Linear recurrence relations with constant coefficients

Transcription

1 Liea ecuece elatios with costat coefficiets Recall that a liea ecuece elatio with costat coefficiets c 1, c 2,, c k c k of degee k ad with cotol tem F has the fom a c 1 a 1 + c 2 a c k a k + F k It follows fom the geeal ecusio theoem that fo evey stig of iitial values a, a 1,, a k 1 thee is exactly oe sequece {a } that satisfies the above ecuece elatio ad matches the give iitial coditios Cosequetly, if o iitial coditios ae imposed, thee will always be a ifiite set of solutios The theoems below will show how to fid all solutios to such a ecuece elatio povided the cotol tem is of the special fom F qs fo some polyomial q ad some costat s Moe geeal cotol tems ae discussed i advaced text books o diffeece equatios, such as A Itoductio to Diffeece Equatios by Sabe N Elaydi Spige Velag 1996 We fist coside the case whee F fo all The ecuece elatio is the called homogeeous Hee is the solutio fomula: Theoem 1 [The homogeeous case: fidig all solutios] Let c 1, c 2,, c k be costats with c k Coside the homogeeous liea ecuece elatio a c 1 a 1 + c 2 a c k a k k Let λ 1, λ 2,, λ t be the distict solutios to the chaacteistic equatio λ k c 1 λ k 1 + c 2 λ k c k The a sequece {a } is a solutio to if ad oly if it is of the fom a p 1 λ 1 + p 2 λ p t λ t with polyomials p 1, p 2,, p t such that the degee of p i is less tha the multiplicity of λ i fo all 1 i t Note: The coefficiets of the polyomials p i ae detemied by a, a 1, a k 1

2 Example The chaacteistic equatio to the ecuece elatio is give by which is equivalet to a a a a a a 5 λ 5 λ λ λ λ , 3λ 2 3 2λ Hece, λ 1 2/3 with multiplicity 3 ad λ 2 1/2 with multiplicity 2 ad the geeic solutio to is give by 2 a α + α 1 + α α 3 + α If oe is give iitial coditios, like a 1, a 1 2, a 2 3, a 3, ad a 4 6, the α,, α 4 ca be detemied by solvig the appopiate system of five liea equatios i five ukows esultig fom takig, 1,, 4 i the fomula defiig a Poof [of Theoem 1] Thee ae may diffeet ways to establish this esult The poof that we will sketch hee uses some kowledge of liea algeba We will illustate the geeal poof at a example of degee 5 Fist suppose that {a } is a solutio to a c 1 a 1 + c 2 a c 5 a 5 5 We ca ewite this i matix fom as follows: a a 1 a 2 a 3 a 4 c 1 c 2 c 3 c 4 c } {{ } A a 1 a 2 a 3 a 4 a 5 So, the matix A effects a shift i ay cosecutive stig of legth 5 take fom the ifiite sequece {a } This allows us to compute all values of {a } fom a, a 1,, a 4 by iteatig the matix A: 1 a +4 a +3 a +2 a +1 a A A A A }{{} -times a 4 a 3 a 2 a 1 a

3 Also otice that the matix A is ivetible, sice c 5 We theefoe ca ceate a histoy a 1, a 2, a 3, fo ou sequece {a } that follows the same ecuece elatio by simply applyig A 1 athe tha A Howeve, the fom that the matix A is cuetly i does ot allow fo coveiet iteatio A chage of basis will help us out hee Recall that if B is ay 5 5 ivetible matix, the the liea tasfomatio epeseted by A i the stadad basis, will be epeseted by J B 1 AB i the basis made up fom the colum vectos of B To fid a good choice fo B, we compute the eigevalues of the matix A by expadig acoss the fist ow: det A λi det c 1 λ c 2 c 3 c 4 c 5 1 λ 1 λ 1 λ 1 λ c 1 λ λ 4 c 2 λ 3 + c 3 λ 2 c 4 λ + c λ 5 c 1 λ 4 c 2 λ 3 c 3 λ 2 c 4 λ c 5 Obseve that the chaacteistic equatio of the matix A is exactly the chaacteistic equatio of the ecuece elatio Hece the ame! To stay withi a specific example, say this polyomial factos as follows: 1 5 λ 5 c 1 λ 4 c 2 λ 3 c 3 λ 2 c 4 λ c λ λ 1 2 λ λ 2 3 The Joda Nomal Fom Theoem ow implies that we ca choose B such that J has the fom λ 1 1 λ 1 J λ 2 1 λ 2 1 λ 2 I geeal, the Joda caoical fom has squae blocks o its diagoal ad zeos elsewhee Each squae block has oe eigevalue o its mai diagoal ad the umbe 1 o its subdiagoal ad is of size equal to the multiplicity of that eigevalue With a bief iductio you ca veify that the matix J ca be iteated quite easily 5: J J J J }{{} -times 1 λ 1 λ 1 λ 1 1 λ 2 λ 1 2 λ 1 2 λ 2 2 λ λ 2

4 Sice A BJB 1 that helps us iteate the matix A: A A A A A BJB 1 BJB 1 BJB 1 BJB 1 BJ B 1 Obseve that the expessio f ,! whe viewed as a polyomial i the vaiable, has degee Also, λ i λ i λ i Theefoe, if we substitute BJ B 1 fo A i 1 ad focus o its last ow we get 3 a α + α 1 λ 1 + α 2 + α 3 + α 4 2 λ 2 5, with some coefficiets α,, α 4 O the face of it, 3 holds oly fo 5 Howeve, we could have used the iitial coditios a 1, a 2,, a 5 i all of the above, athe tha a 4, a 3,, a, ad still aive at a fomula that is of the fom 3 Because such a shift i pespective substitutes + 5 fo i 3, which the tus ito aothe fomula of the same type Hece, fomula 3 actually holds fo all This shows that evey solutio is of the poposed fom Next, we pove that evey sequece of this fom is a solutio As pat of the above discussio we see that the five iitial coditios a,, a 4 of ay solutio {a } to detemie the five coefficiets α,, α 4 i the fomula fo {a }, ad they do so by a liea tasfomatio Moeove, this tasfomatio is ijective, because the α s detemie the etie sequece {a } Now, evey liea tasfomatio fom IR 5 to IR 5 which is ijective, must also be sujective Cosequetly, if we ae give ay umbes α,, α 4, whatsoeve, we ca fid umbes a,, a 4 by the ivese of this tasfomatio, ad the poduce a ifiite sequece {a } by applyig the ecuece ule whose fomula will be a α + α 1 λ 1 + α 2 + α 3 + α 4 2 λ 2 I shot, evey sequece of this fom is a solutio to

5 Next, we will examie how to fid oe paticula solutio to a o-homogeeous ecuece elatio of the specific type F qs : Theoem 2 [Fidig oe paticula solutio] Let costats c 1, c 2,, c k c k be give, alog with a costat s ad a polyomial q The the ecuece elatio a c 1 a 1 + c 2 a c k a k + qs k amog othe solutios always has a solutio of the fom b m ps, whee m is the multiplicity of s i the chaacteistic equatio λ k c 1 λ k 1 + c 2 λ k c k fo the coespodig homogeeous equatio ad p is a polyomial whose degee is less tha o equal to the degee of q Note: This paticula solutio caot accommodate abitay iitial coditios So, it might ot be the solutio you wee lookig fo I this case, you eed to fid all solutios usig Theoem 3 below ad pick the oe you actually wated Example Coside a 11a 1 39a a Sice the chaacteistic equatio to the coespodig homogeeous poblem is λ 3 11λ 2 39λ + 45, which is equivalet to λ 3 2 λ 5, we ca take b 2 β + β 1 + β β The coect coefficiets β,, β 3 ca be detemied by substitutig {b } back ito the oigial o-homogeeous equatio: b 11b 1 39b b This will poduce oe amog ifiitely may solutios To fid them all, icludig the oe you wated, use Theoem 3

6 Poof [of Theoem 2] Deote the degee i the vaiable of the polyomial q by d The fo evey, the fuctio fx q x is a polyomial of degee d i the vaiable x So, by the execise below, d+1 k d + 1 k Multiplyig though with s we see that d+1 k d + 1 k 1 k q k fo all }{{} fk 1 k s k }{{} s k k q ks } {{ } F k fo all But that meas that the sequece {F } {qs } satisfies a homogeeous liea ecuece elatio whose chaacteistic equatio is d+1 k d + 1 k s k λ d+1 k λ s d+1 Now, let {a } be ay solutio to the o-homogeeous ecuece elatio We claim that the sequece {a } also satisfies a homogeeous liea ecuece elatio whose chaacteistic equatio is λ k c 1 λ k 1 c 2 λ k 2 c k λ s d+1 I ode to keep the fomalism to a miimum, we will illustate this with a example that coveys the geeal idea: Suppose ou oigial o-homogeeous ecuece elatio eads a 7a s That is, k 1, c 1 7, d 1, q 3 + 2, ad we ae claimig that {a } satisfies a homogeeous liea ecuece elatio with chaacteistic equatio λ 7λ s 2 The above says that the sequece {F } satisfies a ecuece elatio coespodig to the chaacteistic equatio λ 2 2sλ + s 2 λ s 2, amely F 2sF 1 + s 2 F 2

7 The a 7a 1 F 2sa 1 7a 2 2sF 1 s 2 a 2 7a 3 s 2 F 2 a 7 + 2sa 1 +s 2 7a 2 7a 3, which has chaacteistic equatio λ 7λ s 2 This is best see by substitutig λ i fo a i ow by ow ad factoig out λ 3 λ 7 Hece the claim The we ca apply Theoem 1 to fid a fomula fo {a } We get a α 7 + α 1 + α 2 s if s 7 ad m, O the othe had, a α + α 1 + α 2 2 s if s 7 ad m 1 d α 7 solves the homogeeous pat of ou ecuece, by Theoem 1 Hece, by Theoem 3 below, b a d is agai a solutio to Notice that i both cases, this solutio {b } has the fom m ps, with degee p equal to the degee of q amely 1 The same agumet holds fo ay o-homogeeous liea ecuece elatio of the fom Theoem 3 [The geeal fomula] Let costats c 1, c 2,, c k c k ad a cotol sequece F : IN IR be give Suppose you have foud oe paticula solutio {b } to the o-homogeeous ecuece elatio a c 1 a 1 + c 2 a c k a k + F IN The the geeic solutio {a } to is of the fom a d + b, whee {d } is the geeic solutio to the coespodig homogeeous ecuece elatio a c 1 a 1 + c 2 a c k a k IN

8 Poof If a is of the stated fom, the it satisfies : a d + b c 1 d 1 + c 2 d c k d k + c1 b 1 + c 2 b c k b k + F c 1 d 1 + b 1 + c 2 d 2 + b c k d k + b k + F c 1 a 1 + c 2 a c k a k + F Covesely, if {a } is ay solutio to the the sequece {d } defied by d a b satisfies the homogeeous ecuece elatio : d a b c 1 a 1 + c 2 a c k a k + F c 1 b 1 + c 2 b c k b k + F c 1 a 1 b 1 + c 2 a 2 b c k a k b k c 1 d 1 + c 2 d c k d k Hece, a d + b has the poposed fom

9 Execise a Show that 1 c fo all > ad all costats c [Simply facto out c] b Show that 1 1 fo all 1 [This is the Chaipeso Idetity] c Show that fo evey polyomial fx of degee d we have wheeve d < 1 f, [Use iductio o d Pat a seves as the basic step d Fo the iductive step, assume that d 1 ad that the statemet is tue fo all polyomials of degee less tha d To show that the statemet also holds fo polyomials of degee d, you oly eed to veify that 1 d fo all d < Make use of Pat b ad the iductive hypothesis: 1 d 1 d d d d 1 ]