Journal of Mathematical Analysis and Applications 259, 127 136 (21) doi:1.16/jmaa.2.7399, available online at http://www.idealibrary.com on On Positive Solutions of Boundary Value Problems on the Half-Line Mirosława Zima nstitute of Mathematics, Pedagogical University, Rejtana 16 A, 35-31Rzeszów, Poland E-mail: mzima@atena.univ.rzeszow.pl Submitted by J. Henderson Received July 19, 2 We study the existence of positive solutions of boundary value problems on the half-line for differential equations of second order. The Krasnoselskii fixed point theorem on cone compression and expansion is used. 21 Academic Press 1. NTRODUCTON n this paper, we are concerned with the existence of positive solutions for the following boundary value problem (BVP for short) for second order differential equations x t k 2 x t +f t x t = (1) x = lim t x t = and x t k 2 x t +f t x t x t = x = lim t x t = (2) where t =, k>, and f is continuous and a non-negative function. We establish existence theorems for (1) and (2) in the appropriate function spaces equipped with Bielecki s norm (see 5]). The proofs of these theorems are based on the Krasnoselskii fixed point theorem on cone expansion and compression of norm type (see 1]). For the convenience of the reader, we include here the definitions of a cone and completely continuous operator. 127 22-247X/1 $35. Copyright 21 by Academic Press All rights of reproduction in any form reserved.
128 mirosława zima Definition 1. A nonempty subset K of a Banach space E is called a cone if K is convex, closed, and (i) αx K for all x K and α, (ii) x x K implies x = θ. Definition 2. An operator F E E is said to be completely continuous if F is continuous and maps bounded sets into precompact sets. Next we state the Krasnoselskii fixed point theorem. Proposition 1. Let E be a Banach space and K E be a cone in E. Assume that 1 and 2 are two bounded open sets in E such that θ 1 and 1 2 Let F K 2 \ 1 K be a completely continuous operator such that either (iii) Fx x for x K 1 and Fx x for x K 2 or (iv) Fx x for x K 1 and Fx x for x K 2 is satisfied. Then F has at least one fixed point in K 2 \ 1 For applications of the Krasnoselskii theorem and its modifications for k-set contractions to various BVPs on finite intervals see, for example, 1, 7, 9, 11, 12, 18] and the references therein. The BVPs on infinite intervals have been considered, for example, in 4, 8, 13]. Finally, let us mention a few other techniques used in recent contributions to obtain the existence or multiplicity results for BVPs: Schauder s fixed point theorem 15], Borsuk s antipodal theorem 6], topological degree methods 14], and the Leggett Williams theorem 2]. 2. PRELMNARY RESULTS n this section we will study some properties of the Green s function for the following homogeneous BVP: x t k 2 x t = t (3) x = lim t x t = The Green s function G t s for (3) is given by G t s = 1 e ks e kt e kt t s 2k e kt e ks e ks s t and its partial derivative with respect to t (4) G t t s = 1 e ks e kt + e kt t<s 2 e kt e ks e ks s<t. (5)
positive solutions of boundary value problems 129 t is easy to show that G t s fulfills the inequalities G t s (6) and t s G t s e µt G s s e ks (7) for µ k and t, s. Moreover, for any, δ such that <<δand t δ we have G t s mg s s e ks (8) where s and m = min e kδ e k e k} (9) Denote by G t s + s the right-hand side derivative of (4) at s s. tis easy to verify that for k 1, t δ, and t>swe get where 2kG t s +G t t s m 1 G s s + G t s + s ] e ks (1) m 1 = Moreover, for t s, t>s, and µ k G s s + G t s + s ] e ks 1 2 k k + 1 e kδ G t s + G t t s ] e µt (11) n the case <k<1, we have for t δ and t>s G t s +G t t s m 2 G s s + G t s + s ] e ks (12) where and 1 k m 2 = min 1 + k e kδ e k + k 1 } k + 1 e k G s s + G t s + s ] e ks G t s + G t t s ] e µt (13) for t s, t>sand µ k. The inequalities (1) (13) also hold if we replace G t s + s by the lefthand side derivative G t s s and take t<s. Clearly, the constants m, m 1, m 2 are less than 1.
13 mirosława zima 3. BANACH SPACES WTH BELECK S TYPE NORM Let p be a continuous function. Denote by E the Banach space consisting of all functions x continuous on and satisfying sup x t p t } < equipped with the norm x =sup x t p t } (14) The Arzela Ascoli theorem fails to work in the space E; however, some sufficient conditions of compactness are known (see 3]). n our further considerations we will need the following modification of the compactness criterion from 16]. Proposition 2. Let E. f the functions x are almost equicontinuous on and uniformly bounded in the sense of the norm x q = sup x t q t } where the function q is positive and continuous on and p t lim t q t = then is relatively compact in E. Recall that the functions x are said to be almost equicontinuous on if they are equicontinuous in each interval T <T<. Next, denote by E the Banach space of the functions x continuously differentiable on and such that sup x t + x t ] p t } < with the norm x =sup x t + x t ] p t } (15) n E the following compactness criterion holds. Proposition 3 17]. Let be a subset of E. f the functions x and their derivatives are almost equicontinuous on and uniformly bounded with respect to the norm x q = sup x t + x t ] q t } where q is positive and continuous on and p t lim t q t = then is relatively compact in E.
positive solutions of boundary value problems 131 4. EXSTENCE THEOREM FOR PROBLEM (1) Now we state and prove our result on positive solutions of (1). We will apply Proposition 1 in the space E with p t =e λt, where λ>k. Theorem 1. that: Let G t s be the Green s function given by (4). Suppose (1 ) f is continuous and f t x a t +b t x for t x, where a b are continuous functions, (2 ) the integrals M 1 = e ks a s ds and M 2 = e λ k s b s ds are convergent and M 2 < 2k, (3 ) there exist α>, α M 1 2k M 2 1, δ >, <δ, and t such that t δ x mα αe λδ δ 1 f t x αe λt G t s ds] where m is given by (9). Then the problem (1) has at least one positive solution. Proof. n the space E consider the set } K = x E x t on and min x t m x δ t is clear that K is a cone in E. Define the integral operator F by Fx t = G t s f s x s ds where x K and t. Obviously, every fixed point of F is a solution of (1). We will show that F satisfies the assumptions of Proposition 1. t is easy to prove that F maps K into E. Particularly, by (1 ), (2 ), and (7) we get for x K and t Fx t e λt = e λt G t s f s x s ds 1 2k e ks G s s a s +b s x s ds e ks a s ds + 1 2k x e λ k s b s ds (16) which implies that sup Fx t e λt <. Next, observe that F K K. ndeed, from (1 ) and (6) we see that Fx t on. Moreover, by (7)
132 mirosława zima and (8) for any t δ and s τ we obtain Consequently, min δ Fx t =min δ m G t s f s x s ds G s s e ks f s x s ds me kτ G τ s f s x s ds = me kτ Fx τ min Fx t m Fx δ and we conclude that F maps K into itself. Fix r = M 1 2k M 2 1 and R = α and define 1 = x E x <r and 2 = x E x <R Without loss of generality we may assume that r<r.letx K 2. Proceeding analogously to (16) we can prove that the functions Fx are uniformly bounded with respect to the norm x µ = sup where <µ<λ. The standard arguments show that the Fx are almost equicontinuous on (see 17]). Thus, by Proposition 2, F is completely continuous on K 2.fx K 1 then from (16) and (2 ) e λt Fx t 1 2k M 1 + rm 2 =r hence Fx x. fx K 2 then x t e µt } min x t mα δ and sup x t e λt } = α Thus, for t δ, we have mα x t αe λδ.by(3 ), Fx t = G t s f s x s ds δ G t s f s x s ds δ ( δ 1 G t s G t τ dτ) αe λt ds = αe λt which implies e λt Fx t α. This gives Fx x for x K 2 By Proposition 1 the operator F has at least one fixed point in the set K 2 \ 1, which means that the problem (1) has a positive solution x such that r x R. This completes the proof of Theorem 1.
positive solutions of boundary value problems 133 5. EXSTENCE THEOREMS FOR PROBLEM (2) n this part of our paper we will study the existence of positive solutions of problem (2) in the space E with p t =e λt. First, consider the case k 1. Theorem 2. Suppose that G t s is given by (4) and (4 ) f is continuous and f t x y a t +b t x + y for t x y, where a b are continuous functions satisfying 2 with M 2 < 1 2 (5 ) there exist α>, α 2M 1 1 2M 2 1, <<δ, and t such that δ f t x y 2kαe λt 2kG t s +G t t s ds for all t δ and x y D, where D = ] 1 x y 2kx + y m 1 2 α and x + y αeλδ} Then the problem (2) has at least one positive solution. Proof. Consider the space E and its subset K = x E x t on and min δ 2kx t +x t m } 1 2 x t is easy to check that K is a cone in E.Fort and x K define the operator Fx t = G t s f s x s x s ds From (4 ) and (7) it follows that F maps K into E. We will show that F K K.fx K, then in view of (4 ) and (6), Fx t for t.
134 mirosława zima Furthermore, by (1) and (11) we get for t δ and s τ, 2k Fx t + Fx t ] = 2kG t s +G t t s f s x s x s ds Thus t m 1 G s s + G t s+ s e ks f s x s x s ds + m 1 t ] G s s + G t s s e ks f s x s x s ds 2 e λτ t + t G τ s + G t τ s f s x s x s ds ] G τ s + G t τ s f s x s x s ds m 1 2 e λτ Fx τ + Fx τ ] min δ 2k Fx t + Fx t m 1 2 Fx and in consequence F K K. Put r = 2M 1 1 2M 2 1, R = α, and assume that r<r.let 1 = x E x <r and 2 = x E x <R By (4 ) and Proposition 3, the operator F is completely continuous on the set K 2. Moreover, for x 1 K Fx 2 M 1 + M 2 r =r = x f x 2 K then 2kx t +x t 1 2 m 1α and x t + x t αe λδ for t δ. Thus, by (5 ) 2k Fx 2ke λt Fx t + Fx t ] e λt 2k Fx t + Fx t ] e λt e λt δ δ 2kG t s +G t t s f s x s x s ds 2kG t s +G t t s δ 1 2kG t τ +G t t τ dτ] 2kαe λt ds = 2kα = 2k x Hence Fx x on 2 ) K. By Proposition 1, there exists a fixed point of F in the set K ( 2 \ 1. This completes the proof of Theorem 2.
positive solutions of boundary value problems 135 n a similar way we can prove the following existence theorem for problem (2) with k 1. Theorem 3. Suppose that the assumption (4 ) is satisfied with M 2 < 1 and: (6 ) there exist α>, α M 1 / 1 M 2, <<δ, and t such that δ ] 1 f t x y αe λt G t s +G t t s ds for t δ and x y D, where D = x y x + y m 2 α and x + y αe λδ} Then the problem (2) has at least one positive solution. The proof is similar to that of Theorem 2. t is enough to consider the cone } K = x E x t on and min δ x t +x t m 2 x and use the inequalities (6), (7), (12), and (13). Remark. t is worth pointing out that using the norms (14) and (15) instead of usual supremum one we can impose weaker growth conditions on the function f. Finally, we will give an example of application of Theorem 2. Example 1. Consider the following BVP x t x t +t + 1 3 e 2t x t + x t = (17) x = lim t x t = where t. Fix λ = 2, = t = 1, and δ = 2. Then m 1 = 1 2 e 2, M 1 = 1, M 2 = 1 3,2M 1 1 2M 2 1 = 6, and 2 2G 1 s +G t 1 s ds = 3e2 1 e 1 1 2e 3 t is easy to check that the function f t x y =t + 1 3 e 2t x + y fulfills the assumptions of Theorem 2 with α = 24e6 3e 2 1 e 1 96e 11 3e 2 1 e 1 Evidently, α 6. By Theorem 2, the problem (17) has a positive solution x such that α x 6.
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