Periodic and almost periodic solutions of integral equations

Applied Mathematics and Computation 105 (1999) 121±136 www.elsevier.nl/locate/amc Periodic and almost periodic solutions of integral equations D. O'Regan a b a,b,* , M. Meehan b Department of Mathematics, National University of Ireland, Galway, Ireland Department of Mathematics, National University of Ireland, Dublin, Ireland Abstract Existence principles are presented which guarantee the existence of a continuous periodic or continuous Ralmost periodic solution y of the nonlinear Fredholm integral equation: y t† ˆ h t† ‡ I k t; s†f s; y s†† ds; t 2 I: Ó 1999 Published by Elsevier Science Inc. All rights reserved. 1. Introduction and preliminaries In this paper we consider the nonlinear integral equation Z y t† ˆ h t† ‡ k t; s†f s; y s††ds; t 2 I 1† I with a view to proving the existence of a continuous periodic or continuous almost periodic solution y of Eq. (1). In Section 2 ± the main part of the paper ± several admissibility results and existence principles are presented for Eq. (1). However before proceeding to Section 2 we ®rstly recall some facts about periodic and almost periodic functions, and secondly, we discuss the general approach that will be taken to obtain the various existence principles. Let 0 < x < 1 and suppose I is an interval of R that contains at least one compact subinterval of length x; which we denote by Ix : We de®ne Ax I† to be * Corresponding author. 0096-3003/99/$ ± see front matter Ó 1999 Published by Elsevier Science Inc. All rights reserved. PII: S 0 0 9 6 - 3 0 0 3 ( 9 8 ) 1 0 0 9 5 - 4 122 D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 the subspace of BC I† (bounded continuous functions on I with values in R) consisting of all x-periodic mappings, that is, if y 2 Ax I†; then y is continuous on I and y t ‡ x† ˆ y t†; for all t such that t ‡ x 2 I: The norm on Ax I† is the same as the norm j
j0 on BC I†; and it is clear that for y 2 Ax I† jyj0 ˆ supjy t†j ˆ supjy t†j: t2I t2Ix For I ˆ R; we denote by AP I†; the space of continuous almost periodic functions on I with values in R: A continuous function is said to be almost periodic (in the Bohr sense) if for any  > 0; there exists l † > 0 such that any interval of I of length l † contains an element s such that jx t ‡ s† ÿ x t†j <  for all t 2 R: Continuous almost periodic functions are bounded [1,2], with the norm on AP I† given by j
j0 : Finally we have the following compactness criteria for the space AP I†; the proof of which is given in Ref. [1], p. 143. Theorem 1.1 (Compactness criteria for AP(I)). The necessary and sucient condition that M  AP I† be relatively compact is that the following properties hold true: (i) for any t 2 I; the set of values of functions from M is relatively compact in R; (ii) M is equi-continuous, (iii) M is equi-almost periodic. Our aim in Section 2 will be to present various existence principles which establish the existence of a solution y 2 Ax I† or y 2 AP I† of Eq. (1). In each case the Nonlinear Alternative is instrumental in obtaining our results. Theorem 1.2 (Nonlinear alternative). Let C be a convex subset of a normed linear space E, and let U be an open subset of C, with pH 2 U : Then every compact, continuous map N~ : U ! C has at least one of the following two properties: (i) N~ has a ®xed point, (ii) there is an x 2 oU ; with x ˆ 1 ÿ d†pH ‡ dN~ x for some 0 < d < 1: Consider the operator equation y t† ˆ h t† ‡ KFy t†; 2† D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 123 de®ned on an interval I. The following existence principle for Eq. (2) follows easily from the Nonlinear Alternative, and in fact it is this result that we use in Section 2. Theorem 1.3 (Existence principle). Let E1 and E2 be normed spaces with norms denoted by jj
jjE1 and jj
jjE2 respectively. Suppose that h 2 E1 ; 3† F : E1 ! E2 is a bounded and continuous operator 4† K : E2 ! E1 is a continuous and completely continuous operator 5† and hold. In addition, suppose there exists a constant M > 0; independent of k; with jjyjjE1 6ˆ M for any solution y 2 E1 of y t† ˆ k h t† ‡ KFy t†† for each k 2 0; 1†: Then Eq. (2) has a solution y 2 E1 : Proof. It is easy to check using Eqs. (3)±(5) that the operator N : E1 ! E1 de®ned by Ny t† :ˆ h t† ‡ KFy t† is continuous and completely continuous. Therefore the hypotheses of Theorem 1.2 are satis®ed with C ˆ E ˆ E1 ; U ˆ fy 2 E1 : jjyjjE1 < Mg; pH ˆ 0 and N~ ˆ N : Note that condition (ii) cannot occur. Consequently N has a ®xedpoint y in E1 ; which is equivalent to saying that Eq. (2) has a solution y 2 E1 :  It is clear that by de®ning the linear integral operator Z Ky t† :ˆ k t; s†y s† ds 6† I and the possibly nonlinear operator Fy t† :ˆ f t; y t††; 7† one can rewrite Eq. (1) as the operator equation (2). Therefore if we aspire to establish an existence principle for Eq. (1) by using Theorem 1.3, we must show that the integral operator K as de®ned in Eq. (6) is well-de®ned, continuous and completely continuous. The conditions that the kernel k is required to satisfy for this to be true, we call admissibility conditions. Therefore in Section 2, before stating each existence principle, we ®rst present an admissibility result for the linear integral operator K. 124 D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 2. Periodic and almost periodic, continuous solutions Consider the nonlinear, Fredholm integral equation Z y t† ˆ h t† ‡ k t; s†f s; y s†† ds; t 2 I; 8† I where I is an interval of R (®nite or in®nite), h : I ! R; k : I  I ! R and f : I  R ! R: In this paper we assume that either h 2 AP I† or h 2 Ax I†; and discuss conditions that k and f are required to satisfy in order for Eq. (8) to have a solution y 2 AP I†; or y 2 Ax I†, respectively. Notation. Throughout this paper when discussing the normed space AP I† we assume that I ˆ R; and when considering Ax I† we assume that 0 < x < 1 and I is an interval of R that contains at least one compact subinterval of length x; which we denote by Ix : Obviously by de®ning Z Ky t† :ˆ k t; s†y s† ds; 9† I Fy t† :ˆ f t; y t†† 10† on the interval I, one can write Eq. (8) in the form of the operator equation y t† ˆ h t† ‡ KFy t†; t2I 11† which was discussed in Section 1. Therefore for h 2 Ax I† or h 2 AP I†; our aim is to impose conditions on k and f so that K and F as de®ned above satisfy the hypotheses of Theorem 1.3 with E1 ˆ Ax I† or E1 ˆ AP I†, respectively. We thus ensure that Eq. (11), or equivalently Eq. (8) has a solution y 2 Ax I† or y 2 AP I†, respectively. In particular most of this section will be concerned with illustrating conditions for k : I  I ! R such that the linear integral operator K satis®es K : E2 ! E1 is continuous and completely continuous; 12† where E2 and E1 are certain normed function spaces. In Refs. [3,4] the pair of function spaces E2 ; E1 † is said to be admissible with respect to the operator K if for any y 2 E2 ; we have Ky 2 E1 : Note that no requirement of continuity or complete continuity of K is made in this de®nition, however for convenience we extend the de®nition by saying that E2 ; E1 † is admissible with respect to K if Eq. (12) is satis®ed. Condition Eq. (12) is discussed in the literature [5], Ch. 4 where K is as de®ned in Eq. (9), E2 ˆ Lp ‰0; T Š; 1 6 p 6 1 and E1 ˆ C‰0; T Š: Our ®rst two admissibility results extend the conditions given on k in Ref. [5], so that D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 125 Eq. (12) is now true with E2 ˆ Lp I†; 1 6 p 6 1 and E1 ˆ Ax I† or AP I†: (Note that I may now be an in®nite interval.) Theorem 2.1 (Admissibility result). Let 1 6 p 6 1 be a constant, q be such that 1=p ‡ 1=q ˆ 1; 0 < x < 1 and I be an interval of R that contains at least one compact subinterval Ix of length x: Assume that kt s† :ˆ k t; s† 2 Lq I† for each t 2 I; 13† the map t 7! kt is continuous from I to Lq I† and Z q jkt‡x s† ÿ kt s†j ds ˆ 0 for all t such that t ‡ x 2 I hold: 14† 15† I Then K : Lp I† ! Ax I† is continuous and completely continuous 16† is true. Remark 2.1. From Eq. (14) and the triangle inequality we have that jjkt jjq is continuous on I. In addition, Eq. (15) and the triangle inequality implies that jjkt‡x jjq ˆ jjkt jjq for all t such that t ‡ x 2 I: Hence we see that jjkt jjq 2 Ax I†; 17† and therefore supjjkt jjq ˆ supjjkt jjq ˆ M0 < 1: t2I 18† t2Ix Also notice that Eq. (15) implies for almost every s 2 I that k t ‡ x; s† ˆ k t; s† for all t such that t ‡ x 2 I: 19† Proof of Theorem 2.1. We have immediately from Eqs. (13) and (14) and H older's Inequality that K : Lp I† ! C I† is well-defined; since for t1 , t2 2 I and y 2 Lp I†; Z jKy t1 † ÿ Ky t2 †j 6 q !1=q Z jkt1 s† ÿ kt2 s†j ds I !1=p p jy s†j ds : 20† I Replacing t1 and t2 in Eq. (20) with t ‡ x 2 I and t 2 I, respectively, yields from Eq. (15) jKy t ‡ x† ÿ Ky t†j 6 jjkt‡x ÿ kt jjq jjyjjp ˆ 0 21† 126 D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 and therefore K : Lp I† ! Ax I† is well-defined: From Eq. (18) and H older's Inequality it follows easily that K : Lp I† ! Ax I† is continuous, since if yn ! y in Lp I†; then jKyn ÿ Kyj0 6 supjjkt jjq jjyn ÿ yjjp ! 0: t2I 22† We ®nally show that K : Lp I† ! Ax I† is completely continuous. Let X be a bounded set in Lp I†; that is there exists M1 > 0 such that jjyjjp < M1 for all y 2 X and consider KX: We want to show that KX is relatively compact in Ax I†: 1 1 Let yn †nˆ1 be a sequence in X and hence Kyn †nˆ1 is a sequence in KX: Now for all t 2 Ix jKyn t†j 6 supjjkt jjq jjyn jjp 6 supjjkt jjq M1 < 1; t2Ix t2Ix Kyn †1 nˆ1 is uniformly bounded on Ix : From Eq. (20) with t1 ; t2 2 showing that ela-Ascoli Ix we have that Kyn †1 nˆ1 is also equi-continuous. Therefore the Arz Theorem implies that there exists Ky 2 C Ix † and a subsequence Kynk †1 kˆ1 of 1 Kyn †nˆ1 which converges uniformly on Ix to Ky: Let Ky t† ˆ Ky t ‡ x†; then since Kynk t† ˆ Kynk t ‡ x† we have Kynk ! Ky in Ax I†: Consequently K : Lp I† ! Ax I† is completely continuous and the result is proved.  Remark 2.2. If q ˆ 1 in Eqs. (13)±(15), Theorem 2.1 implies that K : L1 I† ! Ax I† is continuous and completely continuous: 23† Consequently, since Ax I†  BC I†  L1 I†, K : X ! Ax I† is well-de®ned, continuous and completely continuous if X ˆ Ax I†; or BC I†: Our second admissibility result gives conditions on k such that Eq. (12) is true with E2 ˆ Lq I†; 1 6 q 6 1 as before, but now E1 ˆ AP I†: Theorem 2.2 (Admissibility result). Let 1 6 p 6 1 be a constant, q be such that 1=p ‡ 1=q ˆ 1 and I ˆ R. Assume that Eqs. (13) and (14) and for any  > 0; there exists l † > 0 such that any interval of length l † in I contains a s such that !1=q Z jjkt‡s ÿ kt jjq ˆ jkt‡s s† ÿ kt s†jq ds <  for all t 2 R I 24† D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 127 hold. Then K : Lp I† ! AP I† is continuous and completely continuous 25† is true. Remark 2.3. As noted in Remark 2.1, Eq. (14) and the triangle inequality imply that jjkt jjq 2 C I†: Fix  > 0: Then the triangle inequality and Eq. (24) imply that there exists an l † > 0 such that any interval of length l † contains an element s such that for all t 2 R, jjkt‡s jjq ÿ jjkt jjq 6 jjkt‡s ÿ kt jjq < : Consequently, jjkt jjq 2 AP I†: 26† Therefore since almost periodic functions are bounded [1,2] supjjkt jjq < 1: t2I Proof of Theorem 2.2. As seen in the proof of Theorem 2.1, K : Lp I† ! C I† is de®ned well by Eqs. (13) and (14). We now show that K : Lp I† ! AP I† is well-defined: Fix  > 0 and let y 2 Lp I† be such that jjyjjp < M2 : Then since k satis®es Eq. (24), there exists l =M2 † > 0 such that any interval of length l =M2 † contains a s such that  jjkt‡s ÿ kt jjq < for all t 2 R: M2 Hence for  > 0; any interval of length l =M2 † contains an element s such that  M2 ˆ : 27† M2 Therefore for  > 0, any interval of length l =M2 † contains an element s such that for all t 2 R, jKy t ‡ s† ÿ Ky t†j 6 jjkt‡s ÿ kt jjq jjyjjp < jKy t ‡ s† ÿ Ky t†j <  and Ky 2 AP I† for y 2 Lp I†: The continuity of K follows easily from Eq. (26) and H older's Inequality, since if yn ! y in Lp I†; then jKyn ÿ Kyj0 6 supjjkt jjq jjyn ÿ yjjp ! 0: t2I 128 D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 To see that K : Lp I† ! AP I† is completely continuous, let X be a bounded set in Lp I†; that is there exists M3 > 0 such that jjyjjp < M3 for all y 2 X; and consider KX: If KX satis®es (i)±(iii) of Theorem 1.1, the result follows. Using arguments similar to those used in the proof of Theorem 2.1, conditions (i) and (ii) are easily veri®ed. Finally, replacing M2 with M3 in Eq. (27) shows that (iii) is also satis®ed.  Example 2.1. If for some 0 < x < 1 and I an interval of R that contains at least one compact subinterval of length x; k t; s† ˆ a t†b s†; t 2 I; a:e: s 2 I; where a 2 Ax I† and b 2 Lq I†; then trivially k satis®es the hypotheses of Theorem 2.1. Alternatively, if a 2 AP I†; I ˆ R and b 2 Lq I†; then k satis®es the properties of Theorem 2.2. We now present two existence principles which follow from Theorem 1.3 with the help of Theorems 2.1 and 2.2, respectively. Recall ®rst the de®nition of an Lp -Caratheodory function. De®nition 2.1. Let I be an interval in R: A function f : I  R ! R is an Lp Caratheodory function if the following conditions hold: (i) the map t ! f t; y† is measurable for all y 2 R; (ii) the map y ! f t; y† is continuous for almost all t 2 I; (iii) for any r > 0; there exists lr 2 Lp I† such that jyj 6 r implies that jf t; y†j 6 lr t† for almost all t 2 I: Theorem 2.3 (Existence principle). Let 1 6 p 6 1 be a constant, q be such that 1=p ‡ 1=q ˆ 1; 0 < x < 1 and I be an interval of R that contains at least one compact subinterval Ix of length x: Assume that Eqs. (13)±(15) hold along with h 2 Ax I†; 28† f : I  R ! R is an Lp -Carath eodory function: 29† In addition, suppose that there exists a constant M > 0; independent of k; with jyj0 6ˆ M for any solution y 2 Ax I† of ! Z k t; s†f s; y s†† ds ; y t† ˆ k h t† ‡ t2I I for each k 2 0; 1†: Then Eq. (8) has a solution y 2 Ax I†: 30† D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 129 Proof. Let y 2 BC I† be such that jyj0 < r: Then Eq. (29) ensures the existence of lr 2 Lp I† such that jf t; y t††j 6 lr t†; for a.e. t 2 I: Consequently F : BC I† ! Lp I† as de®ned in Eq. (10) is bounded since Z Z p jf t; y t††j dt 6 lpr t† dt < 1: I I Also Eq. (29) and the Lebesgue Dominated Convergence Theorem imply that F : BC I† ! Lp I† is continuous. To see this suppose that yn ! y in BC I† and jyn j0 ; jyj0 < r: Then from Eq. (29) f t; yn t†† ! f t; y t†† pointwise for a:e: t 2 I; jf t; yn t†† ÿ f t; y t††j 6 2lr t†; a:e: t 2 I: Hence from the Lebesgue Dominated Convergence Theorem we obtain Z p jjFyn ÿ Fyjjp ˆ jf t; yn t†† ÿ f t; y t††jp dt ! 0 as n ! 1: I Now since Ax I†  BC I† F : Ax I† ! Lp I† is bounded and continuous: 31† By Theorem 2.1, Eq. (16) is true where K is given by Eq. (9). It is therefore clear that the hypotheses of Theorem 1.3 are satis®ed with E1 ˆ Ax I† and E2 ˆ Lp I† and hence Eq. (8) has a solution y 2 Ax I†:  Theorem 2.4 (Existence principle). Let 1 6 p 6 1 be a constant, q be such that 1=p ‡ 1=q ˆ 1 and I ˆ R: Assume that Eqs. (13), (14), (24) and (29) and h 2 AP I† 32† hold. In addition, suppose that there exists a constant M > 0; independent of k; with jyj0 6ˆ M for any solution y 2 AP I† of Eq. (30) for each k 2 0; 1†: Then Eq. (8) has a solution y 2 AP I†: Proof. Since AP I†  BC I†; we obtain from Theorem 2.3 that F : AP I† ! Lp I† is bounded and continuous; where F is as de®ned in Eq. (10). Theorem 2.2 implies that Eq. (25) is true. Hence the proof follows from Theorem 1.3 with E1 ˆ AP I† and E2 ˆ Lp I†:  Recall from Theorem 2.1 that if k : I  I ! R satis®es Eqs. (13)±(15), or equivalently, Eqs. (13), (14) and (19), then the linear integral operator as given by Eq. (9) satis®es 130 D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 K : Lp I† ! Ax I† is continuous and completely continuous: It seems reasonable that if one were to restrict the domain of de®nition of K to Ax I†; then the periodicity properties of this space could be exploited to ®nd alternative admissibility conditions for K. We ®nd that this is in fact the case. However, not only will the domain of K play a more signi®cant role in determining the conditions we place on k : I  I ! R; our admissibility results will also be sensitive to our choice of interval I. We therefore present two further admissibility results for K as de®ned by Eq. (9). In the ®rst we let I ˆ R; while in the second I ˆ ‰0; T Š where 0 < T < 1 is dependent on the period x: Consider the linear, integral operator Z 33† K1 y t† :ˆ k t; s†y s† ds; t 2 R: R Theorem 2.5 (Admissibility result). Let 0 < x < 1 and assume that Eqs. (13) and (14) hold with I ˆ R and q ˆ 1; along with k t ‡ x; s ‡ x† ˆ k t; s†; for all t 2 R; a:e: s 2 R: 34† Then K1 : Ax I† ! Ax I† is continuous and completely continuous 35† is true. Proof. As in Theorem 2.1, Eqs. (13) and (14) with I ˆ R and q ˆ 1 give that K1 : L1 R† ! C R†is well-defined: However for y 2 Ax R†  L1 R†; Eq. (34) further implies that K1 y 2 Ax R†: To see this note that by making a change of variable in the ®rst integral, using Eq. (34) and the fact that y 2 Ax R†; we obtain for t 2 R Z K1 y t ‡ x† ÿ K1 y t† ˆ Z k t ‡ x; s†y s† ds ÿ R R Z Z k t ‡ x; s ‡ x†y s ‡ x† ds ÿ ˆ R k t; s†y s† ds R Z ˆ k t; s†‰ y s ‡ x† ÿ y s†Š ds ˆ 0: R k t; s†y s† ds D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 131 Now from Eq. (34) we see that by making a change of variable Z Z jjkt‡x jj1 ˆ jk t ‡ x; s†j ds ˆ jk t ‡ x; s ‡ x†j ds R R Z jk t; s†j ds ˆ jjkt jj1 ; ˆ t 2 R: R This fact along with Eq. (14) (with I ˆ R and q ˆ 1) imply that jjkt jj1 2 Ax R† and hence supt2R jjkt jj1 < 1 (see Remark 2.1). The remainder of the proof now follows in a similar fashion to the proof of Theorem 2.1.  Example 2.2. Consider the convolution operator Z ~ t† ˆ k~ t ÿ s†y s† ds; t 2 R: Ky 36† R Then for k~ 2 L1 R†; Eqs. (13) and (14) with I ˆ R and q ˆ 1 are satis®ed with k t; s† ˆ k~ t ÿ s†; a:e: t; s 2 R: 37† ~ In addition, k t; s† ˆ k t ÿ s† satis®es Eq. (34) for any x > 0: Hence Theorem 2.5 implies that for all x > 0; K~ : Ax R† ! Ax R† is continuous and completely continuous 38† is true. Convolution kernels are discussed in detail in Refs. [3,4]. Remark 2.4. If k~ t† ˆ eÿjtj ; then by Example 2.2 k t; s† ˆ k~ t ÿ s† ˆ eÿjtÿsj satis®es the hypotheses of Theorem 2.5. Note however that k t ‡ x; s† ˆ eÿjt‡xÿsj 6ˆ eÿjtÿsj ˆ k t; s† and hence this particular kernel does not satisfy the hypotheses of Theorem 2.1 (with I ˆ R and q ˆ 1). On the other hand, if we recall the kernel k t; s† ˆ a t†b s†; t 2 R; a:e: s 2 R discussed in Example 2.1, and if we assume that b 2 L1 R† is not x-periodic, then we have an example of a kernel that satis®es the hypotheses of Theorem 2.1 (with I ˆ R and q ˆ 1) but not those of Theorem 2.5. In the proof of Theorem 2.5 the fact that K1 is de®ned on R was signi®cant when making the various changes of variable. Therefore one might ask whether for some 0 < T < 1 and kernel k : ‰0; T Š  ‰0; T Š ! R satisfying an analogous condition to Eq. (34), an admissibility result for the linear, integral operator 132 D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 Z K2 y t† :ˆ T 0 k t; s†y s† ds; t 2 ‰0; T Š 39† could be established. The following is one answer to this question. Theorem 2.6 (Admissibility result). Let 0 < x < 1; T ˆ nx for some n 2 N‡ and assume that Eqs. (13) and (14) hold with I ˆ ‰0; T Š and q ˆ 1; along with k t ‡ x; s ‡ x† ˆ k t; s† for all t 2 ‰0; T ÿ xŠ; a:e: s 2 ‰0; T ÿ xŠ; 40† k t ‡ mx; s† ˆ k t; s ‡ n ÿ m†x† for all t 2 ‰0; T ÿ mxŠ; a:e: s 2 ‰0; T ÿ n ÿ m†xŠ; where m 2 f0; 1; . . . ; ng: 41† K2 : Ax ‰0; T Š ! Ax ‰0; T Š is continuous and completely continuous 42† Then is true. Proof. Conditions Eqs. (13) and (14) with I ˆ ‰0; T Š and q ˆ 1 imply that K2 : C‰0; T Š ! C‰0; T Š is well-de®ned. For t 2 ‰0; T ÿ xŠ we see that Z T Z T k t ‡ x; s†y s† ds ÿ k t; s†y s† ds K2 y t ‡ x† ÿ K2 y t† ˆ Z ˆ x 0 Z ‡ T x 0  Z k t ‡ x; s†y s† ds ÿ 0 0 T ÿx Z ‡ T T ÿx  k t; s†y s† ds: 43† Using Eq. (40) and by making a change of variable, we obtain for y 2 Ax ‰0; T Š and t 2 ‰0; T ÿ xŠ that Z T ÿx Z T k t ‡ x; s†y s† ds ÿ k t; s†y s† ds x 0 Z T ÿx Z T k t ‡ x; s†y s† ds ÿ k t ‡ x; s ‡ x†y s† ds ˆ x 0 Z T k t ‡ x; s†‰y s† ÿ y s ÿ x†Š ds ˆ 0; ˆ x thus reducing Eq. (43) to K2 y t ‡ x† ÿ K2 y t† Z ˆ 0 x Z k t ‡ x; s†y s† ds ÿ T T ÿx k t; s†y s† ds; t 2 ‰0; T ÿ xŠ: 44† Since T ˆ nx; some n 2 N‡ ; then once again by making a change of variable, recalling that y 2 Ax ‰0; T Š ˆ Ax ‰0; nxŠ and Eq. (41), we have from Eq. (44) for t 2 ‰0; n ÿ 1†xŠ; D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 K2 y t ‡ x† ÿ K2 y t† Z Z x k t ‡ x; s†y s† ds ÿ ˆ Z ˆ Z ˆ 0 x 0 x 0 Z k t ‡ x; s†y s† ds ÿ nx nÿ1†x nx nÿ1†x 133 k t; s†y s† ds k t; s†y s ÿ n ÿ 1†x† ds ‰k t ‡ x; s† ÿ k t; s ‡ n ÿ 1†x†Šy s† ds ˆ 0: Thus K2 : Ax ‰0; T Š ! Ax ‰0; T Š where T ˆ nx; some n 2 N‡ ; is well de®ned. The continuity and complete continuity of K2 follows using an argument similar to that used in the proof of Theorem 2.1.  Example 2.4. Consider Z 2p sin t ÿ s†y s† ds; Ky t† ˆ 0 t 2 ‰0; 2pŠ: 45† Now k : ‰0; 2pŠ  ‰0; 2pŠ ! R de®ned by k t; s† ˆ sin t ÿ s† satis®es Eqs. (13) and (14) with I ˆ ‰0; 2pŠ and q ˆ 1: Clearly for any 0 < x 6 2p k t ‡ x; s ‡ x† ˆ sin t ‡ x ÿ s ÿ x† ˆ sin t ÿ s† ˆ k t; s†; 0 6 t; s 6 2p ÿ x; and therefore k also satis®es Eq. (40) with T ˆ 2p. Finally for any 0 < x 6 2p that divides 2p; that is, 2p ˆ nx; some n 2 N‡ ; we have for m 2 f0; 1; . . . ; ng that k t ‡ mx; s† ˆ sin t ‡ mx ÿ s† ˆ sin t ‡ mx ÿ s ÿ nx† ˆ sin t ÿ s ‡ n ÿ m†x†† ˆ k t; s ‡ n ÿ m†x† since sin t has period 2p ˆ nx: Hence k satis®es Eq. (41) with T ˆ 2p: Therefore Theorem 2.6 implies that K : Ax ‰0; T Š ! Ax ‰0; T Š where K is as de®ned in Eq. (45), is continuous and completely continuous for any x > 0 that divides 2p: In general k t; s† ˆ k~ t ÿ s† where k~ 2 C‰ÿT ; T Š satis®es Eqs. (13) and (14) with I ˆ ‰0; T Š and q ˆ 1; and Eq. (40) for any 0 < x 6 T : In addition, if k~ is T-periodic and 0 < x 6 T is such that T ˆ nx; some n 2 N‡ ; then k satis®es Eq. (41) since k t ‡ mx; s† ˆ k~ t ‡ mx ÿ s† ˆ k~ t ‡ mx ÿ nx ÿ s† ˆ k~ t ÿ s ‡ n ÿ m†x††; where m 2 f0; 1; . . . ; ng; t 2 ‰0; n ÿ m†xŠ; s 2 ‰0; mxŠ: 134 D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 To complete the section we state two existence principles, the ®rst for Z y t† ˆ h t† ‡ k t; s†f s; y s†† ds; t 2 R; 46† R and the second for Z y t† ˆ h t† ‡ 0 T k t; s†f s; y s†† ds; t 2 ‰0; T Š: 47† Theorem 2.7 (Existence Principle). Let 0 < x < 1 and assume that Eqs. (13) and (14) with q ˆ 1; and Eq. (28) hold with I ˆ R: Suppose also that Eq. (34) and f : R  R ! R is continuous and f t ‡ x; y† ˆ f t; y† for all t; y 2 R 48† are true. In addition, suppose that there exists a constant M > 0; independent of k; with jyj0 6ˆ M for any solution y 2 Ax R† of Eq. (30) with I ˆ R; for each k 2 0; 1†: Then Eq. (46) has a solution y 2 Ax R†: Proof. From Theorem 2.5, Eq. (35) is true, hence if we show that F1 y t† ˆ f t; y t††; t2R 49† is such that F1 : Ax R† ! Ax R† is bounded and continuous; 50† it is clear that the result follows from Theorem 1.3 with I ˆ R; E1 ˆ E2 ˆ Ax R†; K ˆ K1 and F ˆ F1 where K1 and F1 are as de®ned in Eqs. (33) and (49), respectively. Clearly Eq. (48) gives that F : BC R† ! BC R† is bounded and continuous and for y 2 Ax R†; t2R Fy t ‡ x† ˆ f t ‡ x; y t ‡ x†† ˆ f t ‡ x; y t†† ˆ f t; y t†† ˆ Fy t†: Hence Eq. (50) is true.  Theorem 2.8 (Admissibility result). Let 0 < x < 1; T ˆ nx for some n 2 N‡ and assume that Eqs. (13) and (14) with q ˆ 1; and Eq. (28) hold with I ˆ R: Suppose also that Eqs. (40) and (41) and f : ‰0; T Š  R ! R is continuous and f t ‡ x; y† ˆ f t; y† for all t 2 ‰0; T ÿ xŠ; y 2 R 51† D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 135 are true. In addition, suppose that there exists a constant M > 0; independent of k; with jyj0 6ˆ M for any solution y 2 Ax ‰0; T Š of Eq. (30) with I ˆ R for each k 2 0; 1†: Then Eq. (47) has a solution y 2 Ax ‰0; T Š: Proof. Using Theorem 2.6 and an argument similar to the one used in the proof of Theorem 2.7, the result follows from Theorem 1.3 with I ˆ ‰0; T Š; E1 ˆ E2 ˆ Ax ‰0; T Š; K ˆ K2 where K2 is as de®ned in Eq. (39) and Fy t† ˆ F2 y t† ˆ f t; y t††; We omit the detail. t 2 ‰0; T Š:  Remark 2.5. Existence results which ensure the existence of a continuous solution of Eq. (8) are plentiful in the literature [5]. The techniques used in obtaining these results can be adapted to provide existence results for Eq. (8) when a solution y 2 Ax I† or y 2 AP I† is required. Remark 2.6. One could also consider the equation Z y t† ˆ h t† ‡ k t; s†f s; y s†† ds; a:e: t 2 I 52† I where I is a ®nite subinterval of R: Results analogous to those presented for Eq. (8) in this section with I ®nite can be obtained which prove the existence of a solution y 2 Lp I†; 1 6 p < 1 of Eq. (52) such that y t† ˆ y t ‡ x† a:e: t such that t ‡ x 2 I: We omit the detail as the results are readily established using the arguments presented in this paper along with those in Ref. [5], Ch. 4. In Ref. [4] Corduneanu discusses Eq. (52) with I ˆ R and convolution kernel, that is, Z 53† y t† ˆ h t† ‡ k~ t ÿ s†f s; y s†† ds; a:e: t 2 R: R A result is presented which proves the existence of a solution y 2 Px R† of Eq. (53). Here Px R† is de®ned to be the space of locally integrable functions with period x > 0: We refer the reader to Ref. [4] for further details. References [1] C. Corduneanu, Almost Periodic Functions, Interscience, New York, 1968. [2] N. Dunford, J.T. Schwartz, Linear Operators, Part I, Interscience, New York, 1958. [3] C. Corduneanu, Admissibility with respect to an integral operator and applications, SIAM Stud. Appl. Math. 5 (1969) 55±63. 136 D. O'Regan, M. Meehan / Appl. Math. Comput. 105 (1999) 121±136 [4] C. Corduneanu, Integral Equations and Stability of Feedback Systems, Academic Press, New York, 1973. [5] D. O'Regan, M. Meehan, Existence Theory for Nonlinear Integral and Integrodi€erential Equations, Kluwer Academic Publishers, Dordrecht, 1998.