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Hindawi Publishing Corporation Boundary Value Problems Volume 2008, Article ID 123823, 11 pages doi:10.1155/2008/123823 Research Article Existence and Uniqueness of Solutions for Singular Higher Order Continuous and Discrete Boundary Value Problems Chengjun Yuan, 1, 2 Daqing Jiang, 1 and You Zhang 1 1 School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, Jilin, China 2 School of Mathematics and Computer, Harbin University, Harbin 150086, Heilongjiang, China Correspondence should be addressed to Chengjun Yuan, ycj7102@163.com Received 4 July 2007; Accepted 31 December 2007 Recommended by Raul Manasevich By mixed monotone method, the existence and uniqueness are established for singular higher-order continuous and discrete boundary value problems. The theorems obtained are very general and complement previous known results. Copyright q 2008 Chengjun Yuan et al. This is an open access article distributed under the Creative Commons Attribution License, which p ermits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction In recent years, the study of higher-order continuous and discrete boundary value problems has been studied extensively in the literature see 1–17 and their references. Most of the results told us that the equations had at least single and multiple positive solutions. Recently, some authors have dealt with the uniqueness of solutions for singular higher- order continuous boundary value problems by using mixed monotone method, for example, see 6, 14, 15. However, there are few works on the uniqueness of solutions for singular dis- crete boundary value problems. In this paper, we state a unique fixed point theorem for a class of mixed monotone op- erators, see 6, 14, 18. In virtue of the theorem, we consider the existence and uniqueness of solutions for the following singular higher-order continuous and discrete boundary value problems 1.1 and 1.2 by using mixed monotone method. We first discuss the existence and uniqueness of solutions for the following singular higher-order continuous boundary value problem y n tλqtgyhy0, 0<t<1,λ>0, y i 0y n−2 10, 0 ≤ i ≤ n − 2, 1.1 2 Boundary Value Problems where n ≥ 2, qt ∈ C0, 1, 0, ∞, g : 0, ∞ → 0, ∞ is continuous and nondecreasing; h : 0, ∞ → 0, ∞ is continuous and nonincreasing, and h may be singular at y 0. Next, we consider the existence and uniqueness of solutions for the following singular higher-order discrete boundary value problem Δ n yiλqin−1gyin−1 hyin−1 0,i∈ N {0, 1, 2, ,T − 1},λ>0, Δ k y0Δ n−2 yT 10, 0 ≤ k ≤ n − 2, 1.2 where n ≥ 2, N {0, 1, 2, ,Tn}, qi ∈ CN , 0, ∞, g : 0, ∞ → 0, ∞ is continuous and nondecreasing; h : 0, ∞ → 0, ∞ is continuous and nonincreasing, and h may be singular at y 0. Throughout this paper, the topology on N will be the discrete topology. 2. Preliminaries Let P be a normal c one of a Banach space E,ande ∈ P with e≤1,e / θ. Define Q e {x ∈ P | x / θ, there exist constants m, M > 0 such that me ≤ x ≤ Me}. 2.1 Now we give a definition see 18. Definition 2.1 see 18. Assume A : Q e × Q e → Q e . A is said to be mixed monotone if Ax, y is nondecreasing in x and nonincreasing in y,thatis,ifx 1 ≤ x 2 x 1 ,x 2 ∈ Q e implies Ax 1 ,y ≤ Ax 2 ,y for any y ∈ Q e ,andy 1 ≤ y 2 y 1 ,y 2 ∈ Q e implies Ax, y 1 ≥ Ax, y 2 for any x ∈ Q e . x ∗ ∈ Q e is said to be a fixed point of A if Ax ∗ ,x ∗ x ∗ . Theorem 2.2 see 6, 14. Suppose that A: Q e × Q e → Q e is a mixed monotone operator and ∃ a constant α, 0 ≤ α<1, such that A tx, 1 t y ≥ t α Ax, y,forx,y∈ Q e , 0 <t<1. 2.2 Then A has a unique fixed point x ∗ ∈ Q e . Moreover, for any x 0 ,y 0 ∈ Q e × Q e , x n A x n−1 ,y n−1 ,y n A y n−1 ,x n−1 ,n 1, 2, , 2.3 satisfy x n −→ x ∗ ,y n −→ x ∗ , 2.4 where x n − x ∗ o 1 − r α n , y n − x ∗ o 1 − r α n , 2.5 0 <r<1, r is a constant from x 0 ,y 0 . Theorem 2.3 see 6, 14, 18. Suppose that A: Q e × Q e → Q e is a mixed monotone operator and ∃ a constant α ∈ 0, 1 such that 2.2 holds. If x ∗ λ is a unique solution of equation Ax, xλx, λ>02.6 in Q e ,thenx ∗ λ − x ∗ λ 0 →0,λ→ λ 0 . If 0 <α<1/2, then 0 <λ 1 <λ 2 implies x ∗ λ 1 ≥ x ∗ λ 2 , x ∗ λ 1 / x ∗ λ 2 ,and lim λ→∞ x ∗ λ 0, lim λ→0 x ∗ λ ∞. 2.7 Chengjun Yuan et al. 3 3. Uniqueness positive solution of differential equations 1.1 This section discusses singular higher-order boundary value problem 1.1. Throughout this section, we let Gt, s be the Green’s function to −y 0,y0y10, we note that Gt, s ⎧ ⎨ ⎩ t1 − s, 0 ≤ t ≤ s ≤ 1, s1 − t, 0 ≤ s ≤ t ≤ 1, 3.1 and one can show that Gt, tGs, s ≤ Gt, s ≤ Gt, t, for Gt, s ≤ Gs, s, t, s ∈ 0, 1 × 0, 1. 3.2 Suppose that y is a positive solution of 1.1.Let xty n−2 t, 3.3 from y i 0y n−2 10, 0 ≤ i ≤ n−2, and Taylor Formula, we define operator T : C 2 0, 1 → C n 0, 1,by ytTxt t 0 t − s n−3 n − 3! xsds, for 3 ≤ n, ytTxtxt, for n 2 3.4 Then we have x 2 tλft, Txt 0, 0 <t<1,λ>0, x0x10. 3.5 Then from 3.4, we have the next lemma. Lemma 3.1. If xt is a solution of 3.5,thenyt is a solution of 1.1. Further, if yt is a solution of 1.1,implythatxt is a solution of 3.5. Let P {x ∈ C0, 1 | xt ≥ 0, for all t ∈ 0, 1 }. Obviously, P is a normal cone of Banach space C0, 1. Theorem 3.2. Suppose that there exists α ∈ 0, 1 such that gtx ≥ t α gx, 3.6 h t −1 x ≥ t α hx, 3.7 for any t ∈ 0, 1 and x>0,andq ∈ C0, 1, 0, ∞ satisfies 1 0 s n−1 n − 2s −α qsds < ∞. 3.8 Then 1.1 has a unique positive solution y ∗ λ t. And moreover, 0 <λ 1 <λ 2 implies y ∗ λ 1 ≤ y ∗ λ 2 ,y ∗ λ 1 / y ∗ λ 2 . If α ∈ 0, 1/2,then lim λ→0 y ∗ λ 0, lim λ→∞ y ∗ λ ∞. 3.9 4 Boundary Value Problems Proof. Since 3.7 holds, let t −1 x y, one has hy ≥ t α hty. 3.10 Then hty ≤ 1 t α hy, for t ∈ 0, 1,y>0. 3.11 Let y 1. The above inequality is ht ≤ 1 t α h1, for t ∈ 0, 1. 3.12 From 3.7, 3.11,and3.12, one has h t −1 x ≥ t α hx,h 1 t ≥ t α h1,htx ≤ 1 t α hx,ht ≤ 1 t α h1, for t ∈ 0, 1,x>0. 3.13 Similarly, from 3.6, one has gtx ≥ t α gx,gt ≥ t α g1, for t ∈ 0, 1,x>0. 3.14 Let t 1/x, x > 1, one has gx ≤ x α g1, for x ≥ 1. 3.15 Let etGt, tt1 − t, and we define Q e x ∈ C0, 1 | 1 M Gt, t ≤ xt ≤ MGt, t,t∈ 0, 1 , 3.16 where M>1 is chosen such that M>max λg1 1 0 qsds λh1 1 0 s n−1 n − 2s n! −α qsds 1/1−α , λg1 1 0 Gs, s s n−1 n − 2s n! α qsds λh1 1 0 Gs, sqsds −1/1−α . 3.17 First, from 3.4 and 3.16, for any x ∈ Q e ,wehavethefollowing. When 3 ≤ n, 1 M t n−1 n − 2t n! ≤ t 0 1 M Gs, s t − s n−3 n − 3! ds ≤ Txt ≤ t 0 MGs, s t − s n−3 n − 3! ds ≤ M t n−1 n − 2t n! ≤ M, for t ∈ 0, 1, 3.18 Chengjun Yuan et al. 5 when n 2, 1 M t n−1 n − 2t n! ≤ Txtxt ≤ M tn − 2t n! ≤ M, for t ∈ 0, 1, 3.19 then 1 M t n−1 n − 2t n! ≤ Txt ≤ M t n−1 n − 2t n! ≤ M, for t ∈ 0, 1. 3.20 For any x, y ∈ Q e , we define A λ x, ytλ 1 0 Gt, sqsgTxs hTysds, for t ∈ 0, 1. 3.21 First, we show that A λ : Q e × Q e → Q e . Let x, y ∈ Q e , from 3.14, 3.15,and3.20,wehave gTxt ≤ gM ≤ M α g1, for t ∈ 0, 1, 3.22 and from 3.13,wehave hTyt ≤ h 1 M t n−1 n − 2t n! ≤ t n−1 n − 2t n! −α h 1 M ≤ M α t n−1 n − 2t n! −α h1, for t ∈ 0, 1. 3.23 Then, from 3.2, 3.21, 3.22 and 3.23,wehave On the other hand, for any x, y ∈ Q e ,from3.13 and 3.14,wehave gTxt ≥ g 1 M t n−1 n − 2t n! ≥ t n−1 n − 2t n! α g 1 M ≥ t n−1 n − 2t n! α 1 M α g1, hTyt ≥ hMh 1 1/M ≥ 1 M α h1, for t ∈ 0, 1. 3.24 Thus, from 3.2, 3.21 and 3.24,wehave A λ x, yt ≥ λGt, t 1 0 Gs, sqsM −α s n−1 n − 2s n! α g1ds 1 0 Gs, sqsM −α h1ds ≥ 1 M Gt, t, for t ∈ 0, 1. 3.25 So, A λ is well defined and A λ Q e × Q e ⊂ Q e . 6 Boundary Value Problems Next, for any l ∈ 0, 1, one has A λ lx, l −1 y tλ 1 0 Gt, sqs glTxs h l −1 Tys ds ≥ λ 1 0 Gt, sqs l α gTxs l α hTys ds l α A λ x, yt, for t ∈ 0, 1. 3.26 So the conditions of Theorems 2.2 and 2.3 hold. Therefore, there exists a unique x ∗ λ ∈ Q e such that A λ x ∗ ,x ∗ x ∗ λ . It is easy to check that x ∗ λ is a unique positive solution of 3.5 for given λ>0. Moreover, Theorem 2.3 means that if 0 <λ 1 <λ 2 , then x ∗ λ 1 t ≤ x ∗ λ 2 t, x ∗ λ 1 t / x ∗ λ 2 t and if α ∈ 0, 1/2,then lim λ→0 x ∗ λ 0, lim λ→∞ x ∗ λ ∞. 3.27 Next, from Lemma 3.1 and 3.4,wegetthaty ∗ λ Tx ∗ λ is a unique positive solution of 1.1 for given λ>0. Moreover, if 0 <λ 1 <λ 2 , then y ∗ λ 1 t ≤ y ∗ λ 2 t, y ∗ λ 1 t / y ∗ λ 2 t and if α ∈ 0, 1/2,then lim λ→0 y ∗ λ 0, lim λ→∞ y ∗ λ ∞. 3.28 This completes the proof. Example 3.3. Consider the following singular boundary value problem: y n tλ μy a ty −b t 0,t∈ 0, 1, y i 0y n−2 10, 0 ≤ i ≤ n − 2, 3.29 where λ, a, b > 0, μ ≥ 0, max{a, b} < 1/n − 1. Applying Theorem 3.2,letα max{a, b} < 1/n − 1, qt1, gyμy a , hyy −b , then gty ≥ t α gy,h t −1 ≥ t α hy, 1 0 s n−1 n − 2s −α ds < ∞. 3.30 Thus all conditions in Theorem 3.2 are satisfied. We can find 3.29 has a unique positive so- lution y ∗ λ t. In addition, 0 <λ 1 <λ 2 implies y ∗ λ 1 ≤ y ∗ λ 2 ,y ∗ λ 1 / y ∗ λ 2 .Ifα max{a, b}∈0, 1/2, then lim λ→0 y ∗ λ 0, lim λ→∞ y ∗ λ ∞. 3.31 Chengjun Yuan et al. 7 4. Uniqueness positive solution of difference equations 1.2 This section discusses singular higher-order boundary value problem 1.2. Throughout this section, we let Ki, j be Green’s function to −Δ 2 yiui 10, i ∈ N, y0yT 10, we note that Ki, j ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ jT 1 − i T 1 , 0 ≤ j ≤ i − 1, iT 1 − j T 1 ,i≤ j ≤ T 1, 4.1 and one can show that Ki, i ≥ Ki, j,Kj, j ≥ Ki, j,Ki, j ≥ Ki, i T 1 , for 0 ≤ i ≤ T 1, 1 ≤ j ≤ T. 4.2 Suppose that y is a positive solution of 1.2.Let xiΔ n−2 yi, for 0 ≤ i ≤ T 1. 4.3 From Δ i y0Δ n−2 yT 10, 0 ≤ i ≤ n − 2, and Δ m yi − 1Δ m−1 yi − Δ m−1 yi − 1,sowe define operator T,by Txiyi n − 1 i1 l1 C n−2 i−ln−1 xl, for 0 ≤ i ≤ T. 4.4 Then Δ 2 xiλFi n − 1,Txi 0, 0 ≤ i ≤ T − 1,λ>0, x0xT 10. 4.5 Lemma 4.1. If xi is a solution of 4.5,thenyi is a solutionn of 1.2. Proof. Since we remark that xi is a solution of 4.5,ifandonlyif xi T j1 Ki, jλFj n − 1,Txj, for 0 ≤ i ≤ T 1. 4.6 Let Txiyi n − 1, for 0 ≤ i ≤ T. 4.7 From 4.4 we find Δ i y0Δ n−2 yT 10, 0 ≤ i ≤ n − 2, and xiΔ n−2 yi,sothatyi is asolutionof1.2. Further, if yi is a solution o f 1.2,implythatxi is a solution of 4.5. Let P {x ∈ CN , 0, ∞ | xi ≥ 0, for all i ∈ N }. Obviously, P is a normal cone of Banach space CN , 0, ∞. 8 Boundary Value Problems Theorem 4.2. Suppose that there exists α ∈ 0, 1 such that gtx ≥ t α gx, h t −1 x ≥ t α gx, 4.8 for any t ∈ 0, 1 and x>0,andq ∈ CN , 0, ∞. Then 1.2 has a unique positive solution y ∗ λ i. And moreover, 0 <λ 1 <λ 2 implies y ∗ λ 1 ≤ y ∗ λ 2 , y ∗ λ 1 / y ∗ λ 2 .Ifα ∈ 0, 1/2,then lim λ→0 y ∗ λ 0, lim λ→∞ y ∗ λ ∞. 4.9 Proof. The proof is the same as that of Theorem 3.2,from4.12 and 4.13, one has h t −1 x ≥ t α hx,h 1 t ≥ t α h1,htx ≤ 1 t α hx,ht ≤ 1 t α h1, for t ∈ 0, 1,x>0; gtx ≥ t α gx,gt ≥ t α g1, for t ∈ 0, 1,x>0. 4.10 gtx ≥ t α gx,gt ≥ t α g1, for t ∈ 0, 1,x>0. 4.11 Let t 1/x, x > 1, one has gx ≤ x α g1, for x ≥ 1. 4.12 Let eiKi, i/T 1, and we define Q e x ∈ P | 1 M ei ≤ xi ≤ Mei, for 0 ≤ i ≤ T 1 , 4.13 where M>1 is chosen such that M>max λT 1g1 T j1 qj n − 1 j1 l1 C n−2 j−ln−1 α λT 1 1α h1 T j1 K −α j, jqj n − 1 1/1−α ; λg1 T j1 qj n − 1 Kj, j T 1 α λh1 T j1 qj n − 1 j1 l1 C n−2 j−ln−1 −α −1/1−α . 4.14 From 4.4 and 4.13, for any x ∈ Q e ,wehave 1 M ej ≤ Txj j1 l1 C n−2 j−ln−1 xl ≤ Mej j1 l1 C n−2 j−ln−1 , for 0 ≤ j ≤ T. 4.15 Chengjun Yuan et al. 9 For any x, y ∈ Q e , we define A λ x, yiλ T j1 Ki, jqj n − 1gTxj hTyj, for 0 ≤ i ≤ T 1. 4.16 First we show that A λ : Q e × Q e → Q e . Let x, y ∈ Q e , from 4.11 and 4.12,wehave gTxj ≤ g Mej j1 l1 C n−2 j−ln−1 ≤g M j1 l1 C n−2 j−ln−1 ≤M α j1 l1 C n−2 j−ln−1 α g1, for 1≤j ≤ T , 4.17 and from 4.10,wehave hTyj ≤ h 1 M ej ≤ e −α jh 1 M ≤ M α e −α jh1, for 1 ≤ j ≤ T. 4.18 Then, from 4.2 and the above, we have A λ x, yi ≤ λKi, i T j1 qj n − 1gTxj T 1hTyj ≤ eiM α λT 1 g1 T j1 qj n − 1 j1 l1 C n−2 j−ln−1 α h1 T j1 e −α jqj n − 1 ≤ eiM α λT 1 g1 T j1 qj n − 1 j1 l1 C n−2 j−ln−1 α h1 T j1 Kj, j T 1 −α qj n − 1 ≤ Mei, for 0 ≤ i ≤ T 1. 4.19 On the other hand, for any x, y ∈ Q e ,from4.10 and 4.12,wehave gxj ≥ g 1 M ej ≥ e α j 1 M α g1, for 1 ≤ j ≤ T, 4.20 hyj ≥ h Mej j1 l1 C n−2 j−ln−1 ≥ M −α j1 l1 C n−2 j−ln−1 −α h1, for 1 ≤ j ≤ T. 4.21 Thus, from 4.2 and 4.16,wehave A λ x, yi ≥ λei T j0 qj n − 1gTxj T j0 qj n − 1hTyj ≥ λeiM −α g1 T j0 qj Ki, i T 1 α h1 T j0 qj n − 1 j1 l1 C n−2 j−ln−1 −α ≥ 1 M ei, for 0 ≤ i ≤ T 1. 4.22 10 Boundary Value Problems So, A λ is well defined and A λ Q e × Q e ⊂ Q e . Next, for any l ∈ 0, 1, one has A λ lx, l −1 y iλ T j1 Ki, jqj n − 1 gTlxj h T l −1 yj λ T j1 Ki, jqj n − 1 glTxj h l −1 Tyj ≥ λ T j1 Ki, jqj n − 1 l α gTxj l α hTyj ds l α A λ x, yi, for 0 ≤ i ≤ T 1. 4.23 So the conditions of Theorems 2.2 and 2.3 hold. Therefore, there exists a unique x ∗ λ ∈ Q e such that A λ x ∗ ,x ∗ x ∗ λ . It is easy to check that x ∗ λ is a unique positive solution of 4.5 for given λ>0. Moreover, Theorem 2.3 means that if 0 <λ 1 <λ 2 , then x ∗ λ 1 t ≤ x ∗ λ 2 t, x ∗ λ 1 t / x ∗ λ 2 t and if α ∈ 0, 1/2,then lim λ→0 x ∗ λ 0, lim λ→∞ x ∗ λ ∞. 4.24 Next, on using Lemma 3.1,from4.5,wegetthaty ∗ λ Tx ∗ λ is a unique positive solution of 1.2 for given λ>0. Moreover, if 0 <λ 1 <λ 2 , then y ∗ λ 1 t ≤ y ∗ λ 2 t, y ∗ λ 1 t / y ∗ λ 2 t and if α ∈ 0, 1/2,then lim λ→0 y ∗ λ 0, lim λ→∞ y ∗ λ ∞. 4.25 This completes the proof. Example 4.3. 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Corporation Boundary Value Problems Volume 2008, Article ID 123823, 11 pages doi:10.1155/2008/123823 Research Article Existence and Uniqueness of Solutions for Singular Higher Order Continuous and Discrete Boundary. continuous and discrete boundary value problems 1.1 and 1.2 by using mixed monotone method. We first discuss the existence and uniqueness of solutions for the following singular higher- order continuous. theorem for a class of mixed monotone op- erators, see 6, 14, 18. In virtue of the theorem, we consider the existence and uniqueness of solutions for the following singular higher- order continuous
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