; max f 1 < A ðparticularly, min f0 ¼ þ1, max f1 ¼ 0Þ. If there exist four positive numbers a1 < b1 < a2 < b2 such that max f ðt; ai Þ 6 ai A;
06t61
min
1=4 6 t 6 3=4
f ðt; bi =2Þ P bi B; i ¼ 1; 2:
Then problem ðP Þ has at least three symmetric positive solution w1 ; w2 ; w3 2 K satisfying 0 < kw1 k 6 a1 < b1 6 kw2 k 6 a2 < b2 6 kw3 k e i , i ¼ 1; 2; 3, i.e., and there exists 0 < c < a1 such that wi ¼ limk!1 Tnk w
k
e i ÞðtÞ wi ðtÞ ¼ 0; lim max ðTn w k!1 0 6 t 6 1
where e 1 ðtÞ ¼ cqðtÞ; w
e 2 ðtÞ ¼ b1 qðtÞ; w
e 3 ðtÞ ¼ b2 qðtÞ; t 2 ½0; 1: w
Q. Yao / Appl. Math. Comput. 138 (2003) 1–9
9
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