- Email: [email protected]
On the existence of positive solutions of fourth-order difference equations
Applied Mathematics and Computation 161 (2005) 139–148 www.elsevier.com/locate/amc
On the existence of positive solutions of fourth-order difference equations q Zhimin He
a,*
, Jianshe Yu
b
a
b
Department of Applied Mathematics, Central South University, Changsha 410083, Hunan, PR China Department of Applied Mathematics, Hunan University, Changsha 410082, Hunan, PR China
Abstract In this paper, by means of KrasnoselÕskiiÕs fixed point theorem, the existence of positive solutions of fourth-order difference equations is considered. Ó 2003 Elsevier Inc. All rights reserved. Keywords: Positive solution; Difference equation; KrasnoselÕskiiÕs fixed point theorem
1. Introduction For notation, given a < b in Z, we employ intervals to denote discrete sets such as ½a;b ¼ fa;aþ1;...;bg, ½a;bÞ ¼ fa;aþ1;...;b1g, ½a;1Þ ¼ fa;aþ1;...:g, etc. Let T P1 be fixed. In this paper, we are concerned with determining eigenvalues, k, for which there are positive solutions of the fourth-order difference equation D4 uðt 2Þ kaðtÞf ðuðtÞÞ ¼ 0;
t 2 ½2; T þ 2;
ð1Þ
satisfying the boundary conditions uð0Þ ¼ D2 uð0Þ ¼ uðT þ 2Þ ¼ D2 uðT Þ ¼ 0;
q
ð2Þ
This work was supported by Mathematical Tianyuan Foundation (No: A0324621) of PeopleÕs Republic of China. * Corresponding author. E-mail address: [email protected] (Z. He). 0096-3003/$ - see front matter Ó 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2003.12.016
140
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
or uð0Þ ¼ D2 uð0Þ ¼ DuðT þ 1Þ ¼ D3 uðT 1Þ ¼ 0;
ð3Þ
where (A) f : Rþ ! Rþ is continuous (Rþ denotes the nonnegative reals), (B) aðtÞ is a positive valued function defined on ½1; T þ 1, and (C) f0 ¼ limx!0þ f ðxÞ and f1 ¼ limx!1 f ðxÞ exist. x x The motivation for the present work stems from many recent investigations in [1–6]. For the continuous case, boundary value problems analogous to (1) and (2) (or (1) and (3)) arise in various nonlinear phenomena for which only positive solutions are meaningful; see, for example [7,8]. In this paper we will apply the following KrasnoselÕskiiÕs fixed point theorem in a cone to yields positive solutions of (1) and (2) (or (1) and (3)). This theorem can be found in the book by KrasnoselÕskii [9] as well as in the book by Deimling [10]. Theorem 1.1 [9,10]. Let B be a Banach space, and let P B be a cone in B. Assume X1 , X2 are open subsets of B with 0 2 X1 X1 X2 , and let T : P \ ðX2 n X1 Þ ! P be a completely continuous operator such that, either i(i) kTuk 6 kuk; u 2 P \ oX1 , and kTuk P kuk; u 2 P \ oX2 , or (ii) kTuk P kuk; u 2 P \ oX1 , and kTuk 6 kuk; u 2 P \ oX2 . Then T has a fixed point in P \ ðX2 n X1 Þ.
2. Solutions of (1) and (2) in a cone Lemma 2.1. uðtÞ is a solution of the BVP (1) and (2) if and only if uðtÞ ¼ k
T þ1 X
Gðt; sÞ
s¼1
T X
G1 ðs; iÞaðiÞf ðuðiÞÞ;
t 2 ½0; T þ 2;
ð4Þ
i¼2
where Gðt; sÞ ¼
1 T þ2
G1 ðt; iÞ ¼
1 T
sðT þ 2 tÞ; tðT þ 2 sÞ;
1 6 s 6 t 6 T þ 2; 0 6 t 6 s 6 T þ 1;
ð5Þ
and
ðT þ 1 tÞði 1Þ; ðT þ 1 iÞðt 1Þ;
2 6 i 6 t 6 T þ 1; 16t6i6T:
ð6Þ
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
141
Proof. BVP (1) and (2) is equivalent to solving D2 uðt 1Þ ¼ k
T X
G1 ðt; iÞaðiÞf ðuðiÞÞ;
t 2 ½1; T þ 1;
ð7Þ
i¼2
uð0Þ ¼ uðT þ 2Þ ¼ 0;
ð8Þ
and it is also equivalent to solving uðtÞ ¼ k
T þ1 X
Gðt; sÞ
s¼1
T X
G1 ðs; iÞaðiÞf ðuðiÞÞ;
t 2 ½0; T þ 2:
ð9Þ
i¼2
h
The proof of the lemma is complete. From (5)
for ðt; sÞ 2 ð1; T þ 1Þ ð1; T þ 1Þ;
Gðt; sÞ > 0;
ð10Þ
and Gðt; sÞ 6 Gðs; sÞ ¼
sðT þ 2 sÞ T þ2
for 1 6 s 6 T þ 1; 0 6 t 6 T þ 2:
ð11Þ
Merdivinici [3] has shown that, if Y is defined by T þ1 3ðT þ 1Þ Y ¼ t2Z: 6t6 ; 4 4 and if r ¼ min
min Y T þ 2 max Y ; T þ1 T þ1
;
then Gðt; sÞ P rGðs; sÞ ¼
rsðT þ 2 sÞ T þ2
for t 2 Y ; 1 6 s 6 T þ 1:
ð12Þ
Let B ¼ fu : ½0; T þ 2 ! Rjuð0Þ ¼ uðT þ 2Þ ¼ D2 uð0Þ ¼ D2 uðT Þ ¼ 0g; with norm, kuk ¼ maxt2½0;T þ2 juðtÞj, then ðB; k kÞ is a Banach space. Define a cone, P , by P ¼ fu 2 B : uðtÞ P 0 for t 2 ½0; T þ 2; and mint2Y uðtÞ P rkukg:
ð13Þ
Also, we define the number s 2 ½0; T þ 2 by T þ1 X s¼1
Gðs; sÞ
X i2Y
G1 ðs; iÞaðiÞ ¼ max
t2½0;T þ2
T þ1 X s¼1
Gðt; sÞ
X i2Y
G1 ðs; iÞaðiÞ:
ð14Þ
142
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
Theorem 2.1. Assume that conditions (A), (B) and (C) are satisfied. Then, for each k satisfying 1 P P T þ1 r s¼1 Gðs; sÞ i2Y G1 ðs; iÞaðiÞ f1 < k < P
T þ1 s¼1
Gðs; sÞ
1 PT
i¼2
; G1 ðs; iÞaðiÞ f0
ð15Þ
there exists at least one solution of (1) and (2) in P . Proof. Let k be given as in (15). Then, let e > 0 be such that 1 P P þ1 Gðs; sÞ i2Y G1 ðs; iÞaðiÞ ðf1 eÞ r Ts¼1 6k6 P T þ1 s¼1
Gðs; sÞ
PT
1
i¼2
: G1 ðs; iÞaðiÞ ðf0 þ eÞ
Define a summation operator T : P ! B by ðTuÞðtÞ ¼ k
T þ1 X
Gðt; sÞ
s¼1
T X
G1 ðs; iÞaðiÞf ðuðiÞÞ;
u 2 P:
ð16Þ
i¼2
We note that from (5), (6) and (16), if u 2 P , then ðTuÞðtÞ P 0 for t 2 ½0; T þ 2. From the properties of G, Tu satisfies the boundary conditions (2). Moreover, for u 2 P , we have ðTuÞðtÞ ¼ k
T þ1 X
Gðt; sÞ
s¼1
6k
T þ1 X
T X
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2
Gðs; sÞ
s¼1
T X
G1 ðs; iÞaðiÞf ðuðiÞÞ;
t 2 ½0; T þ 2;
i¼2
and so kTuk ¼ max jðTuÞðtÞj 6 k t2½0;T þ2
T þ1 X
Gðs; sÞ
s¼1
T X
G1 ðs; iÞaðiÞf ðuðiÞÞ:
i¼2
Thus, if u 2 P , then minðTuÞðtÞ ¼ min k t2Y
T þ1 X
t2Y
P rk
Gðt; sÞ
T X
s¼1
T þ1 X s¼1
Gðs; sÞ
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2 T X i¼2
G1 ðs; iÞaðiÞf ðuðiÞÞ P rkTuk:
ð17Þ
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
143
Consequently, T : P ! P . It is also easy to check that T : P ! P is completely continuous. Now, turning to f0 , there exits H1 > 0 such that f ðuÞ 6 ðf0 þ eÞu, for 0 < u 6 H1 . Thus, for u 2 P with kuk ¼ H1 , (11) and (16) implies that ðTuÞðtÞ 6 k
T þ1 X
Gðs; sÞ
s¼1
6k
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2
T þ1 X
Gðs; sÞ
s¼1
6k
T X
T X
G1 ðs; iÞaðiÞðf0 þ eÞuðiÞ
i¼2
T þ1 X
Gðs; sÞ
s¼1
T X
G1 ðs; iÞaðiÞðf0 þ eÞkuk 6 kuk;
t 2 ½0; T þ 2:
i¼2
Therefore, kTuk 6 kuk;
for u 2 P \ oX1 ;
ð18Þ
where X1 ¼ fu 2 B : kuk < H1 g: If we next consider f1 , there exists an H 2 > 0 such that f ðuÞ P ðf1 eÞu, for all u P H 2 . Let H2 ¼ maxf2H1 ; r1 H 2 g, and define X2 ¼ fu 2 B : kuk < H2 g: If u 2 P with kuk ¼ H2 , then mint2Y uðtÞ P rkuk P H 2 , and ðTuÞðsÞ ¼ k
T þ1 X
Gðs; sÞ
s¼1
Pk
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2
T þ1 X
Gðs; sÞ
X
G1 ðs; iÞaðiÞðf1 eÞuðiÞ
i2Y
s¼1
P kr
T X
T þ1 X s¼1
Gðs; sÞ
X
G1 ðs; iÞaðiÞðf1 eÞkuk P kuk:
i2Y
Hence, kTuk P kuk;
for u 2 P \ oX2 :
ð19Þ
From (18) and (19), Theorem 1.1 implies that T has a fixed point u 2 P \ ðX2 n X1 Þ. This fixed point u is a solution of (1) and (2) corresponding to the given value of k. The proof of Theorem 2.1 is complete. h Theorem 2.2. Assume that conditions (A), (B) and (C) are satisfied. Then, for each k satisfying
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Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
1 P P T þ1 r s¼1 Gðs; sÞ i2Y G1 ðs; iÞaðiÞ f0 < k < P
T þ1 s¼1
Gðs; sÞ
1 PT
i¼2
; G1 ðs; iÞaðiÞ f1
ð20Þ
there exists at least one solution of (1) and (2) in P . Proof. Let k be given as in (20). Then, we can choose e > 0 such that 1 P P T þ1 r s¼1 Gðs; sÞ i2Y G1 ðs; iÞaðiÞ ðf0 eÞ 6k6 P T þ1 s¼1
Gðs; sÞ
1
: G ðs; iÞaðiÞ ðf þ eÞ 1 1 i¼2
PT
Let T be the cone preserving, completely continuous operator that was defined by (16). Beginning with f0 , there exits H1 > 0 such that f ðuÞ P ðf0 eÞu, for 0 < u 6 H1 . So, for u 2 P with kuk ¼ H1 , we have ðTuÞðsÞ ¼ k
T þ1 X
Gðs; sÞ
s¼1
Pk
T þ1 X
Gðs; sÞ
T X
G1 ðs; iÞaðiÞðf0 eÞuðiÞ
i¼2
T þ1 X
Gðs; sÞ
X
G1 ðs; iÞaðiÞðf0 eÞuðiÞ
i2Y
s¼1
P kr
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2
s¼1
Pk
T X
T þ1 X s¼1
Gðs; sÞ
X
G1 ðs; iÞaðiÞðf0 eÞkuk P kuk:
i2Y
Therefore, kTuk P kuk;
for u 2 P \ oX1 ;
ð21Þ
where X1 ¼ fu 2 B : kuk < H1 g: Using the assumption concerning f1 , there exists an H 2 > 0 such that f ðuÞ 6 ðf1 þ eÞu, for all u P H 2 . There are two cases, (a) f is bounded, and (b) f is unbounded.
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
145
Case (a): Assume M > 0 be such that f ðuÞ 6 M, for all 0 < u < 1. Let ( ) T þ1 T X X H2 ¼ max 2H1 ; Mk Gðs; sÞ G1 ðs; iÞaðiÞ : s¼1
i¼2
Then, for u 2 P with kuk ¼ H2 , we have ðTuÞðtÞ ¼ k
T þ1 X
T X
Gðt; sÞ
s¼1
6 kM
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2
T þ1 X
Gðs; sÞ
s¼1
T X
G1 ðs; iÞaðiÞ 6 kuk;
t 2 ½0; T þ 2:
i¼2
Hence kTuk 6 kuk;
for u 2 P \ oX2 ;
ð22Þ
where X2 ¼ fu 2 B : kuk < H2 g: Case (b): It can be shown without much difficulty that there is a H2 > maxf2H1 ; H 2 g such that f ðuÞ 6 f ðH2 Þ, for 0 < u 6 H2 . Choosing u 2 P with kuk ¼ H2 , ðTuÞðtÞ 6 k
T þ1 X
Gðs; sÞ
T X
s¼1
6k
T þ1 X
Gðs; sÞ
T X
s¼1
6k
G1 ðs; iÞaðiÞf ðH2 Þ
i¼2
T þ1 X
Gðs; sÞ
T X
s¼1
¼k
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2
G1 ðs; iÞaðiÞðf1 þ eÞH2
i¼2
T þ1 X s¼1
Gðs; sÞ
T X
G1 ðs; iÞaðiÞðf1 þ eÞkuk 6 kuk;
t 2 ½0; T þ 2:
i¼2
Therefore kTuk 6 kuk;
for u 2 P \ oX2 ;
ð23Þ
where X2 ¼ fu 2 B : kuk < H2 g: We apply Theorem 1.1 to conclude that T has a fixed point u 2 P \ ðX2 n X1 Þ: Thus in either of the case, Theorem 1.1 (ii) applied to (21) and (22) or (23) yields a fixed point of T which belongs to P \ ðX2 n X1 Þ. This fixed point u is a solution of (1) and (2) corresponding to the given k. The proof of Theorem 2.2 is complete. h
146
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
3. Solutions of (1) and (3) in a cone Lemma 3.1. uðtÞ is a solution of the BVP (1) and (3) if and only if uðtÞ ¼ k
T þ1 X
Kðt; sÞ
T X
s¼1
K1 ðs; iÞaðiÞf ðuðiÞÞ;
t 2 ½0; T þ 2;
ð24Þ
i¼2
where Kðt; sÞ ¼
t; s;
0 6 t 6 s 6 T þ 1; 1 6 s 6 t 6 T þ 2;
ð25Þ
and K1 ðt; iÞ ¼
t 1; i 1;
16t6i6T; 2 6 i 6 t 6 T þ 1:
ð26Þ
Proof. BVP (1) and (3) is equivalent to solving D2 uðt 1Þ ¼ k
T X
K1 ðt; iÞaðiÞf ðuðiÞÞ;
t 2 ½1; T þ 1;
ð27Þ
i¼2
uð0Þ ¼ DuðT þ 1Þ ¼ 0;
ð28Þ
and it is also equivalent to solving uðtÞ ¼ k
T þ1 X
Kðt; sÞ
T X
s¼1
K1 ðs; iÞaðiÞf ðuðiÞÞ;
t 2 ½0; T þ 2:
ð29Þ
i¼2
The proof of the lemma is complete.
h
From (25) Kðt; sÞ > 0;
for ðt; sÞ 2 ½1; T þ 2 ½1; T þ 1;
ð30Þ
and Kðt; sÞ 6 Kðs; sÞ ¼ s Kðt; sÞ P
for 1 6 s 6 T þ 1; 0 6 t 6 T þ 2;
1 1 Kðs; sÞ ¼ s 4 4
for t 2 Y ; 1 6 s 6 T þ 1:
ð31Þ ð32Þ
Let B1 ¼ fu : ½0; T þ 2 ! Rjuð0Þ ¼ DuðT þ 1Þ ¼ D2 uð0Þ ¼ D3 uðT 1Þ ¼ 0g; with norm, kuk ¼ maxt2½0;T þ2 juðtÞj, then ðB1 ; k kÞ is a Banach space.
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
147
Define a cone, P1 , by 1 P1 ¼ u 2 B1 : uðtÞ P 0 for t 2 ½0; T þ 2; and mint2Y uðtÞ P kuk : 4 ð33Þ Also, we define the number c 2 ½0; T þ 2 by T þ1 X
Kðc; sÞ
X
K1 ðs; iÞaðiÞ ¼ max
t2½0;T þ2
i2Y
s¼1
T þ1 X s¼1
Kðt; sÞ
X
K1 ðs; iÞaðiÞ:
ð34Þ
i2Y
Similar to the existence results of the previous section, we have the following two theorems for positive solutions of (1) and (3). Theorem 3.1. Assume that conditions (A), (B) and (C) are satisfied. Then, for each k satisfying P
T þ1 s¼1
Kðc; sÞ
< k < P
T þ1 s¼1
4 P
i2Y
s
PT
K1 ðs; iÞaðiÞ f1 1
i¼2
; K1 ðs; iÞaðiÞ f0
ð35Þ
there exists at least one solution of (1) and (3) in P1 . Theorem 3.2. Assume that conditions (A), (B) and (C) are satisfied. Then, for each k satisfying P
T þ1 s¼1
Kðc; sÞ
< k < P
T þ1 s¼1
4 P
i2Y
s
PT
K1 ðs; iÞaðiÞ f0
i¼2
1
; K1 ðs; iÞaðiÞ f1
ð36Þ
there exists at least one solution of (1) and (3) in P1 . References [1] R.P. Agarwal, J. Henderson, Positive solutions and nonlinear eigenvalue problems for thirdorder difference equations, Comp. Math. Appl. 36 (11–12) (1998) 347–355. [2] J. Henderson, Positive solutions for nonlinear difference equations, Nonlinear Stud. 4 (1) (1997) 29–36. [3] F. Merdivenci, Two positive solutions of a boundary value problem for difference equations, J. Diff. Eqns. Appl. 1 (1995) 263–270. [4] R.P. Agarwal, D. OÕRegan, A fixed-point approach for nonlinear discrete boundary value problems, Comp. Math. Appl. 36 (10–12) (1998) 115–121.
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Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
[5] J. Henderson, H.Y. Wang, Positive solutions for nonlinear eigenvalue problems, J. Math. Anal. Appl. 208 (1997) 252–259. [6] B.G. Zhang, L.J. Kong, Y.J. Sun, X.H. Deng, Existence of positive solutions for BVPs of fourth-order difference equations, Appl. Math. Comput. 131 (2002) 583–591. [7] R.Y. Ma, H.Y. Wang, On the existence of positive solutions of fourth-order ordinary differential equations, Appl. Anal. 59 (1995) 225–231. [8] J.R. Graef, B. Yang, On a nonlinear boundary value problem for fourth order equations, Appl. Anal. 72 (1999) 439–448. [9] M.A. KrasnoselÕskii, Positive Solutions of Operator Equations, Noordhooff, Groningen, 1964. [10] K. Deimling, Nonlinear Functional Analysis, Springer-Verlag, New York, 1985.
On the existence of positive solutions of fourth-order difference equations q Zhimin He
a,*
, Jianshe Yu
b
a
b
Department of Applied Mathematics, Central South University, Changsha 410083, Hunan, PR China Department of Applied Mathematics, Hunan University, Changsha 410082, Hunan, PR China
Abstract In this paper, by means of KrasnoselÕskiiÕs fixed point theorem, the existence of positive solutions of fourth-order difference equations is considered. Ó 2003 Elsevier Inc. All rights reserved. Keywords: Positive solution; Difference equation; KrasnoselÕskiiÕs fixed point theorem
1. Introduction For notation, given a < b in Z, we employ intervals to denote discrete sets such as ½a;b ¼ fa;aþ1;...;bg, ½a;bÞ ¼ fa;aþ1;...;b1g, ½a;1Þ ¼ fa;aþ1;...:g, etc. Let T P1 be fixed. In this paper, we are concerned with determining eigenvalues, k, for which there are positive solutions of the fourth-order difference equation D4 uðt 2Þ kaðtÞf ðuðtÞÞ ¼ 0;
t 2 ½2; T þ 2;
ð1Þ
satisfying the boundary conditions uð0Þ ¼ D2 uð0Þ ¼ uðT þ 2Þ ¼ D2 uðT Þ ¼ 0;
q
ð2Þ
This work was supported by Mathematical Tianyuan Foundation (No: A0324621) of PeopleÕs Republic of China. * Corresponding author. E-mail address: [email protected] (Z. He). 0096-3003/$ - see front matter Ó 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2003.12.016
140
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
or uð0Þ ¼ D2 uð0Þ ¼ DuðT þ 1Þ ¼ D3 uðT 1Þ ¼ 0;
ð3Þ
where (A) f : Rþ ! Rþ is continuous (Rþ denotes the nonnegative reals), (B) aðtÞ is a positive valued function defined on ½1; T þ 1, and (C) f0 ¼ limx!0þ f ðxÞ and f1 ¼ limx!1 f ðxÞ exist. x x The motivation for the present work stems from many recent investigations in [1–6]. For the continuous case, boundary value problems analogous to (1) and (2) (or (1) and (3)) arise in various nonlinear phenomena for which only positive solutions are meaningful; see, for example [7,8]. In this paper we will apply the following KrasnoselÕskiiÕs fixed point theorem in a cone to yields positive solutions of (1) and (2) (or (1) and (3)). This theorem can be found in the book by KrasnoselÕskii [9] as well as in the book by Deimling [10]. Theorem 1.1 [9,10]. Let B be a Banach space, and let P B be a cone in B. Assume X1 , X2 are open subsets of B with 0 2 X1 X1 X2 , and let T : P \ ðX2 n X1 Þ ! P be a completely continuous operator such that, either i(i) kTuk 6 kuk; u 2 P \ oX1 , and kTuk P kuk; u 2 P \ oX2 , or (ii) kTuk P kuk; u 2 P \ oX1 , and kTuk 6 kuk; u 2 P \ oX2 . Then T has a fixed point in P \ ðX2 n X1 Þ.
2. Solutions of (1) and (2) in a cone Lemma 2.1. uðtÞ is a solution of the BVP (1) and (2) if and only if uðtÞ ¼ k
T þ1 X
Gðt; sÞ
s¼1
T X
G1 ðs; iÞaðiÞf ðuðiÞÞ;
t 2 ½0; T þ 2;
ð4Þ
i¼2
where Gðt; sÞ ¼
1 T þ2
G1 ðt; iÞ ¼
1 T
sðT þ 2 tÞ; tðT þ 2 sÞ;
1 6 s 6 t 6 T þ 2; 0 6 t 6 s 6 T þ 1;
ð5Þ
and
ðT þ 1 tÞði 1Þ; ðT þ 1 iÞðt 1Þ;
2 6 i 6 t 6 T þ 1; 16t6i6T:
ð6Þ
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
141
Proof. BVP (1) and (2) is equivalent to solving D2 uðt 1Þ ¼ k
T X
G1 ðt; iÞaðiÞf ðuðiÞÞ;
t 2 ½1; T þ 1;
ð7Þ
i¼2
uð0Þ ¼ uðT þ 2Þ ¼ 0;
ð8Þ
and it is also equivalent to solving uðtÞ ¼ k
T þ1 X
Gðt; sÞ
s¼1
T X
G1 ðs; iÞaðiÞf ðuðiÞÞ;
t 2 ½0; T þ 2:
ð9Þ
i¼2
h
The proof of the lemma is complete. From (5)
for ðt; sÞ 2 ð1; T þ 1Þ ð1; T þ 1Þ;
Gðt; sÞ > 0;
ð10Þ
and Gðt; sÞ 6 Gðs; sÞ ¼
sðT þ 2 sÞ T þ2
for 1 6 s 6 T þ 1; 0 6 t 6 T þ 2:
ð11Þ
Merdivinici [3] has shown that, if Y is defined by T þ1 3ðT þ 1Þ Y ¼ t2Z: 6t6 ; 4 4 and if r ¼ min
min Y T þ 2 max Y ; T þ1 T þ1
;
then Gðt; sÞ P rGðs; sÞ ¼
rsðT þ 2 sÞ T þ2
for t 2 Y ; 1 6 s 6 T þ 1:
ð12Þ
Let B ¼ fu : ½0; T þ 2 ! Rjuð0Þ ¼ uðT þ 2Þ ¼ D2 uð0Þ ¼ D2 uðT Þ ¼ 0g; with norm, kuk ¼ maxt2½0;T þ2 juðtÞj, then ðB; k kÞ is a Banach space. Define a cone, P , by P ¼ fu 2 B : uðtÞ P 0 for t 2 ½0; T þ 2; and mint2Y uðtÞ P rkukg:
ð13Þ
Also, we define the number s 2 ½0; T þ 2 by T þ1 X s¼1
Gðs; sÞ
X i2Y
G1 ðs; iÞaðiÞ ¼ max
t2½0;T þ2
T þ1 X s¼1
Gðt; sÞ
X i2Y
G1 ðs; iÞaðiÞ:
ð14Þ
142
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
Theorem 2.1. Assume that conditions (A), (B) and (C) are satisfied. Then, for each k satisfying 1 P P T þ1 r s¼1 Gðs; sÞ i2Y G1 ðs; iÞaðiÞ f1 < k < P
T þ1 s¼1
Gðs; sÞ
1 PT
i¼2
; G1 ðs; iÞaðiÞ f0
ð15Þ
there exists at least one solution of (1) and (2) in P . Proof. Let k be given as in (15). Then, let e > 0 be such that 1 P P þ1 Gðs; sÞ i2Y G1 ðs; iÞaðiÞ ðf1 eÞ r Ts¼1 6k6 P T þ1 s¼1
Gðs; sÞ
PT
1
i¼2
: G1 ðs; iÞaðiÞ ðf0 þ eÞ
Define a summation operator T : P ! B by ðTuÞðtÞ ¼ k
T þ1 X
Gðt; sÞ
s¼1
T X
G1 ðs; iÞaðiÞf ðuðiÞÞ;
u 2 P:
ð16Þ
i¼2
We note that from (5), (6) and (16), if u 2 P , then ðTuÞðtÞ P 0 for t 2 ½0; T þ 2. From the properties of G, Tu satisfies the boundary conditions (2). Moreover, for u 2 P , we have ðTuÞðtÞ ¼ k
T þ1 X
Gðt; sÞ
s¼1
6k
T þ1 X
T X
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2
Gðs; sÞ
s¼1
T X
G1 ðs; iÞaðiÞf ðuðiÞÞ;
t 2 ½0; T þ 2;
i¼2
and so kTuk ¼ max jðTuÞðtÞj 6 k t2½0;T þ2
T þ1 X
Gðs; sÞ
s¼1
T X
G1 ðs; iÞaðiÞf ðuðiÞÞ:
i¼2
Thus, if u 2 P , then minðTuÞðtÞ ¼ min k t2Y
T þ1 X
t2Y
P rk
Gðt; sÞ
T X
s¼1
T þ1 X s¼1
Gðs; sÞ
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2 T X i¼2
G1 ðs; iÞaðiÞf ðuðiÞÞ P rkTuk:
ð17Þ
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
143
Consequently, T : P ! P . It is also easy to check that T : P ! P is completely continuous. Now, turning to f0 , there exits H1 > 0 such that f ðuÞ 6 ðf0 þ eÞu, for 0 < u 6 H1 . Thus, for u 2 P with kuk ¼ H1 , (11) and (16) implies that ðTuÞðtÞ 6 k
T þ1 X
Gðs; sÞ
s¼1
6k
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2
T þ1 X
Gðs; sÞ
s¼1
6k
T X
T X
G1 ðs; iÞaðiÞðf0 þ eÞuðiÞ
i¼2
T þ1 X
Gðs; sÞ
s¼1
T X
G1 ðs; iÞaðiÞðf0 þ eÞkuk 6 kuk;
t 2 ½0; T þ 2:
i¼2
Therefore, kTuk 6 kuk;
for u 2 P \ oX1 ;
ð18Þ
where X1 ¼ fu 2 B : kuk < H1 g: If we next consider f1 , there exists an H 2 > 0 such that f ðuÞ P ðf1 eÞu, for all u P H 2 . Let H2 ¼ maxf2H1 ; r1 H 2 g, and define X2 ¼ fu 2 B : kuk < H2 g: If u 2 P with kuk ¼ H2 , then mint2Y uðtÞ P rkuk P H 2 , and ðTuÞðsÞ ¼ k
T þ1 X
Gðs; sÞ
s¼1
Pk
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2
T þ1 X
Gðs; sÞ
X
G1 ðs; iÞaðiÞðf1 eÞuðiÞ
i2Y
s¼1
P kr
T X
T þ1 X s¼1
Gðs; sÞ
X
G1 ðs; iÞaðiÞðf1 eÞkuk P kuk:
i2Y
Hence, kTuk P kuk;
for u 2 P \ oX2 :
ð19Þ
From (18) and (19), Theorem 1.1 implies that T has a fixed point u 2 P \ ðX2 n X1 Þ. This fixed point u is a solution of (1) and (2) corresponding to the given value of k. The proof of Theorem 2.1 is complete. h Theorem 2.2. Assume that conditions (A), (B) and (C) are satisfied. Then, for each k satisfying
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Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
1 P P T þ1 r s¼1 Gðs; sÞ i2Y G1 ðs; iÞaðiÞ f0 < k < P
T þ1 s¼1
Gðs; sÞ
1 PT
i¼2
; G1 ðs; iÞaðiÞ f1
ð20Þ
there exists at least one solution of (1) and (2) in P . Proof. Let k be given as in (20). Then, we can choose e > 0 such that 1 P P T þ1 r s¼1 Gðs; sÞ i2Y G1 ðs; iÞaðiÞ ðf0 eÞ 6k6 P T þ1 s¼1
Gðs; sÞ
1
: G ðs; iÞaðiÞ ðf þ eÞ 1 1 i¼2
PT
Let T be the cone preserving, completely continuous operator that was defined by (16). Beginning with f0 , there exits H1 > 0 such that f ðuÞ P ðf0 eÞu, for 0 < u 6 H1 . So, for u 2 P with kuk ¼ H1 , we have ðTuÞðsÞ ¼ k
T þ1 X
Gðs; sÞ
s¼1
Pk
T þ1 X
Gðs; sÞ
T X
G1 ðs; iÞaðiÞðf0 eÞuðiÞ
i¼2
T þ1 X
Gðs; sÞ
X
G1 ðs; iÞaðiÞðf0 eÞuðiÞ
i2Y
s¼1
P kr
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2
s¼1
Pk
T X
T þ1 X s¼1
Gðs; sÞ
X
G1 ðs; iÞaðiÞðf0 eÞkuk P kuk:
i2Y
Therefore, kTuk P kuk;
for u 2 P \ oX1 ;
ð21Þ
where X1 ¼ fu 2 B : kuk < H1 g: Using the assumption concerning f1 , there exists an H 2 > 0 such that f ðuÞ 6 ðf1 þ eÞu, for all u P H 2 . There are two cases, (a) f is bounded, and (b) f is unbounded.
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
145
Case (a): Assume M > 0 be such that f ðuÞ 6 M, for all 0 < u < 1. Let ( ) T þ1 T X X H2 ¼ max 2H1 ; Mk Gðs; sÞ G1 ðs; iÞaðiÞ : s¼1
i¼2
Then, for u 2 P with kuk ¼ H2 , we have ðTuÞðtÞ ¼ k
T þ1 X
T X
Gðt; sÞ
s¼1
6 kM
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2
T þ1 X
Gðs; sÞ
s¼1
T X
G1 ðs; iÞaðiÞ 6 kuk;
t 2 ½0; T þ 2:
i¼2
Hence kTuk 6 kuk;
for u 2 P \ oX2 ;
ð22Þ
where X2 ¼ fu 2 B : kuk < H2 g: Case (b): It can be shown without much difficulty that there is a H2 > maxf2H1 ; H 2 g such that f ðuÞ 6 f ðH2 Þ, for 0 < u 6 H2 . Choosing u 2 P with kuk ¼ H2 , ðTuÞðtÞ 6 k
T þ1 X
Gðs; sÞ
T X
s¼1
6k
T þ1 X
Gðs; sÞ
T X
s¼1
6k
G1 ðs; iÞaðiÞf ðH2 Þ
i¼2
T þ1 X
Gðs; sÞ
T X
s¼1
¼k
G1 ðs; iÞaðiÞf ðuðiÞÞ
i¼2
G1 ðs; iÞaðiÞðf1 þ eÞH2
i¼2
T þ1 X s¼1
Gðs; sÞ
T X
G1 ðs; iÞaðiÞðf1 þ eÞkuk 6 kuk;
t 2 ½0; T þ 2:
i¼2
Therefore kTuk 6 kuk;
for u 2 P \ oX2 ;
ð23Þ
where X2 ¼ fu 2 B : kuk < H2 g: We apply Theorem 1.1 to conclude that T has a fixed point u 2 P \ ðX2 n X1 Þ: Thus in either of the case, Theorem 1.1 (ii) applied to (21) and (22) or (23) yields a fixed point of T which belongs to P \ ðX2 n X1 Þ. This fixed point u is a solution of (1) and (2) corresponding to the given k. The proof of Theorem 2.2 is complete. h
146
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
3. Solutions of (1) and (3) in a cone Lemma 3.1. uðtÞ is a solution of the BVP (1) and (3) if and only if uðtÞ ¼ k
T þ1 X
Kðt; sÞ
T X
s¼1
K1 ðs; iÞaðiÞf ðuðiÞÞ;
t 2 ½0; T þ 2;
ð24Þ
i¼2
where Kðt; sÞ ¼
t; s;
0 6 t 6 s 6 T þ 1; 1 6 s 6 t 6 T þ 2;
ð25Þ
and K1 ðt; iÞ ¼
t 1; i 1;
16t6i6T; 2 6 i 6 t 6 T þ 1:
ð26Þ
Proof. BVP (1) and (3) is equivalent to solving D2 uðt 1Þ ¼ k
T X
K1 ðt; iÞaðiÞf ðuðiÞÞ;
t 2 ½1; T þ 1;
ð27Þ
i¼2
uð0Þ ¼ DuðT þ 1Þ ¼ 0;
ð28Þ
and it is also equivalent to solving uðtÞ ¼ k
T þ1 X
Kðt; sÞ
T X
s¼1
K1 ðs; iÞaðiÞf ðuðiÞÞ;
t 2 ½0; T þ 2:
ð29Þ
i¼2
The proof of the lemma is complete.
h
From (25) Kðt; sÞ > 0;
for ðt; sÞ 2 ½1; T þ 2 ½1; T þ 1;
ð30Þ
and Kðt; sÞ 6 Kðs; sÞ ¼ s Kðt; sÞ P
for 1 6 s 6 T þ 1; 0 6 t 6 T þ 2;
1 1 Kðs; sÞ ¼ s 4 4
for t 2 Y ; 1 6 s 6 T þ 1:
ð31Þ ð32Þ
Let B1 ¼ fu : ½0; T þ 2 ! Rjuð0Þ ¼ DuðT þ 1Þ ¼ D2 uð0Þ ¼ D3 uðT 1Þ ¼ 0g; with norm, kuk ¼ maxt2½0;T þ2 juðtÞj, then ðB1 ; k kÞ is a Banach space.
Z. He, J. Yu / Appl. Math. Comput. 161 (2005) 139–148
147
Define a cone, P1 , by 1 P1 ¼ u 2 B1 : uðtÞ P 0 for t 2 ½0; T þ 2; and mint2Y uðtÞ P kuk : 4 ð33Þ Also, we define the number c 2 ½0; T þ 2 by T þ1 X
Kðc; sÞ
X
K1 ðs; iÞaðiÞ ¼ max
t2½0;T þ2
i2Y
s¼1
T þ1 X s¼1
Kðt; sÞ
X
K1 ðs; iÞaðiÞ:
ð34Þ
i2Y
Similar to the existence results of the previous section, we have the following two theorems for positive solutions of (1) and (3). Theorem 3.1. Assume that conditions (A), (B) and (C) are satisfied. Then, for each k satisfying P
T þ1 s¼1
Kðc; sÞ
< k < P
T þ1 s¼1
4 P
i2Y
s
PT
K1 ðs; iÞaðiÞ f1 1
i¼2
; K1 ðs; iÞaðiÞ f0
ð35Þ
there exists at least one solution of (1) and (3) in P1 . Theorem 3.2. Assume that conditions (A), (B) and (C) are satisfied. Then, for each k satisfying P
T þ1 s¼1
Kðc; sÞ
< k < P
T þ1 s¼1
4 P
i2Y
s
PT
K1 ðs; iÞaðiÞ f0
i¼2
1
; K1 ðs; iÞaðiÞ f1
ð36Þ
there exists at least one solution of (1) and (3) in P1 . References [1] R.P. Agarwal, J. Henderson, Positive solutions and nonlinear eigenvalue problems for thirdorder difference equations, Comp. Math. Appl. 36 (11–12) (1998) 347–355. [2] J. Henderson, Positive solutions for nonlinear difference equations, Nonlinear Stud. 4 (1) (1997) 29–36. [3] F. Merdivenci, Two positive solutions of a boundary value problem for difference equations, J. Diff. Eqns. Appl. 1 (1995) 263–270. [4] R.P. Agarwal, D. OÕRegan, A fixed-point approach for nonlinear discrete boundary value problems, Comp. Math. Appl. 36 (10–12) (1998) 115–121.
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[5] J. Henderson, H.Y. Wang, Positive solutions for nonlinear eigenvalue problems, J. Math. Anal. Appl. 208 (1997) 252–259. [6] B.G. Zhang, L.J. Kong, Y.J. Sun, X.H. Deng, Existence of positive solutions for BVPs of fourth-order difference equations, Appl. Math. Comput. 131 (2002) 583–591. [7] R.Y. Ma, H.Y. Wang, On the existence of positive solutions of fourth-order ordinary differential equations, Appl. Anal. 59 (1995) 225–231. [8] J.R. Graef, B. Yang, On a nonlinear boundary value problem for fourth order equations, Appl. Anal. 72 (1999) 439–448. [9] M.A. KrasnoselÕskii, Positive Solutions of Operator Equations, Noordhooff, Groningen, 1964. [10] K. Deimling, Nonlinear Functional Analysis, Springer-Verlag, New York, 1985.