Finding Positive Solutions of Boundary Value Dynamic Equations on Time Scale

Transcription

1 Marshall University Marshall Digital Scholar Theses, Dissertations and Capstones 2009 Finding Positive Solutions of Boundary Value Dynamic Equations on Time Scale Olusegun Michael Otunuga Follow this and additional works at: Part of the Dynamical Systems Commons, Dynamic Systems Commons, Ordinary Differential Equations and Applied Dynamics Commons, and the Partial Differential Equations Commons Recommended Citation Otunuga, Olusegun Michael, "Finding Positive Solutions of Boundary Value Dynamic Equations on Time Scale" (2009). Theses, Dissertations and Capstones. Paper 734. This Thesis is brought to you for free and open access by Marshall Digital Scholar. It has been accepted for inclusion in Theses, Dissertations and Capstones by an authorized administrator of Marshall Digital Scholar. For more information, please contact

2 FINDING POSITIVE SOLUTIONS OF BOUNDARY VALUE DYNAMIC EQUATIONS ON TIME SCALE Otunuga Olusegun Michael Thesis submitted to the Graduate College of Marshall University in partial fulfillment of the requirements for the degree of Master of Arts in Mathematics Dr. Basant Karna, Ph.D., Committee Chairperson Dr. Bonita Lawrence, Ph.D., Committee Co-chairperson Dr. Judith Silver, Ph.D., Committee Member Department of Mathematics Marshall University Huntington, West Virginia 2009 Keywords: Positive solutions, boundary value problems, eigenvalues, fixed point theorem, Green s function. Copyright 2009 by Otunuga Olusegun Michael

3 FINDING POSITIVE SOLUTIONS OF BOUNDARY VALUE DYNAMIC EQUATIONS ON TIME SCALE. Otunuga Olusegun Michael ABSTRACT This thesis is on the study of dynamic equations on time scale. Most often, the derivatives and anti-derivatives of functions are taken on the domain of real numbers, which cannot be used to solve some models like insect populations that are continuous while in season and then follow a difference scheme with variable step-size. They die out in winter, while the eggs are incubating or dormant; and then they hatch in a new season, giving rise to a non overlapping population. The general idea of my thesis is to find the conditions for having a positive solution of any boundary value problem for a dynamic equation where the domain of the unknown function is a so called time scale, an arbitrary nonempty closed subset of the reals.

4 3 Acknowledgement I will like to thank my advisor, Dr. Basant Karna, for his guidance during my study and research at Marshall University. He was always accessible and willing to help in every way, both academically, socially and morally. I can never forget all your kindness in my life. Words are not enough to express my gratitude. Thank you very much. I was delighted to interact with Dr. Bonita Lawrence by attending one of her classes, attending her Differential analyser workshop and having her as my co-advisor. Her insights to differential equations and its application is second to none. Besides, she sets an example of a world-class researcher for her rigor and passion on research. Dr. Judith Silver deserves a special thanks as my thesis committee members. She is one of the greatest and smartest ladies I ever met in the country. I would regret my Masters year at Marshall University if I had not met Dr. and Mrs Akinsete. They are like father and mother to me. Thank you so much. I will also like to thank Lamina Faye Maynard Queen and the graduate college, Marshall University, for the grant given to me to support my thesis. My deepest gratitude goes to my family for their unflagging love and support throughout my life; this thesis is simply impossible without them. I am indebted to my father, Stephen Sola Otunuga, for his care and love. He worked industriously to support the family and spare no effort to provide the best possible environment for me to grow up and attend school. Although he is no longer with us, he is forever

5 4 remembered. I am sure he shares our joy and happiness in heaven. Also to my mum, Esther Otunuga, I love you so much. I remember her constant support during the good and bad days, giving her children the best education. I am so proud of you. I feel proud of my siblings, Tosin, Aanu, and Seun Otunuga. I love you guys. Lastly, thanks to Jesus Christ for my life in the past two years. You have really shown me that You are all I need. Thanks for your joy of salvation, and for being there for me every time. You are my Lord. I give glory and honor to you Lord.

6 5 Contents. Introduction 7 2. Second Order Boundary Value Problem on R Solution to the Second Order Differential Equation Properties of the Function G(t,s) Definition of Green s Function Existence of Positive Solutions Example Green s Function and Bounds for the 2n (th) Order Boundary Value Differential Equation Finding the Green s Function for the 2n (th) Order DE Bounds for the Green s Function Example Third-Order Boundary Value Problem on R with Green s Function and Bound Solving the Third Order Equation Bounds for the Green s Function Existence of Positive Solution(s) Green s Function and Bound for the 3n (th) Order BVP Bounds for the Green s Function Example Second Order Boundary Value Problems on a Time Scale Time Scales Solution to the Second Order Differential Equation Bounds for the Green s Function Existence of Positive Solution 74

7 Green s Function for 2n th Order BVP on Time Scale Example 8 6. Third Order Boundary Value Problem on Time Scale Solution to the Eigenvalue Boundary Value Problem on T Existence of Positive Solution Example Conclusion 93 References 94

8 7. Introduction This paper focuses on determining eigenvalues λ, for which there exist positive solutions with respect to a cone, of the nonlinear eigenvalue dynamic equation y + λf(t, y σ ) = 0, t T with boundary conditions α y( ) + α 2 y ( ) = 0 α 2 y(σ(t 2 )) + α 22 y (σ(t 2 )) = 0. In this equation, we consider the case where the solution is defined on any closed subset of real numbers (called a time scale) denoted by T, initiated by Stephan Hilger in order to create a theory that unifies discrete and continuous analysis in calculus. This is fully defined in Section 5. For the special case where the time scale is the real numbers, the equation takes the form y + λf(t, y) = 0, t [, t 2 ], subject to the two-point boundary conditions α y( ) + α 2 y ( ) = 0, α 2 y(t 2 ) + α 22 y (t 2 ) = 0.

9 8 Also, we consider the 3rd-order eigenvalue problem y λf(t, y σ ) = 0, t T, with boundary conditions y( ) = β, y(σ(t 2 )) = β 2, on a general time scale T. y(σ 2 (t 3 )) = β 3 In the case where the time scale is real numbers, the equation is of the form y = λf(t, y) t [, t 3 ], subject to the three-point boundary conditions y( ) = β, y (t 2 ) = β 2, y (t 3 ) = β 3. Boundary value problems for higher order differential equations play a role in both theory and applications. The existence of positive solutions for two-point eigenvalue problems, has been studied by many researchers by using the Guo-Krasnosel skii fixed point theorem. We refer readers to [, 2, 3] for some recent results. However, few papers can be found in the literature for third order three-point boundary

10 value problems (BVPs), most papers deal with existence of positive solutions when the nonlinear term f is nonnegative []. In this paper, we deal with the existence of a positive solution for the 2nd and 3rd order BVPs on the general time scale, when the nonlinear term f is nonnegative, by first defining their respective Green s function. This Green s function is then used to derive the Green s function for the 2n th order BVP. The Green s function is also used to derive the condition for which a positive solution of the 2n th order eigenvalue differential equation can be derived. 9 The rest of this paper is organised as follows. In Section 2, we compute the Green s function for the two-point boundary value problem on R and also find the condition under which a positive solution will exist for the two-point problem. In Section 3, we derive the Green s functions for even order BVPs, compute the bounds which are finally used to proof the existence of positive solution(s) for 2n th order BVPs. In Section 4, we find the conditions in which positive solution(s) will exist for the three-point problem. Sections 5 and 6 offer some background on time scales and we derive similar existence results for even order problems and third order problem. Conclusions and future work are discussed in the last section.

11 0 2. Second Order Boundary Value Problem on R 2.. Solution to the Second Order Differential Equation For this section, we are going to consider the second order boundary value eigenvalue problem on the time scale T = R. Consider the second order eigenvalue BVP () y (t) + λf(t, y(t)) = 0, t [, t 2 ] with boundary conditions α y( ) + α 2 y ( ) = 0 (2) α 2 y(t 2 ) + α 22 y (t 2 ) = 0 where f : [, t 2 ] R + R + is continuous, and α, α 2, α 2, α 22 are real constants. We will assume the following condition: A : f : [, t 2 ] R + R + is continuous. We define the nonnegative numbers f 0, f 0, f, and f by f 0 = lim f 0 = lim f = lim f = lim min y 0 + t [,t 2 ] max y 0 + t [,t 2 ] min y t [,t 2 ] max y t [,t 2 ] f(t, y) y f(t, y) y f(t, y) y f(t, y) y

12 and assume that they all exist in the extended reals. Now we are going to find the solution of the second order problem. We shall show that the solution y(t) is of the form y(t) = G(t, s)g(s) ds where G(t, s) is defined below. Writing y (t) = g(t, y(t)) where g(t, y(t)) = λf(t, y(t)) and solving the differential equation () using Laplace transform, we have L(y (t)) = L(g(t)) which implies s 2 L(y(t)) sy(0) y (0) = L(g(t)). That is L(y(t)) = y(0) + y (0) L(g(t)). s s 2 s 2 Taking the inverse Laplace of both side, we have y(t) = y(0) + ty (0) y (t) = y (0) t t g(s) ds. Using the boundary conditions, we have (t s)g(s) ds, and α y( ) + α 2 y ( ) = α (y(0) + y (0)) + α 2 y (0) = 0, which implies, Likewise, implies α 2 (y(0) + t 2 y (0) α y(0) + (α + α 2 )y (0) = 0. α 2 y(t 2 ) + α 22 y (t 2 ) = 0 ( ) (t 2 s)g(s) ds )+α 22 y (0) g(s) ds = 0,

13 2 which implies that α 2 y(0) + (α 2 t 2 + α 22 )y (0) = Solving for y(0) and y (0), we have y(0) = β A D y (0) = α A D (α 2 (t 2 s) + α 22 )g(s) ds. where (3) So, β i = β(t i ) = α i t i + α i2, i =, 2 A = t 2 (β 2 α 2 s)g(s)ds D = α β 2 α 2 β, y(t) = β A D + α D At = Therefore, where t (t s)g(s) ds β D (β 2 α 2 s)g(s) ds + = D t (t s)g(s) ds (4) G(t, s) = (β 2 α 2 s)(α t β )g(s) ds y(t) = G(t, s)g(s) ds α D t(β 2 α 2 s)g(s) ds t (t s)g(s) ds. (β D α s)(α 2 t β 2 ) if s t t 2 ; (β D 2 α 2 s)(α t β ) if t s t 2.

14 3 Throughout this section, we will require the following conditions: A 2 : α > 0, α 2 > 0; A 3 : m t 2 m 2, where m i = β(t i) α i = β i α i, i =, 2 Note: β α implies that β α 0 which implies that α 2 0 and β 2 α 2 t 2 implies that β 2 α 2 0 which implies that α Now, we establish some preliminary results that will be used later Properties of the Function G(t,s) We give some Lemmas on the above function G(t,s). Lemma 2.. G(t, s) > 0 for (t, s) [, t 2 ] [, t 2 ]. Proof. For s t t 2 using conditions A and A 2, we have β α s t β 2 α 2 so that D = α β 2 α 2 β > 0 and G(t, s) = D (α 2t β 2 )(β α s) > 0. Also, for t s t 2, we have β α t s β 2 α 2 so that G(t, s) > 0. Therefore, G(t, s) > 0 for (t, s) [, t 2 ] [, t 2 ]. Lemma 2.2. The function G(t, s) satisfies the homogeneous differential equation y = 0 and the boundary conditions (2) for fixed s. Proof. Since G(t, s) is a polynomial of degree one, then it satisfies d 2 t 2 G(t, s) = 0 for all (t, s) [, t 2 ] [, t 2 ]. Note that differentiation is with respect to t. For t s t 2, d dt G(t, s) = D α (β 2 α 2 s) so that α G(, s) + α 2 G (, s) = 0.

15 4 Also for s t t 2, G (t, s) = D α 2(β α s) so that α 2 G(t 2, s) + α 22 G (t 2, s) = 0. Lemma 2.3. For any fixed s [, t 2 ], the function G(t, s) is continuous for every t [, t 2 ]. Proof. Clearly, G(t, s) is continuous everywhere on [, t 2 ] [, t 2 ] since it is continuous at the point t = s. Hence, the proof is complete. d Lemma 2.4. G(t, s) has a jump discontinuity with a jump of factor dt at the point t = s. Proof. Here, we show that the limit of d G(t, s) as t approaches s from dt above differ from its limit as t approaches s from below by. G (s +, s) G (s, s) = lim t s + G (t, s) lim t s G (t, s) = D (α 2β α 2 α s α β 2 + α α 2 s) = D (α 2β α β 2 ) =. Lemma 2.5. Define { G(t, s) (5) γ = min G(s, s), G(t } 2, s), G(s, s) then 0 < γ <.

16 5 Proof. Since G(t, s) > 0 for all (t, s) [, t 2 ] [, t 2 ], γ > 0. { } Case (i) If s =, γ = min, G(t 2, ) G(,, which implies ) Case (ii) If s = t 2, then γ = G(t 2, ) G(, ) = α 2t 2 β 2 α β <. { } γ = min, G(,t 2 ) G(t 2,t 2 which implies γ = G(, t 2 ) G(t 2, t 2 ) <. Hence, the proof is complete. Theorem 2.6. Assume that conditions A A 3 holds then, γg(s, s) G(t, s) G(s, s) where 0 < γ = min{ G(, s) G(s, s), G(t 2, s) G(s, s) } <. Proof. Case (i) For s t t 2, G (t, s) = α 2 D (β α s) < 0, which implies that G(t, s) is a decreasing function of t so that G(t, s) G(s, s) and also for t t 2, G(t,s) G(t 2,s) G(s,s) G(s,s) γg(s, s) G(t, s). γ which implies Case (ii) For t s t 2, G (t, s) = D α (β 2 α 2 s) > 0 implies that G(t, s) is an increasing function of t so that G(t, s) G(s, s) and also for t, G(t, s) G(s, s) G(, s) G(s, s) γ

17 6 and so we have γg(s, s) G(t, s). Therefore, γg(s, s) G(t, s) G(s, s) for t, s t Definition of Green s Function Consider the linear homogeneous differential equation (6) n a i y (i) = 0, t [, t n ] i=0 subject to the homogeneous boundary conditions (7) α i y(t i ) + α i2 y (t i ) + + α in y n (t i ) = 0, i =, 2..., n. For each fixed s [, t 2 ], a function H(t,s) with the property that (i) (ii) H(t,s) satisfies the differential equation H(t,s) satisfies the homogeneous boundary conditions (iii) H (i) (t, s), i = 0,,2,..., n-2 is a continuous function of t on [, t n ] (iv) H (n ) (s +, s) H (n ) (s, s) = a n(s) at t = s is called the Green s function of (6) satisfying (7). So, from this definition, we can conclude from Lemmas 2.2, 2.3 and 2.4 that the function G(t, s) is the Green s function for the equation y (t) = 0, t [, t 2 ] with boundary conditions α y( ) + α 2 y ( ) = 0 α 2 y(t 2 ) + α 22 y (t 2 ) = 0.

18 Existence of Positive Solutions In this section,we find the range of λ for which there exist a positive solution for () satisfying (2). Definition 2.7. Let X be a Banach space. A non empty closed convex set κ is called a cone of X, if it satisfies the following conditions: (i) α u + α 2 v κ, u, v κ and α, α 2 0, (ii) u κ and u κ, implies u = 0. Let y(t) be the solution of the BVP () satisfying (2), given by (8) y(t) = λ G(t, s)f(s, y(s)) ds. Define with norm X = {u u C[, t 2 ]}, u = max t [,t 2 ] u(t). Then, (X,. ) is a Banach space. Define a set κ by (9) κ = {u X : u(t) 0 on [, t 2 ] and min u(t) γ u } t [,t 2 ] where γ is defined in (5). In the next Lemma, we show that κ defined above is a cone. Lemma 2.8. The set κ is a cone in X, where κ is defined in (9). Proof. Let {u n } κ,n N, be such that u n u 0 0 as n, where u 0 X. Then u n (t) 0 on [, t 2 ], and min{u n (t)}

19 8 γ u n, n N. Thus, given ɛ > 0, there exist N N such that ɛ < u n (t) u 0 (t) < ɛ, t [, t 2 ], n N and so 0 u n (t) u 0 (t) + ɛ, t [, t 2 ], n N. Hence u 0 (t) 0 on [, t 2 ], then lim min{u n (t)} n γ lim u n and min {u n n(t)} γ u 0, t [, t 2 ], implies u 0 κ and t [,t 2 ] so κ is closed. Now let u,v κ and α, α 2 0. Then α u(t) + α 2 v(t) 0, t [, t 2 ], and min {α u(t) + α 2 v(t)} α min {u(t)} + α 2 t [,t 2 ] t [,t 2 ] α γ u + α 2 γ v min {v(t)} t [,t 2 ] γ α u + α 2 v. Therefore, α u + α 2 v κ. Hence the proof is complete. Define the operator T : κ X by (0) (T y)(t) = λ G(t, s)f(s, y(s))ds, for all t [, t 2 ]. If y κ is a fixed point of T, then y satisfies (8); hence y is a positive solution of the BVP () - (2). We seek a fixed point of the operator T in the cone κ. Now, we show that the operator defined in (0) preserves the cone. Lemma 2.9. The operator T, as defined in (0), preserves κ. That is, T : κ κ. Proof. Let y κ. Since G(t, s) > 0 for all t [, t 2 ], we have (T y)(t) 0 for all t [, t 2 ]. Then from Lemma 2.6,

20 9 (T y)(t) = λ G(t, s)f(s, y(s)) ds λ λγ γg(s, s)f(s, y(s)) ds max G(t, s)f(s, y(s)) ds, t [,t 2 ] γλ max t [,t 2 ] = γ Ty. G(t, s)f(s, y(s)) ds, Therefore, min (Ty)(t) γ Ty. t [,t 2 ] So, (T y)(t) κ. Hence T : κ κ. Now we need to show that the operator T is completely continuous on the cone κ. Lemma 2.0. The operator T is completely continuous, where T is define in (0). Proof. Let y κ and ɛ > 0 be given. By the continuity of f, there exists δ > 0 such that for any u [0, ) with y(t) u < δ, t [, t 2 ], then f(t, y(t)) f(t, u) < ɛ. Let w κ with y w < δ, then w(t) y(t) < δ, for all t [, t 2 ]. So we have, (T y)(t) (T w)(t) = λ G(t, s) f(s, y(s)) f(s, w(s)) ds ɛλ G(t, s) ds

21 20 Thus, (T y)(t) (T w)(t) ɛλ t 2 G(t, s) ds and T is continuous. Now, let {y n } be a bounded sequence in κ. Since f is continuous, there exists N > 0 such that f(t, y(t)) N for all n where t [, t 2 ]. For each n, (T y n )(t) = λ G(t, s)f(s, y n (s))ds λ Nλ G(s, s) f(s, y n (s)) ds G(s, s)ds. By choosing successive subsequences, there exists a subsequence {T y nj } which converges uniformly on [, t 2 ]. Hence T is completely continuous. To establish the eigenvalue intervals where a fixed point exists, we will employ the following Fixed Point Theorem due to Guo and Krasnosel skii. Theorem 2. (Guo-Krasnosel skii Fixed Point Theorem). Let X be a Banach space, κ X be a cone, and suppose that Ω, Ω 2 are open subsets of X with 0 Ω Ω 2 and Ω Ω 2. Suppose further that T : κ (Ω 2 \Ω ) κ is completely continuous operator such that either (i) T u u, u κ Ω and T u u, u κ Ω 2, or (ii) T u u, u κ Ω and T u u, u κ Ω 2, holds. Then T has a fixed point in κ (Ω 2 \Ω ). We are going to present our first existence result.

22 Theorem 2.2. Assume that conditions (A )-(A 3 ) are satisfied. Then, for each λ satisfying 2 () [γ 2 t 2 < λ < G(s, s) ds]f [ t 2 G(s, s) ds]f, 0 there exist at least one positive solution of the BVP ()-(2) in κ, where f and f 0 are as define in Section 2.. Proof. Let λ be given as in (). Now, let ɛ > 0 be chosen such that [γ 2 t 2 λ t G(s,s) ds](f ɛ) [ t 2 G(s,s) ds](f 0 +ɛ). Let T be the cone preserving, completely continuous operator defined in (0). By definition of f 0, there exists H > 0 such that f(t, y) max t [,t 2 ] y (f 0 + ɛ), for 0 < y H. It follows that, f(t, y) (f 0 + ɛ)y, for 0 < y H. Choose y κ with y = H. Then, we have from the boundedness of G(t, s) and the nature of λ, that (T y )(t) = λ G(t, s)f(s, y (s)) ds λ λ λ y. G(s, s)f(s, y (s)) ds G(s, s)(f o + ɛ)y (s) ds G(s, s)(f o + ɛ) y ds

23 22 Consequently, T y y. So, if we define Ω = {u X : u < H }, then, (2) T y y, for y κ Ω. By definition of f, there exists H 2 > 0 such that It follows that Let and let f(t, y) min t [,t 2 ] y (f ɛ), for y H 2. f(t, y) (f ɛ)y, for y H 2. H 2 = max{2h, γ H 2}, Ω 2 = {u X : u < H 2 }. Now, choose y 2 κ Ω 2 with y 2 = H 2, so that Then, min t [,t 2 ] y 2(t) γ y 2 H 2. (T y 2 )(t) = λ G(t, s)f(s, y 2 (s)) ds λ λγ γg(s, s)f(s, y 2 (s)) ds G(s, s)(f ɛ)y 2 (s) ds

24 23 γ 2 λ G(s, s)(f ɛ) y 2 ds y 2. Thus, (3) T y y, for y κ Ω 2 Applying Theorem 2.(i), from (2) and (3) we have that T has a fixed point y(t) κ (Ω 2 \Ω ). This fixed point is the positive solution of the BVP ()-(2) for the given λ. Another existence result applying Theorem 2.(ii) is as follow: Theorem 2.3. Assume that conditions (A )-(A 3 ) are satisfied. Then, for each λ satisfying (4) [γ 2 t 2 < λ < G(s, s) ds]f 0 [ t 2 G(s, s) ds]f there exist at least one positive solution of the BVP ()-(2) in κ. Proof. Let λ be given as in (4). Now, let ɛ > 0 be chosen such that [γ λ 2 t 2 t G(s,s) ds](f 0 ɛ) [. t 2 t G(s,s) ds]u(f +ɛ) Let T be the cone preserving, completely continuous operator defined in (0). By definition of f 0, there exists J > 0 such that f(t,y) min (f t [,t 2 ] y 0 ɛ), for 0 < y J.

25 24 It follows that, f(t, y) (f 0 ɛ)y, for 0 < y J. Choose y κ with y = J. Then, we have from the boundedness of G(t,s) that (T y)(t) = λ G(t, s)f(s, y(s))ds λ γλ γ 2 λ y. γg(s, s)f(s, y(s))ds G(s, s)(f o ɛ)y(s)ds G(s, s)(f o ɛ) y ds Consequently, T y y. So, if we define Ω = {u X : u < J }, then (5) T y y, for y κ Ω. It remains for us to consider f. By the definition of f, there exists an J 2 > 0 such that It follows that f(t,y) max (f + ɛ), for y J t [,t 2 ] y 2. f(t, y) (f + ɛ)y, for y J 2. There are two cases to consider. Case (i). The function f is bounded. Suppose L >0 is such that f(t, y) L, for all 0 < y <.

26 25 Let J 2 = max{2j, Lλ G(s, s) ds}, Then, for y 2 κ with y 2 = J 2, we have (T y 2 )(t) = λ G(t, s)f(s, y 2 (s)) ds λ λl y 2. G(s, s)f(s, y 2 (s)) ds G(s, s) ds Thus T y y. So, if we define Ω 2 = {u X : u < J 2 }, then (6) T y y, for y κ Ω 2. Case (ii). The function f is unbounded. Let J 2 > max{2j, J 2 } be such that f(t, y) f(t, J 2 ), for 0 < y J 2. Let y 2 κ with y 2 = J 2. Then, (T y 2 )(t) = λ G(t, s)f(s, y 2 (s)) ds λ λ G(s, s)f(s, y 2 (s)) ds G(s, s)f(s, J 2 ) ds which implies

27 26 (T y 2 )(t) λ G(s, s)(f + ɛ)j 2 ds J 2 = y 2. Thus, T y y. For this case, if we define Ω 2 = {u X : u < J 2 }, Then (7) T y y, for y κ Ω 2. Thus, in either of the cases, Theorem 2. in light of (5),(6) and (7) yields that T has fixed point y(t) κ (Ω 2 \Ω ). This fixed point is the solution of the BVP ()-(2) for the given λ. Let s consider the example with boundary conditions 2.5. Example y( + 200y) y (t) + λ + y = 0, t [0, ] y(0) y (0) = 0 2y() + 3y () = 0 The green s function is given by G(t, s) = ( s)( 5 + 2t) if 0 s t; 7 (5 2s)( + t) if 0 t s. 7

28 We found that γ = 2, f = 200, and f 0 =. So, employing (), there is a positive solution for all λ in the range ( 3 25, 6 5 ). 27

29 28 3. Green s Function and Bounds for the 2n (th) Order Boundary Value Differential Equation Our interest in this section is finding positive solutions to all differential equation of the form (8) ( ) n 2 y (n) + λf(t, y(t)) = 0 for even n, with boundary conditions α y (2k) ( ) + α 2 y (2k+) ( ) = 0 (9) α 2 y (2k) (t 2 ) + α 22 y (2k+) (t 2 ) = 0, k = 0,, 2,... n. 2 Before we can do this, we need to be able to generate the Green s function of the homogeneous boundary value problem which we do in the following subsection. 3.. Finding the Green s Function for the 2n (th) Order DE In this section, we will derive Green s function for 2nth order homogeneous differential equation (2) satisfying (22). Theorem 3.. Suppose that G 2 (t, s) is the Green s function satisfying y (t) = 0 with boundary conditions α y( ) + α 2 y ( ) = 0 α 2 y(t 2 ) + α 22 y (t 2 ) = 0

30 29 Then, (20) G n (t, s) = is the Green s function for G 2 (t, w)g n 2 (w, s) dw n {2k + 2 : k N} (2) ( ) n 2 y n (t) = 0, n {2k + 2 : k N}, with boundary conditions α y ( 2k)( ) + α 2 y ( 2k + )( ) = 0 (22) α 2 y ( 2k)(t 2 ) + α 22 y ( 2k + )(t 2 ) = 0, k = 0,, 2,... n. 2 Proof. Suppose G 2 (t, s) is the Green s function satisfying y (t) = 0, then so that y (t) = g y(t) = G 2 (t, s)g(s) ds y (t) = g (y ) = g which implies y (t) = G 2 (t, s)g(s) ds = H(t). Then, y(t) = = = = = G 2 (t, w)h(w) dw G 2 (t, w) { { { } G 2 (w, s)g(s) ds dw } G 2 (t, w)g 2 (w, s)g(s) ds dw } G 2 (t, w)g 2 (w, s) dw g(s) ds G 4 (t, s)g(s) ds

31 30 where G 4 (t, s) = From definition of G 2 (t, s), G 4 (t, s) = t 2 G 2 (t, w)g 2 (w, s) dw. G 2 (t, w)g 2 (w, s) dw implies G 4(t, s) = G 2 (t, s) which in turn implies that y satisfies the boundary conditions (2). That is, α y ( ) + α 2 y ( ) = 0, α 2 y (t 2 ) + α 22 y (t 2 ) = 0. Likewise, G 4 (t, s) satisfies boundary conditions (2) so that y(t) satisfies the BC α y( ) + α 2 y ( ) = 0 α 2 y(t 2 ) + α 22 y (t 2 ) = 0 α y ( ) + α 2 y ( ) = 0 α 2 y (t 2 ) + α 22 y (t 2 ) = 0. So, G 4 (t, s) is the Green s function for the equation y (t) = 0, satisfying the BCs α y( ) + α 2 y ( ) = 0 α 2 y(t 2 ) + α 22 y (t 2 ) = 0

32 3 α y ( ) + α 2 y ( ) = 0 α 2 y (t 2 ) + α 22 y (t 2 ) = 0. g(t) For n=6, we have y (6) (t) = g(t) or, equivalently, (y ( ) ) ( ) (t) = which implies y (t) = so that y(t) = = = = G 4 (t, s)g(s) ds = H(t) G 2 (t, w)h(w) dw G 2 (t, w) { { } G 4 (w, s)g(s) ds dw } G 2 (t, w)g 4 (w, s)g(s) dw ds G 6 (t, s)g(s) ds, where (23) G 6 (t, s) = By definition of G 2 (w, s), (23) implies G 2 (t, w)g 4 (w, s)g(s) dw. G 6(t, s) = G 4 (t, s) which means that y we have satisfies the boundary conditions above. That is,

33 32 (24) α y ( ) + α 2 y ( ) = 0 α 2 y (t 2 ) + α 22 y (t 2 ) = 0 α y ( ) + α 2 y ( ) = 0 α 2 y (t 2 ) + α 22 y (t 2 ) = 0 Also, G 6(t, s) satisfies the boundary conditions (2) so that y(t) satisfies the BC α y( ) + α 2 y ( ) = 0 α 2 y(t 2 ) + α 22 y (t 2 ) = 0 α y ( ) + α 2 y ( ) = 0 α 2 y (t 2 ) + α 22 y (t 2 ) = 0 α y ( ) + α 2 y ( ) = 0 α 2 y (t 2 ) + α 22 y (t 2 ) = 0. Continuing in this way, we find out that G n (t, s) = is the Green s function for G 2 (t, w)g n 2 (w, s) dw n {2k + 2; k N} ( ) n 2 y n (t) = 0, n {2k + 2; k N} with boundary conditions α y (2k) ( ) + α 2 y (2k+) ( ) = 0 α 2 y (2k) (t 2 ) + α 22 y (2k+) (t 2 ) = 0, k = 0,, 2,... n 2.

34 Bounds for the Green s Function Here, we find bound for the Green s function for 2nth order problem. Theorem 3.2. Assuming conditions (A)-(A3),then γ n 2 Gn (s, s) G n (t, s) G n (s, s) for n {2k; k N} Proof. From previous theorem, γg 2 (s, s) G 2 (t, s) G 2 (s, s) (t, s) [, t 2 ]. So, G 4 (t, s) = G 2 (t, x)g 2 (x, s) dx G 2 (s, x)g 2 (x, s) dx = G 4 (s, s). Therefore, G 4 (t, s) G 4 (s, s). Also G 4 (t, s) = G 2 (t, x)g 2 (x, s) dx γg 2 (s, x)g 2 (x, s) dx and since 0 < γ <, we have γg 2 (s, x)γg 2 (x, s) dx = γ 2 G 4 (s, s). Therefore, G 4 (t, s) γ 2 G 4 (s, s), so that γ 2 G 4 (s, s) G 4 (t, s) G 4 (s, s).

35 34 Similarly, G 6 (t, s) = G 2 (t, x)g 4 (x, s) dx G 2 (s, x)g 4 (x, s) dx = G 6 (s, s). Therefore, G 6 (t, s) G 6 (s, s). Also, G 6 (t, s) = G 2 (t, x)g 4 (x, s) dx γg 2 (s, x)γ 2 G 4 (s, s) dx and since G 4 (t, s) G 4 (s, s), we have γg 2 (s, x)γ 2 G 4 (x, s) dx = γ 3 G 6 (s, s). Therefore G 6 (t, s) γ 3 G 6 (s, s) so that γ 3 G 6 (s, s) G 6 (t, s) G 6 (s, s). Continuing in this, we have that γ n 2 Gn (s, s) G n (t, s) G n (s, s) for n {2k + 2; k N} The following theorem gives us the eigenvalue interval for which there exists positive solution(s) for even order problems. Theorem 3.3. For n {2k; k N}, assuming that conditions (A )- (A 3 ) is satisfied, then for each λ satisfying

36 35 (25) [γ n t 2 < λ < G n (s, s)ds]f 0 [ t 2 G n (s, s)ds]f, there exist at least one positive solution of the BVP (26) ( ) n 2 y n (t) = λf(t, y(t)). with boundary conditions α y (2k) ( ) + α 2 y (2k+) ( ) = 0 α 2 y (2k) (t 2 ) + α 22 y (2k+) (t 2 ) = 0, k = 0,, 2,... n 2. Proof. The proof follows by using Theorem 2. and changing γ to be γ n 2 in () and (4). Doing this, we have [γ n t 2 < λ < G n (s, s) ds]f o [ t 2 G n (s, s) ds]f Example Using equation (2), we can easily generate the Green s function for the case where n = 4, 6, 8, 0, and so on. Below is one of such computed Green s function using mathematica. For the case where n=4, G 4(t, s) = (β α s)(s t)( β 2 +α 2 t)(3β ( 2β 2 +α 2 (s+t))+α (3β 2 (s+t) 2α 2 (s2 +st+t 2 ))) 6D 2 + (β α s)(β α t)((β 2 α 2 t)3 +( β 2 +α 2 t 2 ) 3 ) 3α 2 D 2 + (β 2 α 2 s)(β 2 α 2 t)(( β +α s)3 +(β α ) 3 ) 3α D 2 if s t t 2; (β 2 α 2 s)(s t)(β α t)( 3β ( 2β 2 +α 2 (s+t))+α ( 3β 2 (s+t)+2α 2 (s 2 +st+t 2 ))) 6D 2 + (β α s)(β α t)((β 2 α 2 s)3 +( β 2 +α 2 t 2 ) 3 ) 3α 2 D 2 + (β 2 α 2 s)(β 2 α 2 t)(( β +α t)3 +(β α ) 3 ) 3α D 2 if t s t 2.

37 36 is the Green s function for y (4) = 0 with boundary conditions α y( ) + α 2 y ( ) = 0 α 2 y(t 2 ) + α 22 y (t 2 ) = 0 (27) α y ( ) + α 2 y ( ) = 0 α 2 y (t 2 ) + α 22 y (t 2 ) = 0. Considering the equation with boundary conditions y (4) y( + 200y) (t) + λ = 0, t [0, ], + y the Green s function is G 4(t, s) = y(0) y (0) = 0 2y() + 3y () = 0 y (0) y (0) = 0 2y () + 3y () = 0 47 (8 + ( s)3 )(5 2s)(5 2t) ( s)( t)(25 + ( 5 + 2t)3 ) ( s)(5 2t)(s t) ( 5(s + t) + 4(s 2 + st + t 2 ) + 3( 0 + 2(s + t)) ) if s t t 2; 47 (5 2s)(8 + ( t)3 )(5 2t) ( s)(25 + ( 5 + 2s)3 )( t) 294 (5 2s)( t)(s t) ( 5(s + t) + 4(s 2 + st + t 2 ) 3(0 2(s + t)) ) if t s t 2. We found that γ = 2, f = 200, and f 0 =. Employing (), we get the eigenvalue interval < λ < , for which there exists a positive solution.

38 4. Third-Order Boundary Value Problem on R with Green s Function and Bound For this section, we are going to consider the third order eigenvalue problem on R. We are going to consider nonhomogeneous boundary conditions. In this section, we assume f(t, y(t)) to be as defined in Section Solving the Third Order Equation Consider the boundary value problem (28) y (t) = λf(t, y(t)), t [, t 3 ] with boundary conditions (29) y( ) = ρ y (t 2 ) = ρ 2 y (t 3 ) = ρ 3 have Defining g(t) λf(t, y(t)) and taking the Laplace transform of 28, we L(y (t)) = L(g) which implies s 3 L(y) s 2 y(0) sy (0) y (0) = L(g). This implies that L(y) = s y(0) + s 2 y (0) + s 3 y (0) + s 3 L(g). Taking the inverse Laplace gives

39 38 y(t) = y(0) + ty (0) + 2 t2 y (0) + L { s 3 L(g)} = y(0) + ty (0) + 2 t2 y (0) + 2 t (t s) 2 g(s) ds. Then, t y (t) = y (0) + ty (0) + t y (t) = y (0) + g(s) ds Using the boundary conditions, we have that (t s)g(s) ds, and y(0) = ρ ρ 2 2 (t2 2 t 2 )ρ (t 2 s)g(s) ds, y (0) = ρ 2 t 2 ρ 3 + y (0) = ρ 3 g(s) ds, so that t 2 g(s) ds (t 2 2 t 2 )g(s) ds (t 2 s)g(s) ds, y(t) = y(0) + ty (0) + 2 t2 y (0) + 2 t (t s) 2 g(s) ds = ρ + (t )ρ (t2 2tt 2 t t 2 )ρ 3 (t 2 2tt 2 t t 2 )g(s) ds 2 + (t 2 s)( t)g(s) ds + t (t s) 2 g(s) ds 2

40 39 That is, Defining y(t) = ρ + (t )ρ ( (t t2 ) 2 (t 2 ) 2) ρ 3 ((t t 2 ) 2 (t 2 ) 2 )g(s) ds 2 (t 2 s)(t )g(s) ds + t (t s) 2 g(s) ds. 2 (30) z(t) ρ + (t )ρ ((t t 2) 2 (t 2 ) 2 )ρ 3, we have y(t) = z(t) ( (t t2 ) 2 (t 2 ) 2) g(s) ds 2 (t 2 s)(t )g(s) ds + t (t s) 2 g(s) ds, 2 where z(t) is the solution of the homogeneous boundary value differential equation y (t) = 0, with boundary conditions Also, y( ) = ρ y (t 2 ) = ρ 2 y (t 3 ) = ρ 3.

41 40 (3) G(t, s) = 2 (s ) 2 if s t t 2 < t 3 ; [ (s t ) 2 (s t) 2] if t s t 2 < t 3 ; [ (t2 ) 2 (t 2 t) 2] if t t 2 s t 3 ; [ (t2 ) 2 (t t 2 ) 2 + (t s) 2] if < t 2 s t t 3 ; [ (t2 ) 2 (t t 2 ) 2] if < t 2 t s t 3 ; 2 (s ) 2 if s t 2 t t 3. is the Green s function for the equation (32) y (t) = 0, with boundary conditions (33) y( ) = 0 y (t 2 ) = 0 y (t 3 ) = 0. From above, z(t) as defined in (30) has zeroes t and t where (34) t t = (ρ 3t 2 b 2 )+ A ρ 3, = (ρ 3t 2 b 2 ) A ρ 3, and A = [ρ 3 ( t 2 ) + ρ 2 ] 2 2ρ ρ 3. We assume the following conditions on, t 2, t 3 and ρ, ρ 2, ρ 3 throughout this section: B : t 2 > +t 3 2, t 3 < t

42 4 B2 : ρ > 0, ρ 3 < 0, ρ 3 (t 2 ) < ρ 2 < ρ 3 (t 2 t 3 ). Note: B is derived from the fact that G(t 3, s) must be nonnegative on the interval < t 2 t s t 3 and we choose t 3 < t so that (, t 3 ) (t, t ). B2 is derived such that < t 2 ρ 2 ρ 3 < t 3 where t 2 ρ 2 ρ 3 is the maximum point of z(t). Also, we make ρ 3 < 0 because we want z(t) to be concave down and ρ > 0 since we want a positive solution for y(t) Bounds for the Green s Function In this section, we find the bounds for the Green s function (3). Theorem 4.. Given that condition B holds, G(t, s) > 0 for (t,s) (, t 3 ] (, t 3 ]. Proof. For s t t 2 < t 3, G(t, s) > 0 since s. For < t < s t 2 < t 3, since < t < s, we have s > s t > 0 and so G(t, s) = [ 2 (s t ) 2 (s t) 2] > 0. Also, if t = s, then G(t, s) = 2 (s ) 2 > 0. Therefore G(t, s) > 0. For < t < t 2 s t 3, since < t < t 2, we have t 2 > t and so G(t, s) = [ 2 (t2 ) 2 (t 2 t) 2] > 0. Also, if t = t 2, then G(t, s) = 2 (t 2 ) 2 > 0. Therefore G(t, s) > 0. For < t 2 s t t 3, since t 2 > +t 3 2, we have t 2 > t 3 t 2 > t t 2. So, G(t, s) = [ 2 (t2 ) 2 (t t 2 ) 2 + (t s) 2] > 0. For < t 2 t s t 3, G(t, s) > 0 since t 2 > +t 3 2. Lastly, for < s t 2 t < t 3, G(t, s) > 0 since s. In the next theorem, we find the bounds for the Green s function 3. This bound is later used to find the range of λ values for which (28) satisfying (29) has a positive solution.

43 42 Theorem 4.2. For a fixed s, G(t, s) 2 (s ) 2 for all (t,s) (, t 3 ] (, t 3 ]. G(t, s) 2 ((t 2 ) 2 (t 3 t 2 ) 2 ) for all (t,s) [t 2, t 3 ] [t 2, t 3 ]. Proof. For t < s < t 2 < t 3, G (t, s) = s t > 0 which implies that G(t, s) is an increasing function of t. So, G(t, s) < G(s, s) for t < s. For t t 2 s t 3, G (t, s) = t 2 t 0 which implies that G(t, s) is a nondecreasing function of t. So, G(t, s) G(t 2, s) = 2 (t 2 ) 2 2 (s ) 2 for t t 2 s. Likewise, for < t 2 s t t 3, G (t, s) = t 2 s 0, so G(t, s) is a nonincreasing function of t so that G(t, s) G(s, s) = [ 2 (t2 ) 2 (s t 2 ) 2] 2 (t 2 ) 2 2 (s ) 2. For < t 2 t s t 3, t < t 3 so that (t t 2 ) 2 > (t 3 t 2 ) 2 which implies G(t, s) = ( 2 (t2 ) 2 (t t 2 ) 2) ( 2 (t2 ) 2 (t 3 t 2 ) 2). Likewise on < t 2 s t t 3, from above, G(t, s) = ( (t2 ) 2 (t t 2 ) 2 + (t s) 2) 2 2 ( (t2 ) 2 (t 3 t 2 ) 2) 2 ( (t2 ) 2 (t t 2 ) 2) 4.3. Existence of Positive Solution(s) In this subsection, we find the range of λ for which (28) satisfying (29) has positive solution. Let y(t) be the solution of the BVP (28)- (29), given by (35) y(t) = z(t) + λ G(t, s)f(s, y(s)) ds

44 43 Defining equation (35) can be re-written as v(t) y(t) z(t), (36) v(t) = λ G(t, s)f(s, v(s)) ds, which is the solution of the homogeneous boundary value differential equation (37) v (t) = λf(t, v(t)), t [, t 3 ], with boundary conditions (38) v( ) = 0 v (t 2 ) = 0 v (t 3 ) = 0. Also G(t, s) is the Green s function for the differential equation v (t) = 0, t [, t 3 ] with boundary conditions Define a set, X, by v( ) = 0 v (t 2 ) = 0 v (t 3 ) = 0. X = {u u C[, t 3 ]} with norm u = max t [,t 3 ] u(t),

45 44 Then (X,. ) is a Banach space. Let (39) m = min { min t 2 s t { } (t2 ) 2 (t 3 t 2) 2 + (t 3 s) 2 (t 2 ) 2 + (t 2 s) 2, (t2 } t)2 (t 3 t 2) 2. (t 2 ) 2 We first show that 0 < m <. Since for < t 2 s t t 3, we have G (t, s) = t 2 s 0, so G(t, s) is a decreasing function of t and G(t 3, s) < G(t 2, s). Also, for < t 2 t s t 3, we have G (t, s) = t 2 t 0, so G(t, s) is a decreasing function of t and G(t 3, s) < G(t 2, s). Define a set κ by κ = {u X : u(t) 0 on [, t 2 ] and Then by Lemma 2.8, κ is a cone. Using condition B2, min u(t) m u }. t [t 2,t 3 ] z(t) > 0 for t (t, t ), where t and t are as define in (34). From the fact that z(t ) = 0 and z( ) = ρ > 0, we conclude that t < since z(t) is concave down. Also, since t 3 < t then (, t 3 ) (t, t ). So, we conclude that z(t) 0 for t [, t 3 ]. Define the operator T : κ X by (40) (T v)(t) = λ G(t, s)f(s, v(s))ds, t [, t 3 ] From Lemma 2.9, T preserves κ. If v κ is a fixed point of T, then v satisfies (37) and hence v is a positive solution of the BVP (37) - (38). We seek a fixed point of the operator,t, in the cone κ. From Lemma 2.9, the operator T as defined in (40) preserves κ. Now, we find the range of λ that gives a positive solution for (36)

46 Theorem 4.3. Assume that conditions (B),(B2) is satisfied. Then, for each λ satisfying 45 (4) [m t 3 t 2 2 ((t < λ < 2 ) 2 (t 3 t 2 ) 2 ) ds]f [ t 3 2 (s ) 2 ds]f, 0 there exist at least one positive solution of the BVP (37)-(38) in κ where m is defined in (39). Proof. Let λ be given as in (4). Now, let ɛ > 0 be chosen such that [m λ t 3 t 2 2 ((t 2 ) 2 (t 3 t 2 ) 2 ) ds](f ɛ) [. t 3 t G(s,s) ds](f 0 +ɛ) Let T be the cone preserving, completely continuous operator defined in (40). By definition of f 0, there exist H > 0 such that f(t, v) max (f 0 + ɛ), for 0 < v H. t [,t 3 ] v It follows that, f(t, v) (f 0 + ɛ)v, for 0 < v H. So choosing v κ with v = H. Then, we have from the boundedness of G(t, s) that (T v )(t) = λ λ λ λ G(t, s)f(s, v (s)) ds 2 (s ) 2 f(s, v (s)) ds 2 (s ) 2 (f 0 + ɛ)v (s) ds 2 (s ) 2 (f 0 + ɛ) v ds v. Consequently, T v v. So, if we define Ω = {u X : u < H },

47 46 Then (42) T v v, for v κ Ω. By definition of f, there exists an H 2 > 0 such that f(t,v) min v (f ɛ), for v H 2. t [,t 3 ] It follows that f(t, v) (f ɛ)v, for v H 2. Let H 2 = max{2h, m H 2}, and let Ω 2 = {u X : u < H 2 }. Now choose v 2 κ Ω 2 with v 2 = H 2, so that Consider, T (v 2 )(t) = λ λ λ min t [,t 2 ] v 2(t) m v 2 H 2. t 2 t 2 mλ v 2. G(t, s)f(s, v 2 (s)) ds 2 ((t 2 ) 2 (t 3 t 2 ) 2 )f(s, v 2 (s)) ds 2 ((t 2 ) 2 (t 3 t 2 ) 2 )(f ɛ)v 2 (s) ds t 2 2 ((t 2 ) 2 (t 3 t 2 ) 2 )(f ɛ) v 2 ds Thus, (43) T v v, for v κ Ω 2

48 Applying Theorem 2. to (42) and (43) yields a fixed point for T v(t) κ (Ω 2 \Ω ). This fixed point is the positive solution of the BVP (37)- (38) 47 for the given λ. Next, we prove the other range for λ for which a positive solution exists. Theorem 4.4. Assume that conditions (B)-(B2) is satisfied. Then, for each λ satisfying (44) [m t 3 t 2 2 ((t < λ < 2 ) 2 (t t 2 ) 2 ) ds]f 0 [ t 3 2 (s ) 2 ds]f there exist at least one positive solution of the BVP (37)-(38) in κ. Proof. Let λ be given as in (44). Now, let ɛ > 0 be chosen such that [m t 3 t 2 2 ((t 2 ) 2 (t 3 t 2 ) 2 ) ds](f 0 ɛ) λ [ t 3 2 (s ) 2 ds]u(f + ɛ). Let T be the cone preserving, completely continuous operator defined in (2.9). By definition of f 0, there exist J > 0 such that f(t, v) min (f 0 ɛ), for 0 < v J. t [,t 3 ] v

49 48 It follows that, f(t, v) (f 0 ɛ)v, for 0 < v J. So choosing v κ with v = J. we have from the boundedness of G(t,s) that (T v )(t) = λ λ λ λ t 2 t 2 t 2 mλ v. G(t, s)f(s, v (s)) ds G(t, s)f(s, v (s)) ds 2 ((t 2 ) 2 (t 3 t 2 ) 2 )f(s, v (s)) ds 2 ((t 2 ) 2 (t 3 t 2 ) 2 )(f o ɛ)v (s) ds t 2 2 ((t 2 ) 2 (t 3 t 2 ) 2 )(f o ɛ) v ds Consequently, T v v. So, if we define Ω = {u X : u < J }, Then (45) T v v, for v κ Ω. It remains for us to consider f. By the definition of f, there exists an J 2 > 0 such that It follows that There are two cases. f(t, v) max (f + ɛ), for v J 2. t [,t 3 ] v f(t, v) (f + ɛ)v, for v J 2. Case (i): The function f is bounded. Suppose L >0 is such that f(t, v) L, for all 0 < v <.

50 49 Let Then for v 2 κ with v 2 = J 2, J 2 = max{2j, λl 2 (s ) 2 ds}, (T v 2 )(t) = λ λ λl G(t, s)f(s, v 2 (s)) ds 2 (s ) 2 f(s, v 2 (s)) ds 2 (s ) 2 ds J 2 = v 2. Thus, T v v. So, if we define then Ω 2 = {u X : u < J 2 }, (46) T v v, for v κ Ω 2. Case (ii): The function f is unbounded. Let J 2 > max{2j, J 2 } be such that f(t, v) f(t, J 2 ), for 0 < v J 2. Let v 2 κ with v 2 = J 2. Then (T v 2 )(t) = λ λ G(t, s)f(s, v 2 (s)) ds 2 (s ) 2 f(s, v 2 (s)) ds λ 2 (s ) 2 f(s, J 2 ) ds λ 2 (s ) 2 (f + ɛ)j 2 ds J 2 = v 2. Thus, T v v. For this case, if we define Ω 2 = {u X : u < J 2 },

51 50 then (47) T v v, for v κ Ω 2. In either of the cases, an application of part (ii) of Theorem 2. to (45), (46) and (47) yields a fixed point for T v(t) κ (Ω 2 \Ω ). This fixed point is the solution of the BVP (37)-(38) for the given λ Green s Function and Bound for the 3n (th) Order BVP Our interest in this section is to find positive solutions to all differential equations of the form (48) y (n) + λf(t, y(t)) = 0 subject to some boundary conditions (49) y (3k) ( ) = ρ y (3k+) (t 2 ) = ρ 2 y (3k+2) (t 3 ) = ρ 3, k = 0,, 2,..., n 2. Before we can do this, we need to be able to generate the Green s function of the homogeneous boundary value problem. The following theorem offers us a method. Theorem 4.5. Suppose that G 3 (t, s) is the Green s function of y (t) = 0 satisfying the boundary conditions

52 5 (50) then, y( ) = 0 y (t 2 ) = 0 y (t 3 ) = 0 (5) G n (t, s) = is the Green s function for G 3 (t, w)g n 3 (w, s) dw n {3k + 3 : k N} (52) y n (t) = 0, n {3k + 3 : k N}, with boundary conditions (53) y (3k) ( ) = 0 y (3k+) (t 2 ) = 0 y (3k+2) (t 3 ) = 0, k = 0,, 2,..., n 2. Proof. Suppose G 3 (t, s) is the Green s function of y (t) = 0, satisfying the boundary conditions y( ) = 0, y (t 2 ) = 0, y (t 3 ) = 0 then, the solution of y (t) = g satisfying the above BCs is y(t) = G 3 (t, s)g(s) ds. So, if y (6) (t) = g, that is, (y (3) ) (3) = g, then,

53 52 which implies y (3) (t) = G 3 (t, s)g(s) ds H(t) y(t) = = = = = G 3 (t, w)h(w) dw G 3 (t, w) { { } G 3 (w, s)g(s) ds dw G 3 (t, w)g 3 (w, s)g(s) ds dw } G 3 (t, w)g 3 (w, s) dw g(s) ds G 6 (t, s)g(s) ds, where G 6 (t, s) = By definition of G 3 (t, s), G 6 (t, s) = G 3 (t, w)g 3 (w, s) dw. G 3 (t, w)g 3 (w, s) dw implies that which implies that y G 6 (t, s) = G 3 (t, s) satisfies the boundary conditions for the equation (50), that is, (y )( ) = 0 (y ) (t 2 ) = 0 (y ) (t 3 ) = 0.

54 Likewise, G 6 (t, s) satisfies the boundary conditions (50) so that y(t) satisfies 53 the boundary conditions y( ) = 0 y (t 2 ) = 0 y (t 3 ) = 0. So, G 6 (t, s) is the Green s function for the equation y (6) (t) = 0, with boundary conditions y( ) = 0 (54) Similarly, implies that y (t 2 ) = 0 y (t 3 ) = 0y ( ) = 0 y (4) (t 2 ) = 0 y (5) (t 3 ) = 0. y (9) (t) = g(t) which gives us (y (3) ) (6) (t) = g(t) so that y (3) (t) = G 6 (t, s)g(s) ds = H(t)

55 54 y(t) = = = = G 3 (t, w)h(w) dw G 3 (t, w) { { } G 6 (w, s)g(s) ds dw } G 3 (t, w)g 6 (w, s)g(s) dw ds G 9 (t, s)g(s)ds, where G 9 (t, s) = By definition of G 3 (w, s) G 3 (t, w)g 6 (w, s)g(s) dw. G 9 (t, s) = G 6 (t, s) which means that y satisfies the boundary conditions (54), that is y (3) ( ) = 0 y (4) (t 2 ) = 0 y (5) (t 3 ) = 0 y (6) ( ) = 0 y (7) (t 2 ) = 0 y (8) (t 3 ) = 0. Also, G 9 (t, s) satisfies the boundary conditions (38) and y(t) satisfies the boundary conditions

56 55 y( ) = 0 y (t 2 ) = 0 y (2) (t 3 ) = 0 y (3) ( ) = 0 y (4) (t 2 ) = 0 y (5) (t 3 ) = 0 y(6)( ) = 0 y (7) (t 2 ) = 0 y (8) (t 3 ) = 0. Continuing in this way, we find that G n (t, s) = is the Green s function for G 3 (t, w)g n 3 (w, s)dw, n {3k + 3; k N}, y n (t) = 0, n {3k + 3; k N}, with boundary conditions y (3k) ( ) = 0 y (3k+) (t 2 ) = 0 y (3k+2) (t 3 ) = 0, k = 0,, 2,... n Bounds for the Green s Function In this section, we find the bounds for Green s function, G n (t, s).

57 56 Theorem 4.6. Assuming conditions (B),(B2),then for n {3k; k N}, ( t3 2 G n (t, s) 3 ) n 3 ( (t2 ) 2 (t 3 t 2 ) 2) n 3 G n (t, s) for all (t, s) [t 2, t 3 ] [t 2, t 3 ]. ( ) n 3 (t3 ) n 3 (s ) 2 for all (t, s) [, t 3 ] [, t 3 ]. 6 Proof. From Theorem (4.2), G 3 (t, s) 2 (s ) 2 for all (t, s) [, t 3 ] [, t 3 ], and G 3 (t, s) 2 ((t 2 ) 2 (t 3 t 2 ) 2 ) for all (t, s) [t 2, t 3 ] [t 2, t 3 ]. So, G 6 (t, s) = G 3 (t, x)g 3 (x, s) dx 2 (x ) 2 2 (s ) 2 dx = 3 ( ) 2 (s ) 2 (t 3 ) 3. 2 Therefore, G 6 (t, s) 3 Also for all (t, s) [t 2, t 3 ] [t 2, t 3 ], ( ) 2 (s ) 2 (t 3 ) 3. 2 G 6 (t, s) = G 3 (t, x)g 3 (x, s) dx ( { (t2 ) 2 (t 3 t 2 ) 2}) 2 dx 2 ( { (t2 ) 2 (t 3 t 2 ) 2}) 2 (t 3 ). 2 Therefore, G 6 (t, s) ( ) 2 2 ((t 2 ) 2 (t 3 t 2 ) 2 ) (t 3 ). Similarly,

58 57 G 9 (t, s) = = ( 3 G 3 (t, x)g 6 (x, s) dx ( ) 2 (s ) 2 (t 3 ) 3 dx 2 2 (x ) 2 3 ) 2 ( ) 3 (s ) 2 (t 3 ) 6. 2 Also, G 9 (t, s) = = G 3 (t, x)g 6 (x, s) dx [ 2 ((t 2 ) 2 (t 3 t 2 ) 2 ) 2 ( 2 ((t 2 ) 2 (t 3 t 2 ) 2 ) { (t2 ) 2 (t 3 t 2 ) 2}] 2 (t 3 ) dx ) 3 (t 3 ) 2. Therefore, G 9 (t, s) Continuing in this sense, we have that G n (t, s) G n (t, s) ( ) n ( ) 3 2 ((t 2 ) 2 (t 3 t 2 ) 2 ) (t 3 ) 2. ( ) n 3 (t3 ) n 3 (s ) 2 for all (t, s) [, t 3 ] [, t 3 ], 2 ( ) n 2 ((t 2 ) 2 (t 3 t 2 ) 2 3 ) (t3 ) n 3 3 for all (t, s) [t 2, t 3 ] [t 2, t 3 ], By defining the two functions F n (s, s) = E n (s, s) = ( ) n ( ) n 3 (t3 ) n 3 (s ) 2, 2 ( ) n 2 ((t 2 ) 2 (t 3 t 2 ) 2 3 ) (t3 ) n 3 3,

59 58 going by (4) and (44), we can state the following theorems: Theorem 4.7. Assume that conditions (B),(B2) are satisfied. Then, for each λ satisfying [m t 3 < λ < t 2 E n (s, s) ds]f [ t 3 F n (s, s) ds]f, 0 there exist at least one positive solution of the BVP (48)-(49) in κ. Proof. The proof is similar to that of Theorem 4.3. Theorem 4.8. Assume that conditions (B),(B2) are satisfied. Then, for each λ satisfying [m t 3 < λ < t 2 E n (s, s) ds]f 0 [ t 3 F n (s, s) ds]f there exist at least one positive solution of the BVP (48)-(49) in κ. Proof. The proof is similar to that of Theorem Example Consider the third order boundary value problem y (t) + λy( e 7y ) = 0, t [0, ], with boundary conditions, y() = y (2.6) = 0 y (4) =

60 59 The Green s function is given by 2 (s )2 if s t 2.6 < 4; G(t, s) = For this particular example, 2 ( + 2s t)( + t) if t s 2.6 < 4; 2 (4.2 t)(t ) if t 2.6 s 4; 2 ( s t 2st) if < 2.6 s t 4; 2 (4.2 t)(. + t) if < 2.6 t s 4; 2 (s )2 if s 2.6 t 4. z(t) = + 2 (2.56 (t 2.6)2 ), m = , f = 200, f 0 = 2. Using (4), positive solution exists for all λ in the interval ( , ).

61 60 5. Second Order Boundary Value Problems on a Time Scale In this section, we will find positive solutions for the two boundary value problem discussed in Section 2 and Section 3, on a general time scales. First, we will discuss what a time scale is. 5.. Time Scales The calculus of time scales was introduced by Stefan Hilger in his Ph.D. thesis (Universität Würzburz,988) in order to unify the discrete and continuous analysis. The definitions and the theorems in this subsection are from [2]. A time scale is an arbitrary non-empty closed subset of the real numbers. It is usually denoted by T. Thus R, Z, N, N 0 are some examples of time scales. But Q, R Q {irrationals}, C and (0, ),i.e., the rational numbers, the irrational numbers, the complex numbers, and the open interval between 0 and, are not time scales. We move through the time scale using forward and backward jump operators. The gaps in the time scale are measured by a function µ, defined in terms of forward jump operator, σ. Definition 5.. Forward jump operator Let T be a time scale. For t T we define the forward jump operator σ : T T, by σ(t) := inf {s T : s > t} Backward jump operator : An operator ρ : T T is given by ρ(t) := sup {s T : s < t} Note : If σ(t) > t, we say that t is right scattered, while if ρ(t) < t we say that t is left scattered. The points which are both right-scattered and left-scattered are called isolated.

62 6 Note 2 : If t < supt and σ(t) = t, then t is called right-dense. Note 3 : If t > inft and ρ(t) = t, then t is called left-dense. The forward jump operator defined on t, σ(t), is not always equal to t. The difference between σ(t) and t is called graininess. Graininess Function: The Graininess of a time scale,t, µ : [0, ) is defined by µ(t) = σ(t) t for all t T. Note 4: If T has a left-scattered maximum m, then T k = T {m}. Otherwise T k = T. That is, T (ρ(sup T), supt] if supt < T k = T if sup T =. Note 5: Let f : T R be a function, then we define the function, f σ : T R, by f σ (t) = f(σ(t)) for all t T i.e., f σ = f σ. Using σ we define the delta derivative of a function f in a natural way. Definition 5.2. Differentiation: Assume f : T k R is a function and let t T k. Then we define f (t) to be the number(provided it exists) with the property that given any ɛ > 0 there exists a neighborhood U = (t δ, t+δ) T of t for some δ > 0 such that [f(σ(t)) f(s)] f (t)[σ(t) s] ɛ σ(t) s for all s U where f (t) is called delta derivative of f at t. Using the limit definition, Assume f : T R is continuous and let t T k. Then we define f f(σ(t)) f(s) (t) = lim, s t σ(t) s provided the limits exist. We will introduce the delta derivative f for a function f defined on T. It

63 62 is expressed as (i) f = f (is the usual derivative) if T = R and (ii) f = f (is the forward difference operator) if T = Z. Theorem 5.3. Assume f : T R is a function and let t T k. Then we have the following. (i) If f is differentiable at t, then f is continuous at t. (ii) If f is continuous at t, then f is differentiable at t with f (t) = f(σ(t)) f(t) µ(t) (iii) If t is right- dense, then f is differentiable at t iff the limit f(t) f(s) lim s t t s exists as a finite number. In this case f f(t) f(s) (t) = lim. s t t s (iv) If f is differentiable at t,then f(σ(t)) = f(t) + µ(t)f (t). Now we introduce the most powerful fundamentals of derivatives: the sum rule, product rule, quotient rule and the transformation of the sigma function in terms of original function and its derivative. Theorem 5.4. Assume f, g: T R are differentiable at t T k. Then: (i) The sum f + g : T R is differentiable at t with (f + g) (t) = f (t) + g (t).

64 63 (ii) For any constant α, αf : T R is differentiable at t with (αf) (t) = αf (t). (iii) The product fg : T R is differentiable at t with (fg) (t) = f (t)g(t) + f(σ(t))g (t) = f(t)g (t) + f (t)g(σ(t)). (iv) If f(t)f(σ(t)) 0, then f (v) If g(t)g(σ(t)) 0 then f g is differentiable at t with { } (t) = f (t) f f(t)f(σ(t)) is differentiable at t and { } f (t) = f (t)g(t) f(t)g (t) g g(t)g(σ(t)) In addition to the differentiability we need couple of more conditions for integrability of the function. Definition 5.5. A function f : T R is called regulated, provided its rightsided limits exists(finite) at all right-dense points in T and its left-sided limits exists(finite) at all left-dense points in T. The set of such function is denoted by R. A function f : T R is called rd continuous provided it is continuous at right-dense points in T and its left-sided limits exist at left-dense points in T. It is denoted by C rd = C rd (T) = C rd (T, R). A continuous function f : T R is called pre-differentiable in the region of differentiation D, provided D T k, T k D is countable and contains no right-scattered elements of T, and f is differentiable at each each t

65 64 D. Assume f : T T is regulated function. Any function F is called a pre antiderivative of f if F (t) = f(t). Theorem 5.6. Existence of Pre-Antiderivative Let f be regulated. Then there exists a function F which is pre-differentiable with region of differentiation D such that F (t) = f(t) holds for all t D. The indefinite integral of a regulated function f is given by f(t) t = F (t) + C where C is an arbitrary constant and F is a pre-antiderivative of f. We define the Cauchy integral by: s r f(t) t = F (s) F (r) for all r, s T. A function F : T R is called an antiderivative of f : T R provided F (t) = f(t) holds for all t T k. Z TABLE: Time scale derivative and Antiderivative for T = R or T = Time T symbol R Z Backward jump operator ρ(t) t t Forward jump operator σ(t) t t + Graininess µ(t) 0 Derivative f (t) f (t) f(t) b Integral a f(t) t b a f(t)dt b f(t)(if a < b) Rd-continuous f continuous f any f t=a