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POSITIVE SOLUTIONS OF NEUTRAL FUNCTIONAL DIFFERENTIAL EQUATIONS

1

Huang Zhenxun (jf.f,J.,Jfh) Dept. of Mathematics, Fudan University, Shanghai 200433, China.

Gao Guozhu ( ;t; 00 it. ) Basic Sciences Department, China Textile University, Shanghai 200051, China.

Abstract This paper is to obtain sufficient conditions under which the neutral functional differential equation d

dx [x(t)

+

jt c

x( 8)d.p.(t, 8)] +

jt c

f( t, x( 8»d.11(t, 8) = 0, t ~ to ~ c

(1)

has a positive solution on [c, +00). Some results in [1] are generalized. Then we apply our results to functional differential equations of special form and obtain sufficient conditions for those equations to have a positive solution.

Key words Positive solutions, neutral functional differential equations.

1 Introduction Consider the first order neutral functional differential equation d

dx [x(t) +

jt x(s)d.JL(t, s)] + jt f(t, x(s))d'TJ(t, s) = c

c

0, t

?: to ?: c

(1)

where -00 < c ~ to < 00 and the function f,J-t, and 1] satisfy the following conditions: (i) f : [to, +(0) x R -+ R is a continuous function for which there exist a constant fJ > 0 and a continuous function a(t) > 0 for t 2: to such that

o < f(t, x) ~ a(t)x,

for t

2: to and 0 < x < 6,

(ii) 1] : [to, +(0) x [c, +00) -+ R is a function such that for fixed t ~ to, 1/(t, s) is nondecreasing on c ~ s ~ t and 1/(t, t) - 1/(t, c) ~ 0 for t > to. Moreover the function Jet y(t, s )d$1](t, s) is continuous on [to, +00) whenever y is continuous on {(t, s) : t 2: to, c ~ s ~ t}. (iii) 1£ : [to, +00) x [c,+oo) -+ R is a function such that for any interval [e,r] C [c, +00 ), p( t, s) is a function on [e, r] of bounded variation and V 1£(t, t+.) -+ 0 as s -+ 0+ [--,,0]

where the mark V f(.) means the total variation of f on interval [a, b]. [a,b]

1 Received

Dec. 3, 1994; revised Dec.l, 1995.

.

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322

Under the above conditions there exists a solution of(1) for any (to, ¢) E [c,

+00)XB[13].

B is the Banach space of bounded continuous functions mapping the interval [c, to] into R

with the superemum norm. J-L satisfies the following conditions yet: (iv) If h(t) > 0, t ~ to, h(.) E C([c, +00), R) and ¢(s) > 0, s E [c, to], the solution x(to, ¢, h)(t) on [to, +00) of the difference equation

x(t) + Jet x(s)d$J-L(t, s) { x(s) = ¢(s)

= h(t)

t 2: to 2:

c

(2)

c

is positive. (v) The solution x(s) of the equation (2) satisfies

KII¢II + K

Ix(to, ¢, h)(t)1 ~ where K is a positive constant,

II¢I! =

D(t,

sup h(u)

(3)

to~tt~t

1¢(s)l. Obviously, the uniformly stable operator

sup e~"~to

The oscillatory behavior of the solutions of neutral functional differential equations has been studied, see [3,4,5,6]. Recently, the existence of positive solutions for the neutral delay differential equations have been investigated by many authors [1,9,10,11,12]. In this paper, the nonoscillatory problem of (1) are considered and some results of [1] are extended. As is customary, a solution is said to be oscillatory if it has arbitrarily large zeros and nonoscillatory if it is eventually positive or eventually negative. A differential equation is called oscillatory if all of its solutions oscillate, otherwise it is called nonoscillatory.

2 Theorems and Proofs Our main results are the following. Theorem 1 Suppose that there exist a continuous function lJ : [to, +00) positive constant d such that

d E (0,0*),0* = min{o,

k(3

t

+ 2 Je d"J-L(t o, s))

i r t (exp

+

a(u )11( u)du)d.TJ(t, s)

Jmax(",to)

e

b to (exp d(1 + Je d"J-L(to, s)

it to

a(u)lJ(u)du)

}.

R and a

(4)

0

TJ(to, to) - TJ(to, c) < 1I(to )(1 + and

o

~

i

c

to

d. p,( to, s))

it [e,,,] e

V J-L(s·)d,,1J(t, s) :::; lJ(t),

(5)

(6)

Vt ~ to, where the constants k and b are as the above. Then equartion (1) has a positive solution x on [c, +00) such that 0 < x(t) < S, t ~ to and

x(t o) +

i

c

tux(s)d"JL(t

o, s) exp( -

it a(u)v(u)) < x(t) + it X(S)d,4JL(t, s), to

e

(7)

323

Huang & Gao: SOLUTIONS OF DIFFERENTIAL EQUATIONS

No.3

Theorem 2 Suppose that there exist a continuous function v : [to, +(0) positive constant d such that the conditions (4) and (5) hold and

d(l

+

lt o

e

d$Jl(t o,s)(l- exp(-

it to

a(u)v(u)du)) > fJ

it to

-+

R and a

a(r)(T/(r,r) -T/(r,c))dr,

(8)

then equation (1) has a positive solution x on [c, +(0) such that 0 < x(t) < fJ, t ~ to and the estimate (7) holds. Remark If Jl = 0 Theroem 1 becomes Theorem 1 in [1]. The following lemma will be useful in the proofs of the above theorems. Lemma[l] Suppose that the functions Yi(t, s)(i = 1,2) are continuous on {(t, s)lt ~ to, c ~ s ~ t} and the condition (ii) holds. If 0 < YI(t, s), Y2(t, s) for t > to and c ~ s t, then

o<

it

it

(Yl(t, s)d.'l(t, s) <

:s

Y2(t, s)d."l(t, s), t > to.

Proof of Theorem 1 Consider the initial value (to,

x(t)

i

+

t

x(s )d&Jl(t, s)

c

= y(t) exp( -

therefore

y'(t) clearly, y(t)

= - i t f(t, x(s))d."l(t, s) exp e

r

lto

lmax(t,to)

a(u)v(u)du, c

to

~

t ::; t l.

a(u)v(u)du + a(t)v(t)y(t), to

~ t < tl-

= d(1 + J: dlfJl(t, s)), c ~ t ::; to. From the conditions (i) and (5) it follows that y'(to)

= - let o f(t o, d)dlfT/(t O, s) + a(tO)v(tO)y(tO) ~ a(tO)v(tO)y(t O) - J:o da(to)dlfT/(t O' S) . =da(to)[v(to)(1 + fet o d$Jl(to, S)) - (T/(to, to) -T/(to, c))] > o.

By (5) and the nondecreasing property of T/(to, s) in s, we obtain (1 + + Thus for any t > to near to, we have

d(l:-

l

c

t

o

d.p,(to;s)) < y(t) < 2d(1 +

lto c

r

f:

o

d, Jl(to, s)) >

d.p,(to,s))exP(lto a(u)v(u)du).

o.

(9)

We will show that the inequality (9) holds for any t E (to, tl). Otherwise we have the following two possible cases:

r

(a) There is a t 2 E (to, t l) such that (9) holds for to < t < t2 and

y(t 2 )

= 2d(1 +

l

o

t

dlfJl(to, s)) exp(

(b) There is a t 3 E (to, t l ) such that (9) holds for to

y(t3) = d(l +

a(u)lI(u)du).

to

e

l

c

to

< t < t3 and

d.p,(to, s)).

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ACTA MATHEMATICA SCIENTIA

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In the case (a), by (9) we obtain

t: a(u)v(u)du

d(1 + J:o d~J-L(to, s ) ~ (X(t) + Jet x(t)dsJ-L(t, s» exp

< 2d(1 + J:o dsJ-L(t o, s ) exp J/o a(u)v(u)du, to ~ t < t2; and

o < x(t) +

i

c

t

x(s)d.j.t(t, s) < 2d(1 +

jto c

d.j.t(t, s)), to

~

t < t2'

(10)

From (iv),(v),(4) and (10) we have

o<

x(t) ~

KII¢oll + K

::; Kd + +2Kd(1 +

sup (x(u)

+ J:~ x(s)d~J-L(u, s)

t: d~J-L(to, s)) < 6, to ::; t < t2. to~1t~t

That is

o < x(t)

< fJ, to

~

t < t 2.

(11)

From (i), we have

y'(t)

~

a(t)v(t)y(t), to

and by the Gronwall inequality

yet)

~ y(to) exp

it

~

t < t2

a(u)v(u)du, to

to

~ t < tz

which leads to the contradiction:

to y(t2) ~ y(to) exp ft:2 a(u)v(u)du < 2d(1 + fe d~J-L(to, s ) exp ftt02 a(u)v(u)du

= y(t2).

Consider now the case (b). Similarly to the case (a),

o < ~ (t)

< s, to

~

t < t3.

We will show that

y'(t»O for to~t

(12)

Since y'(to) > 0 and y'(t) is continuous it is clear that (12) holds for any t 2: to near to. If (12) does not hold there is a t 4 E (to, t 3) such that y'(t) > 0 in (to, t 4 ) and y'(t 4 ) = o. From the equation of y'(t) we obtain

J: a(t)x(s)d~TJ(t, s) exp t: a(u)v(u)du + a(t)v(t)y(t) = -(exp to a(u)v(u)du)U: a(t)y(s) exp( - ft7 ax o)a(u)v(u)du)d.1](t, s)] h a(u)v(u)du) J: a(t)(Je~ x(r)drJ-L(s, r»d~TJ(t, s) + a(t)v(t)y(t)

y'(t) 2: -

(. ,t

t

+(exp

=a(t){ -

o

J: y(s)(exp J~ax(~,to) a(u)v(u)du)d~TJ(t,s) t

+expJt o a(u)v(u)du)f: f; x(r)drJ-L(s,r)d~TJ(t,s)+ v(t)y(t)}.

Huang & Gao: SOLUTIONS OF DIFFERENTIAL EQUATIONS

No.3

325

Thus

J:4 y(s)(exp J~:.x(.,tO) a(U)V(u)du)d.1](t4, s)

y'(t4) ~ a(t 4){ -

+[exp ftt4 a(u)v(u)du) f: 4 f: x(r)drJ-L(s, r)d$TJ(t 4, s) + V(t4)y(t 4)} 0

f:

4 > a(t4)y(t4){ (ex p f~:.x($,to) a(u)v(u)du)d$TJ(t4, s) + V(t4) t 4 + (exp ft a(u)v(u)du)Y(~4) f: 4 f: x(r)drJ-L(s, r)d$TJ(t4, s)} 0

J:4(exp J::ax(.,t o) a(u)v(u)du)d.1](t4' s) 1'4 a(u)v(u)du) J'4 V 1l(&,.)d.'1(t4,$)

~ a(t4)y(t4){V(t4) 6(exp

to

C

[Ct.)

}

~o

(by the condition (6)), that is, y'(t 4) > 0 which contradicts the definition of t 4. Therefore [12) holds and y(t3) > y(to). This contradicts the definition of t4.. Since both cases (a) and (b) lead to a contradiction, for all t in the interval (to, tl) d(l

+

t

d$J-L(t o, s) < yet) < 2d(1 +

c

t

d$J-L(to, s)) exp

c

it to

a(u)v(u)du.

Similarly to the case (a), we obtain

o < x(t) < 6, to ::; t < t l. From (i) and (ii), t l = +00 (see [13]) and so (7) holds. The proof of the theorem is complete. Proof of Theorem 2 As the proof of Theorem 1, we will establish that (9) holds for any t E (to, tl)' Otherwise we have two possible cases (a) and (b) as in the proof of Theorem 1. To discuss the case (a) is the same as in the proof of Theroem 1. Now consider the case (b). That is to say that there exists t3 E (to, t l ) such that (9) holds for any t E (to, t 3) and y(t 3 ) = d(l + 0 d8 J.L (t o, s)). Similarly to (a), we obtain

1:

Thus

y'(t) 2: -

jt a(t)x(s)d.1](t, s) exp(i a(u)v(u)du) + a(t)v(t)y(t), t

to

c

to ::; t < t3. By the Gronwall inequality we have yet) 2 y(to) exp It: a(u)v(u)du

- It: I~4 a(s)x(r)dr'T](s, r)(exp It: a(u)v(u)du)(exp I8t a(u)v(u)du)ds ~ y(to) exp It: a(u)v(u)du - t; a(s)('T](S, s) -'T](S, c))ds6 exp It: a(u)v(u)du, to ~ t < t3.

and

d(l

+ J:o d.Jl(to, s)) ~ d(l + J:o d.Jl(to, s)( exp itto'a(u)v(u)du) -6(exp

Itto a(u)v(u)du) Itto a(s)(TJ(S, s) _OTJ(S, c))ds, 3

3

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ACTA MATHEMATICA SCIENTIA

that is,

d(l

+ J:o d~p,(to, s»)(exp J/

3 0

~ 6(exp

Vol.16

a(u)v(u)du - 1)

ftt a(u)v(u )du) ftt a(s)( 1]( s, s) - 1]( s, c) )ds. 3

03

0

This contradicts the condition (8). Since both cases (a) and (b) lead to a contradiction, (9) holds for all t E (to, t 1 ) . Similarly to the proof of Theroem 1, we have a positive solution x(t) of (1) such that 0 < x(t) < 6 for all t E [to, +00) and (7) holds. The proof of Theorem 2 is complete. Corollary 1 Consider the first order neutral differential equation d

dt [xCt) + AX(t - r)] + I(t, x[g(t)]) := 0, t ? to

(13)

where get) is continuous, get) ~ t'tlim g(t) = 00 and f satisfies condition (i), r ? 0, A E -+00

(-1,0]. If e

exp(--, 1+ A

it

max(g(t),to)

a(r)dr)

e . + ( 41AIA)2 ex p(--,.

1+

1+

1 t

to

A

e

a(r)dr) ~ - 1' ,t ? to

+A

(14)

then (13) has a positive solution x(t) on [c,oo), where c = min{t o, -r, inf g(t)} and x t~to

satisfies

(x(to)

+ AX(to -

Proof Let 1](t, s)

1(u)

k

={

0, U

1,u

= 1~..\ [2.7].

<

°.

e r» exp( - ~

= l(s -

+

get»~,

p,(t,s)

It to

a(r)dr) < x(t)

+ AX(t -

r), t ~ to,

(15)

= Al(s - t + r), t ~ to, s ~ c, where

Then 1] and p. satisfy the assumptions (ii), (iii), (iv) and (v), moreover,

~ 0

We take d

= !o(1 + A) and v(t) = 1~..\' then d and v satisfy (4) and (5). = exp( 1~..\ J~ax(g(t),to) a(r)dr)

The left side of (6)

6

+( 16(1 + ..\) 2 exp( 1~..\ 4

it: a(r)dr

V

[c,g(t)]

JL(g(t), .)

= exp(l~..\ f~ax{g(t).to} a(r)dr) + (1~~\2 exp(l~..\a(r)dr) ~ 1~~ (by(14» = vet), That is, (6) of Theorem 1 holds and the result follows from theorem 1.

=

Remark If A 0, Corollary 1 becomes Corollary 1 of [1]. Corollary 2 Consider the first cr-Ier linear neutral differential eqauation

~ [x(t) + AX(t -

r)]

+ a(t)x[g(t)]

= 0, t

~ to

(16)

where a : [to, +00) --+ R is continuous and g(t) as in Corollary 1. If the condition (14) is satisfied then (16) has a positive solution x(t) on [c, +00), where c is as in Corollary 1, and x(t) satisfies

x(t)

~ z(to)[IAl

t

O

- ,.t

+ 1 + (1-IAI~)exp( __ e-

1+A

r a(r)dr],t ~ to.

Jt o

(17)

327

Huang & Gao: SOLUTIONS OF DIFFERENTIAL EQUATIONS

No.3

Proof From Corollary 1 it follows that (16) has a positive solution x(t) on [e, +00). Take fJ = 4 and d = ~fJ(1 + A) = 1 + A. By [7] we have for t ~ to, to + nr ~ t ~ to + (n + 1)r n

x(t) = (_1)n+1Ant l¢o(t - to - (n + 1)r) + L( _1)i Ai[X(t - ir) + AX(t - ir - r)], i=O

where n is a nonnegative integer. Since A < 0, ¢o(s)

x(t)

n

= !A\n+1(1 + A) + L: IAli[x(t ~

IAl n+l(1 + A) +

= IAln+l(1

= d on

ir) + AX(t - ir - r)]

i=O

i: IAli[x(to) + AX(to -

i=O

+ A) + (1 + A)2

[e, to] and (15), we obtain

r)) exp(-

s:

1~.\

t IAl exp( -1~.\ s;: a(r)dr)

a(r)dr)]

0

i

i=O

0

~ X(tO)[IAI~+l + (1 -IAI~+l) exp( -1~.\) J,'o a(r)dr)] ,t ~ to x(to) = 1 + A. The proof of the Corollary 2 is complete. Corollary 3 Consider the neutral differential equation with several delays d

dt [x(t) + AX(t - r)]

n

+ ~Pi(t)X(t - ri(t)) = 0, t 2: to

where pi(t), ri(t) are positive and continuous functions, lim (t-ri(t» = Let c

=

min {inf (t - ri(t)} and K(t)

it

1~I~n t~to

e exp(-l\

+

A

n

LPi(r)dr) + (1

k(t) i=1

=

t-+oo

(18)

+oo(i =

1,2,··· ,n).

min max{t - ri(t), to}. If

1~I~n

41AI exp e A)2 ( -1 ' + +

A

1 t

e

n

LPi(r)dr) S - 1"t 2:: to

+

to i=O

(19)

A

then (18) has a positive solution x(t) on [e, +(0) and x(t) satisfies

x(t)

~ x(to)[IAI

o t-,.t

Proof Let 1J(t,s) =

+1

+ (1 -IAI t-,.t

Q

n

L: Pi(t)I(s -- [t -

i=1

)

exp( -1 e . : tp;(r)dr)], t

+

A

ri(t)]), p,(t, s)

to

i=1

= A1(s - t + r)

~ to.

and f(t,x)

(20)

=

X,

where I is the uni' step function. Then j",1] and JL satisfy the assumptions (i)-(iv). Take d

= 6(1:).)

n

and v(t)

side of (6) is equal to n

t

Then (4) and (5) in Theorem 1 hold. Moreover, the left

i=1

~ Pi(t) exp(!maX(t-Ti(t),tQ)

1=1

.

= 1~..\ L: p(t).

~

en 1+).

ten

.L: Pi (r)dr) + (1+).)2 exp(h o 1+).

1=1

n

~ Pi(r)dr) ~ Pi(t)

1=1

.=1

~ 1=1 tpi(t)[exp(l~.\J~(t) 1=1 tpi(r)dr) + (1~~)' exp(l~.\ Jt: tpi(r)dr) 1=1 n

~

L: Pi(t) 1~..\ (by(19» = v(t), t 2:: to,

i=1

that is, (6) holds. Therefore, from Theorem 1 we have a positive solution x(t) on [e, +00) of (18) and

(x(t o) + AX(to - r)) exp(

-1:

1: A

~Pi( r)dr) < x(t) + AX(t -

r), t

~ to.

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ACTA MATHEMATICA SCIENTIA

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Let 6 = 4 and xo(s) = d on [c, to], and d = (1 + A). Similarly to Corollary 2, we obtain

x(t)

~ x(tO)[IAI~+l + (1 - IAI~) exp( -1 e

+

. : tpi(r)dr), t to i=l

A

~ to,

x(t o) = 1 + A. This completes the proof of the corollary. Remark If A 0, Corollary 3 becomes Corollary 3 in [1]. Example 1 Consider the equation

=

d

1

dt [x(t) - 2 x(t - r)] + 20(2

111 1 + t)2 x(t - 3") + 30(1 ~. t)2 x(t -- 2)

= 0, t ~ O.

(21)

Then c ~ min {inf(t - ri(t))} = -~, k(t) = min max{t - ri(~), O} = max{t - ~,O}A l<.

=

=

1/(t, s)

=L 2

=

Pi (t)I( s

=

- [t - ri(t)]), p,(t, s)

1

=- -I( s - t + r), 2

i=l

rl(t)

=!, r2(t) =! 3 2

=z,

and f(t, x)

Thus (4) holds and q(O, 0) - q(O, the left side of (6) is equal to

-t) =

8~

where I is as in the proof of Corollary l.

< 1(1-

t) = t

t

2

=

that is, (5) is valid. Moreover,

2

t

i~ p,(t) exp fmax{t-Ti(t),O} v(r)dr) + 8 i~ pi(t) exp fo v(r)dr ~

2

t

. t"

(E pi(t)) exp(h_ J. v(r)dr) + 8 exp fo v(r)dr» i=l

"

2

= (1 + t)(t:h" + 8){-~ ~\2~t)2 + 30(1~t)~) ~ 10(t + 1)( 20(1~t)2 + 30(1~t)2) ~ l~t

= vet), t ~ o.

That is, (6)" of Theorem 1 holds. Therefore, from Theorem 1 it follows that equation (21) has a positive solution x(t) on [-~, +00) and similarly to the proof of Corollary 2, we have

1

t

x(t)~x(0)[(2);:

+1

1

1

i

+ t+1(1-(2»·]),t~0.

Example 2 Consider the equation d -d [x(t) + Ax(t - r)]

t

1

+ (1 +t )2 x f3(t -

r(t»

= O.t >- 0

(22)

where A E (-1, O],{3 > 1, ret) is a positive and continuous func..on such that lim (t ret)~

= +00.

f(t, x)

Let c

= t>o inf{t ~ ret)}, 1/(t, s) = I[s -

= (1';t)2 x f3 , such that

1

{3 1

(t - ret))], p,(t, s)

= AI(s -

( O

t-+oo

t + r) and

Huang & Gao: SOLUTIONS. OF DIFFERENTIAL EQUATIONS

NQ.3

where fJ satisfies fJ{3-1

329

1 < -(1 + A)2 .4

Set d =~15(1 +A), a(t) = (~~;2 and v(t) = 6~:'1 (1 +t) for all t ~. O. By the proof of Corollary 1, (4) holds and from the definition of fJ we have 1](0,0) - 1](0, c) ~ 1 <

1 fj{3-1

(1 + A)

.

= v(O)(l + A)

that is, (5) holds. From the definition of d, aCt) and vet), the condition (8) of Theorem 2 becomes 1 -(1+A)2(1-exp(4

it 0

1 --dr)) «

l+r

s":'

it 0

(1 1 p(1J(r,r)-1J(t,c))dr,t> 0 +r

The definition of "1 leads to {3-1

e

ft

Jo

1

(1+r)2(1J(r,r)-1J(t,c)) dr

{3-1

r (l+r)2dr-fJ 1

:s o Jo

_

(23)

{3-1 _ _ t_

(l+t)'

so (23) holds. Therefore (8) of Theorem 2 is valid and by Theorem 2 equation (22) has a positive solution x(t) on [c, +00) such that 0 < x(t) < s for all t 2: 0 and fJ{3-1 < ~(1 + A)2 and the estimate (7) holds. Similarly to the proof of Corollary 2, we obtain t+l

x(t) ~ x(to)[I>'I"

1

+ 1 + t (1 - IAI")], t t

~

o.

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