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©BEYKENT UNIVERSITY

ON THE LINEAR

DELAY-DIFFERENTIAL QUATIONS

Zekeriya GÜNEY

Muğla Üniversitesi Eğitim Fakültesi,

Orta Öğretim Fen ve Matematik Alanlar Eğitimi Bölümü zguney@mu.edu.tr

ABSTRACT

In this study, solutions has been generated for some special forms of delay linear differential equations system that it conclude of first order n linear delay differential equations and stone equation y'(t) = c y(t-r), that is to form a base for these solutions has been investigated. Also, solutions of this equation on some available time scales has been generated.

Key words: Delay Differential Equations

ÖZET

Bu çalışmada, birinci mertebeden n lineer gecikmeli diferansiyel denklemden oluşan gecikmeli lineer diferansiyel denklem sistemlerinin bazı özel biçimleri için çözümler üretilmiş ve bu çözümler için temel oluşturan, y'(t) = c y(t-r) çekirdek denklemi incelenmiştir. Ayrıca bu denklemin uygun bazı zaman skalalaları üzerinde çözümleri üretilmiştir.

I. PRELIMINARY ON THE DELAY DIFFERENTIAL

EQUATIONS

I.1. INTRODUCTION

A general delay differential equation can be written in the form

F (t, y ( f „ ( f ) ), ..., y( flm(t ) ), y ' (&(/) ), ..., y ' ( f u t ) ), ••• , y("'v(. ( 1 )

fnl(t) ), .... y(n-1>(fUt) ),y( n )(t) ) = 0 ( 1 )

(2)

F( t, y(l-l>( t - j t ) ), y(n>(t) ) = 0 (2)

where t is the independent variable, f y (t) = t- ry (t), ry (t) > 0, given functions for i = l,... ,n,

j = 1,...,m ; y is unknown function. if it is possible to solve y(n>(t) from (2),

especially

y(nn (t ) = <P ( t, y(i-l) ( t - ry(t) ) (3) n m

y(n> ( t ) = 2 S a'j(t) y(i-l) ( t - ry(t) ) -A t ) (4)

i = 1 j = 1

in (4), ay(t>'s are given variable coefficients, ry(t>'s are given variable delays and f(t) is a given function. By the transformation yi (t> = y(l~l>(t>, equations

(3) and (4) may be reduced into the systems of equations that are written in vector form which contain n unknown function of the first order

Y (t> = F( t, Y (t-ry(t))) (5)

and

m

Y (t) = S Ay (t) Y (t-ry(t)) + H (t) (6)

i = 1

where Y(t> = [ y(>, ... , yn(t> ]T and H(t>= [ hi(t>, ... , hn(t> ]T are given

vector functions.

When doing solvability analysis for the equation (5), it is meant for the solution that Y ensures the condition

te[y, to ] ^ Y (t) = 6(t)

te [to- P ] => Y(t)=F(t,Y(t-r,m

in the interval [y. Pi ), where to, P e Rn, Pj e (tQ, P ] , j =1,... ,m : to< t< p y

< t-r/U < t0 and 8 : [y,to ] K" initially function, if F and y

continuous,

Y (t) is solution on [y,p ] o Y (t)

6 (t) , for te[y, to ]

t (8)

6 (t<,) + I F (s, Y( s-ys)) ds, for te [tJto o, P],

The characteristic equations that play important role in the analysis of solutions, solvability and nature of solutions, of the linear homogen differential equations with fixed delay and fixed coefficient, are obtained by applying Laplace transform to the both sides of delay differential equation, or by making use of the relation E =eD that is between differrential and shifting

(3)

operators or also by suggesting a solution in the exponantial form y=est . If we

assume that coefficient and delays are fixed in (4),(6), the characteristic equations of these equations are

n m

sn = H s1"1 e-srj (9)

i = 1 j=1 and

m

det [ si - 2 Aİ e"srj ] (10)

j=1

respectively. In (10) I is the unit matrix. These equations are in the class of quasi-polinomial.

They are transandantal equations and generally have infinite complex zeros. In the scala case for each zero s e C . there is a solution of delay differential equation in the form y(t)= est. If s e R then corresponding solution is

nonoscillating. For example if there is relation ad2 = bcd+ b2

between coefficient of the second-order linear homogen delay differential equation

y" (t) = ay(t) + by(t-r) + cy'(t) +dy' (t-r) (11)

with fixed coefficient and fixed delays, then the charesteristic equation s2 = a+ es + (b+ds)e-rs

has a real zero in the form s=-b/d and also equation (11) has a nonoscillating solution in the form y= e- if the set of all zeros of equation (9) is { s J k e I }

then the function

y(t) = 2 ck eskt

kel

is the solution for equation (4) where ck are scalars. [3],[4],[8]

II. THE ANALYSIS OF THE EQUATION y ' (t) = a y(t) + b y (t-r)

II.1. REDUCTION

The equation y ' (t) = a y(t) + b y (t-r), can be reduced the form

x ' (t) = c x (t-r), applying the transformation y(t) = eatx(t), where a,b,ceR\

{0}, reR+ :

y (t) = a y(t) + b y (t-r) 1

r ^ a eat x(t) + eat x '(t) = a eat x(t) + b ea(t-r) x(t-r)

(4)

^ x (t) = b e-ar x(t-r)

^ x (t) = c x(t-r), c = b ear

Because of this, to study with the equation y (t) = c y(t-r), is sufficient.

II.2. INITIAL VALUE PROBLEM

Theorem:if v(t) = 0 (t) be given integrable initial function for le \-r, 0\, kel , / be an index set in a countable cardinality, and sk e {51 s-aesr 0, s e C

than the solution of the equation

y ' ( t ) = a y (t-r) , a e R\ {0}, r > 0 (12) according to 8 is

0

y(t) = S eskt ( ae-skr J e-sku y(u) du) + 8(0 ) .

- r

Proof. Applying the Laplace transformation to both sides of (14) we have L \y(t)] = J e-st y(t) dt = z ( s ), L [y'(t)] = s z (s)-y ( 0 ) ^

0

0

s z ( s ) - y( 0 ) = a e- J e- y( u ) du + ae'r z ( s ) ^ -r

0

z (s) = (y(0) + ae-rs J e-su y(u) du ) ( s- a e-sr)'' , -r

f(s) = s - a e-sr is a non-zero analytic function and all its roods are isolated.

So, provided that I is an index set that is equivalent to the set of roots, for sk e

{ 5 /s-a'sr 0, .v e C ¡ and k e I, from Mittag-Leffler Theorem it can written

as follows

( s-a e-sr ) = ^ Ck/ (s-sk ) + h(s) kel

Here, it is accepted that h(s) is a entire function, that ck =1/ d/ds [( s- a e-sr ) -1s=sk end that sk is a single . A non-zero function should have a singularity at the

infinity. But, since h(s)=0 as s ^ w it should be h( s ) = 0 than. 0

z (s) = ( y(0) + ae-rs J e-su y(u) du ) £ ck / (s-sk )

- r ke l

and from inverse Laplace transformation and Cauchy formula we get

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KJ

Î

C + f »

[ Sck( >-(0) + ae-rs I e-su y(u) du)est) /s-sk] ds • C — f » J - r 0 Ck eskt ( | aesk (r+uU y(u) du + y ( 0 ) ) -r 0 ; eskt (ae-skr | e-sk u y(u) du ) / 1+ rsk +0(0)). = S Example:

y ' (t) = e-1 y(t) + e y (t-e), y(t) = et/e x(t)

^ x ' ( t ) = x ( t-e)

V ^ s = es

x(t) = c est J ^ ^

e-1 e { s | s-e-se = 0, s £ C }

^ a solution is x(t) = c et/e, y(t) = c e2t/e

II.3. SOLUTION BY STEP METHOD

Depending on an initial function v(t) = 80(t) in the interval [ tQ-r, to ], we can

find a solution in the interval [to, a\ for Vae K \{tQ} a delay differential

equation in the form

y(n) (t) = 9 (t,y(l_1)(t),y(l_1) (t-r) ), i = 1 , ^ , n (13)

which is formed by taking the coefficients constant, m=2, r1=0, r2 = r in

equation (3). Since

t e [ to, to+r ] ^ t-r e [ to-r, to ] we have t e [ to, to+r ] ^ y(i-1) (t-r) ) = 9(l-1) (t-r) )

and for t e [ to, to+r ] , equation (13) becomes the ordinary differential equation

y(n) (t) = 9 (t, y(l_1) (t), 6 o( i - 1 ) (t-r) ), i=1,...,n (14)

if y(t)= 81 (t) is taken as a new initial solution of (14) in the interval [ to, to+r ] ,

y(t)= 82 (t) will be the solution of

y(n) (t) = 9 (t, y(i-1) (t), e j( i - 1 ) (t-r) ), i = 1 , . , n

in the interval [ to+r , to+2r ]. Continuing in the same way, solution of (12).

Restricted to the interval [ to+(n-1)r , to+nr ] is found in n steps. Thus in the

interval . [ to-r , to+nr ], we get

' e o (t) , t £ [ to-r, to ] e1 (t) , t £ [ to, to+r ]

y(t) = J e2 (t) , t e [ to+r, to+2r ]

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as solution. Since n e N can be taken forV r > 0,so that a < nr for n = inin{m | a < mr, mei^J}, a solution is found in the interval [ to, a ] after n steps. Here, it is accepted that differential equations y(n) (t) = 9 (t, y(l-1) (t), 6(l-1) (t-r) )

can be solved for Vi=1,.. ,,n. This method is called step method. [4]

Example: l . J b r a = r =1, 6(t) =1, the solution of (14) in the interval [-1,5] is

t e [-1,0] t e [0,1] t e [1,2] t e [2,3] t e [3,4] t e [4,5] r y(t) = 1, t + 1, t2/2 + 3/2 . t3/6 t4 - t3/3 + 7t2/4 - 5t/2 + 85/24, t2/2 + 2t +1/6 . t5/120 t4/8 - t3 - 13t2/12 + 19t/6 - 1213/40 by step method.

II.4.A CRITERIAN RELATED NONOSCILLATING SOLUTIONS:

Theorem: Equation (12) has one nonoscillating solution if a e {-1/er }u (0,ro ) and two nonoscillating solutions if a e (-1/er, 0 ). For a e (-ro, -1/er ) all the roots of the characteristic equation are complex.

Proof.

y ' ( t ) = a y (t-r)

\ ^ X - a e-Xr = 0,

y = c eXt J

( f: R^-R ) ( f(X) = X - a e-Xr ) ( a > 0 )

^ ( f continuous ) [ lim f (X ) = -ro ] [ lim f (X ) = + ro ]

[ f(X) < X ] [ f (X) = 1+ ra e-Xr > 0 ] '

=i> I { XI X - a e_Aj = 0, XeR } I = 1, (i)

( f: M->]HL ) ( f(X) = X - a e ^ ) ( a < 0 )

^ ( f continuous ) [ lim f (X ) = lim f (X ) = + œ ] X^ - œ X^ + œ

[ f (X) = 1+ ra e-Xr = 0 ^ X = ln (-ar)1/r < (1/r ) + ln (-ar)1/r = f (ln

(-ar)1/r ) = f (X ) ]

[ f ' (X) = -a r e_Aj > 0 ]

=> (1/r ) + ln (-or)1/r < 0 -1/er < a < 0 =i> | { X | X - a e_Aj = 0, Xe®. } = 2,

(1/r ) + ln (-ar)1/r = 0 ^ a = -1/er ^

I { XI X - a e_Aj = 0, XeR } | = | { ln (-or)1/r } | =1,

(1/r )+ ln (-or)1/r>0=> a < -1/er =i> | { X | X - a e_Aj = 0, Xeffi. } | = 14> I = 0 (ii),

(i),(ii) ^ [ a e {-1/er }u (0,œ ) ^ one nonoscillating solutions ] [ a e (-1/er, 0 ) ^ two nonoscillating solutions ] [ a e (-œ, -1/er ) ^ no nonoscillating solution ] .

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Example

The equations y' (t) = 5y(t-1) and y' (t) = -1/2e y(t-2) have one nonoscillating solutions, in the form y=ce ; the equation y' (t) = -1/6 y(t-2) has two nonoscillating solutions, in the form y=ceM ; the equation y' (t) = -5

y(t-2) has not nonoscillating solutions, in the form y=ceM .

II.5. Y'(T) = C Y(T-R) EQUATION ON SOME TIME SCALES Preliminary:

T time scale :<=> ( c | ) ^ X c I i t ) ( \ X e U (usual topology on K ) ) CT forward jump operator :<=> ( CT : X —>• X ) [ CT (t) := inf {si s > t } ] p backward jump operator.^ ( p : T —> X ) [ p (t) := sup {si s < t }] [ n : T - > [ 0 , o o ) ] [ n(t) = a ( t ) - t ] a e Sr := { t I ct (t) > t e X } a e Si := { t I p (t) < t e X } :<=> a e I := { t I p (t) < t < ct (t), t e X } :<=> a e Dr := { t I ct (t) = t < sup X , t e X :<=> a e Di := { t I p (t) = t > inf X , t e X t £TK= T \ { {max T } n Si } :=>

f (t) the delta derivative of f at the point t ( Ve > 0 ) ( 38 > 0 )

[ se ( t - 8 , t +8 ) n T =i> I [ f (ct (t))- f (s) ] - f* (t) [ c r ( t ) - s ] | < s I cr (t) - s I ]

fis (delta) differentiable on TK :<=>/e D [TK ] : = { g | [ g: X - J ]

graininess function a right-scattered a left- scatteredt a isolated a right-dense a left- dense f . T [ t e TK^ 3 gA (t) ]} continuous } / continuous :CÎ> / e C [T] := { g | g: X f regulated f e Re :={ g |( g: T ^ I R ) [ t e Dr=>lim g ( s ) < ® ] [ t e D , => lim g(s)< oo ] } s^t+ f rd- Continuous f £ C r d = { g I ( g | D r £ C ( D r ) [ t £ D i ^ lim g (s) < <» ] } s^t

-f di-f-ferentiable and which derivate is rd- Continuous

f £ C r d1 = { g I ( g£ D [TK ] ) ( gA £ C r d ) }= D [TK ] n C r d

/ pre- differentiable with D :ci>

f £ DD = {G| ( G£ C ) ( D c TK )(| Tk \ D | < K0) ( X \SR C TK \ D ) ( g£D [D ] ) } F is a pre-antiderivative of f ( f £ Re (T) ) [ t £ D ^ Fa (t) = f t ) ] Indefinite integral of f £ Re (T) : J f t ) At = F (t) + C r, s £ X Cauchy Integral : f t ) At = F (s) - F (r) F: T—>R is a an ti derivative o f f :<=> [ t £ TK => FA (t) = f(t) ] / regressive f e R (T) = {g | g: X te TK =s> l+ju(t) g (t) * 0 }

Generalized exponential function : ( s , t £ X ) ( p £ R (T)) :=>

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-ep (t,s) = exp ( f ( p(r) )Ar

Js

First order dynamic equation: yA = / ( t, y, y° ) , f: X x l2- > IE

v is a solution of yA = / ( t, y, ya ):<»( y: T IE ) [ teTK => yA (t) = / ( t, y(t), ya(t) ]

General solution of yA = / ( t , y, y° ) :ci>

S[ yA = / ( t , y , ya) ] = {yl (y: X ^ R ) [ t e TK =^yA(t) =/"(t,y(t)},yCTCt) ] }

First order linear dynamic equation:

yA = g(t)y+h(t) v yA = g(t) y° +h(t), &h : T - > ] R

Adjoint equation of first order nonhomogenous dynamic equation yA = g(t) y +h(t) :

xA = -g(t)xfJ+h(t)

Initial value problem: to6 T , yDe IE, y(to) = yQ, yA = / ( t , y, y° )

y is a solution of the initial value problem yA=/( t, y, ya) t0e X ,y0eIE,y(t0)=yc„ :<»

(y: X - > R) [ t e TK \ {tD } yA (t) = / ( t. y(t). y°(t) ] [y(t0) = yD ]

THEOREMS

1. ( t e Sr ) ( f is continuous at t ) ^ f (t) = [ f (a (t) )- f (t) ] / |(t) ( If t right-scattered and f is continuous at t , thanfA (t) = [ f (a (t) )- f (t) ] / |(t) )

2. f is differentiable at t ^ f (a (t) ) = f (t) + | ( t ) fA (t)

. ( if f is differentiable at t, than f (a (t) ) = f (t) + |(t) f (t) ) 3. f e R ^ ( 3 F e D D ) [ t e D ^ FA (t) = f t ) ]

( If f regressive than there exist a regulated F such that FA (t) = f t ) for

each t e D )

4. / e Crd => ( 3 F: X —>• IE) [ t e TK ^ FA( t ) = f t ) ]

( I f / r d - Continuous than there exist a F: X IE such that FA (t) = /(t)

for each t e TK )

5.

(a;b E T ) ( f e Qi )=>

{ If a b in lime scala T and /rd- Continuous lhan )

(i) T =R =• £ f ( t ) A t = £ f(Odt

{

Ltekb) fi(t) fit), a < b

0, a = b

-Step^M© fit) : a > b

(iii) (h > 0 ) (t ehZ ) f ( t ) A t =

T ^ ' *- 1 f f k h )h . K b

(9)

6. [yA = p ( t ) y ] [ p e R ( T ) ] (toe T )

ep (,,s) is unique of the initial value problem yA = p(t)y, y(tQ) = 1 on T

7. [ p e R (T) ] ( tDe T ) ( y0e K ) ^

y(t) = ep( t,tQ )yQ is unique solution of the initial value problem yA = p(t)y,

y(t0) = yD on T

8. [ p e R (T) ] ( tDe T ) ( x0e R )

x(t) = e0 p ( t,to )xo is unique solution of the initial value problem xA = -p(t)x° , X(to) = xo

9. [ p e R (T) ] ( f e Crd ) ( tDe T ) ( y0e l ) y(t) = ep( ttD )yD+ f ep( Jto

t, CT(r) ) f (r) Ar

is unique solution of the initial value problem yA = p(t)y+ f(t), y(tQ) = yQ

10. [ p e R (T) ] ( f e Crd ) ( tDe T ) ( x „ e l ) ^

x(t) = e0 p ( t,to )xo + I e0 p( t, r ) f (r) Ar is unique solution of the initial to

value problem

xA = -p(t)x° + f(t), x(to) = xo Examples

Lets firstly not that, delay dynamic equations should not be formulated with the delay of the form t-s as t,s e T for some time scale T does not necessarilly imply t-s e T. For these, only some available time scales has been studied 1. For the time scale T

=hNo

= { hn | n e No } = { 0, h, 2h,...}, e solution

of the initial value problem, xA (t) = ax ( t-h ), x(0) = 1 , t > h, Sr := T,

ct

(t) := t +h T =hN0 => XA (t) = [x{t+h> x(t) ] ' xA(t) = ix (t-h } x(t-h)-x(t) = ha x (t-h )" x<t) = c est c ez(Hii _c = cha => esr- 1 =iaea r => s = In [ (l-(l+4ha)L- ) / 2 ]L 1 x (0) = 1 !> => x(t) = [ (l+(l+4ha.)1;'2 ) /2 T4 x(t) =c est

J

2.

T =

hN

0

,

XA (t) = X ( t ) => [x(t-h>- x ( t ) ] / h = x(t) x(0) =1 U x(t) =21°a <W4< h = (|+h)Wi. = ^ o ) x(t>=c:st J

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3. T =2NoT=2N0, t >2,

xA (t) = x (t-2 ) => [x{t+2> x(t) ] / 2 = x(t-2) "1

x(0)

=1

=>

s(t) = = ei(t,0),

x(t)

=c

2

st

J

Indeed ( 2 t 2 = [ 2 - 2t 2 ] ¡1 = 2®"*°.

4. For the time scale T

=

e

N

0 { en | neNo } = { 0, e, 2e,...}, e solution of

the initial value problem, xA (t) = x ( t-e ), x(0) = 1, t > e,

T =eN0 =• xA (t) = [x(t+e> x(t) ] " e xA (t) = x ( t-e } x(t+e)- x(t) = e x ( t-e ) x(t) =c est => c e5^ _ c e=t - c e => e=a- 1 = ee"5 3 => s — In [(l+(l+4e)l a ) / 2 ]' -x (0) = 1 => x(t) = [ (l+(l+4e)l f l ) /2 f ° x(t) =c est

J

5. T = N „ = t > 1 = s4© = x ( t - l ) => x(t+l)= x(t)+ x(t - 1 } x(0)= 0, x ( l ) =1 x (t) = a1 x{t) = ( [ ( 1 - 5l 2 >2 f - [ (1- 5m yi f ) / 5l f l x [ «0] = { 0. 1. 1. 2. 3. 5. 8. 13=...} Fibonacci numbers

III. SOLUTION OF THE EQUATION Y = A Y (t-r) + H (t)

The linear delay differential equation system

y'i (t) = an yi (t) + y2 (t-r) + ...+ am yn (t-r) + hi(t)

y'2 (t) = a2i yi (t) + a22 y2 (t-r) + ...+ a2n yn (t-r) + h2(t)

y ' n (t) = a n i y i (t) + a „ 2 y2 (t-r) + ...+ am y n (t-r) + hn(t)

that includes n non-homogenaus equations of degree one, with constant coefficient and with

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constant r delay is written on the matix form as

f

>

r

f

f

y'ı (t) y'2 (t) = 3-11 a 12 3-21 3.22 • -- ain • • a2n >1 (t-r)

Yl

(t-r) -lii(t) h2(t) y ' n ( t ) V.

J

anı a„i .

J

L

J

yn (t-r) Vhn(t)

J

shortly as or

Y

= A Y (t-r) + H (t)

We want to solve this equation. Let the eigen values of the matrix A be .. .,Xn and its modal matrix be M. Then

M ' A M = 0 .. 0 yi(0 xi(t) 0 À.Ï . . 0 = Ai for Y(t) = . X © = 0 0 . . /ijji O ^ i . by the transformation we get found (15) Y(t) = M X(t) (16) (17) M X (t) = AMX (t) + H (t)

By the matrix multiplication of (17) with M-1 from left,

M-1 M X (t) = M-1 A M X (t-R) + M-1 H (t) ^

X (t) = A1 X (t-r) + H 1 (t) ^ X1 (t) = X X1 (t-r) + k1 (t) X2 (t) = X X1 (t-r) + k2 (t) Xn (t) = Xn X1 (t-r) + kn (t)

which are of (14) form in the homogenaus part form (14). The sum of the general solutions of homogenaus part and special solutions of

(12)

(t) is obtained and also from Y (t) = M X (t), Y (t) is obtained. For example, for the equation

(18) y 1 ( t ) 1 4 V i ( t - 1 ) -t y 2. ( t ) 1 1 V 2 ( t - 1 ) 2t we have 1 4 -2 2 -1/4 1/2 -1 0 A= 1 1 : M = 1 1 : M"1 = 1/4 1/2 = Ai = 0 3 or idantically x ' 1 (t) = - x j (t-1) + 3/4 t x ' 2 (t) = 3 X2 (t-1) + 5/4 t

for the homogenaus parts of the above equations using we obtain „ - s k ^ ? o „ - s ' k

sk = - e , s k = 3 e

xi(t) = 2 Ck eskt, X2(t) = 2 c'k es'kt,

KgI KGI and we obtain, for non-homogeneaus equations the special solutions

xj(t) = 3/4 t, X2(t) = -5t /12 -5/9 and for (19) the solution

X ( t ) =

3t ,4 - £ e**,

1=1

-5t /12 -5/9 - 2

(19)

and also for (18) the solution

f -7t / 3 - 10 / 9 - £ ( - ct e3tt + 2 c\ e ^) "

Y(t) = M X (t) =

t / 3 - 5 / 9 + 2 < cte 3 1 t + c ' k ^ ) ITiJ

IV.SOLUTION OF THE EQUATION Y = A Y (t) + B Y(t-r)

it is possible to diogonalize two matrixes at the same time if some conditions are satisfied. if this is satisfied for the matrixes A and B the equation

(13)

can be transformed to n scalar equations of the form (14). Let M-1A M =Ai,

M-1B M =B1; diagonal and Y (t) = M Z (t). From (20)

M

Z

(t) = AMZ (t) + BMZ (t-r)

M"1 M

Z

(t) = M-1 AMZ (t) + M-1 BMZ (t-r) ^

Z

(t) = Ai Z (t) + Bi Z (t-r) ^ z'i (t) = aii Zi (t) + bii Zi (t-r), z (t) = e aii 4 Xi (t), ai = bu e aii r, i=1, ... , n ^

x'i (t ) = aiXi (t-r), i =1,...,n

we obtain n equations in the form (14). Solving these and going the inverse direction, Y(t) is found. For example, for the equation

we have 1 -Js

i V?

"3 4 " — 1 2 ' . 5 0 2 1 4 -3 - 2 1 0 5 1 -2 1 0 1 2 1 0 2 1 0 1 -2 1 0 1

by this method Y (t) is obtained as

2 ( ck e (sk-5) t + 2c'k e(s'k +5 )t )

Y( t ) =

(14)

IX. CONCLUSION

As it's seen insections IV-V, equations (12) is a starting point for more complex equations. Thus, solutions can be found for some equations of the n th order and systems of equations in the form

Y = A Y (t) + B Y(t-r) + H (t).

Because the characteristic equation of (12) has generally infinite number of complex roots, the problem is the impossibility of determining these roots exactly. There exist some localization theorems given by Borel and Krall. Here have been gave a criterian for the roots that give non-oscillating solutions. This criterion can be generalized for the systems, by taking the intersections. Also, for any delay differential equation on some available time scales, by to benefit from concepts in [1] and [2], similar solution analysises could be construct.

REFERENCES

[1 ]. Bohner, M, Peterson, A. ; Dynamic Equations on Time Scales, Birkhauser Boston, (2001)

[2]. Bohner, M, Peterson, A. Martin Bohner, Allan Peterson, Advances In Dynamic Equations on Time Scales, Birkhauser Boston, ( 2002 )

[3]. Bole,G.; Calculus of Finite Differences, Chelsra Company, New York, N.Y. (1980) [4]. Driver, R.D.; Ordinary and Delay Differential Equations, Kington, (1976)

[5]. Korreil, K., Grove, E.A. Lads, G. " Delay Differential Equations with Positive and Negative Coefficients" Applicaple Analysis, Vol 27, pp 181-197, Great Britain (1988) [6]. Kulenoviç, M.R.C. Ladas, G, Sficas, Y.G.; Oscillations of second order Linear Delay Differential Equations, Applicaple Analysis, Vol 27, pp 109-123, Great Britain (1988)

[7]. Reza, F.; Linear Spaces in Engineering, Ginn Company, London, (1971) [8]. Saaty, T.L.; Modern Nonlinear Equations, Dover Publications, New York (1981). [9]. Uluçay, C.; Fonksiyonlar Teorisi ve Riemann Yüzeyleri, Ankara Üniv. Basımevi, (1978 )

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