Comparable solutions to nth-order linear difference equations

MATHEMATICAL COMPUTER MODELLING Mathematical and Computer Modelling 32 (2000) 549-564

PERGAMON

www.elsevier.nl/locate/mcm

Comparable Solutions to nth-Order Linear Difference Equations R. KRUEGER Department of Mathematical Sciences, Coe College Cedar Rapids, IA 52402, U.S.A. [email protected] Abstract-We

consider the nth-order linear difference equation

for t E [o, 61 where q(t) is a real-valued function defined on [a, b]. We define the (formal) adjoint operator 1; of 1, by

l;%(t) = L:z(t) + (-l)nq(t)z

(t + I;\)

1

for t E [a, b]. We compare boundary value solutions of Ing(t) = 0 to similar solutions of the adjoint equation l;z(t) = 0. @ 2000 Eisevier Science Ltd. All rights reserved. Keywords-Difference

equation, Generalized zero, Adjoint, Polya factorization.

1. INTRODUCTION Using [zJ to denote

greater

than

the greatest

integer

or equal to z, we consider

less than or equal to z and ]z] to denote the nth -order linear difference equation

blv(t> = Ly(t) + dt)Y

(t+ I;]) =0,

for t in the discrete

(1)

interval [a, b] := {a, a + 1,. . I , b} where q(t) is a real-valued on [a, b]. We define the nth- order linear difference operator L, by ~&y(t) where pi(t)

are real-valued

:= y(t + n) + pl(t)y(t functions

the least integer

function

defined

+ 12 - 1) + . . * + h(t>y(t),

(2)

on [a, b + n] with

(3) We start with

a few definitions.

Given

that

y(t) is a real-valued

function

on [a, b + n], we say

to = a is a generalized zero for y(t) if y(a) = 0 and that to > a is a generalized zero for y(t) if - k)y(to) > 0 either y(to) = 0 or there exists an integer k, 1 5 k 5 t 0 - a, such that (-l)“y(tc and if k > 1, y(to - k + 1) = -. * = y(to - 1) = 0. The difference equation Zny(t) = 0 is said to be 0895-7177/00/$ - see front matter @ 2000 Elsevier Science Ltd. Ail rights reserved. PII: SO895-7177(00)00152-7

Typeset by dn/ts-Tl$

550

R. KRUEGER

disconjugate on [a, b -t n] if no nontrivial solution has n or more generalized zeros (or real zeros) in [a, b + n]. For this paper, we will consider only the real zeros of y(t). An open question in this area is whether or not the following results hold for generalized zeros. In 1978, Hartman [l] was interested in the nth -order linear difference equation &y(t) = 0. This was the discrete version of the nth-order differential equation studied by Polya in [2]. In this important paper, he proved that for a fixed n E IV, if the difference equation &y(t) = 0 with (3) is disconjugate on [a, b + n], then there are solutions VI(~), . . . , yn(t) such that the Wronskian (Casoratian) defined by YN

Yl(t) Yl(t + 1)

yz(t

+

*** 1)

Y/c@I yk (t + 1)

-a*

*. yl(t + k - 1) on [a,b+n-k+l] factorization of L,y

for

yz(t + k - 1)

..s

> 0,

yk(t + k - 1)

< k I n, and for any y(t) defined on [a, b + n], we obtain the Polya

b/(t) = pntt)A tpn-l(t>A t+.. A bo(Mt)) +- ,)>I

(5)

for t E [a, b], where pa(t) =

Pi(t) = Pratt)=

-L >0, Yltt>

t E [a,b+n],

w(t) w(t + 1) W-1(t

+ l)W+1@)

%(t + 1) w,_l(t

-I-

1) ’ O,

>

o

’

t

[a,b + n -

E

i],

1 2 i I n - 1,

[t&b].

t E

This was an important result because it gave a new form for disconjugate nth-order equations. In addition, this factorization suggests that many of the results for a disconjugate nth-order linear difference equation are the same as for the simple nth -order difference equation Any(t) = 0. In this paper, we will assume that &y(t) = 0 is disconjugate on [a, b + n]. We would like to note that even though Lny(t) = 0 is disconjugate, I&t) = 0 may or may not be disconjugate. For some of our main results, it will be necessary that Imy = 0 is not disconjugate. We will define the (formal) adjoint for the operators L, and 1,. Our definition of the adjoint will depend upon the choice of the Polya factorization. For this reason, we fix a Polya factorization for L,. Note that we could find a Trench factorization [3,4] on [a, co) and then restrict our domain to [a, b]. Our objective in this work is to compare solutions of Zny(t) = 0 with what we will define to be a (n - k, k)-pair of zeros to solutions of the adjoint equation lzz(t)= 0 with a (k, n - k)-pair of zeros. Much of this work is a generality or extension of the work done in [5-g]. It is convenient to extend the domain of definitions of the pi(t) by Pi(t) := Pi@),

for t < a and for 0 I i I n and pi(t) := pi@ + n - i),

for t 2 b + n - i and for 0 5 i I n. DEFINITION. The (formal) adjodnt, L;,

of L, for any function z(t) defined on [a, b + n] is defined

by

Lb@)= PO(tfor t E [a, 61.

It1f n)A (PI(t - IF1+n -

1) A - - - A (p” (t - [zl)

z(t)) . . .) ,

(6)

Comparable

Solutions

551

DEFINITION. The (formaZ) adjoint, l;, of 1, for any function z(t) defined on [a, b + n] is defined by I;+)

= L;z(t)

+ (-l)“q(t)z

(t + [;I)

)

(7)

for t E [a, b].

DEFINITION. The quasidi;tferences, for 0 5 j 5 n, for any functions y and z defined on [a, b + n] are defined by

Ljytt) = pj@>A(~~-ltW. . . A (pott)~l(t)).. .I,

(8)

(t - Is1+j)A (P~_.~+~ (t - [zl x A...A (pn(t - [;I) z(t)). . .) ,

L;Z(t)= h-j

fort6

[a,b+n-

+ j _ 1)

(9)

j].

DEFINITION. Tlie Lagrange Bra&et {z; y} of the functions 2 and y defined on [a, b + n] is defined for n even by n-1

{z(t); y(t)} = C(-l)'L-i-1Y

(t + [;J)@(t+

p-;-‘i>,

(10)

i=O and for n odd by n-1

{%(t);y(t)} =

C(-l)'L-i-IY

( [;j) LZ%(t+p$J),

(11)

t +

i=o fort E [a,b+

11.

THEOREM 1.1. LAGRANGE IDENTITY. If y and z are defined on [a, b + n], then E

(t +

[;I)

I,y(t) + (-I)++/

(t + [;I)

$2(t)

= A {4t);y(t))

7

(12)

for t E [a,b].

The proof of the previous theorem is a complicated repetition of the product rules for difference equations and other standard procedures used in finding an adjoint. I’v’oticethat in the n even case, if we summed both sides of (12) from a to b, we would obtain the adjoint if the Lagrange bracket evaluated at b + 1 and at a were zero. Thus, since we cannot always guarantee these criterion, we refer to our adjoint as a (formal) adjoint. The following definition will generate a linearly independent set of solutions for &(t) = 0 and fzz(t)= 0. DEFINITION. For 0 5 j 5 n - 1, let yj(t, s) be the solution of the “initial value problem”

(13)

and let Zj(t, s) be the solution of the “initial value problem”

when n is even

l;%(t) = 0, Lrtj

Oliln-1, (,+ L”-l-‘J,S)/tz8=6ij.

(14)

552

Ft.

KRUEGER

and when n is odd

(15) It is important to know the signs of the functions Y;(t, s) and zj(t, s) at the points before and after the zeros. The following remarks will classify those values. REMARK 1.2. For each fixed s, the function YP(t,s), 0 5 p I n - 1, is a solution of I,Y = 0 with p consecutive zeros starting at t = s -I- [(n - p)/2J. REMARK 1.3. If n - p is even, then n-p-2

(

Yp s+

9s

2 n-p-2

YP s+ (

>O,

7.3 co, )

n+p -,s 2

,(s+

>

) >o,

2

when n is even, when n is odd,

(16) and

(17)

for all n.

(16)

CO,

when n is even,

(19)

>O,

when n is odd,

REMARK 1.4. If n - p is odd, then YP

Yp

(

s+

(

s+

(

n-p-3

9s

2 n-p-3

9s

2 nfp-1

>

>

and

(20)

for all n. >O, (21) > REMARK 1.5. For each fixed s, the function zP(t,s) is a solution of lzz = 0 with p consecutive zeros starting at YP s+

9s

2

2n-p

t=s+ I

)

when n is even,

(22)

when n is odd.

(23)

when n is even,

(24)

J

and

REMARK 1.6. If n - p is even, then

when n is odd,

and

(25)

for all n.

(26)

when n is even,

(27)

when n is odd,

(26)

REMARK 1.7. If n - p is odd, then n-p-3

2 n-p-l 2 n+p-1 2

when n is even,

n+p+l 2

when n is odd.

and

(29) (30)

ComparableSolutions

553

The following is an important relationship between the solutions {yj(t, tl)}?:: the solutions {Zj (t, tz)}Tgi

of l,y = 0 and

of the adjoint equation 1i.z = 0.

THEOREM 1.8. For 0 2 i,j 5 n - 1, when n is even,

(s+[;J,t- [‘“-;-‘J),

S) = (-l)i+jL~_j_l%n_i_l

Li?Jj(t,

(31)

and when n is odd,

for all t and s. PROOF. Recall the Lagrange Identity (12), 2

(t + [;I)

hdt) +

(t+ [;I) Gdt)

C-l)"-'Y

= A {dt); y(t)),

for t E [a,b]. Fix tr and tz. Fix i,j for 0 I i, j 2 n - 1. Since yj(t, tl) is a solution to (13) and z+.i_l(t, t2) is a solution to (14) when n is even and (15) when n is odd, we get

A {.GA--I (4 tz>; yj (t, tl)} = 0, for t E (a, b]. Hence, there is a constant c E R such that {&-i-l

(tj t2) i Yj (t, tl)}

=

G

for t E [a, b + 11. Specifically, {&-i-l

(t

l,t2)

;Yj (tl,tl)}

=

(h-i-1

(t29t2)

;Yj

(33)

(t2,tl)I*

CASE 1. n even. By (lo), n-1

RHS (33) = C(-l)kL,--k-_lyj k=O

Using the initial conditions of z,_i_l(t,t2),

we get only the term when k = n - i - 1,

RHS (33) = (-l)n-i-lL,-(,_i-l)-lyj

( t2 + i”-;-‘i

A),

which simplifies to RHS (33) = (-l)‘+iLiyj

( tg + I”-;-‘]

Jl).

Similarly, using (lo), we get n-l

LHS (33) = ~(-l)k&_&ryj

(

tr +

k=O

Using the initial conditions of yj(t, tl), we only get the term when k = n - j - 1, LHS (33) = (-l)n-j-lL;_j_rz,_i_l

R. KRUEGER

554

Hence, when n is even, we get that (-l)n-i-%iyj

( t2 + 1” -;-‘1

,tl> = (-l)n-j-lL~_j_lzn_i_l

(t, + [;J ,b)

which gives the desired result where t = t2 + [(n - i - l)/ZJ and s = tl. arbitrary, we get the desired result (31).

)

Since tl and t2 are

CASE 2, n odd. We proceed similarly by (11) to get

Using the initial conditions of z,++l(t,

tz), we get only the term when k = n - i i 1,

RI-IS (33) = (-l)n-%#j

( t2 + 1y-y

A).

Similarly, using (ll), we get n-l

LHS (33) = c(-l)“L,+~yj k=O

Using the initial conditions of yj(t, tl), we only get the term when k = n - j - 1, LHS (33) = (-l)n-+L;_j_Iz,+i_l

(t*+[+A).

Hence, when n is odd, we get that (-l)“-i-‘Liyj

( t 2 + yJ,tl)

= (-l)n-j-lL~_j_lZn_i_l

(1t +

which gives the desired result where t = t2 -t- [(n - i - 1)/2] and s = tl. arbitrary, we get the desired result (32).

2. PRELIMINARY

p$+2)>

Since tl and t2 are I

RESULTS

DEFINITION.Let p and q be positive integers, then we say that a nontrivial solution y(t) of Zny(t) = 0 has a (p,q)-pair of zems at tl and tz if y(tl f i) = 0 for 0 5 i 5 p - 1, y(tz+ i) = 0 forO
is a linearly independent

PROOF. By (13), we know, for each fixed tl, yj(t, tl) is a solution to I,,y(t) = 0 for 0 5 j 5 n- 1. c,+l are constants such that Assumem,q,..., coy0 (t, t1) + ClYl @,h>

for all t E [a,b+n].

+ . * * + cn-lYn-1

@,h)

= 0,

(34)

ComparableSolutions

555

CASE 1. n even. By (16) in Remark 1.3, we have yo(tl+(n_2)/2,tl) # 0. Using Remark 1.2, yl(t, tl) has a zero at tl+ [(n- 1)/2J. Since [(n - 1)/2] = (n - 2)/2 for n even, y~(t~ + (n - 2)/2, tl) = 0. Similarly, y~(t, tl) has two consecutive zeros starting at tl + L(n - 2)/2]. Thus, yz(tl + (n - 2)/2, tl) = 0. Continuing this process, we see by Remark 1.2 that yi(t1-t (n - 2)/2,tl) = 0 for i = 1,. . . , n - 1. Hence, letting t = tl + (n - 2)/2 in (34), we get that ~0 = 0. Now since n - 1 is odd, we use (21) in Remark 1.4 to see that yl(tl +n/2,tl) # 0. Using the same process as above, by Remark 1.2, we see yi(tl +n/2,

tl) = 0 for i = 2,. . . , n - 1. Hence, letting t = tl + n/2 in (34), we get that cl = 0.

Continuing in this fashion, we get ci = 0 for 0 5 i < n - 1. Therefore, we have {yj((t, tl)}~~~

is

a linearly independent set of solutions for Iny(t) = 0 on [a, b + n] for n even. CASE 2. n odd.

By (21) in Remark 1.4, we have y,~(tl+ (n- 1)/2, tl) # 0. Using Remark 1.2, yl(t, tl) has a zero at tl -t [(n - 1)/2j I Thus, yl(tl

= 0. Similarly, y~z(t,TV) has two consecutive zeros starting at tl+[(n-2)/2). Since \(n-2)/2J +I = (n-1)/2 for n odd, yz(tl+(n-1)/2,tl) = 0. Continuing this process, we see by Remark 1.2 that yi(t1-t (n - 1)/2, tl) = 0 for i = 1, . . . , n - 1. Hence, letting t = tl + (n- 1)/2 in (34), we get that co = 0. Now since n- 1 is even, we use (17) in Remark 1.3 to see that yl(tl +(n-3)/2, tl) # 0. Using the same process as above, by Remark 1.2, we see yi(tl + (n - 3)/2, tl) = 0 for i = 2,. . . , n - 1. Hence, letting t = tl f (n - 3)/2 in (34), we get that cl = 0. Continuing in this fashion, we get ci = 0 for 0 < i 2 n - 1. Therefore, we have {y3 (t, tl)}y;t is a linearly independent set of solutions for Iny(t) = 0 on [a, b + n] for n odd. 1 LEMMA 2.2.

+ (n - 1)/2,tl)

There is a nontrivial solution y(t) of Iny(t) = 0 with n-k

consecutive zeros starting

at tl + [k/2] 8 ?dt)

=

%-l&Z-l

(tttl) +

cn-251~3-2

(6

tl) + ‘. . + Cn-t;Ywk (t, tl) ,

(35)

where the G, n - k < i 5 n - 1, are constants not all equal to zero. PROOF. (+)

By Lemma 2.1, any solution y(t) of Iny(t) = 0 has the form y(t) = coy0 (t, h) + cry1 (t, h) + *. . + Cn-lYn-1

(6 t1).

(36)

CASE 1. n even. Note that by (16) in Remark 1.3, yo(tl + (n - 2)/2,tl) # 0 and by Remark 1.2, y,i(tl + (n - 2)/2, tl) = 0 for i = 1,. . . , n - 1. Thus, by letting t = tl + (n - 2)/2 in (36), we get that Q = 0. Now, by (21) in Remark 1.4, yl(tl + n/2, tl) # 0 and by Remark 1.2, yi(tl + n/2, tl) = 0 for i = 2,. . . , TZ- 1. Thus, by letting t = tl + n/2 in (36), we get cl = 0. Continuing in this fashion, we get that Q = cl = . . . = cn-_lc-2 = 0. Finally, to show that &_k_l = 0, we consider the following two cases (noting that n - (n - k - 1) = k -t 1). SUBCASE 1.1. k + 1 even. By (16) in Remark 1.3, y+k_l(tl+(k-1)/2, tl)#O. But by Remark 1.2, yi(tl+(k-1)/2,tl)=O for i = n - k, . . . , n - 1. Hence, setting t = tl -t (k - 1)/2 in (36), we get the desired result c,+l~_~ = 0 for k + 1 even. SUBCASE 1.2. k + 1 odd. By (21) in Remark 1.4, we see that yn_!+l(tl + (2n - k - 2)/2, tl) # 0. But by Remark 1.2, yi(tl + (2n - k - 2)/2, tl) = 0 for i = n - k,. . . , n - 1. Again, setting t = tl + (2n - k -- 2)/2 in (36), we get the desired result f?,+k__l = 0 for k + 1 odd. Hence, when n is even, we get that (35) holds. CASE 2. n odd.

The proof in this case is similar to the proof of Case 1 and will be omitted. (+) Assume y(t) is of the form (35) where the ci, n-k < i <_ n- 1, are constants not all equal to zero. Since y(t) is a nontrivial linear combination of linearly independent solutions of Zny(t) = 0,

R.

656

KRUEGER

then y(t) is a nontrivial solution of Inu(t) = 0. By Remark 1.2, yi(t,tr) has i consecutive zeros starting at t = tr + [(n - i)/2J. So when i = n-k, y,+k(t, tl) has n -k consecutive zeros starting has n - k + 1 consecutive zeros at t = tr + [k/ZJ as desired. When i = n - k + 1, y,+k+r(t,tr) starting at t = tl + [(k - 1)/2J. S o we have one more zero either starting at the same place if k is odd or one place back if k is even. Thus, they have the same n - k zeros in common starting at t = tl + Lk/Zj. Similarly, when i = n - k + 2, y,.+k+a(t, tl) has n - k + 2 consecutive zeros starting at t = tl + I( k - 2)/2J. Thus, we have two more zeros starting one place back, so they have the same n - k zeros in common. Therefore, at each step, we move a half of a step back but add a zero. So we can continue thii process to see that yi(t,tr) for n - k 5 i 5 n - 1 has n - k consecutive zeros starting at t = tr + [k/2J. Therefore, since y(t) is a linear combination of yi(t,tr) for n - k 5 i 5 n - 1, we have that y(t) has n - k consecutive zeros starting at t = tr + lk/2J. I DEFINITION. For the functions zcr(t), Wronskian wk(t)

. . . , Uk(t)

defined on [a, b + n], we define the generalized

by wk(t)

[Ul(t>,

= wk

Jwl

. * * ,Uk(t)]

L&!(t) LluZ(t)

(t)

J%w(t)

=

*** -’* ‘.

LOUk (t> LlUk(t)

, Lk-lUk(t)

fortE[a,b+n-k+l],l_
(t, ‘h) = wk

[&a-l

(6 h)

, * * . , !/n-k

(6 tl)]

*

THEOREM 2.3. There is a nontrivial solution of Zny(t) = 0 with a (n - k, k)-pair of zeros at tl and t2, (tz > tl + n - k) if? wk(tz.tl-

[;j)

=o.

PROOF. Assume there exists a nontrivial solution y(t) with a (n-k, k)-pair of zeros at tl and t2. Since y(t) is a nontrivial solution of l,y = 0 that has n - k consecutive zeros starting at tl, we have, by Lemma 2, y(t) =

Ga-lY,-1

( t, t1 -

I;])

+.*.

where ~-1,. . . , c+_l~ are not all zero. Since y(t) has k consecutive zeros starting at %-l&z-I

(t,,tl

-

[;I)

+‘%-k’y,-k

(,,t,

-

L;j),

t2,

+“‘+C,-k%z-k

(t2,tl

-

I;])

=o,

(37) (38)

cn--1Yn-I

tz + k - 1, tl -

(39)

;

If we multiply (37) by PO(h), by (S), we have c?a-I’%&%-1 (,,,tl

-

I;])

+‘*‘+Cn-kLOh-.k

(t2,tl

-

[;I)

=o.

Comparable

Solutions

If we multiply (38) by ~o(tz + l), (37) by -me, pi(&),

557

take their sum, and multiply the sum by

by (8), we have

cn-lLIYn--l

(t,,t,- [;J)+...+Cn-kLIYn_k (t,,tl- Lsl> =0.

Continuing the process, it follows that

G%-lLoYn-l

(t2,t1 - 19 +...+cn-kloyn_k (,,,t,-[;J)=o,

~-l~l~~-l(tz,tl-~~J)+...+~_xLl~,*(t2,tl-~~J

Since c,_i,

. . . , c,_k

are not all zero, we get that

The converse is essentially reversing the above proof. LEMMA

2.4.

independent PROOF.

l;z(t)

For each fixed tar the collection of functions set of solutions for Zz.z(t) = 0 on [a, b + n].

I {zj(t,

[j/2])}:;:

tz -

By (14) and (15), we know, for each fixed t2, that Zj(t,tZ - lj/2]) = 0 for 0 < j < n - 1. Assume do, dl, . . . , d,_r are constants such that

d,,zo (t, tz) + dlzl

(t, tz) + dm

(t, tz + 1) +. . a+ dn-lz,-1

(,.t,

-

p$])

is a linearly is a solution to

= 0,

(40)

for all t E [a, b + n]. CASE 1. n even. By (24) in Remark 1.6, zo(t2 + (n - 2)/2, t2) # 0 and by (22) in Remark 1.5, zi(t2 + (n -. 2)/2, tg - [i/21) = 0 for i = 1,. . . , n - 1. Hence, letting t = t2 + (n - 2)/2 in (40), we get that dP = 0. Now since n - 1 is odd, we use (27) in Remark 1.7 to see that zl(t2 + (n - 4)/2,t2) # 0 and by (22) in Remark 1.5 to see that zi(t2 + (li - 4)/2, t2 - [i/2]) = 0 for i = 2,. . . ,n - 1. Hence, letting t = tz + (n - 4)/2 in (40), we get that dr = 0. Now since n - 2 is even, we use (24) in Remark 1.6 to see zz(t2 + (n - 4)/2 - l,t2 - 1) # 0 and by (22) in Remark 1.5 to see that ri(t2 + (n - 4)/2 - 1, t2 - [i/2j) = 0 for i = 3,. . . , n - 1. Hence, letting t = t2 -t (n - 4)/2 - 1 in (40), we get dz = 0. Continuing in this fashion, =re get di = 0 for 0 < i 5 n - 1. Therefore, we have {q(t, tz - tj/2])}7=: for n even.

is a linearly independem -3t of solutions for Zgz(t) = 0 on [a, b + n]

2. n odd. Once again the proof is similar to Case 1 and will be omitted.

CASE

LEMMA 2.5. If z(t) is a nontrivial

solution

of lcz(t)

I

= 0 with n - k consecutive

zeros ending at

tz -t [(n - 1)/2J iff z(t) = dn-lzn_l

(,,t,-

p$j) +.*.+ dn-kzn-k

where the di, n - k 5 i 5 n - 1, are constants

(,.,,

not all equal to zero.

-

191)

,

(41)

R.

558

KRUEGER

PROOF. (+) By Lemma 2.4, any solution z(t) of &r(t) = 0 has the form z(t)

= doze (t, t2) + 4.21 (t, t2) + * * * +d,_lz,_l

(,,t,

-

13).

(42)

CASE 1. n even. Note that ze(t2 + (n - 2)/2,t2) # 0 by (24) and zi(tz + (n - 2)/2, t2 - [i/2]) = 0 for i = 1,.**, n - 1 by (22). Thus, letting t = t2 + (n - 2)/2 in (42), we get that dc = 0. Also, .zr(t2 + (n - 4)/2, t2) # 0 by (27) and zi(t2 + (n - 4)/2, t2 - li/2J) = 0 for i = 2,. . . , n - 1 by (22). Thus, by letting t = t2 -I- (n - 4)/2 in (42), we get dr = 0. Continuing in this fashion, we get that do = dr = . . . = dn-k-2 = 0. Finally, to show that dn-k-i = 0, we consider the following two cases (noting that n - (n - k - 1) = k + 1). SUBCASE 1.1. k + 1 even. BY (241,&k-l@2 + (k - 1)/2 - L(n - k - 1)/2J,t2 - [(n - k - 1)/2J) # 0. But by (22), .zi(tz + (k - 1)/2 - [(n - k - 1)/2J, t2 - L(n - i)/2J) = 0 for i = n - k, . . . , n - 1. Hence, setting t = tz + (k - 1)/2 - L(n - k - 1)/2] in (42), we get d,+k_i = 0 for k + 1 even.

SUBCASE 1.2. k + 1 odd. BY (271,%-k-l@2 + (k - 2)/2 - l(n - k - 1)/2J, tz - \(n - k - 1)/2J) # 0 and by (22), ti(t2 + (k - 2)/2 - [(n - k - 1)/2J, tz - [i/2j) = 0 for i = n - k, . . . ,n - 1. Finally, setting t = t2 + (k - 2)/2 - l(n - k - 1)/2J in (42), we get the desired result d,+k-1 = 0 for k + 1 odd.

Hence, when n is even, we get that (41) holds. CASE 2. n odd.

The proof is again very similar as in Case 1 (x=) Assume z(t) is of the form (41) where equal to zero. Since z(t) is a nontrivial linear It.z(t) = 0, then z(t) is a nontrivial solution of

and will be omitted. the di, n - k < i 5 n - 1, are constants not all combination of linearly independent solutions of lzz(t) = 0.

CASE 1. n even. By (22) in Remark 1.5, zi(t,ta - li/2J) h as i consecutive zeros starting at t = t2 - ii/21 + lb - i)/ZJ. So when i = n - k, zn_k(t, t2 - [(n - k)/2J) has n - k consecutive zeros starting at

Thus, z&t,)!2

- [(n - k)/2J)

h as n - k consecutive zeros ending at

t=ta-

[qj

+ I%] +n-k-l.

If k is even, then

=t2--

as desired.

n-k

k

2n-2k-2

2

+5+

2

Comparable Solutions

559

If k is odd, then

= t2 =t2+2

n-k-l

k-l

2

2n - 2k - 2

+2+

2

n-2 n-l

=t2+

I J --y

as desired. Similarly, when i = n - k + 1, .z,+k+l(t,

t:! - [(n - k + l)/ZJ)

11a.sn - k + 1 consecutive zeros

starting at

Thus, z+k+l(t,

t2 - [(n - k + 1)/2J) has n - k + 1 consecutive zeros ending at t=t,-I”-;+‘j+[y]+n-k.

If k is even, then

=t2_=t2+=tz+

n-k 2 n-2

+

k-2+2n-2k 2

2

2 n-l --y-

I 1

as desired. If k is odd, then t=t2-

= t2 -

Ln-2k+l]+ [k&L] n-k+1 2

k-1+2n-2k -+ 2

+n_k

2

n-2 =t2+y-

as desired. Continuing this process, we can see that at each step we add a consecutive zero but the i consecutive zeros of zi(t, t2 - [i/2J) for n - k 5 i < n - 1 will always end at t = t2 + L(n - 1)/Z]. Thus, the i consecutive zeros of zi(t, t2 - [i/Z]) for n - k 5 i 2 n - 1 will end with the same n - k consecutive zeros. Therefore, since z(t) in a linear combination of zi(t, t2 - Lila]) for n - k 5 i < n - 1, we have that z(t) has n - k consecutive zeros ending at t2 + [(n - 1)/2]. CASE 2. n odd. The proof is similar to Case 1 and will be omitted.

I

560

FL KRUEGER

DEFINITION. For the functions

Wmnstian w,(t)

W;(t)

s(t),

. . . , ok

defined on [a, b + n], we define the generalized

for n even by

= W,* [VI(t), . * - 1WC(t)]

(,+p$J) LTV1 (,+ [yJ) . L;_,Vl l?J) Ll;?Jl

=

(,I

fort E [a,b+n-

w;(t)

k + 1 - L(n - k)/2J],

= w, [4),

* * ’ ,w&)]

Wl

(t + [;j)

=

1 I k 5 n, and for n odd by

L$& (t + l$J)

*. ’

LTV2 (t + 191)

LTV1 (t + 191)

Gja

**’

(t + pj J)

LiQ (t + 191)

**

9

J) fort E [a, b + n - k -I-1 -

[(n - k + 1)/2J],

1 I k 5 n.

We will also use the notation

w,*(t, t2) THEOREM 2.6.

t1 +

l(n -

=

w,+[ %n_] (t, t2 -

[yJ> ,..., (t,t2-pg.

‘I%ere is a nontrivial solution of l$

%n-k

= 0 with a (k,n - k)-pair of zeros at

)/2J and t2 + [(n - 1)/2J iff

PROOF. We will use the notation: fi = li/2J. Assume there exists a nontrivial solution z(t) with a (k, n - k)-pair of zeros at tl + fn-l and t2 f fn-l. Since z(t) is a nontrivial solution of 1;~ = 0 that has n - k consecutive zeros ending at t2 + fn_l, we have by Lemma 2.5

z(t) = dn-ltn-1

(t, tz - fra-1) + * * * + &-i&-k

(t,tz

-

he-k),

where the di, n - k 5 i 5 n - 1, are constants not all equal to zero. Since z(t) has k consecutive zeros ending at t 1 + fn_l, dn-lz,-l

(h + &-I

- k + 1, tz - &-.I) + .. . + &-k&-k

d,p~z,_l

(tl + fn-1 - k + 2, tz - &-I)

+ ‘. ’ + &-k&-k

(tl

+ fn-I

(tl + fn-1

dn-lzn-1 (h + fn--1,tz - fn-I) + *** + dn-k&-k

(tl

k + 1, tz - fn-k)

= 0,

- k + 2, t2 - fn-k)

= 0,

+ h-1,

= 0.

-

tz -

h-k)

The [k/2jth equation above is

+...

(43) ++_k~-k(tl+~~-~-k+r$],t2-fn-k)=O.

Comparable Solutions

561

It is easy to prove ‘that (44)

k= Using (44) in (43)) we get &-lh-1

@l + fn-1 - fk,

t2

+ . . ’ + ‘&-&n-k

fn-1)

-

(tl + fn-1 -

fk,

t2 -

fn-k)

=

(45)

0.

If we multiply (45) by pn(tr + fn_i - fk - [n/21), then, using (9), we obtain d+iL;FzW-i

(ti + &Z--l -

The ([k/21 dn-izn-1

fk,t2

-

fn-I)+’

* .+d,+kh;&_k

(tl

+ &_I

-

fk,

t2 -

&_k)

=

0.

(46)

+ l)st equation above is

(

ti +.&-I

> 11 +..* +&-k&k tl+fn-1 - 11

-k+

+l,t2-fn-1

5

k +

f

+ 1,t2 - fn_k

>=0.

By (44), we have dn-rzn--l +

(tl + fn-1 -

* * * + &-k&-k

(tl

+ fn-.1

fk -

+

l,t2

fk

f

-

fn-1)

1, t2 -

fn_k)

=

(47)

0.

Now if we multiply (47) by pn(ti + fn-i - fk - [n/21 + l), multiply through (45) by pn(tr + fn_r - fk - m/21), and subtract the second expression from the first, we have A-iA

[p- (tr + fn-1 +

*** + dn-kA

fk

-

[;1>

[Pm (tl

&-I + fn-1

(tl -

+ fn-1 .fk -

Then multiplying the above equation by p+l(tl dn-iL;Zn-1

(ti + fn-1 -

CASE 1. n even. Since f,+_r = f,+s A-l-Gz,-1

(tl

fk,

t2 -

-

[;I)

fk,t2

-

&-I,]

&a-k

(tl

+ fn-I

-

.fk,t2

-

j-n-k)]

=

0.

- fk - [n/21 f l), using (9), we get

+ fn-l

fn-I)+*.

‘ f&-kL;%-k

(tl

+ fn-1

-

fk,

t2 -

fn-k)

=

0.

(48)

-

fk,

t2 -

fn-k)

=

0.

(49)

for n even, we can substitute into (48) to obtain

-t fn-2

-

fk,t2

-

&-I)+*.

.+&-kL;G-k

(tl

+ fn-2

This is the second equation we need in system (50) below. The first is equation (46). Repeatedly using a similar process as we used to find equations (46) and (49), we can obtain the system of equations &-r-&-r (ti + k-1 - .fk,tZ - fn-1) +...

+...

+ dn-kL&-k

+

(tl

+ fn-1

-

fk,

h

-

fn-k)

dn--1L;zw-l

(h

+ fn-z

-

fk,

t2 -

fw-1)

dn-kL;&-k

(h

+

fn-2

-

fk,t2

-

fn-k)

(tl + fn-k

-

.fk,t2

-

fn-1)

(h +

-

fk,

t2 -

fn-k)

A.--1L;-1+1 +... Since d,+k,.

. . , d,_l

CASE 2. n odd.

+ dn-&-l&-k

fn-k

= 0,

= 0,

= 0.

are not all zero, we have the desired result for n even.

(50)

R. KRUEGER

562

Since fn_r = fn when n is odd, we can substitute (tl + fn - fk, t2 - fn-1) + . * * +

&a-IL&a-l

into equation (46) to get

ha-k’+w-k

(h

+ fn

-

fk,

t2 -

&a-k)

=

0.

(51)

This is the first of the desired equations in system (52) below. The second equation is (48). Repeatedly using the process we used to find equations (51) and (48), we can obtain the system of equations A-lL&-1 (tl + fn - .fk, t2 - fn-1) + . ’ ’ + &-k-@n-k

(h

&--lL;&a-1 + *. * + dn-kL;Ga-k

A-&-1+-l

+ fn

-

fk,

t2 -

fn-k)

(h + ha-1 -

fk,

t2 -

ha-l)

(tl

fk,

t2 -

fn-k)

+ fn-1

-

(tl + h-k+1

+ ’ * ’ + &-/&~%-k

=

0,

=

0,

=

0.

(52)

- fk, t2 - &,-I)

(tl + ha-k+1 -

fk,

t2 -

fn-k)

Since d,,_k, . . . , d,_l

are not all zero, we have the desired result for n odd, The proof of the converse is essentially reversing the steps above, so it will be omitted.

3. MAIN

1

RESULTS

One of our main results is the following theorem. The analogous result for the continuous case is given by Hinton [lo, Theorem 4.21. Our objective was to state this result as close as we could to Hinton’s result. Thus, much of the work leading up to this theorem took on its form to give us the present theorem. THEOREM 3.1.

For each t and s,

(-l)(n-l)k,,$$,

Wk(t, s) = PROOF.

t).

Assume that n is even. Consider wk(t,

3)

=

[Yn-l(t, s>, . . . ,%a-k(t,

wk

LoYn-

l@, 8)

LlYn-l(t,

s)

s)]

LoYyn-2(t, 8)

*. .

LO&-k(t,

JhYn-2(t, 8)

..

LlYn-k (6 s)

.

s)

*. Lk-lYn-l(t,

8)

Lk-lYn--a(&

s)

*..

Lk-l%a--k(t,S)

By (31), we have

(-l)“-lL;;zn-r (-l)nL;z,_z

t-11n+k-2L&-k

(s + [r+J (s + 191

(S +

l+J

, t - py)

(-l)“-QL&_r

(s+

, t - pyJ>

(-l)“-lLiZn-z

(S + [VJ

, t - 191)

191

,t-

(--l)n+k-3Li%n_k (S'+ l?J

***

(-l)n-kL;_&_r

* +*

(-l)n-k+‘L;_,%,_s

(s + 191 (s + [YJ

pgj)

, t - 191)

,t - [YJ)

, t - p+J) , t - 1-J)

Comparable Solutions

..*

563

L;_+-k

(S + [%$I ,t - [=$j)

By the definition of determinant, we have

.. .

L;_lzn_lc (s + 191

,t - [+J)

Since the determinant of a matrix is equal to the determinant of its transpose, L&z*_1 (s + [y]

) t - 191)

L;;zn-2 (s + l?J

)

LiZn-l

,t

LiZn-2

,

(S +

191

-

l+])

(S +

L;_rr,_z

= (-lpl)“W;(s

[qJ

(s -t 191

...

L;&-k

...

L;_l%,-k

(a + 1’

(s + lqj

,t - 191)

, t).

Thus, we have the result for n even. The proof for n odd is very similar and will be omitted.

I

Now with all these preliminary results, we can prove the following theorem.

THEOREM 3.2, The difference equation Zny(t) = 0 has a nontrivial solution with an (n -’ k, k)pair of zeros at tl and tz, (t2 > tl +n - k) iff there is a nontrivial solution of the adjoint equation Izz(t) = 0 with a (k, n - k)-pair of zeros at tr + [(n - 1)/2] and t2 + L(n - 1)/2J. Note by Theorem 2.3, the difference equation l,y = 0 has a nontrivial solution with a (n - k, k)-pair of zeros at tr and t2 iff

PROOF.

564

R. KRUEGER

But by Theorem 3.1, this equation is true iff

w+-[SJ,t2)=o. Finally, by Theorem 2.6, this equation holds iff there is a nontrivial solution of l;z = 0 with a I (k,n - k)-pair of zeros at tl + [(n - 1)/2] and t2 + [(n - 1)/2J.

REFERENCES

5.

6. 7. 8. 9. 10. 11.

P. Hartman, Difference equations: Disconjugacy, principal solutions, Green’s functions, complete monotonicity, Frans. Amer. Math. Sot. 246, l-29, (1978). G. Polya, On the mean-value theorem corresponding to a given linear homogeneous differential equation, ‘1Pans.Amer. Math. Sot. 24, 312-324, (1922). B. Harris, R. Krueger and W. Trench, Trench’s canonical form for a disconjugate nth-order linear difference equation, PanAmerican Journal 8 (3), 55-71, (1998). W. Trench, Canonical forms and principal systems for general disconjugate equations, tins. Amer. Math. Sot. 189, 319-327, (1974). C. Ahlbrandt and J. Hooker, Disconjugacy criteria for second order linear difference equations, In Qualitative Properties of Di&nzntial Equations, Proc. of 1984 Edmonton Conference, University of Alberta, Edmonton, Alberta, Canada, pp. 15-26, (1987). C. Ahlbrandt and A. Peterson, Discrete Hamiltonian Systems, Kluwer Academic, Boston, MA, (1996). J. Henderson and A. Peterson, Disconjugacy for a third-order linear difference equation, Computers Math. Applic. 28 (l-3), 131-139, (1994). M. Morelli, Disconjugacy of a third-order linear difference equation, Doctoral Thesis, University of Nebraska, (1997). T. Peil, Disconjugacy for nth order linear difference equations, J. Math. Anal. AppZ. 160, 132-148, (1991). D. Hinton, Disconjugate properties of a system of differential equations, Journal of Differential Equations 2, 420-437, (1966). W. Kelley and A. Peterson, Difference Equations: An Introduction with Applications, Academic, San Diego, (1991).