INEQUALITIES FOR POSITIVE
SOLUTIONS OF
n
THE EQUATION ẏ(t) = − i=1 (ai + bi /t)y(t − τi )
JOSEF DIBLÍK AND MÁRIA KÚDELČÍKOVÁ
The equation ẏ(t) = − ni=1 (ai + bi /t)y(t − τi ), where ai ,τi ∈ (0, ∞), i = 1,2,...,n, and
is considered when t → ∞ under supposition that the transcendenbi ∈ R are constants,
tal equation λ = ni=1 ai eλτi has two real and different roots. The existence of a positive
solution is proved as well as its asymptotic behaviour.
Copyright © 2006 J. Diblı́k and M. Kúdelčı́ková. This is an open access article distributed
under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
We consider equation
ẏ(t) = −
n i =1
ai +
bi y t − τi ,
t
(1.1)
where ai ,τi ∈ R+ := (0, ∞), i = 1,2,...,n, and bi ∈ R are constants. The case when there
exist positive solutions is studied. In the supposition of existence of two real (positive)
different roots λ j , j = 1,2, λ1 < λ2 , of the transcendental equation
λ=
n
i =1
ai eλτi ,
(1.2)
we prove the existence of a positive solution y = y(t) having for t → ∞ asymptotic behaviour
y(t) ∼ e−λ1 t t r1
(1.3)
with an appropriate number r1 . Corresponding below and upper inequalities are given
for y(t), too.
Hindawi Publishing Corporation
Proceedings of the Conference on Differential & Difference Equations and Applications, pp. 341–350
342
Solutions of the equation ẏ(t) = −
n
i=1 (ai + bi /t)y(t − τi )
2. A nonlinear theorem
Let Rn be equipped with the maximum norm. With Rn≥0 (Rn>0 ) we denote the set of all
componentwise nonnegative (positive) vectors v in Rn , that is, v = (v1 ,...,vn ) with vi ≥ 0
(vi > 0) for i = 1,...,n. For u,v ∈ Rn , we say u ≤ v if v − u ∈ Rn≥0 , u v if v − u ∈ Rn>0 ,
= v.
u < v if u ≤ v and u Let C([a,b],Rn ), where a,b ∈ R, a < b, be the Banach space of the continuous mappings from the interval [a,b] into Rn equipped with the supremum norm ψ = supθ∈[a,b]
|ψ(θ)|, ψ ∈ C([a,b],Rn ). We will denote this space as Cr if a = −r < 0 and b = 0.
Let us consider a system of functional differential equations
ẏ(t) = f t, yt ,
(2.1)
where f : Ω → Rn is a continuous quasibounded functional which satisfies a local Lipschitz condition with respect to the second argument and Ω is an open subset in R × Cr .
We assume that the derivative in (2.1) is at least right-sided.
If σ ∈ Rn , A ≥ 0, and y ∈ C([σ − r,σ + A],Rn ), then for each t ∈ [σ,σ + A] we define
yt ∈ Cr by means of relation yt (θ) = y(t + θ), θ ∈ [−r,0].
In accordance with [2], a function y is said to be a solution of the system (2.1) on
[σ − r,σ + A) with A > 0 if y ∈ C([σ − r,σ + A),Rn ), (t, yt ) ∈ Ω for t ∈ [σ,σ + A) and
y(t) satisfies (2.1) for t ∈ [σ,σ + A). For given σ ∈ R, ϕ ∈ Cr , we say y(σ,ϕ) is a solution
of (2.1) through (σ,ϕ) ∈ Ω if there is an A > 0 such that y(σ,ϕ) is a solution of system
(2.1) on [σ − r,σ + A) and yσ (σ,ϕ) = ϕ. In view of the above conditions, each element
(σ,ϕ) ∈ Ω determines a unique solution y(σ,ϕ) of system (2.1) through (σ,ϕ) ∈ Ω on its
maximal interval of existence which depends continuously on the initial data [2].
For given k ∈ Rn>0 , let us consider the integro-functional inequalities
L1 (t) ≤ −λ(t) − I k,L1 (t)
L2 (t) ≥ −λ(t) − I k,L2 (t)
−1 −1 f t,I k,L1
f t,I k,L2
t
,
(2.2)
t
on [t0 , ∞), t0 ∈ Rn , for L j ∈ C([t0 − r, ∞),Rn ), j = 1,2, where
I : Rn>0 × C t0 − r, ∞ ,Rn −→ C t0 − r, ∞ ,Rn
(2.3)
is defined by
I i (k,L)(t) := ki exp −
t
t 0 −r
λi (s)ds +
∞
t
Li (s)ds ,
(2.4)
i = 1,...,n, t ∈ [t
0 − r, ∞) with a fixed function λ ∈ C([t0 − r, ∞),Rn ), provided that im∞
proper integrals Li (s)ds exist.
We consider an operator equation
L(t) = (TL)(t) := −λ(t) − I(k,L)(t)
−1 f t,I(k,L)t ,
(2.5)
J. Diblı́k and M. Kúdelčı́ková 343
where λ is a fixed function on interval [t0 − r, ∞). It is easy to verify that system (2.1) is
connected with operator equation (2.5) by the substitution
y(t) = I(k,L)(t).
(2.6)
A function L is said to be a solution of operator equation (2.5) on interval [σ − r,σ + A)
with A > 0 if L ∈ C([σ − r,σ + A),Rn ), (t,Lt ) ∈ Ω for t ∈ [σ,σ + A) and L(t) satisfies (2.5)
for t ∈ [σ,σ + A). The next theorem which is necessary for our investigation and taken
from [1] indicates conditions under which there exists solution of system (2.1).
Theorem 2.1. Let us suppose that
(i) for any M ≥ 0, ϑ ≥ t0 , there are K1 , K2 such that
(TL)(t) ≤ K1 ,
(TL)(t) − (TL)(t ) ≤ K2 |t − t |
(2.7)
for any t,t ∈ [t0 ,ϑ] and any function L ∈ C([t0 − r,ϑ),Rn ) with |L| ≤ M;
(ii) there are k ∈ Rn>0 and functions L j ∈ C([t0 − r, ∞),Rn ), j = 1,2, with convergent
∞
integrals Lij (s)ds, j = 1,2;i = 1,...,n, satisfying L1 (t) ≤ L2 (t) on [t0 − r, ∞) and
the inequalities (2.2) on [t0 , ∞), that is, the inequalities:
L1 (t) ≤ TL1 (t),
L2 (t) ≥ TL2 (t);
(2.8)
(iii) there is a Lipschitz continuous function ϕ : [t0 − r,t0 ] → Rn satisfying ϕ(t0 ) = 0 and
on [t0 − r,t0 ] inequalities
L1 (t) ≤ TL1 t0 + ϕ(t),
L2 (t) ≥ TL2 t0 + ϕ(t);
(2.9)
(iv) for any functions Λ j (t) ∈ C([t0 − r,ϑ),Rn ), j = 1,2, with Λ1 (t) ≤ Λ2 (t) for t ∈ [t0 −
r,ϑ), ϑ ≥ t0 , we have
TΛ1 (t) ≤ TΛ2 (t),
t ∈ t0 , ∞ .
(2.10)
Then there exists a solution y of the system (2.1) on [t0 − r, ∞) satisfying
I k,L1 (t) ≤ y(t) ≤ I k,L2 (t)
(2.11)
on t ∈ [t0 − r, ∞).
3. Linear corollary of the theorem
Let us apply Theorem
2.1 to the investigation of delayed linear equation of the type (2.1)
n
with f (t, yt ) := − i=1 ci (t)y(t − τi (t)), that is, to the equation
ẏ(t) = −
n
i=1
ci (t)y t − τi (t)
(3.1)
344
Solutions of the equation ẏ(t) = −
n
i=1 (ai + bi /t)y(t − τi )
with a positive Lipschitz continuous bounded coefficients ci on [t0 , ∞) and Lipschitz continuous bounded positive delays τi (t) ≤ r, r ∈ R+ , where we suppose differences t − τi (t)
increasing on [t0 , ∞). Let us define τ(t) := max{τi (t)}, i = 1,2,...,n. Using the substitution (2.6) for linear case we obtain the operator (TL)(t) defined by formula (2.5) in the
form
(TL)(t) := −λ(t) +
n
i =1
ci (t)exp
t
t −τi (t)
λ(s) + L(s) ds .
(3.2)
Now we apply Theorem 2.1 to linear equation (3.1). We omit corresponding (technically cumbersome) proof.
Theorem 3.1. Let us suppose that there are continuous functions L j : [t0 − τ(t0 ), ∞) → R,
j = 1,2, and a Lipschitz continuous function ϕ(t) : [t0 − τ(t0 ),t0 ) → R satisfying ϕ(t0 ) = 0,
∞
L1 (t) ≤ L2 (t) for t ∈ [t0 − τ(t0 ), ∞) and integrals L j (s)ds, j = 1,2, exist. Let a Lipschitz
continuous bounded function λ : [t0 − τ(t0 ), ∞) → R be given such that also the following
inequalities are satisfied:
λ(t) + L1 (t) ≤
n
i =1
λ(t) + L2 (t) ≥
n
i =1
ci (t)exp
ci (t)exp
t
λ(s) + L1 (s) ds ,
t −τi (t)
t
(3.3)
λ(s) + L2 (s) ds
t −τi (t)
(3.4)
on interval [t0 , ∞), and
λ(t0 ) + L1 (t) ≤
n
i=1
λ t0 + L2 (t) ≥
ci t0 exp
n
i=1
ci t0 exp
t0
t0 −τi (t0 )
t0
t0 −τi (t0 )
λ(s) + L1 (s) ds + ϕ(t),
(3.5)
λ(s) + L2 (s) ds + ϕ(t)
(3.6)
on interval [t0 − τ(t0 ),t0 ].
Then there exists a solution y = y(t) of (3.1) on [t0 − τ(t0 ), ∞), such that
∞
exp
t
t
∞
L1 (s)ds ≤ y(t) · e t0 −τ(t0 ) λ(s)ds ≤ exp
t
L2 (s)ds .
(3.7)
4. Investigation of (1.1)
We consider delayed equation of the type (3.1) with ci (t) := (ai + bi /t), τi (t) := τi = const,
where ai ,τi ∈ R+ and bi ∈ R, i = 1,2,...,n, that is, (1.1).
J. Diblı́k and M. Kúdelčı́ková 345
We will try to find so-called approximative solutions corresponding to (1.1) in the
form
y as (t) ∝ e−λt t r 1 +
A
,
t
(4.1)
where r and A are coefficients (specified below in Theorem 4.3) and λ is one from two
real positive different roots λ1 < λ2 of the transcendental equation (1.2), the existence of
which is supposed. We show that these solutions satisfy formally for t → ∞ (1.1) with the
order of accuracy O(e−λt t r −2 ).
We define the auxiliary function f (λ) := λ − ni=1 ai eλτi . Then
f (λ) = 1 −
n
i=1
ai τi eλτi ,
f (λ) = −
n
i =1
ai τi 2 eλτi .
(4.2)
Lemma 4.1. Let positive constants ai ,τi , i = 1,2,...,n, be given and let just two real different
positive roots λ j , j = 1,2, λ1 < λ2 of (1.2) exist. Then f (λ1 ) > 0 and f (λ2 ) < 0.
Proof. The second derivative (4.2) of the function f is on R negative. Therefore the first
derivative f is a decreasing function, f (0) = 1 and f (+∞) = −∞. Since (1.2) has just
two different positive roots, it means that there exists just one point (extremal point)
λ ∈ (λ1 ,λ2 ) for which it holds f (λ ) = 0. At λ the auxiliary function f (λ) reaches its
maximum and f (λ ) > 0. Since that f (λ1 ) > 0 (function f (λ) increases to its maximum)
and f (λ2 ) < 0 (function f (λ) decreases from its maximum).
The following lemma, the proof of which can be made easily using the binomial formula and the method of induction and therefore is omitted, will be used in the proof of
next theorem.
Lemma 4.2. Let r ∈ R be given. Then the asymptotic representation
(t − τ)r = t r 1 −
rτ r(r − 1)τ 2
1
+o 2
+
t
2t 2
t
(4.3)
holds for t → ∞.
Theorem 4.3. Let positive constants ai ,τi , i = 1,2,...,n, be given and (1.2) has just two
real different roots λ j , j = 1,2, λ1 < λ2 . Then there exist two approximative solutions of (1.1)
having the form (4.1):
−λ j t r j
t 1+
y as
j (t) ∝ e
Aj
,
t
j = 1,2,
(4.4)
with
rj = −
Aj =
n
i=1
bi e
λ j τi
−1
f λj
,
(4.5)
n
r j r j − 1 1
λ j τi
−
f
λ
r
b
τ
e
.
−
j
j
i i
f λj
2
i=1
(4.6)
346
Solutions of the equation ẏ(t) = −
n
i=1 (ai + bi /t)y(t − τi )
Proof. At first, note that the coefficients r j , A j , j = 1,2, are well defined since, due to
= 0. Now substituting the approximative solution (4.1) into (1.1), we
Lemma 4.1, f (λ j ) expect that
− λe−λt t r + At r −1 + e−λt rt r −1 + A(r − 1)t r −2
n r
r −1 b
∝−
ai + i e−λ(t−τi ) t − τi + A t − τi
.
(4.7)
t
i=1
With the aid of Lemma 4.2 and after some necessary computations, we get
λA r A(r − 1)
+ +
t
t
t2
n
1
∝ − eλτi ai + ai A − ai rτi + bi
t
i =1
−λ−
(4.8)
+
1
1
ai Aτi + A − ai rτi + bi + ai rτi 2 (r − 1) − bi rτi
t2
2
.
Now comparing the coefficients at the members with the same powers t 0 , t −1 , and t −2 of
t, we obtain
λ=
n
i=1
−λA + r = −
n
n
i =1
ai eλτi ,
eλτi ai A − ai rτi + bi ,
(4.9)
1
A(r − 1) = − e τi ai A − ai Ar + bi A + ai r(r − 1)τi − bi r .
2
i =1
λτi
When using the first relation, from the second one, we get
r=
n
1
bi eλτi
−
f (λ)
i=1
1−
n
i=1
ai τi eλτi
−1 −
n
i=1
bi eλτi .
(4.10)
From the third equation (using above relations) after some simplifications, we have
n
1
r(r − 1) −
A= f (λ) − r bi τi eλτi .
f (λ)
2
i =1
(4.11)
Now identifying the corresponding λ, r, A with λ1 , r1 , A1 or with λ2 , r2 , A2 , we get the
pair of approximative solutions y1as (t) and y2as (t). Thus the proof is finished.
J. Diblı́k and M. Kúdelčı́ková 347
Now we define with the aid of coefficients of approximative solutions function λ(t)
and functions L1 (t) and L2 (t), L1 (t) ≤ L2 (t), corresponding to the root λ = λ1 of (1.2) as
r1
,
t A1 − ε1
,
L1 (t) :=
t2 A1 + ε1
,
L2 (t) :=
t2
λ(t) := λ1 −
(4.12)
with a constant ε1 ∈ (0,1).
Theorem 4.4. Suppose that (1.2) has just two real different roots λ j , j = 1,2, λ1 < λ2 . Then
for every ε1 ∈ (0,1) there exist a t0 ∈ R, t0 > τ, τ = max{τi }, i = 1,2,...,n, and a positive
solution y ∗ (t) of (1.1) on [t0 − τ, ∞) satisfying inequalities
e −λ 1 t t r 1 1 +
A1 − ε1
A1 + ε1
,
≤ y ∗ (t) ≤ e−λ1 t t r1 1 +
t
t
(4.13)
where coefficients r1 , A1 are defined by formulas (4.5), (4.6).
Proof. We employ Theorem 3.1. It is supposed that ε1 is a fixed positive number and t0
is large enough to indicate the asymptotic relations and inequalities are valid. Let λ(t),
L1 (t), L2 (t) be defined by formulae (4.12). At first we show that functions L1 (t), L2 (t)
really satisfy inequalities (3.3) and (3.4). We have to verify that
λ1 −
n
2
A1 − ε1
b t
r1
≤
ai + i e t−τi (t) (λ1 −r1 /s+(A1 −ε1 )/s )ds
+
2
t
t
t
i =1
(4.14)
holds. Let us simplify the right-hand side (denote it by ᏾). After integration we obtain
᏾=
n i=1
=
n i=1
ai +
1
1
bi
t
exp λ1 τi − r1 ln
− A1 − ε1
−
t
t − τi
t t − τi
b
t − τi
ai + i e λ 1 τ i
t
t
r1
(4.15)
· Ᏹ τi
with
τi A1 − ε1
.
Ᏹ τi := exp
t t − τi
(4.16)
It is easy to see that for sufficiently large t,
Ᏹ(τi ) = exp τi A1 − ε1
1
t2
1+
1
τi
+o
t
t
= 1+
τi A1 − ε1
1
+o 2 .
t2
t
(4.17)
Solutions of the equation ẏ(t) = −
348
n
i=1 (ai + bi /t)y(t − τi )
With the aid of Lemma 4.2, we get
᏾=
n i=1
=
n i=1
b
r τ r1 r1 − 1 τi 2
1
ai + i e λ 1 τ i 1 − 1 i +
+o 2
t
t
2t 2
t
ai +
· Ᏹ τi
A1 − ε1 τi r1 r1 − 1 τi 2
bi λ1 τi
rτ
1
e
1− 1 i +
+
+o 2
t
t
t2
2t 2
t
1
ai r1 r1 − 1 τi 2
1
1
+o 2
= e · ai + bi − ai r1 τi + 2 ai A1 τi − ai ε1 τi − bi r1 τi +
t
t
2
t
1
= λ1 + − r1 f λ1 + r1 f λ1 − 1
t
λ1 τ i
+
n
1 1
1
+o 2 .
− r1 r1 − 1 f λ1 − r1 bi τi eλ1 τi + A1 − ε1 1 − f λ1
2
t
2
t
i =1
(4.18)
Now we compare coefficients at the members with the same powers (t 0 , t −1 , and t −2 ) of t
at left-hand and right-hand sides of the inequality (4.14). Left-hand sides are denoted as
ᏸ and right-hand sides as ᏾ with corresponding indices. We obtain
ᏸ0 = λ1 ,
᏾0 = λ1 ,
ᏸ−1 = −r1 ,
᏾−1 = −r1 ,
ᏸ−2 = A1 − ε1 ,
1 ᏾−2 = − r1 r1 − 1 f
2
λ1 − r1
n
i =1
(4.19)
bi τi eλ1 τi + A1 − ε1 1 − f λ1 .
Obviously ᏸ0 = ᏾0 and ᏸ−1 = ᏾−1 is valid. Since, in view of (4.2), (4.5), and (4.6),
᏾−2 = A1 f λ1 + A1 − A1 f λ1 − ε1 + ε1 f λ1 = A1 − ε1 + ε1 f λ1 ,
(4.20)
then, for the validity of inequality ᏸ−2 < ᏾−2 , the inequality f (λ1 ) > 0 is sufficient. This
is true (see Lemma 4.1). That means (3.3) is fulfilled. Inequality (3.4) for L2 (t) holds by
the same arguments.
Let us show that also inequalities (3.5) and (3.6) hold on [t0 − τ,t0 ] with Lipschitz
continuous function
ϕ(t) := L1 (t) − L1 t0 .
(4.21)
At first we verify validity of (3.5), that is, validity of
λ1 −
n
2
b t0
A −ε
r1 A1 − ε1 A −ε
+
≤
ai + i e t0 −τi (λ1 −r1 /s+(A1 −ε1 )/s )ds + 1 2 1 − 1 2 1 ,
2
t0
t
t
t
t0
0
i =1
(4.22)
J. Diblı́k and M. Kúdelčı́ková 349
which holds automatically since the inequality
λ1 −
n
2
b t0
r1 A1 − ε1 +
≤
ai + i e t0 −τi (λ1 −r1 /s+(A1 −ε1 )/s )ds
2
t0
t
t0
0
i=1
(4.23)
is a special case of (3.3) with t = t0 . To show (3.6), we have to verify if
λ1 −
n
2
b t0
A −ε
r1 A1 + ε1 A −ε
+
≥
ai + i e t0 −τi (λ1 −r1 /s+(A1 −ε1 )/s )ds + 1 2 1 − 1 2 1
2
t0
t
t
t
t0
0
i =1
(4.24)
holds on [t0 − τ,t0 ]. After integration, some simplifications, using (4.2), (4.6), and above
computation for ᏾, we can see that it is necessary to verify the inequality
1 A1 − ε1 2ε1
+ 2 ≥ 2 A1 − ε1 1 − f λ1 + A1 f λ1 ,
2
t
t0
t0
(4.25)
which can be simplified to 2t02 ≥ f (λ1 )t 2 or (since t0 ≥ t) to the inequality f (λ1 ) ≤ 2
which holds obviously. Hence (3.6) is fulfilled on [t0 − τ,t0 ].
All conditions of Theorem 3.1 are valid. It follows from inequalities (3.7) that there
exists a solution y = y(t) such that
y1 (t) ≤ y(t) ≤ y2 (t)
(4.26)
with y1 (t) := I(k,L1 )(t) and y2 (t) := I(k,L2 )(t). After some simplifications, we get
y j (t) = I k,L j (t) = exp −
= exp −
t
λ1 −
t 0 −τ
t
t 0 −τ
λ(s)ds +
r1
ds +
s
∞
∞
t
t
∞ s
t
t
A −ε
+ 1 1
= exp − λ1 t − t0 + τ + r1 ln
t0 − τ
t
= e−λ1 (t−t0 +τ)
= e−λ1 (τ −t0 ) t0 − τ
1
t0 − τ
r1
t
= K j e −λ 1 t t r 1 1 +
exp
A1 − ε1
ds
s2
t
A1 − ε1
= exp − λ1 s − r1 lns t0 −τ + −
L j (s)ds
A1 − ε1
t
(4.27)
A1 − ε1
1
r1 e−λ1 t t r1 1 +
+o
A1 − ε1
1
+o
t
t
,
t
t
j = 1,2,
with
K j := e−λ1 (τ −t0 ) t0 − τ
−r1
,
j = 1,2.
(4.28)
Put y = y ∗ (t) := y(t)/K1 . Since K1 = K2 , inequality (4.13) holds due to the linearity of
(3.1).
350
Solutions of the equation ẏ(t) = −
n
i=1 (ai + bi /t)y(t − τi )
Acknowledgments
The first author was supported by the Grant A 1163401 of Grant Agency of the AS CR
and by the Council of Czech Government MSM 0021630503. The second author was
supported by the Grants nos. 1/0026/03 and 1/3238/06 of the Grant Agency of Slovak
Republic (VEGA).
References
[1] J. Diblı́k and M. Kúdelčı́ková, Inequalities for positive solutions of the equation ẏ(t) = −(a0 +
a1 /t)y(t − τ1 ) − (b0 + b1 /t)y(t − τ2 ), Studies of the University of Žilina. Mathematical Series 17
(2003), no. 1, 27–46.
[2] J. K. Hale and S. M. Verduyn Lunel, Introduction to Functional-Differential Equations, Applied
Mathematical Sciences, vol. 99, Springer, New York, 1993.
Josef Diblı́k: Department of Mathematics, Faculty of Electrical Engineering and Communication,
Brno University of Technology, Technická 8, 61600 Brno, Czech Republic
E-mail address: [email protected]
Mária Kúdelčı́ková: Department of Mathematical Analysis and Applied Mathematics,
Faculty of Science, Žilina University, Hurbanova 15, 01026 Žilina, Slovak Republic
E-mail address: [email protected]