Applications Of Leggett Williams Type Fixed Point Theorems To A Second Order Difference Equation

Transcription

1 Eastern Kentucky University Encompass Online Theses and Dissertations Student Scholarship January 013 Applications Of Leggett Williams Type Fixed Point Theorems To A Second Order Difference Equation Charley Lockhart Seelbach Eastern Kentucky University Follow this and additional works at: Part of the Mathematics Commons Recommended Citation Seelbach, Charley Lockhart, "Applications Of Leggett Williams Type Fixed Point Theorems To A Second Order Difference Equation" (013). Online Theses and Dissertations This Open Access Thesis is brought to you for free and open access by the Student Scholarship at Encompass. It has been accepted for inclusion in Online Theses and Dissertations by an authorized administrator of Encompass. For more information, please contact Linda.Sizemore@eku.edu.

2 APPLICATIONS OF LEGGETT WILLIAMS TYPE FIXED POINT THEOREMS TO A SECOND ORDER DIFFERENCE EQUATION By Charley L. Seelbach Thesis Approved: Chair, Advisory Committee Member, Advisory Committee Member, Advisory Committee Dean, Graduate School

3 STATEMENT OF PERMISSION TO USE In presenting this thesis in partial fulfillment of the requirements for a M.S. degree at Eastern Kentucky University, I agree that the Library shall make it available to borrowers under rules of the Library. Brief quotations from this thesis are allowable without special permission, provided that accurate acknowledgment of the source is made. Permission for extensive quotation from or reproduction of this thesis may be granted by my major professor, or in his absence, by the Head of Interlibrary Services when, in the opinion of either, the proposed use of the material is for scholarly purposes. Any copying or use of the material in this thesis for financial gain shall not be allowed without my written permission. Signature Date

4 APPLICATIONS OF LEGGETT WILLIAMS TYPE FIXED POINT THEOREMS TO A SECOND ORDER DIFFERENCE EQUATION By Charley L. Seelbach Bachelor Of Science University of Kentucky Lexington, Kentucky 010 Submitted to the Faculty of the Graduate School of Eastern Kentucky University in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE May, 013

5 Copyright c Charley L. Seelbach, 013 All rights reserved ii

6 DEDICATION This thesis is dedicated to my parents for all their support. iii

7 ACKNOWLEDGEMENTS I would like to thank my thesis chair, Dr. Jeffrey Neugebauer, for his guidance in this research. I would also like to thank the other committee members, Dr. Steve Szabo, and Dr. Patrick Costello, for their time. I would like to express thanks to my parents for always encouraging me, and providing whatever help they could. I would like to thank my fellow graduate students Ranthony Clark, Gregory Chandler and Ryan Whaley for always being upbeat, and positive. I would finally like to thank by brother and sister, Louis and Elizabeth Seelbach. iv

8 ABSTRACT Using extensions of the LeggettWilliams fixed-point theorem, we prove the existence of solutions for a class of second-order difference equations with Dirichlet boundary conditions. We present these fixed point theorems and then show what conditions have to be met in order to satisfy the theorem. Finally, we provide specific examples to show the hypotheses of the theorems do not contradict one another. v

9 Contents Contents v 1 Introduction 1 Preliminaries 3.1 Definitions The Green s Function Application of the First Fixed Point Theorem The Fixed Point Theorem Preliminaries Positive Symmetric Solutions to (1.1),(1.) Example Application of the Second Fixed Point Theorem The Fixed Point Theorem Preliminaries Positive Symmetric Solutions to (1.1),(1.) Example Bibliography 5 vi

10 Chapter 1 Introduction For years, fixed point theory has found itself at the center of study of boundary value problems. Many fixed point theorems have provided criteria for the existence of positive solutions or multiple positive solutions of boundary value problems. Some of these results can be seen in the works of Guo [1], Krosnosel skii [14], Leggett and Williams [15], and Avery et al. [4, 8]. Applications of the aforementioned fixed point theorems have been seen in works dealing with ordinary differential equations [3, 7, 11] and dynamic equations on time scales [10, 16, 19]. Most relevant to this research, these theorems have been utilized for results that involve finite difference equations [6, 9, 13, 18]. In this paper, we give an application of two recent Avery et al. fixed point theorems to obtain at least one positive solution of the difference equation u(k) + f(u(k)), k {0, 1,..., N}, (1.1) with boundary conditions u(0) = u(n + ) = 0, (1.) Here f : R [0, ) is any continuous function and is the second-difference operator defined by ( u)(k) = u(k + ) u(k + 1) + u(k). 1

11 The original Leggett-Williams theorem states the following. Theorem 1.1 (Leggett-Williams [15]). Let K be a cone in a Banach space E and define the sets K ɛ1 := {x K : x ɛ 1 } and S(β, ɛ, ɛ 3 ) := {x K : ɛ β(x) and x ɛ 3 } for ɛ 1 > 0 and ɛ 3 > ɛ > 0 and any concave positive functional β defined on the cone K, with β(x) x. Suppose that c b > a > 0, α is a concave positive functional with α(x) x and A : K c K is a completely continuous operator such that (i) {x S(α, a, b) : α(x) > a}, and α(ax) > a if x S(α, a, b), (ii) Ax K c if x S(α, a, c), and (iii) α(ax) > a for all x S(α, a, c) with Ax > b. Then A has a fixed point in S(α, a, c). The Leggett-Williams fixed point theorem has been modified by many [1,, 4, 5, 17] in order to generalize the class of boundary value problem the fixed point theorem can be applied to. In the two fixed point theorems used here, the conditions involving the norm in the Leggett-Williams fixed point theorem are replaced by a more general convex positive functional. Also, only subsets of the cone are required to map inward and outward. In Chapter, we provide some background definitions. In Chapter 3, we present the first fixed point theorem and show how this fixed point theorem can be applied to show the existence of a positive symmetric solution of (1.1),(1.). A nontrivial example is then provided. In Chapter 4, similar results are obtained while using a second fixed point theorem.

12 Chapter Preliminaries.1 Definitions Definition.1. Let E be a real Banach space. A nonempty closed convex set P E is called a cone provided: (i) u P, λ 0 implies λu P; (ii) u P, u P implies u = 0. Definition.. A map α is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E if α : P [0, ) is continuous and α(tu + (1 t)v) tα(u) + (1 t)α(v), for all u, v P and t [0, 1]. Definition.3. Similarly, the map β is a nonnegative continuous convex func- 3

13 tional on a cone P of a real Banach space E if β : P [0, ) is continuous and β(tu + (1 t)v) tβ(u) + (1 t)β(v), for all u, v P and t [0, 1].. The Green s Function It is well known that the Green s function for u = 0 satisfying the boundary conditions (1.) is given by H(k, l) = 1 N + k(n + l), k {0,..., l}, l(n + k), k {l + 1,..., N + }, So u solves (1.1),(1.) if and only if u(k) = H(k, l)f(u(l)). Notice that H(k, l) 0. Lemma.1. For (k, l) {0,..., N +} {0,..., N +}, H(N + k, N + l) = H(k, l). 4

14 Proof. For (k, l) {0,..., N + } {0,..., N + }, H(N + k, N + l) = 1 (N + k)(n + (N + l)), 0 N + k N + l N + 1, N + (N + l)(n + (N + k)), 1 N + l N + k N +, = 1 l(n + k), 1 l k N +, N + k(n + l), 0 k l N + 1, = H(k, l). Lemma.. H(k, l) has the property that H(y,l) H(w,l) y w } with w y. for all l, w, y {0,..., N + Proof. When y w l, When y l w, 1 H(y, l) H(w, l) = (y(n + l)) N+ 1 (w(n + l)) = y w. N+ When l y w, 1 H(y, l) H(w, l) = (y(n + l)) N+ y(n + w) (l(n + w)) 1 N+ w(n + w) = y w. 1 H(y, l) H(w, l) = (l(n + y)) N+ 1 (l(n + w)) 1 y w. N+ 5

15 Chapter 3 Application of the First Fixed Point Theorem 3.1 The Fixed Point Theorem Definition 3.1. Let α and ψ be nonnegative continuous functionals on a cone P and δ and β be nonnegative continuous convex functionals on P; then, for nonnegative real numbers a, b, c, and d we define the sets A := A(α, β, a, d) = {x P : a α(x) and β(x) d}, B := B(α, δ, β, a, b, d) = {x A : δ(x) b}, and C := C(α, ψ, β, a, c, d) = {x A : c ψ(x)}. Theorem 3.1 (Anderson, Avery, Henderson []). Suppose P is a cone in a real Banach space E, α and ψ are nonnegative continuous concave functionals on P, β and δ are nonnegative continuous convex functionals on P, and for nonnegative real numbers a, b, c, and d, the sets A, B, and C are defined as above. Furthermore, suppose that A is a bounded subset of P, that T : A P is a completely continuous operator, and that the following conditions hold: 6

16 (A1) {x A : c < ψ(x) and δ(x) < b} and {x P : α(x) < a and d < β(x)} = ; (A) α(t x) a for all x B; (A3) α(t x) a for all x A with δ(t x) > b; (A4) β(t x) d for all x C and, (A5) β(t x) d for all x A with ψ(t x) < c. Then T has a fixed point x A. 3. Preliminaries Define the Banach space E to be E = {u : {0,..., N + } R} with the norm Define the cone P E by u = max u(k). k {0,1,...,N+} P := {u E : u(n + k) = u(k), u is nonnegative and nondecreasing on {0, 1,..., N+ N+ }, and wu(y) yu(w) for w y with y, w {0, 1,..., }}. Define the operator T : E E by T u(k) := H(k, l)f(u(l)), where H(k, l) is the Green s function for u(k) = 0 satisfying the boundary conditions (1.). So if u is a fixed point of T, u solves (1.1),(1.). Lemma 3.1. The operator T : E E is completely continuous. 7

17 Proof. We use the Arzelá Ascoli theorem to show that T is a compact operator, and thus completely continuous. So we must show T is continuous, uniformly bounded, and equicontinuous. First, note that H(k, l) is bounded, so there exists a K > 0 such that H(k, l) K for all k, l {0,..., N + } {0,..., N + }. Let ɛ > 0. Let u E. Since f is continuous, f is uniformly continuous on [ u 1, u + 1]. So there exists a δ > 0 with δ < 1 such that for all x, y [ u 1, u +1], f(x) f(y) < ɛ/(n +1)K. So for all v E with u v < δ, u(k), v(k) [ u 1, u + 1] and u(k) v(k) < δ for all k {0,..., N + }. Thus for all k {0,..., N + }, f(u(k)) f(v(k)) < ɛ/(n + 1)K. Thus for k {0,..., N + } T u(k) T v(k) H(k, l) f(u(l)) f(v(l)) ɛ < K (N + 1)K = ɛ. So T u T v < ɛ and therefore T is continuous. Now let {u n } be a bounded sequence in E with u n K 0 for all n. Since f is continuous, there exists a K 1 > 0 such that f(u n (k)) K 1 for all k {0,..., N + } and for all n. So for k {0,..., N + } T u n (k) H(k, l) f(u n (l)) KK 1 = (N + 1)KK 1 for all n. So {T u n } is uniformly bounded. Lastly, choose δ < 1. So if k 1, k {0,..., N + } with k 1 k < δ, k 1 = k. Thus for all n, T u n (k 1 ) T u n (k ) = 0 < ɛ. So if k 1 k < δ, T u n (k 1 ) T u n (k ) < ɛ. So T is equicontinuous. Hence by the Arzelá Ascoli theorem, T is compact, and thus uniformly continuous. Lemma 3.. The operator T acting on the set A maps A to P. That is T : A 8

18 P. Proof. Let u A. We first need to show T u(n + k) = T u(k). By Lemma.1 H(N + k, N + l) = H(k, l). Now T u(n + k) = Substitute r = N + l. So H(N + k, l)f(u(l)). T u(n + k) = So T u(n + k) = T u(k). r=1 H(N + k, N + r)f(u(n + r)) = H(k, r)f(u(r)) = T u(k). r=1 Next we need to show T u(k) is nonnegative and nondecreasing on {0, 1,..., N+ }. Since H(k, l) 0 for k, l {0,..., N + } and f : [0, ) [0, ), T u(k) is nonnegative for all k {0,..., N + }. To show that T u(k) is nondecreasing on {0, 1,..., N+ }, we show T u(k) is nonnegative on {0, 1,..., N+ }. Now k H(k, l) = H(k + 1, l) H(k, l) = 1 N + N + l, k {0,..., l}, l, k {l,..., N + 1}. 9

19 So T u(k) = k 1 = k 1 = k 1 = k 1 = Since k {0, 1,..., N+ }, T u(k) = k H(k, l)f(u(l)) l N + f(u(l)) + l=k l N + f(u(l)) + l=k N+ k l N + f(u(l)) + r=1 N+ k l N + f(u(l)) + = So T u(k) is nondecreasing. k 1 N+ k l=k N+ k l N + f(u(l)) + N + l N + f(u(l)) N + l f(u(n + l)) N + r N + f(u(r)) l N + f(u(l)). l f(u(l)) 0. N + l N + f(u(l)) Lastly, since by Lemma., H(k, l) has the property that H(y,l) H(w,l) y w l and for w y, wt u(y) yt u(w). Thus T : A P. For u P, define the concave functionals α and ψ on P by for all α(u) := ψ(u) := and the convex functionals δ and β on P by min k {τ,..., N+ } u(k) = u(τ), min k {µ,..., N+ } u(k) = u(µ), δ(u) := max u(k) = u(ν), k {0,...,ν} β(u) := max u(k) = u( N+ ). k {0,..., N+ } 10

20 3.3 Positive Symmetric Solutions to (1.1),(1.) Theorem 3.. Assume τ, µ, ν {1,..., N+ } are fixed with τ µ < ν, that d and m are positive real numbers with 0 < m < dµ and f : [0, ) [0, ) is a N+ continuous function such that (i) f(w) (N + )d (ν τ)(3 + N τ ν) N+ for w [ τd νd N+, N+ (ii) f(w) is decreasing for w [0, m] and f(m) f(w) for w [m, d], and (iii) µ l N+ ( ) ml N + f d µ ( f(m) 1 N+ ) ( N+ ) ( N+ µ µ N+ ). Then T has a fixed point x A. Thus (1.1), (1.) has at least one positive symmetric solution u A(α, β, Proof. Let a = τd N+, b = νd N+ τd N+, d). µd, c = N+. By Lemma 3.1, T is completely continuous. By Lemma 3., T : A P. Let u A. Then β(u) = u ( N+ ) d. But u achieves its maximum at N+, so A is bounded. First, we show (A1) holds. Let u P and let β(u) > d. Then ], α(u) = u(τ) τ N+ N+ u( ) = τ N+ β(u) > τd N+ = a. So {u P : α(u) < ( a and d < β(u)} =. ) d(n + ) d(n + ) Now let K N+ (3N + µ),. Define N+ (3N + ν) 11

21 u K (k) = K H(k, l) = Also, Kk (3N + k). Now (N + ) Kτ α(u k ) = u k (τ) = (3N + τ) (N + ) dτ(3n + τ) > N+ (3N + µ) τd N+ = a. So u k A. Since β(u k ) = u k ( N+ N+ K ) = N+ (3N + (N + ) ) < N+ N+ d(3n + ) (3N + ν) N+ N+ d N+ = d. and Kµ ψ(u k ) = u k (µ) = (3N + µ) (N + ) dµ(3n + µ) > N+ (3N + µ) = µd N+ = c, Kν δ(u k ) = u k (ν) = (3N + ν) (N + ) dν(3n + ν) < N+ (3N + ν) = νd N+ = b, {u A : c < ψ(u) and δ(u) < b}. Therefore (A1) holds. 1

22 Next, we show (A) holds. Let u B with δ(u) < b. By (i), α(t u) = So (A) holds. ν l=τ+1 H(τ, l)f(u(l)) H(τ, l)f(u(l)) (N + )d (ν τ)(3 + N τ ν) N+ τd N+ = a. τ(ν τ)(3 + N τ ν) (N + ) We will now show (A3) holds. Let u A with δ(t u) > b. Then So (A3) holds. α(t u) =T u(τ) H(τ, l)f(u(l)) = τ H(ν, l)f(u(l)) ν = τ δ(t u) ν > τ ν b = dτ N+ = a. Now we show (A4) holds. Let u C. By the concavity of u and since c = µd for all k {0, 1,..., µ},, N+ u(k) ck µ mk µ. 13

23 So, by (ii) and (iii), we have β(t u) = H ( N+, l) f(u(l)) = N+ µ l ( N + N+ ) f(u(l)) N + l ( N+ ) N + f(u(l)) + N+ l=µ+1 l ( N+ ) N + f(u(l)) µ l ( N+ ) ( ( )) ml N + f N+ l ( N+ u + ) µ N + f(m) l=µ+1 ( ) ( ) ( ) 1 N + N + N + d f(m) µ µ N + ( ) ( ) ( ) 1 N + N + N + + f(m) µ µ N + =d. So (A4) is satisfied. Last, we show (A5) is satisfied. Let u A with ψ(t u) < c. So β(t u) = H( N+, l)f(u(l)) < N+ µ N+ µ N+ c µ ψ(t u) = d. H(µ, l)f(u(l)) Therefore T has a fixed point and (1.1), (1.) has at least one positive symmetric solution u A. 14

24 3.4 Example Example 3.1. Let N = 18, τ = 1, µ = 9, ν = 10, d = 5, and m = 4.4. Notice that 0 < τ µ < ν 10 = N+, and 0 < m = = dµ N+. Define a continuous function f : [0, ) [0, ) by f(w) = 45 w 500, 0 w 40 1, w Then, (i) for w [ 1, 5], f(w) f(5) = 5 > 5 63 = 0 5 (10 1) ( )(10), (ii) f(w) is decreasing for w [0, 4.4] and f(m) f(w) for w [4.4, 5], and ( ) (iii) 9 10l 4.4l 0 f = < = 5 f(4.4)( 1 )(10)(10 9)( ). 0 So the hypotheses of Theorem 3. are satisfied. Therefore, the difference equatione u(k) + f(u(k)), k {0, 1,..., 18}, with boundary conditions u(0) = u(0) = 0, has a positive symmetric solution u with u(1) 1 and u(10) 5. 15

25 Chapter 4 Application of the Second Fixed Point Theorem 4.1 The Fixed Point Theorem Definition 4.1. Let ψ and δ be nonnegative continuous functionals on a cone P; then, for positive real numbers a, and b we define the sets P (ψ, b) := {x P : ψ(x) b}, and P (ψ, δ, a, b) := {x P : a ψ(x) and δ(x) b}. Theorem 4.1 (Anderson, Avery, Henderson [4]). Suppose P is a cone in a real Banach space E, α is a nonnegative continuous concave functional on P, β is a nonnegative continuous convex functional on P, and T : P P is a completely continuous operator. Assume there exist nonnegative numbers a, b, c, and d such that: (A1) {x P : a < α(x) and β(x) < b} ; (A) if x P with β(x) = b and α(x) a, then β(t x) < b; 16

26 (A3) if x P with β(x) = b and α(t x) < a, then β(t x) < b; (A4) {x P : c < α(x) and β(x) < d} ; (A5) if x P with α(x) = c and β(x) d, then α(t x) > c; (A6) if x P with α(x) = c and β(t x) > d, then α(t x) > c. If (H1) a < c, b < d, {x P : b < β(x) and α(x) < c}, P(β, b) P(α, c), and P(α, c) is bounded, then T has a fixed point x in P(β, α, b, c). If (H) c < a, d < b, {x P : a < α(x) and β(x) < d}, P(α, a) P(β, d), and P(β, d) is bounded, then T has a fixed point x in P(α, β, a, d). 4. Preliminaries Define the Banach space E to be E = {u : {0,..., N + } R} with the norm Define the cone P E by u = max u(k). k {0,1,...,N+} P := {u E : u(n + k) = u(k), u is nonnegative and nondecreasing on {0, 1,..., N+ N+ }, and wu(y) yu(w) for w y with y, w {0, 1,..., }}. 17

27 Define the operator T : E E by T u(k) := H(k, l)f(u(l)), where H(k, l) is the Green s function for u(k) = 0 satisfying the boundary conditions (1.). So if u is a fixed point of T, u solves (1.1),(1.). By Lemma 3.1, T is completely continuous Lemma 4.1. The operator T : P P. Proof. The proof of this lemma is very similar to the proof of Lemma 3., so it is omitted. For u P, define the concave functional α on P by α(u) := and the convex functional β on P by min k {τ,..., N+ } u(k) = u(τ), β(u) := max u(k) = u( N+ ). k {0,..., N+ } 4.3 Positive Symmetric Solutions to (1.1),(1.) Theorem 4.. If τ {1,..., N+ } is fixed, b and c are positive real numbers with 3b < c, and f : [0, ) [0, ) is a continuous function such that: [ N+ c(n + ) c (i) f(w) > for w c, ], τ(n + 1 τ)( N+ τ) τ [ ] ( ) (ii) f(w) is decreasing for w [ ] bτ w N+, b, b N+ N+, bτ with f bτ N+ f(w) for 18

28 (iii) and τ b f l( N+ ) N + f ( bτ N+ ) ( ) bl 1 N+ ( N+ N+ N+ )( τ)(τ ), then T has a fixed point x in P(β, α, b, c). Thus the discrete right-focal problem (1.1), (1.) has at least one positive symmetric solution u P(β, α, b, c). Proof. Note that by Lemma 4.1, T : P P. By Lemma 3.1, T is completely continuous. First, we let a = bτ N+ and d = c N+. Then, we have a = τ bτ N+ < cτ 3 N+ < c and b < c 3 = dτ 3 N+ < d. ( We proceed to verify properties (A1) and (A4). ) First, for b(n + ) b(n + ) K (3N + τ) N+, (3N + N+ N+ ), define the function u L by u L (k) := LH(k, l) = Lk (3N + k). (N + ) Since and α(u L ) = u L (τ) = β(u L ) = u L ( N+ Lτ (3N + τ) > a, (N + ) ) = L N+ (N + ) N+ (3N + ) < b, u L {u P : a < α(u) and ( β(u) < b}. ) c(n + ) Similarly, for J τ(3n + τ), c(n + ) τ(3n + N+ ), define the function u J by Since and u J (k) := α(u J ) = u J (τ) = JH(k, l) = Jk (3N + k). (N + ) Jτ (3N + τ) > c, (N + ) β(u J ) = u J ( N+ ) = J N+ (N + ) (3N + N+ ) < c N+ τ = d, 19

29 u J {u P : c < α(u) and β(u) < d}. Hence we have {u P : a < α(u) and β(u) < b} = and {u P : c < α(u) and β(u) < d} =. Therefore conditions (A1) and (A4) hold. Turning to (A), let u P with β(u) = b and α(u) a. By the concavity of u, for l {0,..., τ}, we have ( u(τ) u(l) τ and for all l {τ,..., N+ }, we have Hence by (ii) and (iii), it follows that ) l bl N+. bτ u(l) b. N+ β(t v) = H ( N+, l) f(u(l)) = N+ τ τ b f =b. + f l( N+ ) N + f(u(l)) l( N+ ) N + f(u(l)) + l( N+ ) N + f ( ( bτ N+ bτ N+ ) ) ( ) bl + 1 N + 1 N + ( N+ ( N+ N+ l=τ+1 N+ l=τ+1 l( N+ ) N + f(u(l)) l( N+ ) ( N + f bτ N+ ) N+ N+ )( τ)(τ ) N+ N+ )( τ)(τ ) So (A) is satisfied. Next, we establish (A3) of theorem 3.1, and so we let u P with β(u) = b 0

30 and α(t u) < a. By the properties of H(k, l), β(t u) = H( N+, l)f(u(l)) N+ H(τ, l)f(u(l)) τ = N+ α(t u) τ N+ a < = b, τ so (A3) holds. In dealing with (A5), let u P with α(u) = c and β(u) d. Then for l {τ,..., N + }, we have c u(l) d = c N+ τ. Hence by Property (i), α(t u) = = H(τ, l)f(u(l)) τ( N+ l=τ+1 τ) N + l=τ+1 f(u(l)) > H(τ, l)f(u(l)) = l=τ+1 c N + 1 τ = c, and so (A5) is valid. Now we address (A6). So, let u P with α(u) = c and β(t u) > d. Again, by the properties of H, α(t u) = H(τ, l)f(u(l)) τ N+ H( N+, l)f(u(l)) = = τ N+ β(t u) > τd N+ = c 1

31 and so (A6) of Theorem.3 also holds. ( b c Last, we show (H1) holds. Let K N+, 3 N+ ). Then define So u K (k) = K H(k, l) = Kk (3N + k). (N + ) β(u K ) = > N+ K N+ (3N + (N + ) ) b N+ (3N + ) b, (N + ) and α(u K ) = Kτ (3N + τ) cτ < + τ) c. 3(N + ) N+ (3N Thus {u P : b < β(u) and α(u) < c}. If u P(β, b), then α(u) β(u) b < c, and hence P(β, b) P(α, c). Lastly, if u P(α, c), then τ α(u) c, N+ β(u) and so u = β(u) c N+ τ. Therefore P(α, c) is bounded. So (H1) holds. Thus T has a fixed point u P(β, α, b, c).

32 4.4 Example Notice the previous example fails for these new conditions. For N = 18 and τ = 1, by (i), f(w) > 10c 81 for w [c, 9c]. If 9c 40, f(9c) > 10c, implying c < Thus, since 3b < c, b < Then 81 ( ) bτ 1 N+ N+ N+ (iii) fails, since b f ( )( τ)(τ ) < 0. N+ N+ If 9c > 40, then for (i) to hold, > 10c Thus a new example is needed. > Therefore (i) does not hold. Example 4.1. Example: Let N = 10, τ =, b =, and c = 7. Notice that 3b < c. Define a continuous f : [0, ) [0, ) by f(w) = 1 w 6 0 w < w < w w. Then, (i) f(w) > = 7/6 for w [7, 1], (ii) f(w) is decreasing on [ 1, ], and 3 3 f() f(w) for w [, 1], and 3 3 (iii) lf(l) = 0 1 = f( 3 ) Therefore by Theorem 3.1, the right focal boundary value problem, u(k) + f(u(k)), k {0, 1,..., 10}, 3

33 with boundary conditions u(0) = u(1) = 0, has a positive symmetric solution u with u (6) and u () 7. 4

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