Universidad Nacional de Chimborazo
NOVASINERGIA, 2020, Vol. 3, No. 2, junio-diciembre, (67-79)
ISSN: 2631-2654
https://doi.org/10.37135/ns.01.06.06
Research Article
Existence of bounded solutions for retarded equations with
infinite delay, impulses, and nonlocal condition
Existencia de soluciones limitadas para ecuaciones retardadas con
retardo infinito, impulsos y condición no local
Génesis Carrillo , Carlos Chipantiza , Hugo Leiva *
School of Mathematical and Computational Sciences, Department of Mathematics, San Miguel de Urcuquí,
Ecuador, 100119; genesis.carrillo@yachaytech.edu.ec, carlos.chipantiza@yachaytech.edu.ec
* Correspondence: hleiva@yachaytech.edu.ec
Recibido 22 octubre 2020; Aceptado 04 noviembre 2020; Publicado 01 diciembre 2020
Abstract:
In this work, we study the existence of bounded solutions for a semilinear retarded
equation with infinite delay, impulse, and non-local conditions. We also show that under
some conditions this bounded solution is unique, periodic, or almost periodic depending
on the conditions imposed on the terms involving the equation. Through this work, we
shall assume that the associated linear equation has an exponential dichotomy, allowing
us to find a formula for the bounded solutions, and from this formula, we are able to
apply Banach Fixed Point Theorem to prove the existence of such bounded solutions.
Keywords:
Exponential dichotomy, bounded solution, unique, periodic, almost periodic.
Resumen:
En este trabajo, estudiamos la existencia de soluciones acotadas para una ecuación
retardada semilineal con retardo infinito, impulso y condiciones no locales. También
mostramos que bajo algunas condiciones esta solución acotada es única, periódica o
casi periódica dependiendo de las condiciones impuestas a los términos que involucran
la ecuación. A través de este trabajo, asumiremos que la ecuación lineal asociada tiene
una dicotomía exponencial, lo que nos permite encontrar una fórmula para las soluciones
acotadas, y a partir de esta fórmula, podemos aplicar el Teorema del punto fijo de
Banach para demostrar la existencia de tales soluciones acotadas.
Palabras clave:
Cuasi periódica, dicotomía exponencial, periódica, solución acotada, única.
1 Introduction
There are many works on the existence of
bounded solutions without impulses, non-local
conditions and delay simultaneously, to men-
tion we have the works done in((Leiva, 1999a,
2000, 1999b; Leiva & Sivoli, 2003; Leiva & Se-
quera, 2003; Leiva & Sivoli, 2018; Liu, 2000; Liu
et al., 2006)). Recently, in (Ayala et al., 2020)
the existence of solutions for retarded equations
with infinite delay, impulses, and non-local con-
ditions has been proved using Karakosta’s fixed
point theorem. In (Abbas et al., 2020), the exis-
tence of periodic mild solutions of infinite delay
evolution equations with non-instantaneous im-
pulses has been studied, by using Poincare map,
measure of non-compactness and Darbo fixed
point theorem. Compared with these works,
in addition we have non-local conditions, and
first, we prove the existence of bounded solu-
tions, and under son conditions these bounded
solutions are stable, periodic or almost periodic
depending on the conditions impose to the lin-
ear and non-linear term. Without further ado,
in this work we shall study the existence of
bounded solutions for the following semi-linear
non-autonomous retarded equation with infinite
delay, impulses and non-local condition:
http://novasinergia.unach.edu.ec
z
0
= A(t)z + f(t,z
t
), t > 0,t 6= t
k
,
z(s) + g(z
τ
1
,··· ,z
τ
q
)(s) = φ(s), s ∈ (−∞,0) = R
−
,
z(t
+
k
) = z(t
−
k
) + J
k
(t
k
,z(t
k
)), k = 1,2,··· ,P,
where A(t) is a continuous n × n matrix defined on R, φ ∈ PW the space defined as follows
PW = {φ : (−∞,0] → R
n
: φ is bounded and continuous except in a finite number
of point, s
φk
,k = 1,2,... , p, where the side limits exists φ(s
−
φk
), φ(s
+
φk
) = φ(s
φk
)}
endowed with the norm
k φ k
PW
= sup
s∈R
−
k φ(s) k,
where the side limits are defined as follows φ(s
+
φk
) = lim
s→s
+
φk
φ(s) and φ(s
−
φk
) = lim
s→s
−
φk
φ(s).
Here, 0 < t
1
< t
2
< ··· < t
p
, 0 < τ
1
< τ
2
< ... < τ
q
, and the functions
g : (PW)
q
→PW, J
K
: R × R
n
→ R
n
, f : R × PW → R
n
are smooth enough such that the problem (1) admits only one solution z(t) (see (Ayala et al., 2020))
given by
z(t) = U(t,0)[φ(0) − g
z
τ
1
,··· ,z
τ
q
(0)]
+
R
t
0
U(t,s)f (s,z
s
)ds + Σ
0<t
k
<t
U (t,t
k
)J
k
(t
k
,z (t
k
)), t ∈ [0,τ]
z(s) = φ(s) − g(z
τ
1
,··· ,z
τ
q
)(s), s ∈ R
−
.
The space (PW)
q
is endowed with the usual norm, and U(t,s) = Φ(t)Φ
−1
(s), where Φ(·) is the
fundamental matrix of the linear system
z
0
(t) = A(t)z(t),t ∈ R (1)
i.e.,
Φ
0
(t) = A(t)Φ(t)
Φ(0) = I
2 Preliminaries
In this section, we shall choose the space where this problem will be set. To this end, we shall define
the following Banach space:
PW
b
(R;R
n
) = {z : R → R
n
:| .z |
R
−
∈ PW and z
R
+
is bounded and continuous except
at the point t
k
, k = 1,2,..., p, where z(t
+
k
),z(t
−
k
) exist and z(t
+
k
) = z(t
k
)},
endowed with the norm
k z k
b
= sup
t∈R
k z(t) k, z ∈ PW
b
.
Now, we shall assume the following hypotheses:
H1) The linear system (1) admits an exponential dichotomy on R. That is to say, there exist a
continuous projection P (t), t ∈ R, β > 0 and M ≥ 1 such that
i) U(t,s)P (s) = P (t)U(t, s), t,s ∈ R,
ii) k U(t, s)(I − P (s)) k6 Me
β(t−s)
, t > s,
iii) k U(t, s)P (s) k6 Me
β(t−s)
, t 6 s
H2) There exists γ > 0 and ` > 0 such that
k g (z
1
,z
2
,...z
q
) k
PW
<
`
6
, z ∈ PW,
k g (z
1
,...,z
q
) − g (w
1
,··· ,w
q
) k
PW
< γsup
t>a
kz(t) − w(t)k
q
http://novasinergia.unach.edu.ec 68
with
k z(t) − w(t) k
q
:=
q
X
i=1
k z
i
(t) − w
i
(t) k
R
n
.
H3) The function f satisfies the following local Lipschitz condition:
Given an interval [a,b] and a ball B
γ
(0) ⊂ PW, there exists a constant K > 0 such that
k f (t,z
1
) − f (t,z
2
) k
R
n
6 K|t − s|+ k z
1
− z
2
k
PW
, z
1
,z
2
∈ B
γ
(0), t,s,∈ [a,b].
Also, there exists a constant L
f
> 0 such that
k f(t, 0) k
R
n
6 L
f
, t ∈ R.
H4) There are constants S
k
,L
k
> 0, k = 1, 2, ··· ,p, such that
k J
K
(t,z
1
) − J
K
(t,z
2
) k
R
n
6 S
k
k z
1
− z
2
k
R
n
, ∀z
1
,z
2
∈ R
n
, ∀t ∈ R;
and
k J
K
(t,0) k
R
n
< L
k
, k = 1,2,··· ,p, t ∈ R.
Lemma 1. Under the hypotheses H1)-H4). A function z belonging to PW
b
is a solution of (1) if,
and only if, z is a solution of the following integral equation
z(t) =
Z
∞
−∞
G(t,s)f (s,z
s
)ds +
X
0<t
k
6t
p
G(t,t
k
)J
k
(t
k
,z (t
k
)), (2)
z(s) = φ(s) − g(z
τ
1
,··· ,z
τ
q
)(s), s ∈ (−∞, 0] = R
−
,
where G(t,s) is the Green function defined by
G(t,s) =
U(t,s)(I − P (s)), t > s,
−U(t,s)P (s), t 6 s.
(3)
Proof. Suppose that for some ρ > 0, z ∈ B
b
ρ
(0) ⊂ PW
b
, where B
b
ρ
(0) is the ball of center zero and
radius ρ > 0 in PW
b
, i.e.,
B
b
ρ
(0) = {z ∈ PW
b
:k z k
b
< ρ}.
Let L
ρ
be the Lipschitz constant of f in B
b
ρ
(0). On the other hand, we have that
z(t) = U(t,0)
φ(0) − g
z
τ
1
,··· ,z
τ
q
(0)
+
Z
t
0
U(t,s)f (s,z
s
)ds +
X
0<t
k
<t
U (t,t
k
)J
k
(t
k
,z (t
k
))
= U(t, t
0
)
(
U(t
0
,0)
φ(0) − g
z
τ
1
,··· ,z
τ
q
(0)
+
Z
t
0
0
U(t
0
,s)f (s,z
s
)ds +
X
0<t
k
<t
0
U (t
0
,t
k
)J
k
(t
k
,z (t
k
))
)
,
+
Z
t
t
0
U(t,s)f (s,z
s
)ds +
X
t
0
<t
k
<t
U (t,t
k
)J
k
(t
k
,z (t
k
)),
for t
0
< t. So,
z(t) = U(t,t
0
)z(t
0
) +
Z
t
t
0
U(t,s)f (s,z
s
)ds +
X
t
0
<t
k
<t
U (t,t
k
)J
k
(t
k
,z (t
k
)).
http://novasinergia.unach.edu.ec 69
Hence,
(I − P (t))z(t) = (I − P (t))U (t,t
0
)z(t
0
) +
Z
t
t
0
(I − P (t))U (t, s)f (s,z
s
)ds
+
X
t
0
<t
k
<t
(I − P (t))U (t, t
k
)J
k
(t
k
,z (t
k
))
= U(t, t
0
)(I − P (t
0
))z (t
0
) +
Z
t
t
0
U(t,s)(I − P (s))f (s,z
s
)ds
+
X
t
0
<t
k
<t
U (t,t
k
)(I − P (t
k
))J
K
(t
k
,z (t
k
)).
On the other hand,
k U (t,t
0
)(I − P (t
0
))z (t
0
) k6 M k z (t
0
) k e
−β(t−t
0
)
, t
0
6 t.
But, k z k
b
< ρ. Then,
k U (t,t
0
)(I − P (t
0
))z (t
0
) k6 Mρe
−β(t−t
0
)
.
Passing to the limit as t
0
→ −∞, we obtain that
lim
t
0
→−∞
k U (t,t
0
)(I − P (t
0
))z (t
0
) k= 0.
Therefore, we get that
(I − P (t))z(t) =
Z
t
−∞
U(t,s)(I − P (s))f(s, z)ds +
X
0<t
k
<t.
(U (t,t
k
)(I − P (t
k
))J
k
(t
k
,z (t
k
)) . (4)
Now, let us prove that this improper integral converges.
k
Z
t
−∞
U(t,s)(I − P (s))f(s, z
s
)ds k 6
Z
t
−∞
k U(t, s)(I − P (s))f (s, z
s
)ds
6
Z
t
−∞
Me
−β(t−s)
k f(s, z
s
) k ds
=
Z
t
−∞
Me
−β(t−s)
k f(s, z
s
) − f(s,0) + f(s,0) k ds
6
Z
t
−∞
Me
−β(t−s)
(L
ρ
ρ k z
s
k +L
f
)ds
=
M(L
ρ
ρ + L
f
)
β
< ∞.
Now, we shall suppose that t
0
> t. Then,
http://novasinergia.unach.edu.ec 70
P (t)z(t) = P (t)U(t,0)
φ(0) − g(t(z
τ
1
,z
τ
2
,··· ,z
τ
q
)(0)
+
Z
t
0
P (t)U(t,s)f (s,z
s
)ds +
X
0<t
k
<t
P (t)U (t,t
k
)J
k
(t
k
,z (t
k
))
= P (t)U(t, t
0
(
U (t
0
,0)
φ(0) − g
z
τ
1
,z
τ
2
,··· ,z
τ
q
(0)
+
Z
t
0
0
U (t
0
,s)f (s,z
s
)ds +
X
0<t
k
<t
0
U (t
0
,t
k
)J
k
(t
k
,z(t
k
))
)
+
Z
t
t
0
U(t,s)P (s)f (s,z
s
)ds +
X
0<t
k
<t
U (t,t
k
)P (t
k
)J
k
(t
k
,z (t
k
))
− P (t)U (t,t
0
)
X
0<t
k
<t
0
U (t
0
,t
k
)J
k
(t
k
,z(t
k
))
= P (t)U (t,t
0
)z (t
0
) +
Z
t
t
0
U(t,s)P (s)f(s,z
s
)ds
+
X
0<t
k
<t
U (t,t
k
)P (t
k
)J(t
k
,z(t
k
)) −
X
t<t
k
<t
0
U (t,t
k
)P (t
k
)J
k
(t
k
,z (t
k
))
= U (t,t
0
)P (t
0
)z (t
0
) +
Z
t
t
0
U(t,s)P (s)f (s,z
s
)ds
−
X
t<t
k
<t
0
U (t,t
k
)P (t
k
)J
K
(t
k
,z (t
k
)).
From hypothesis H1) − iii), we get that
lim
t
0
→+∞
k U (t,t
0
)P (t
0
)z (t
0
) k= 0.
Hence,
P (t)z(t) = −
Z
∞
t
U(t,s)P (s)f (s,z
s
)ds −
X
t<t
k
6t
p
U (t,t
k
)P (t
k
)J
k
(t
k
,z (t
k
)).
Let us prove that this improper integral converges.
k −
Z
∞
t
U(t,s)P (s)f (s,z
s
)ds k 6
Z
∞
t
k U(t, s)P (s)f (s,z
s
) k ds
6
Z
∞
t
Me
β(t−s)
k f (s,z
s
) − f(s,0) + f(s,0) k ds
6
Z
∞
t
Me
β(t−s)
L
ρ
k z
s
k +L
f
ds
6 M
L
ρ
ρ + L
f
Z
∞
t
e
β(t−s)
ds
= M
L
ρ
ρ + L
f
"
e
β(t−s)
−β
#
∞
t
=
M
L
ρ
ρ + L
f
β
< ∞.
http://novasinergia.unach.edu.ec 71
On the other hand,
z(t) = (I − P (t))z(t) + P (t)z(t)
=
Z
∞
−∞
G(t,s)f (s,z
s
)ds +
X
0<t
k
6t
p
G(t,t
k
)J
k
(t
k
,z (t
k
)).
Now, suppose that z is a solution of the integral equation (2). Then,
z(t) =
Z
t
−∞
U(t,s)(I − P (s))f (s,z
s
)ds −
Z
∞
t
U(t,s)P (s)f (s,z
s
)ds
+
X
0<t
k
<t
U (t,t
k
)(I − P (t
k
) J
k
(t
k
,z (t
k
)) −
X
t<t
k
<t
p
U (t,t
k
)P (t
k
)J
K
(t
k
,z (t
k
)).
Therefore,
z
0
(t) =
Z
t
−∞
A(t)U(t,s)(I − P (s))f (s,z
s
)ds −
Z
∞
t
A(t)U(t,s)P (s)f (s,z
s
)ds
+ (I − P (t)) f (t,z
t
) + P (t)f (t,z
t
) +
X
0<t
k
<t
A(t)U (t,t
k
)(I − P (t
k
))J
k
(t
k
,z(t
k
))
−
X
t<t
k
<t
p
A(t)U (t,t
k
)(P (t
k
))J
k
(t
k
,z(t
k
)).
Hence,
z
0
(t) = A(t)z(t) + f (t,z
t
), t > 0.
3 Existence of bounded solutions
In this section, we shall prove the existence of bounded solutions for the system (1), and under some
conditions, we prove the uniqueness of such a bounded solution. Also, under additional conditions,
we prove the stability of these bounded solutions as well.
Theorem 1. Assume the hypotheses H1)-H4). Let B
b
ρ
be the ball of center zero and radius ρ in
PW, and L
ρ
the Lipschitz constant of f in B
2ρ
. If the following estimate holds
ρ
1 − M
2L
ρ
+ βS
β
> M
2L
f
+ β
e
L
β
!
(5)
where S =
P
0<t
k
<t
s
k
and
e
L =
P
P
K=1
L
k
, then the system (1) admits one, and only one, bounded
solution z
b
with k z
b
(t) k6 ρ, t ∈ R. Moreover, if additionally we assume that P (t) ≡ 0 and
5
6
+
3MLρ
β
+ 3SM < 1 and ` 6 S (6)
this bounded solution is locally stable.
Proof. From Lemma 1, it is enough to prove that the operator
K : PW
b
−→ PW
b
t 7−→ (Kz) (t) =
R
∞
−∞
G(t,s)f (s,z
s
)ds +
P
0<t
k
6 t
p
G(t,t
k
)J
k
(t
k
,z (t
k
)),
http://novasinergia.unach.edu.ec 72
has a fixed point in B
b
ρ
. For z ∈ B
b
ρ
, we have the following estimate
k (Kz)(t) k
R
n
≤
Z
∞
−∞
k G(t,s) kk f (s,z
s
) k ds +
X
0<t
k
6t
p
k G(t,t
k
) k J
k
(t
k
,z (t
k
)) k .
From the definition of the Green function, we obtain that
k G(t,s) k6 Me
−β|t−s|
, t,s ∈ R.
Therefore,
k (Kz(t) k 6
Z
∞
−∞
Me
−β|t−s|
k f (s,z
s
) − f(s,0) + f(s,0) k ds
+
X
0<t
k
<t
p
Me
−β|t−t
k
|
k J
k
(t
k
,z (t
k
)) − J
k
(t
k
,0) − J
k
(t
k
,0) k
6
Z
∞
−∞
e
(β|t−s|
L
ρ
k z
s
k +L
f
ds +
P
X
k=1
M {S
k
k z (t
k
) k +L
k
}
6 M
L
ρ
ρ + L
f
Z
∞
t
e
β(t−s)
+
Z
t
−∞
e
−β(t−s)
+ Mρ
P
X
k=1
s
k
+ M
P
X
k=1
L
k
6 M{L
ρ
ρ + Lρ}
1
β
+
1
β
+ M
ρ
S + M
e
L
=
2M
L
ρ
ρ + L
f
β
+ M{ρS +
e
L}.
From (5), we get that
k (Kz)(t) k< ρ =⇒ k Kz k
b
< ρ ⇐⇒ K(B) ⊂ B
b
ρ
.
k ((Kz)(t) − (K˜z)(t) kk 6
Z
∞
−∞
Me
β|t−s|
k (f, z
s
) − f (s, ez
s
) k
ds
+
P
X
k=1
M k J
k
t
k,
,z (t
k
)
− J
k
(t
k
, ˜z (t
k
)) k
6 ML
ρ
k z − ez k
Z
∞
∞
e
−β|t−s|
ds
+ M
P
X
k=1
S
k
k z (t
k
) − ez (t
k
) k
6
2ML
ρ
β
k z − ez k +MS k z − ez k
=
2ML
ρ
+ βM S
β
k z − ez k
= M
2L
ρ
+ βS
β
k z − ez k .
From (5), we know that
M
2L
ρ
+ βS
β
< 1,
which implies that K is a contraction. Then, applying Banach fixed point Theorem, we get that K
has a unique fixed point in the ball B
b
ρ
, i.e., there exists z
b
∈ B
b
ρ
, such that
z
b
= Kz
b
.
http://novasinergia.unach.edu.ec 73
Hence,
z
b
(t) =
Z
∞
−∞
G(t,s)f(s,z
b
s
)ds +
X
0<t
k
6t
p
G(t,t
k
)J
t
k
,z
b
(t
k
)
.
To prove that z
b
(t) is locally stable, we consider any other solution z(t) of (1) such that
z(t
0
) − z
b
(t
0
)
< ρ/2. Then k z(t
0
) < 2ρ. As long as kz(t)k remains less than 2ρ, we get the fol-
lowing estimate
k z(t) − z
b
(t) k 6k U (t,t
0
) k
h
k z(t
0
) − z
b
(t
0
) k + k g
z
τ
1
,...
τ
q
(t
0
) − g
z
b
τ
1
,···
,z
b
τ
q
(t
0
) k
i
Z
t
t
0
k U(t, s) kk f (s,z
s
) − f
s,z
b
s
k ds
+
X
0<t
k
<t
k U (t,t
k
) kk J
K
(t
k
,z (t
k
)) − J
k
(t
k
,z (t
k
)) k .
Since P (t) ≡ 0, then
k U(t, s) k6 Me
−β(t−s)
, t > s.
Therefore,
k z(t) − z
b
(t) k 6 M k z(t) − z
b
(t) k + k g
z
τ
1
,...
τ
q
− g
z
b
τ
1
,···
,z
b
τ
q
k
+ M
Z
t
t
0
e
−β(t−s)
k f (s,z
s
) − f
s,z
b
s
k ds
+
X
0<t
k
<t
Me
−β(t−t
k
)
k J
k
(t
k
,z (t
k
)) − J
k
t
k
,z
b
(t
k
)
k .
Let t
1
= sup{t > t
0
: kz(t)k < 2ρ}. Then either t
1
= ∞ or kz(t
1
)k = 2ρ. Then, from the above estimate
one can get that
k z(t) − z
b
(t) k <
ρ
2
+
2`
6
+
M
β
L
ρ
sup
S∈[t
0
,t
1
]
k z(s) − z
b
(s) k +M
X
0<t
k
<t
1
S
k
k z (t
k
) − z
b
(t
k
) k
<
ρ
2
+
2`
6
+ 3ρ
M
β
L
ρ
+ 3ρMS
<
1
2
+
1
3
+
3ML
ρ
β
+ 3MS
ρ
=
5
6
+
3ML
ρ
β
+ 3S
ρ.
Thus,
ρ <
5
6
+
3ML
ρ
β
+ 3MS
ρ,
which is a contradiction. Therefore, t
1
= ∞ and z(t) ∈ B
b
2ρ
, t > t
0
. Now, define
k z − z
b
k
+
= sup
t>t
0
k z(t) − z
b
(t) k
Then,
k z − z
b
k
+
6k z (t
0
) − z
b
(t
0
) k + k g
z
τ
1
,...
τ
q
− g
z
b
τ
1
,···
,z
b
τ
q
k +
L
ρ
M
β
k z − z
b
k
+
+SM k z − z
b
k
+
6k z (t
0
) − z
b
(t
0
) k +γ k z(·) − z
b
+
+
M
Lρ
β
k z − z
b
k
+
+SM k z − z
b
k
+
which implies that
1 − Θ −
ML
ρ
β
− MS
k z − z
0
k6k z(t
0
) − z
b
(t
0
) k
http://novasinergia.unach.edu.ec 74
By putting Θ = γ +
ML
ρ
β
+ MS, we obtain that
k z − z
b
1
k
+
6
1
1 − Θ
k z(t
0
) − z
b
(t
0
) k .
This implies the stability.
To prove the uniqueness of the bounded solution globally, we need the following additional hypothesis:
H5) The function f is globally Lipschitz, i.e., there exists a constant L > 0 such that
k f (t,z
1
) − f (s,z
2
) k< L{|t − s|+ k z
1
− z
2
k
PW
} ∀t,s,∈ R, ∀z
1
,z
2
∈ PW.
Theorem 2. Suppose the hypotheses H1),H2),H3),H5 hold and
0 <
ML
β
+ SM <
1
6
.
Then the equation (1) admits one, and only one, bounded solution z
b
(t) for t ∈ R. Moreover, if
condition (6) holds, then this bounded solution is globally uniformly stable.
Proof. Let L > 0 be the Lipschitz constant of f. Then, there exists ρ
1
> 0 such that
1 −
6ML
β
− 6MS
ρ
1
>
ML
β
+
e
LM
⇐⇒
1 −
6ML − 6βMS
β
ρ
1
>
ML + β
e
LM
β
.
Then, applying Theorem 1, for each ρ > ρ
1
, we obtain the existence of an unique bounded solution
of system (1) in the ball B
b
ρ
. Hence the problem (1) has one, and only one, globally bounded solution
z
b
. To prove the uniform stability, we assume that P (t) ≡ 0, consider other solution z(t) of (1), and
the following estimate
k z − z
b
k
+
6
1
1 − Θ
k z (t
0
) − z
b
(t
0
) k,
where
Θ =
γ +
ML
β
+ MS
.
Since Θ does not depend on ρ and t
0
, the stability is globally uniform.
4 Periodic and Almost periodic solutions
In this section, we shall prove that under some additional conditions, the bounded solutions give by
Theorems 1 and 2 are periodic or almost periodic. To this end, in order to prove the periodicity of
the bounded solution z
b
(·), we shall assume the following hypotheses:
H6) f(t,φ) = f(t + T, φ), t ∈ R, φ ∈ PW.
H7) A(t + T ) = A(t), t ∈ R.
From the Floquet Theory, there exists a continuous periodic matrix D(t) and a constant matrix L
such that for t ∈ R
D(t + T ) = D(t), and Φ(t) = D(t)e
Lt
.
From here we get that
U(t + T,s + T ) = D(t + T )e
L(t+T )
e
−L(s+T )
D
−1
(s + T ) = D(t)e
L(t−s)
D
−1
(s) = U(t, s).
Lemma 2. Under the hypotheses H6)-H7) the unique bounded solution z
b
(·) given in Theorem 1 and
Theorem 2 is also T-periodic for t > t
p
.
http://novasinergia.unach.edu.ec 75
Proof. Let z
b
be the unique solution of (1) in the ball B
b
ρ
. Now, we shall prove that z(t) = z
b
(t + T )
b
is also a solution of (1) in the ball B
b
ρ
for t > t
0
> t
p
. Observe that
z
b
s+T
(u) = z
b
(s + u + T ) = z(s + u) = z
u
(s).
Let t > t
0
> t
p
, and consider
z
b
(t) =
Z
∞
−∞
G(t,s)f(s,z
b
s
)ds +
X
0<t
k
6t
p
G(t,t
k
)J(t
k
,z
b
(t
k
))
=
Z
t
−∞
U(t,s)(I − P (s))f(s, z
b
s
)ds −
Z
∞
t
U(t,s)P (s)f(s,z
b
s
)ds
+
X
0<t
k
6t
p
U(t,t
k
)(I − P (t
k
))J
k
(t
k
,z
b
(t
k
)) −
X
t<t
k
6t
p
U(t,t
k
)P (t
k
)J
k
(t
k
,z
b
(t
k
))
=
Z
t
0
−∞
U(t,s)(I − P (s))f(s, z
b
s
)ds +
Z
t
t
0
U(t,s)(I − P (s))f(s, z
b
s
)ds
−
Z
t
t
0
U(t,s)P (s)f (s,z
s
)ds +
Z
t
0
−∞
U(t,s)P (s)f (s,z
s
)ds
+
X
0<t
k
6t
p
U (t,t
k
)(I − P (t
k
))J
k
(t
k
,z
b
(t
k
))
=
Z
∞
−∞
G(t,s)(I − P (s))f
s,z
b
s
ds
+
Z
t
t
0
G(t,s)f
s,z
b
s
ds +
X
0<t
k
6t
p
U (t,t
k
)(I − P (t
k
))J
k
(t
k
,z
b
(t
k
))
= U(t, t
0
)[
Z
∞
−∞
G(t
0
,s)f(s,z
b
s
)ds +
X
0<t
k
6t
p
U (t
0
,t
k
)(I − P (t
k
))J
k
(t
k
,z
b
(t
k
))]
+
Z
t
t
0
G(t,s)f(s,z
b
s
)ds.
Therefore, for t > t
0
> t
p
, we have that
z
b
(t) = U (t,t
0
)z
b
(t
0
) +
Z
t
t
0
G(t,s)f
s,z
b
s
ds. (7)
Hence,
z
b
(t + T ) = U (t + T, t
0
)z
b
(t
0
) +
Z
t+T
t
0
G(t + T,s)f
s,z
b
s
ds
= U(t + T, t
0
)z
b
(t
0
) +
Z
t
t
0
−T
G(t + T,s + T )f
s + T,z
b
s+T
ds
= U (t + T, t
0
+ T )U (t
0
+ T,t
0
)z
b
(t
0
) +
Z
t
t
0
−T
G(t,s)f (s,z
s
)ds
= U (t,t
0
)U (t
0
+ T,t
0
)z
b
(t
0
) +
Z
t
0
t
0
−T
G(t,s)f (s,z
s
)ds +
Z
t
t
0
G(t,s)f (s,z
s
)ds
= U (t,t
0
)
U (t
0
+ T,t
0
)z
b
(t
0
) +
Z
t
0
t
0
−T
G(t
0
,s)f (s,z
s
)ds
+
Z
t
t
0
G(t,s)f (s,z
s
)ds
= U (t,t
0
)z
0
+
Z
t
t
0
G(t,s)f (s,z
s
)ds,
http://novasinergia.unach.edu.ec 76
which implies that
z(t) = U(t,t
0
)z
0
+
Z
t
t
0
G(t,s)f (s,z
s
)ds,
where
z
0
= U(t
0
+ T,t
0
)z
b
(t
0
) +
Z
t
0
t
0
−T
G(t
0
,s)f (s,z
s
)ds.
Therefore,
z(t) = z
b
(t + T )
is a solution of (1) in the ball B
b
ρ
(0). Hence by the uniqueness of the fixed point in this ball we get
that
z
b
(t) = z
b
(t + T ), t > t
p
.
Now, we shall prove that the bounded solution given by Theorem 1 and Theorem 2, under some
conditions, is also almost periodic.
Let us assume the following hypotheses:
H8) J
k
= g = P = 0 and the initial function φ ∈ PW is almost periodic.
H9) A(t) = A and f : R × PW → R
n
is almost periodic in the first variable, uniformly in
φ ∈ PW, and globally Lipschitz in φ.
We recall the following definition and a Theorem from (Toka, 2017).
Definition 1. A jointly continuous function f : R × PW → R
n
is almost periodic uniformly in φ ∈
S ⊂ PW, where S is a bounded set, if for any > 0 there exists `() > 0 such that for any interval of
the form (α,α + `()) contains η with the property
k f(t + η,φ) − f(t,φ) k< ε, ∀t ∈ R, φ ∈ S.
Theorem 3. Let f : R × PW → R
n
be almost periodic in t ∈ R, uniformly in φ ∈ S ⊂ PW, where S
is bounded. Suppose that f is globally Lipschitz in φ ∈ PW. If ζ : R → PW is almost periodic, the
function
Γ : R × PW → R
n
, defined by Γ(t) = f(t,ζ(t)),
is almost periodic.
Proposition 1. Let z ∈ PW
b
be an almost periodic function. Then, the function
π : R −→ PW
t 7−→ π(t) = z
t
,
is almost periodic.
Proof. Since z is almost periodic, then for every > 0 there exists `() > 0 such that any interval
(α,α + `()) contains η such that
k z(t + η) − z(t) k< ε, ∀t ∈ R.
Hence
k z(t + η + s) − z(t + s) k< ε, ∀t, s ∈ R.
So,
k π(t + η) − π(t) k= sup
S∈R−
k z
t+η
(s) − z
t
(s) k< ε, ∀t ∈ R.
In consequence π is almost periodic.
http://novasinergia.unach.edu.ec 77
Now, for a function ξ ∈ PW
b
, we consider the set
H(ξ) = {ξ
t
: t ∈ R},
the closure in the uniform convergence topology, it is called the Hull of ξ, and it is well known
((Toka, 2017)) that: ξ is almost periodic if, and only if, H(ξ) is compact in the uniform convergence
topology.
Also, the following statement holds:
For ρ > 0, the set
A
ρ
= {z ∈ B
b
ρ
: z almost periodic}
is closed.
Theorem 4. Under the hypotheses H8)-H9), the bounded solution z
b
given by Theorems 1 and 2 is
also almost periodic.
Proof. In this case the bounded solution z
b
can be written as follows
z
b
(t) =
Z
t
−∞
e
A(t−s)
f (s,z
s
)ds, t > 0
z
b
(s) = φ(s), s ∈ R
−
.
Now, consider the operator K : A
ρ
→ B
b
ρ
given by (7). From proposition 1, we have that
ξ(t) = f (t,z
t
), z ∈ A
ρ
is almost periodic. On the other hand,
(Kz)(t) =
Z
t
−∞
e
A(t−s)
ξ(s)ds.
Next, we will show that H(Kz) is compact in the uniform convergence topology. In fact, consider a
sequence {(Kz)
η
n
} in H(Kz), where (Kz)
η
n
(t) = (Kz)(t + η
n
). Since ξ is almost periodic, there exists
a convergent sub-sequence {h
η
n
j
}. Now, we have that
(Kz)
η
n
j
(t) = (Kz)
t + η
n
j
=
Z
t+η
n
j
−∞
e
A(t+η
n
j
−s)
ξ(s)ds
=
Z
t
−∞
e
A(t−s)
ξ(s + η
n
j
)ds.
Then,
k (Kz)
η
n
j
(t) − (K)
η
n
j
(t) k6
M
β
k ξ
η
n
j
− ξ
η
n
i
k
b
.
Thus, {(Kz)
η
n
j
} is a Cauchy sequence in PW
b
, which implies that {(Kz)
η
n
j
} converges. Hence
H(Kz) is compact, and (Kz) is almost periodic function. so K(A
ρ
) ⊂ A
ρ
. Therefore, the only fixed
point of K on B
b
ρ
is in A
ρ
. Hence, z
b
(·) is almost periodic.
http://novasinergia.unach.edu.ec 78
5 Conclusion and Final
Remark
In this work, we prove the existence of bounded
solutions for retarded equations with infinite de-
lay, impulses, and non-local conditions. This
is achieved assuming that the associated lin-
ear system has an exponential dichotomy and
applying Banach’s fixed point theorem. Then,
under certain conditions, we prove that this
bounded solution is stable; next, under the ad-
ditional conditions, we prove that this bounded
solution is periodic after the last time impulse
t
p
; in the same way, under certain conditions,
we prove that this bounded solution is almost
periodic. We believe that these results can
be extended to evolution equations in infinite-
dimensional Banach spaces; in fact, this consti-
tutes our next research work in this direction.
Acknowledgment
Many thanks to the anonymous referee for his
comments and suggestions to improve the pre-
sentation of this work.
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