First-Order Recurrence Relations on Isolated Time Scales

Transcription

1 Firs-Order Recurrence Relaions on Isolaed Time Scales Evan Merrell Truman Sae Universiy Rachel Ruger Linfield College Jannae Severs Creighon Universiy Some Preliminaries on ime scales A recen cover sory in New Scienis, [3] repors of many imporan applicaions concerning dynamic equaions on ime scales. Some of hese applicaions include a model of he Wes Nile virus, of elecrical aciviy in he hear, of he sock marke, of combusion in engines, of bulimia and of populaion models ha vary in coninuous ime and discree ime, as well as he sudy of he inerfaces of nanoscale srucures embedded in oher maerials. A ime scale T is an arbirary nonempy closed subse of he real numbers R. Examples include R, he inegers Z, he harmonic numbers {H n } 0 where H 0 = 0 and H n = n i=, and i qz = {0} {q n n Z} where q >. On any ime scale T we define he forward and backward jump operaors by: (.) σ() := inf{s T s > }, ρ() := sup{s T s < }, where inf := sup T and sup := inf T. A poin T, > inf T, is said o be lef-dense if ρ() =, righ-dense if < sup T and σ() =, lef-scaered if ρ() < and righ-scaered if σ() >. A ime scale T is said o be an isolaed ime scale provided given any T, here is a δ > 0 such ha ( δ, +δ) T = {}. The graininess funcion for a ime scale T is defined by () := σ(). The se T κ is derived from he ime scale T as follows: If T has a lef-scaered maximum m, hen T κ = T {m}. Oherwise, T κ = T. Noe ha in he case T = R we have σ() = ρ() =, () 0 This work was done during he summer of 2003 a he REU sie Universiy of Nebraska-Lincoln, NSF gran number , under he guidance of Dr. Allan Peerson.

2 2 for T = Z we have σ() = +, ρ() =, (), for he case q N = {q n n N} we have σ() = q, ρ() =, () = (q ), q and for he ime scale N b 0 := {nb n N 0 } where b N, b ( ) σ() = ( b + ) b, ρ() = ( b b ) b, () = k b. k For a funcion f : T R he (dela) derivaive f () a T κ is defined o be he number f () (provided i exiss) wih he propery such ha given any ɛ > 0, here exiss a neighborhood U of such ha [f(σ()) f(s)] f ()(σ() s) ɛ σ() s for all s T U. The dela derivaive is given by (.2) f () = f(σ()) f() () if f is coninuous a and is righ-scaered. If is righ-dense hen he derivaive is given by (.3) f f(σ()) f(s) f() f(s) () = lim = lim, s s s s provided his limi exiss. This definiion can be generalized o he case where he range of f is any Banach space. A funcion f : T R is said o be righ dense coninuous (rd-coninuous) if i is coninuous a each righ dense poin and here exiss a finie lef limi a all lef dense poins, and f is said o be differeniable if is derivaive exiss. A useful formula is (.4) f σ := f σ = f + f. For a, b T, and a differeniable funcion F, he Cauchy inegral of F = f is defined by b a f() = b a F () = F (b) F (a). We shall make paricular use of he formula b (.5) f() = f(s)(s), a s [a,b) (see []) which holds for isolaed ime scales. a < b

3 We say ha a funcion p : T R is regressive provided + ()p() 0, T. I urns ou ha he se of all regressive funcions on a ime scale T forms an Abelian group (see Theorem 2.7 and Exercise 2.26 in []) under he addiion defined by p q := p + q + pq and he inverse p of he funcion p is defined by p := p + p. We denoe he se of all f : T R which are rd coninuous and regressive by R. If p R, hen we can define he exponenial funcion by ( ) e p (, s) := exp ξ (τ) (p(τ)) τ for T, s T k, where ξ h (z) is he cylinder ransformaion, which is given by { Log(+hz), h 0, ξ h (z) = h z, h = 0. where Log denoes he principal logarihm funcion. Alernaely, for p R one can define he exponenial funcion e p (, 0 ), o be he unique soluion of he IVP x = p()x, x( 0 ) =. We define R + := {f R : + ()f() > 0, T}. We shall be making heavy use of he properies () e 0 (, s), e p (, ) (2) e p (σ(), 0 ) = [ + ()p()]e p (, 0 ) (3) = e e p(, 0 ) p(, 0 ) = e p ( 0, ) (4) e p (, 0 )e q (, 0 ) = e p q (, 0 ) (5) e p (, r)e p (r, s) = e p (, s) where p, q R (see Bohner and Peerson []). Also if p R, hen e p (, s) is real-valued and nonzero on T. If p R +, hen e p (, 0 ) is always posiive. For α R, on he respecive ime scales, he exponenial e α (, 0 ) is R : e α( 0) hz : ( + αh) 0 h q Z : s [ 0, ) [ + (q )αs], > 0 ( {H n } 0 : n+α 0 ) n 0, = n k= k. s 3

4 4 2. General Resuls We will be principally concerned wih he dynamic equaion (2.) y σ p()y = r(), T κ on isolaed ime scales, where p() 0, T κ. In paricular, soluions of he corresponding homogeneous problem u σ p()u = 0 can be used o find explici forms for generalized exponenial funcions on a paricular ime scale. We begin by finding a variaion of consans formula for (2.). Theorem 2.. Variaion of Consans for Firs Order Recurrence Relaions Assume p() 0, T κ. Then he unique soluion o he IVP is given by y σ p()y = r(), y( 0 ) = y 0 y() = e p (, 0 )y 0 + e p 0 (, σ(s)) r(s) (s) s. We shall now give wo proofs, one approaching he problem as a firs order dynamic equaion, and a second following a mehod used in he sudy of difference equaions. Proof. Using formula (.4), we may rewrie he corresponding homogeneous equaion as follows: u σ = p()u u + ()u = p()u ()u = (p() )u u = p() u () and so by he definiion of he generalized exponenial as he soluion of an IVP, we ge ha (2.2) u() = e p (, 0 )u 0 where u 0 = u( 0 ). Using he variaion of consans formula found in Bohner and Peerson [], we find he general soluion o equaion (2.) is (2.3) y() = e p (, 0 )y 0 + e p 0 (, σ(s)) r(s) (s) s where he argumens have been supressed in he subscrips, or alernaely ( ) r(s) (2.4) y() = e p (, 0 ) y (s)p(s)e p (s, 0 ) s.

5 We now derive he variaion of consans formula using a mehod analogous o ha used in he difference equaions case. Firs we prove a lemma. Lemma 2.. The exponenial funcion e p (, 0 ) is given by { e p (, 0 ) = τ [ 0, ) p(τ), 0 τ [, 0 ), <. p(τ) 0 Proof. Since he exponenial e p (, 0 ) is he unique soluion o u σ = p()u wih u( 0 ) =, we noe ha τ [ 0, 0 ) p(τ) = using he convenion ha an empy produc is he ideniy, and for > 0 one may simply ierae he formula u σ = p()u. The case < 0 is similar. Consider he dynamic equaion y σ p()y = r() where p, r C rd and p() 0, T, and le u be a nonzero soluion of he corresponding homogeneous equaion u σ = p()u which by Lemma 2. is τ [ 0, ) p(τ). Le us assume ha y is a soluion o (2.) and now divide (2.) by u σ ()() o ge y σ () u σ ()() p()y() u σ ()() = r() u σ ()(). Hence y σ () u σ ()() p()y() u σ ()() = y σ () y() u σ () u() () = ( ) y() = r() u() u σ ()(). Now, we may inegrae boh sides from 0 o o find ha y() u() y 0 u 0 = 0 r(s) u σ (s)(s) s and hus we have ( ) r(s) y() = u() y 0 + s 0 u σ (s)(s) which, upon noing u() = e p (, 0 )u 0, is exacly (2.3). See Examples 3., 3.2, 3.4, and 3.7 for simple applicaions of he variaion of consans formula. For he case T = Z, we know in he sudy of difference equaions (see [2]) ha he soluion o he problem y( + ) p()y() = r() is (2.5) u() ( y 0 + r() u( + ) where u() is a nonzero soluion o u( + ) = p()u(). However, by he definiion of he generalized exponenial, his funcion u() is a consan muliple of he exponenial e p (, 0 ), and p()e p (, 0 ) = e σ p (, 0 ). Since ) 5

6 6 on Z we have, he Cauchy inegral is exacly an indefinie sum, and hus ( u() y 0 + ) r() = e p (, 0 ) y 0 + r(s) u( + ) e σ p (s, 0 ) r(s) = e p (, 0 ) y 0 + (s)e σ p (s, 0 ) s ( ) r(s) = e p (, 0 ) y 0 + (s)p(s)e p (s, 0 ) s. Theorem 2.2. Facorizaion A soluion o he dynamic equaion u σ = p()u, where p() = p ()p 2 ()... p n () is u() = e p p 2...pn (, 0 ) = e p (, 0 )e p 2 Proof. We noe ha a b = ab for any a, b, and by inducion ha e p p 2...pn (, 0 ) = e p = e p p 2 (, 0 )e p 2 pn (,... 0 ) A basic applicaion is given in Example 3.3. (, 0 )... e pn (, 0 ). (, 0 )... e pn (, 0 ). 2.. Series Soluion. A useful ool for looking a isolaed ime scales is he couning funcion n defined n (, s) := s τ (τ). Noe ha using (.5) he values of his funcion are inegers, where he value of he funcion n couns he number of poins in he half open inerval [s, ) for s and is he negaive of he number of poins in [, s) for < s. On an isolaed ime scale, n (σ k (s), s) = k, wih he convenions σ k () = ρ k () and σ 0 () =. For example, for he ime scale N b, b > 0, n (, s) = b b s and for q Z he couning funcion is given by n (, s) = ln ln s ln q = log q ( s ).

7 The couning funcion provides an enumeraion of he ime scale which can be quie useful in formulas for exponenial funcions, paricularly when considering he produc formula for exponenials given in Lemma 2.. We use his couning funcion in he following heorem. Theorem 2.3. Series Soluion The sum (2.6) n (sup T y(sup T) κ, ) p()p σ ()p σ2 ()... p(sup T κ ) + r σk () p()p σ ()p σ2 ()... p σk () is a soluion o (2.) where sup T = max T R, and if sup T =, hen a soluion is given by he infinie series r σk () (2.7) y() = p()p σ ()p σ2 ()... p σk () {}}{ whenever his series converges, where y σk denoes y( σ(σ(... (σ( ))...))). k imes Proof. Observe ha equaion y σ p()y = r() can be re-expressed in he form y = r()+yσ, and ha y σ = rσ ()+y σσ, and hus y = p() p σ () Coninuing his process, y() = r rσ()+ σ2 ()+ r() + when sup T exiss, and formally we ge p() rσ()+ r() + y() = +y(sup T) p σ2 () p σ () r σ2 ()+ rσ3 ()+ p σ3 () p σ2 () p σ () p(), 7 r σ r+ +y σσ p σ. p if sup T =. Consider firs he case where sup T is finie. The ascending fracion can be expanded ino he sum r() p() + rσ () p()p σ () r(sup T κ ) p()p σ ()... p(sup T κ ) + y(sup T) p()p σ ()... p(sup T κ ). Since here is one summand wih he funcion r() for each poin in he inerval [, sup T), he number of such summands is n (sup T, ) and hus saring our index a zero, hese erms combine as n (sup T, ) r σk n (sup T () κ, ) p()p σ ()p σ2 ()... p σk () = r σk () p()p σ ()p σ2 ()... p σk ().

8 8 For sup T =, formally expanding he ascending coninued fracion yields he infinie series r σk. pp σ p σ2 p σ3... p σk Assume ha his infinie series converges for all T and denoe i by w(). Then r σk () w() = p()p σ ()p σ2 ()p σ3 ()... p σk () r σk () p()w() = r() + p σ ()p σ2 ()p σ3 ()... p σk () k= r σk () p()w() + r() = p σ ()p σ2 ()p σ3 ()... p σk (). Hence we see ha k= p()w() + r() = = and w() is a soluion o (2.). k= = w σ () r σk () p σ ()p σ2 ()p σ3 ()... p σk () r σk+ () p σ ()p σ2 ()p σ3 ()... p σk+ () We now derive some alernae formulas for he soluions given in Theorem 2.3. We noe ha m p σk () = e p (σ m+ (), ) = p σm ()e p (σ m (), ) by Lemma 2., and hus along wih (.5), we see ha n (, ) r σk () σ( ) p σ ()p σ2 ()p σ3 ()... p σk () = r(s) (s)p(s)e p (s, ) s. We may herefore re-express he summaions in our heorem wih he inegrals sup T r(s) e p (, 0 ) (s)p(s)e p (s, 0 ) s

9 9 for sup T finie, and e p (, 0 ) r(s) (s)p(s)e p (s, 0 ) s for sup T =. Thus we find he iniial condiion mus be sup T w 0 = 0 when sup T exiss, else r(s) (s)p(s)e p (s, 0 ) s + y(sup T) e p (sup T, 0 ) w 0 = r σk ( 0 ) p( 0 )p σ ( 0 )p σ2 ( 0 )... p σk ( 0 ) = 0 r(s) (s)p(s)e p (s, 0 ) s. On he ime scale Z wih p() = his leads o a facorial series (see Example 3.4 in [2]). I is also of ineres o consider his series soluion on T = R, where all poins are righ-dense. Thus he equaion y σ p()y = r() becomes simply y p()y = r() and y = r() p(). The series is hen y() = r() p k () = r() p k () = r() p() which is a geomeric series which converges for p() >, exacly as we expeced. 3. Examples In his secion we give several examples following from our previous resuls. The firs wo are direc applicaions of he variaion of consans formula. Example 3.. Solve he dynamic equaion (3.) y σ qy = q on he ime scale T = q N. In his case p() = q, r() = q, and () = (q ). By he variaion of consans formula given in Theorem 2., we

10 0 obain Noe ha Therefore y() = e (q ) (, 0 )y 0 + (q ) = e (, 0 )y 0 + = e = e 0 e (, 0 )y 0 + e (, 0 ) (, 0 )y 0 + e (, 0 ) 0 e (q ) (q ) (q ) (, σ(τ)) τ(q ) τ (, 0 )e ( 0, σ(τ)) τ τ 0 e 0 qe ( 0, σ(τ)) τ τ (τ, 0 ) τ τ. = + (q ) = + q = q. 0 qe (τ, 0 ) τ τ = 0 qτ e (τ, 0 ) τ = [ e (τ, 0 )] 0 = [ e (, 0 ) + ], and we ge ha y() = e = e (, 0 )y 0 + e (, 0 )y 0 + e (, 0 ). = (y 0 + )e (, 0 )[ e (, 0 ) + ] (, 0 ) By Lemma 2., for 0 e (, 0 ) = τ [ 0,) q = qn q n 0 = 0. So y() = C, where C = y 0+ 0, is a general soluion o (3.). Example 3.2. Consider he dynamic equaion (3.2) y σ () ( + ) y = on he so-called ime scale T = {H n } n=0 of Harmonic numbers []. In his case p() = () and r() =. By he variaion of consans formula (+)

11 (Theorem 2.), we obain for 0 y() = e () (+) () (, 0 )y 0 + = e () (+) (, 0 )y 0 + ()(+) = e (+) (, 0 )y 0 + () e () (+) 0 () e () (+) ()(+) 0 0 e (+) () (, σ(τ)) (τ) τ (, σ(τ)) (τ) τ (, σ(τ)) (τ) τ. Bu () = n+, so y() = e (n+2) (n+) (, 0 )y 0 + e (n+2) (n+) (, σ(τ)) 0 (τ) τ = e (, 0 )y 0 + e (, σ(τ)) 0 (τ) τ = e (, 0 )y 0 + (s)e (, σ(s)) (s) s [ 0,) = e (, 0 )y 0 + e (, σ(s)) s [ 0,) = e (, 0 )y 0 + s [ 0,) e (, 0 )e ( 0, σ(s)) = e (, 0 )y 0 + e (, 0 ) e (σ(s), 0 ) s [ 0,) = e (, 0 ) y 0 +. e (σ(s), 0 ) s [ 0,)

12 2 If ( 0 = 0, hen we can use he fac (see page 74 in []) ha e α (, 0 ) = n+α 0 ) n 0 where = n i= so ha i ( ) n + 0 y() = y 0 + ( n 0 n+2 0 ) s [ 0,) n+ 0 = (n + ) y 0 + n + 2 s [ 0,) ( ) n = (n + ) y 0 + k + 2 ( ) n = (n + ) y 0 + k + k= = (n + ) (y 0 + σ() ). Hence a general soluion of equaion (3.2) is y() = (n + ) (y 0 + σ() ). Example 3.3. For an example of Theorem 2.2, consider he homogeneous dynamic equaion (3.3) u σ (q 2 q)u = 0 on he ime scale T = q N 0. In his case p() = (q 2 q) and () = (q ). Bu he original equaion can be wrien in he form u σ q( 2 )u = 0. Hence, we can apply Theorem 2.2. Thus i is sufficien o solve for u () where p () = q and u 2 () where p 2 () = 2. So firs consider u σ qu = 0. Using equaion (2.2), we obain u () = e q (, 0 )u 0 = e q (, 0 )u 0 = e (, 0 )u 0. () (q ) By Example 3., u () = u 0 0. Nex consider u σ 2 (2 )u 2 = 0. As before, by equaion (2.2) we obain Bu by Lemma 2., u 2 () = e ( 2 ) (, 0 )u 0 = e 2 2 (, 0 )u 0. () (q ) u 2 () = e 2 2 (, 0 )u 0 = u 0 (q ) τ [ 0,) (τ 2 ).

13 Thus, by Theorem 2.2, we can wrie he general soluion of equaion (3.3) as u() = u 0 e q 2 q (, 0 ) = u 0 (τ 2 ). (q ) 0 τ [ 0,) Example 3.4. By solving iniial value problems, in some cases we are able o find formulas for exponenial funcions. Consider he nonhomogeneous IVP (3.4) y σ qy = q +, y(0) = 0 on he ime scale T = Z. In his case p() = q, r() = q + and () =. From he variaion of consans formula equaion (2.2), we obain y() = 0 = e q (, 0) = e q (, 0) = e q (, 0) e q (, σ(τ))q τ+ τ e q (σ(τ), 0) qτ+ τ qe q (τ, 0) qτ+ τ q e (q )(τ, 0)q τ+ τ. 3 Noe ha and Hence, (q ) = (q ) + q = q q ( e q (, 0) = + q ) = q q y() = e q (, 0) = e q (, 0) = e q (, 0) = e q (, 0) = ( + q ) = q ( ). q ( ) τ q τ+ τ q q ( ) τ+ q τ+ τ q τ

14 4 is he soluion o he IVP (3.4). In he nex wo examples we consider homogeneous IVP s in order o find a formula for he respecive exponenial funcions. Example 3.5. Consider he IVP (3.5) u σ + u = 0, u() = on he ime scale T = N 2 := { = n 2, n N}. In his case p() = + and () = 2 +. From equaion (2.2), we obain for T = N 2 u() = e + () = e (, ) (, ) = e (, ). 2+ Consider he original equaion u σ = + u. By plugging in he appropriae values of we find u(2 2 ) = + = 2 In general, Hence, Example 3.6. Solve he IVP u(3 2 ) = = 3 u(4 2 ) = = 4 u(5 2 ) = = 5. u() =. e (, ) =. 2+ (3.6) u σ 22 + u = 0, y() =, 2 2

15 on he ime scale T = N 2 = { = n 2 n N}. In his case p() = From equaion (2.2), we obain for T = N 2 u() = e () (, )() = e (, ) (2 2 )() = e 2 (2 2 )() (, ). Bu consider he original equaion u σ () = 22 +u(). By plugging in he 2 2 appropriae values of we find Hence, u(2 2 ) = 2() + 2() () = 3 u(3 2 ) = 2(2) + 2(2) (3) = 5 3 (3) = 5 u(4 2 ) = 2(3) + 2(3) (5) = 7 5 (5) = 7 u(5 2(4) + 2 ) = 2(4) (7) = 9 (7) = 9. 7 e 2 (2 2 ) = 2 2. By varying he ime and ineres rae of an IRA we are able o apply our variaion of consans formula (Theorem 2.) o find ou how much money will be in he IRA afer ime T. Example 3.7. Suppose we open an IRA accoun, iniially a ime =, inves $2,000 and add an addiional $2,000 every ime ineres is compounded. Le T = q N 0 where 0 = q 0 =. The ineres is compounded every = q n years and a a rae of %. How much will we have in he IRA a ime? How much would we have a ime = (.5) 5? Since y() = 0, le y() be he amoun of money afer ime = q n. So y σ () = y() (y() ) y σ () = ( +.0)y() + ( +.0)2000 (3.7) y σ () ( +.0)y() = ( +.0)

16 6 Hence we ge an equaion of he form y σ p()y = r() where p() = +.0 and r() = ( +.0)2000, and we can solve for y() by he variaion of consans formula given in Theorem 2. o obain y() = Since T = q N 0, () = (q ). Thus y() = ( +.0τ)2000 e (+.0) (, σ(τ)) τ. () (τ) = 2000 q ( +.0τ)2000 e.0 (, σ(τ)) τ (q ) (q )τ +.0τ τ e.0 (, σ(τ)) τ. q Bu.0 is a consan and so e.0 (, ) = ρ().0 q q s= [+(q )( )s] = ρ() q s= [+.0s] (see page 74 of []). Hence y() = 2000 q +.0τ τ s [σ(τ),) [ +.0s] τ. Therefore y() = τ [ +.0s] τ q τ s [σ(τ),) = 2000 (i) +.0i [ +.0s] q i i [,) s [σ(i),) = 2000 (q )i +.0i [ +.0s] q i i [,) s [σ(i),) = 2000 ( +.0i) [ +.0s]. i [,) s [σ(i),) So he soluion of he IVP for he IRA problem is ρ() ρ() y() = 2000 ( +.0i) [ +.0s]. i= s=σ(i) If q =.5 and n = 5 we calculae how much money is in our IRA afer

17 = (.5) , so approximaely 6.59 years afer our iniial invesmen, we ge y((.5) 5 ) = 2000 ( +.0i) [ +.0s] i [,(.5) 5 ) which is approximaely $, i [σ(i),(.5) 5 ) Example 3.8. We now consider an example for he Series Soluion Theorem 2.3. On he ime scale T = q Z = {0} {q n n Z}, his series soluion can be useful where p() and r() in (2.) are boh monomials. On his ime scale, we noe ha r σk () = r(q k ) and hus advancing a monomial erm k imes will muliply he erm in he original polynomial r() by q raised o k imes he order of he erm, i.e. (αx β ) σk = α(q k x) β = αx β q kβ. We consider he case p() = α β and r() = A c, where α, β, A, c R. Thus we find he series soluion o y σ α β y = A c for 0 as follows: y() = = = r σk () p()p σ ()p σ2 ()p σ3 ()... p σk () Aq kc c (α β )(αq β β )(αq 2β β )... (αq kβ β ) Aq kc c α k q ( k)β kβ Aq kc c = α k q (k+)k 2 β kβ ( = A c q c α β q (k+) 2 β ) k. Now, we noe ha his series converges by he raio es for all 0 as ( ) k+ q c q k+2 2 lim β α ( β q c k q 2 +k β 2 ) k k = lim k α β q k2 +3k+2 β = lim q c k 2 α β q (k+)β = 0. q c q k+ 2 β α β 7

18 8 Thus for 0 we have as a soluion y() = r() q kc q (k+)k 2 β (α β ) k for any monomial r() on he ime scale q Z, and for = 0, we noe ha 0 is a dense poin in his ime scale and hus y σ (0) = y(0), and y σ () α β y() = r() becomes simply y(0) = r(0). Example 3.9. For he ime scale N b = {n b n N} where b R, b > 0, we find ha for p() a monomial, he corresponding soluion o (2.) is ( ) bβ Γ( (, 0 ) = α n b ) e α β Γ( b 0 ) which is a soluion o u σ () = α β u(), u( 0 ) = as he reader can easily verify. Given his soluion o he homogeneous problem, we can now solve corresponding nonhomogeneous equaions. Example 3.0. Solve he nonhomogeneous problem ( ) bβ y σ α β y = (α β Γ( )α n b ). Γ( b 0 ) Since from he previous problem we have found he exponenial ( ) bβ Γ( (, 0 ) = α n b ) e α β Γ( b 0 ) we may use our variaion of consans formula (2.4) o find ( ) bβ Γ( y() = α n b ) αs β e α β (s, 0 ) y Γ( 0 + b 0 ) 0 (s)αs β e α β (s, 0 ) ( ) s = e α β (, 0 ) y (s) s = e α β (, 0 )(y 0 + n ). This moivaes he following corollary o he variaion of consans heorem: Corollary 3.. The soluion o he IVP y σ p()y = f()e q (, 0 ), y( 0 ) = y 0

19 9 where q R, p() 0, T κ is given by ( ) f(s)e q p (s, 0 ) y() = e p (, 0 ) y 0 + s. p(s)(s) 0 Proof. By Theorem 2., we have he soluion of he given IVP is ( ) f(s)e q (s, 0 ) y() = e p (, 0 ) y (s)p(s)e p (s, 0 ) s, and since e q (, 0 ) e p (, 0 ) = e q(, 0 )e p we have our desired resul. (, 0 ) = e q ( p )(, 0) = e q p (, 0 ) 4. Enumeraions By an enumeraion on an isolaed ime scale, consider he mapping n (, 0 ) : T Z given by he couning funcion, and name each poin of he ime scale wih an index of he value of he couning funcion a ha poin. Thus, our anchor poin 0 says he same, σ() becomes, σ 2 () becomes 2, and so on, again wih he convenion ha negaive exponens indicae use of he backward jump operaor ρ raher han he forward jump operaor σ. For he problem y σ = (n + c)y, c R, enumerae he ime scale and consider he funcion values as he elemens in a sequence, where k = σ k () and y k = y( k ). The soluion o his recurrence relaion is hen y j = { Γ(j+c) y Γ(c) 0, j 0 ( ) j Γ( c) y Γ( c j) 0, j < 0. For c = 0 on Z, he soluion breaks a = 0, bu considering he soluion on he poin 0 we have an alernaing sequence of he reciprocals of facorials, and for 0 we have he facorials. We now consider a generalizaion of our main problem, namely y σk p()y = r(). Viewing his as a recurrence relaion, we noice ha only every k h poin is conneced from any given iniial value; for k = 2, his formula would only relae poins of even index o poins of even index, and poins of odd index o poins of odd index: he soluions on hese wo ses of poins will be independen of each oher.

20 20 Theorem 4.. Recurrence Soluion on Pariioned Time Scale Consider he recurrence relaion (4.) p ()y σa + p 2 ()y σa p n ()y σan + q()y = r(), q() 0 and le g = gcd{a i }, k = max{a i }. Pariion he ime scale T ino he ime scales T n := { n (, 0 ) n(mod g)} for 0 n < g. Then solving his recurrence relaion is equivalen o solving an IVP of order k on each of he g ime scales T n where he funcion is replaced by := + σ + σ σg. Proof. Firs, we noe ha he funcion values y p, y q are relaed by his recurrence relaion iff σ r ( p ) = σ r 2 ( q ) for some r, r 2 which are linear combinaions of he a i. To see his, we noe ha by looking a (4.) wih = p ha y p mus obey a relaionship wih all poins y p+ai, and hen repeaedly applying (4.) yields ha y p is relaed o values y p+c a +c 2 a c na n, where he c i s are ineger consans, including negaive inegers from looking back along he ime scale. Now, if here is some poin q such ha y q+d a +d 2 a d na n = y p+c a +c 2 a c na n hen hese values are relaed hrough (4.). Since he smalles posiive value of a linear combinaion of he a i s is g hen relaed funcion values can be no less han g jumps apar, and since here exiss some linear combinaion of he a i wih sum g hen poins g jumps apar are relaed by our formula. Thus, we mus have g ses of funcion values for y which are no relaed o one anoher by he equaion, and hese independen ses of soluion poins are he values on he ime scales T n. Now, since one use of he jump operaor on T n is equivalen o g uses of he jump operaor on T, he order of he recurrence relaion (4.) is reduced by a facor of g on all ime scales T n. The converse is obvious. Thus we find ha he soluion o y σk p()y = r(), p 0 is y([t]) = y([t n ]) where y Tn = e p (, n ) ( n {0,,2,...,g } y n + n r(s) (s)p(s)e p Example 4.. Solve he sixh order recurrence relaion (4.2) y σ6 y σ3 y = 0 (s, n ) s ).

21 on he ime scale q N 0 wih 0 = and iniial condiions y() =, y(q) = 5, y(q 2 ) = , y(q 3 ) =, y(q 4 ) = 9, y(q 5 ) = Noing ha gcd{3, 6} = 3, we firs pariion q N 0 ino hree ime scales using he couning funcion n (, 0 ) = log q (/ 0 ) = log q (), forming he cells T 0 = q 3N 0, T = q 3N0+, and T 2 = q 3N0+2. On each of hese ime scales we consider he recurrence relaion y σ2 y σ y = 0, which we recognize as he relaion generaing he Fibonacci sequence. Thus, on T 0 wih iniial values and, we have he classic Fibonacci sequence, on T he iniial condiions 5 and 9 will give a Fibonacci ype sequence wih all erms negaive, and on T 2, our iniial condiions y(q 2 ) = , y(q 5 ) = will give Fibonacci sequence values muliplied by one googol. 5. Recurrence Relaions and he Couning Funcion Noe ha for f() 0 for all T ha for (5.) u σ = f σ f u a general soluion o (5.) is u() = cf(). Now, in paricular, if in (5.) we consider f(n ) raher han f() hen we can view he problem as a simple relaion beween funcions of consecuive inegers or erms in a sequence, as f σ (n ) = f(n + ). Now, we may herefore approach problems of he form y σ = p()y + r() by convering p() ino a raional funcion of he form f(n +) f(n ). We noe ha any isolaed ime scale can be enumeraed by use of he couning funcion s n (, 0 ) = (s). Using his approach, we find ha many of our examples can be solved an easier way: u σ = qu = qn+ u = u = cq n q n wih q n = 0 on q Z,, 0 0, as Example 3., wih () = n + on {H n} 0 0 u σ = () σ () u = u = c, as in Example 3.2, u σ = 2 + n + n u = u = c(n + ) 2

22 22 and in paricular, on N 2 where n (, ) = u σ = + u = u = c as in Example 3.5, and on N 2, wih n (, ) = 2 we see ha u σ = u = (2n (, ) + ) σ 2n (, ) + u = u = c(2n (, ) + ) as in Example 3.6. This seems o sugges ha i may be of value o consider recurrence relaions on ime scales as sequences. The closure of he image of he inerior of any isolaed inerval of a ime scale under a map which is increasing and injecive will again be a ime scale, as he image of he open inerval beween poins will be an open inerval beween poins in he image. Thus any porion of a sequence can be ransformed o be on any isolaed inerval of a ime scale, so long as he cardinaliies mach. The couning funcion always provides a map back o he inegers, which seems o sugges ha undersanding he ransformaions beween ime scales depends on sudying he lenghs of he images of inervals. References [] M. Bohner and A. Peerson, Dynamic Equaions on Time Scales: An Inroducion wih Applicaions, Birkhäuser, Boson, 200. [2] W. Kelley and A. Peerson, Difference Equaions: An Inroducion wih Applicaions, 2nd Ediion, Harcour Academic Press, San Diego, 200. [3] V. Spedding, Taming Naure s Numbers, New Scienis, 9 July 2003, 28 3.