1. No category

Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 100, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
PERIODIC SOLUTIONS FOR NONLINEAR NEUTRAL DELAY
INTEGRO-DIFFERENTIAL EQUATIONS
AZZEDDINE BELLOUR, EL HADI AIT DADS
Abstract. In this article, we consider a model for the spread of certain infectious disease governed by a delay integro-differential equation. We obtain the
existence and the uniqueness of a positive periodic solution, by using Perov’s
fixed point theorem in generalized metric spaces.
1. Introduction
The existence of positive solutions to the integral equation with non constant
delay
Z t
x(t) =
f (s, x(s))ds,
(1.1)
t−σ(t)
was considered in [4, 13, 18, 23]. This equation is a mathematical model for the
spread of certain infectious diseases with a contact rate that varies seasonally. Here
x(t) is the proportion of infectious in population at time t, σ(t) is the length of
time in which an individual remains infectious, f (t, x(t)) is the proportion of new
infectious per unit of time (see, for example, [5, 9, 21]).
Ait Dads and Ezzinbi [4] and Ding et al [10] studied the existence of a positive
pseudo almost periodic solution. Ezzinbi and Hachimi [13], Torrej´on [23], Xu and
Yuan [24] showed the existence of a positive almost periodic solution. The existence
of a positive almost automorphic solution was studied in [12, 15, 18]. Bica and
Mure¸san [7, 8] studied the existence and uniqueness of a positive periodic solution
of (1.1) by using Perov’s fixed point theorem in the case of f depends also on x0 (t)
and σ(t) = σ is constant.
Ait Dads and Ezzinbi [3] considered the existence of a positive almost periodic
solution, Ding et al [11] studied the existence of positive almost automorphic solutions for the neutral nonlinear delay integral equation
Z t
x(t) = γx(t − τ ) + (1 − γ)
f (s, x(s))ds,
(1.2)
t−τ
where 0 ≤ γ < 1. We refer to [2, 17] for the meaning of (1.2) in the context of
epidemics.
2010 Mathematics Subject Classification. 45D05, 45G10, 47H30.
Key words and phrases. Neutral delay integro-differential equation; periodic solution;
Perov’s fixed point theorem.
c
2015
Texas State University - San Marcos.
Submitted December 14, 2014. Published April 14, 2015.
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2
A. BELLOUR, E. AIT DADS
EJDE-2015/100
In this paper, we consider the more general equation
Z t
x(t) = γx(t − σ(t)) + (1 − γ)
f (s, x(s), x0 (s))ds,
(1.3)
t−σ(t)
Equation (1.3) includes many important integral and functional equations that arise
in biomathematics (see for example [1, 6, 8, 9, 14, 16, 18, 19, 21]).
We would to use Perov’s fixed point theorem to obtain conditions for the existence and uniqueness of a positive periodic solution to (1.3). This work is motivated
by the work of Wei Long and Hui-Sheng Ding [18]. Moreover, the results obtained
in this paper generalize several ones obtained in [6, 8, 19, 21], and the main goal in
this work is to study the existence and uniqueness of solutions when σ(t) is not
constant in (1.3).
2. Preliminaries and generalized metric spaces
In this section, we recall the following notation and results in generalized metric
spaces.
Definition 2.1 ( [20]). Let X be a nonempty set and d : X × X → Rn be a
mapping such that for all x, y, z ∈ X, one has:
(i) d(x, y) ≥ 0Rn and d(x, y) = 0Rn ⇔ x = y,
(ii) d(x, y) = d(y, x),
(iii) d(x, y) ≤ d(x, z) + d(z, y),
where for x = (x1 , x2 , . . . , xn ) and y = (y1 , y2 , . . . , yn ) from Rn , we have
x ≤ y ⇔ xi ≤ yi , for i = 1, n.
Then d is called a generalized metric and (X, d) is a generalized metric space.
Definition 2.2 ( [22]). Let (E, k · k) be a generalized Banach space, the norm
k · k : E → Rn has the following properties:
(i) kxk ≥ 0 for all x ∈ E and kxk = 0 if and only if x = 0.
(ii) kλxk = |λ| kxk for all λ ∈ K = R or C and for all x ∈ E.
(iii) kx + yk ≤ kxk + kyk for all x, y ∈ E (the inequalities are defined by
components in Rn ).
Remark 2.3. A generalized Banach space is a generalized complete metric space.
Definition 2.4 ( [20]). If (E, d) is a generalized complete metric space and T :
E → E which satisfies the inequality
d(T x, T y) ≤ Ad(x, y)
for all x, y ∈ E,
where A is a matrix convergent to zero (the norms of it its eigenvalues are in the
interval [0, 1)). We say that T is a Picard operator or generalized contraction.
We recall the following Perov’s fixed point theorem.
Theorem 2.5 ( [20]). Let (E, d) be a complete generalized metric space. If T :
E → E is a map for which there exists a matrix A ∈ Mn (R) such that
d(T x, T y) ≤ Ad(x, y),
∀x, y ∈ E
and the norms of the eigenvalues of A are in the interval [0, 1), then T has a unique
fixed point x∗ ∈ E and the sequence of successive approximations xm = T m (x0 )
converges to x∗ for any x0 ∈ E. Moreover, the following estimation holds
d(xm , x∗ ) ≤ Am (In − A)−1 d(x0 , x1 ),
∀m ∈ N∗ .
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PERIODIC SOLUTIONS
3
3. Main result
In this section, we study the existence and uniqueness of a positive and periodic
solution for the equation (1.3). We consider the following functional spaces
P (ω) = {x ∈ C(R) : x(t + ω) = x(t), ∀t ∈ R}
P 1 (ω) = {x ∈ C 1 (R) : x(t + ω) = x(t), ∀t ∈ R}
K + (ω) = {x ∈ P 1 (ω) : x(t) ≥ 0, ∀t ∈ R}
and denote by E the product space E = K + (ω) × P (ω) which is a generalized
metric space with the generalized metric dC : E × E → R2 , defined by
dC ((x1 , y1 ), (x2 , y2 )) = (kx1 − x2 k + kx01 − x02 k, ky1 − y2 k)
where kuk = max{|u(t)| : t ∈ [0, ω]} for any u ∈ P (ω). Before stating the main
result, we need the following lemma.
Lemma 3.1. (E, dC ) is a complete generalized metric space.
Proof. Let (xn ) = (xn , yn ) be a Cauchy sequence, then for any = (1 , 2 ) > 0,
there exists n0 ∈ N such that for all n, m ≥ n0 , we have dC ((xm , ym ), (xn , yn )) ≤ .
Hence, for all n, m ≥ n0 , kxm − xn k + kx0m − x0n k ≤ 1 and kym − yn k ≤ 2 . Then,
(xn ), (x0n ) and (yn ) are Cauchy sequences in P (ω). It is clear that (P (ω), k · k) is
a Banach space, hence there exists y ∈ P (ω) such that
lim kyn − yk = 0
n→+∞
(3.1)
and there exist x, w ∈ P (ω) such that limn→+∞ kxn −xk = limn→+∞ kx0n −wk = 0.
Now, since for all n ≥ n0 , and all t ∈ R+ ,
Z t
xn (t) = xn (0) +
x0n (s)ds.
0
Then, by Lebesgue’s Dominated Convergence Theorem, x0 (t) = w(t) for all t ∈ R+ .
Therefore, for all n ∈ N and all t ∈ R, xn (t) ≥ 0. Then for all t ∈ R, x(t) ≥ 0. As
a consequence, x ∈ K + (ω) and
lim (kxn − xk + kx0n − x0 k) = 0 .
n→+∞
(3.2)
Finally, we deduce, by (3.1) and (3.2), that (xn ) converges to (x, y) ∈ E and (E, dC )
is a complete generalized metric space.
Equation (1.3) will be studied under the following assumptions:
(H1) f ∈ C(R × R+ × R, (0, +∞)) and there exists ω > 0 such that
f (t + ω, x, y) = f (t, x, y) ,
∀(t, x, y) ∈ R × R+ × R.
(H2) There exist α, β > 0 such that
|f (t, u, v) − f (t, u0 , v 0 )| ≤ α|u − u0 | + β|v − v 0 |,
for all t ∈ R and all u, u0 ∈ R+ , for all v, v 0 ∈ R.
(H3) σ ∈ P 1 (ω) and inf t∈[0,ω] σ(t) = σ0 > 0. Let σ1 = supt∈[0,ω] σ(t), σ2 =
supt∈[0,ω] |σ 0 (t)| and assume that γ(1 + σ2 ) < 1.
Under the hypothesis (H1)–(H3), we will use Perov’s fixed point theorem to prove
the following main result.
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A. BELLOUR, E. AIT DADS
EJDE-2015/100
Theorem 3.2. If the hypotheses (H1)–(H3) hold, and if
p
γ(1 + σ2 ) + (1 − γ)β(2 + σ2 ) + L + ζ < 2,
where, L = max(γ(1 + σ2 ), γ + (1 − γ)α(2 + σ1 + σ2 )) and
ζ = γ 2 σ2 (2 + σ2 ) + (1 − γ)2 β(2 + σ2 )(β(2 + σ2 ) + 4α(2 + σ1 + σ2 )) + (L − γ)2
+ 2γβ(1 − γ)(1 + σ2 )(2 + σ2 ) − 2L(β(1 − γ)(2 + σ2 ) + γσ2 ).
Then, (1.3) has a unique solution in K + (ω).
Proof. If we differentiate (1.3) with respect to t and denoting x0 (t) = y(t), for all
t ∈ R, we obtain
h
y(t) = γ(1 − σ 0 (t))y(t − σ(t)) + (1 − γ) f (t, x(t), y(t))
i
− (1 − σ 0 (t))f (t − σ(t), x(t − σ(t)), y(t − σ(t))) ,
which leads to
Z
t
x(t) = γx(t − σ(t)) + (1 − γ)
f (s, x(s), y(s))ds,
t−σ(t)
h
y(t) = γ(1 − σ 0 (t))y(t − σ(t)) + (1 − γ) f (t, x(t), y(t))
i
− (1 − σ 0 (t))f (t − σ(t), x(t − σ(t)), y(t − σ(t))) .
Let T : E → C(R) × C(R) be the map defined by
T1 (x, y)(t)
T (x, y)(t) =
,
T2 (x, y)(t)
where
Z
t
T1 (x, y)(t) = γx(t − σ(t)) + (1 − γ)
f (s, x(s), y(s))ds,
t−σ(t)
and
h
T2 (x, y)(t) = γ(1 − σ 0 (t))y(t − σ(t)) + (1 − γ) f (t, x(t), y(t))
i
− (1 − σ 0 (t))f (t − σ(t), x(t − σ(t)), y(t − σ(t))) .
(3.3)
From Conditions (H1) and (H3), one has that T1 (E) ⊂ C 1 (R). Hence, from the
condition that f is ω-periodic with respect to t, it follows that T1 (E) ⊂ K + (ω).
indeed
Z t+ω
T1 (x, y)(t + ω) = γx(t + ω − σ(t + ω)) + (1 − γ)
f (s, x(s), y(s))ds
t+ω−σ(t+ω)
Z
t
= γx(t − σ(t)) + (1 − γ)
f (s − ω, x(s − ω), y(s − ω))ds
t−σ(t)
= T1 (x, y)(t),
∀t ∈ R, ∀(x, y) ∈ E.
In addition in the same way, one has T2 (x, y)(t + ω) = T2 (x, y)(t). Consequently,
T (E) ⊂ E. Moreover, from Conditions (H2),
|T1 (x1 , y1 )(t) − T1 (x2 , y2 )(t)| + |T10 (x1 , y1 )(t) − T10 (x2 , y2 )(t)|
≤ γ|x1 (t − σ(t)) − x2 (t − σ(t))|
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PERIODIC SOLUTIONS
Z
t
[α|x1 (s) − x2 (s)| + β|y1 (s) − y2 (s)|]ds
+ (1 − γ)
+ γ(1 −
5
t−σ(t)
σ 0 (t))|x01 (t
− σ(t)) − x02 (t − σ(t))|
+ (1 − γ)(α|x1 (t) − x2 (t)| + β|y1 (t) − y2 (t)|)
+ (1 − γ)|1 − σ 0 (t)|α|x1 (t − σ(t)) − x2 (t − σ(t))|
+ (1 − γ)|1 − σ 0 (t)|β|y1 (t − σ(t)) − y2 (t − σ(t))|
≤ L(kx1 − x2 k + kx01 − x02 k) + (1 − γ)β(1 + σ1 + σ2 )ky1 − y2 k
where
L = max(γ(1 + σ2 ), γ + (1 − γ)α(2 + σ1 + σ2 )).
{z
}
| {z } |
r1
r2
Similarly, one has
|T2 (x1 , y1 )(t) − T2 (x2 , y2 )(t)|
≤ (2 + σ2 )(1 − γ)αkx1 − x2 k + [γ(1 + σ2 ) + (1 − γ)β(2 + σ2 )]ky1 − y2 k
≤ (2 + σ2 )(1 − γ)α(kx1 − x2 k + kx01 − x02 k)
+ [γ(1 + σ2 ) + (1 − γ)β(2 + σ2 )]ky1 − y2 k.
So
dC (T (x1 , y1 ), T (x2 , y2 )) ≤ A
kx1 − x2 k + kx01 − x02 k
,
ky1 − y2 k
where

L
(1 − γ)β(2 + σ1 + σ2 )

.
A=
(2 + σ2 )(1 − γ)α γ(1 + σ2 ) + (1 − γ)β(2 + σ2 )
The eigenvalues of this matrix are
p
1
λ1 = [γ(1 + σ2 ) + (1 − γ)β(2 + σ2 ) + L + ζ],
2
p
1
λ2 = [γ(1 + σ2 ) + (1 − γ)β(2 + σ2 ) + L − ζ]
2
where
ζ = γ 2 σ2 (2 + σ2 ) + (1 − γ)2 β(2 + σ2 )(β(2 + σ2 ) + 4α(2 + σ1 + σ2 )) + (L − γ)2
+ 2γβ(1 − γ)(1 + σ2 )(2 + σ2 ) − 2L(β(1 − γ)(2 + σ2 ) + γσ2 ).
In what follows, we show that the eigenvalues of the matrix A are nonnegative real
numbers (λ1 , λ2 ∈ R+ ).
Step 1: We show that λ1 , λ2 are real numbers. It suffices to prove that ζ ≥ 0.
We have two cases:
Case 1: If L = r1 , then
ζ = γ 2 σ2 (2 + σ2 ) + (1 − γ)2 β(2 + σ2 )(β(2 + σ2 ) + 4α(2 + σ1 + σ2 )) + γ 2 σ22
+ 2γβ(1 − γ)(1 + σ2 )(2 + σ2 ) − 2γ(1 + σ2 )(β(1 − γ)(2 + σ2 ) + γσ2 )
= γ 2 σ2 (2 + σ2 ) + (1 − γ)2 β(2 + σ2 )(β(2 + σ2 ) + 4α(2 + σ1 + σ2 )) + γ 2 σ22
− 2γ 2 σ2 (1 + σ2 )
= (1 − γ)2 β(2 + σ2 )(β(2 + σ2 ) + 4α(2 + σ1 + σ2 )) ≥ 0.
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A. BELLOUR, E. AIT DADS
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Case 2: If L = r2 , then
ζ = γ 2 σ2 (2 + σ2 ) + (1 − γ)2 β(2 + σ2 )(β(2 + σ2 ) + 4α(2 + σ1 + σ2 ))
+ (1 − γ)2 α2 (2 + σ1 + σ2 )2 + 2γβ(1 − γ)(1 + σ2 )(2 + σ2 )
− 2(γ + (1 − γ)α(2 + σ1 + σ2 ))(β(1 − γ)(2 + σ2 ) + γσ2 )
= γ 2 σ2 (2 + σ2 ) + (1 − γ)2 β(2 + σ2 )(β(2 + σ2 ) + 2α(2 + σ1 + σ2 ))
+ (1 − γ)2 α2 (2 + σ1 + σ2 )2 + 2γβ(1 − γ)σ2 (2 + σ2 )
− 2γσ2 (γ + (1 − γ)α(2 + σ1 + σ2 ))
= γ 2 σ22 + (1 − γ)2 β(2 + σ2 )(β(2 + σ2 ) + 2α(2 + σ1 + σ2 ))
+ (1 − γ)2 α2 (2 + σ1 + σ2 )2 + 2γβ(1 − γ)σ2 (2 + σ2 )
− 2γσ2 (1 − γ)α(2 + σ1 + σ2 )
= (1 − γ)2 β(2 + σ2 )(β(2 + σ2 ) + 2α(2 + σ1 + σ2 )) + 2γβ(1 − γ)σ2 (2 + σ2 )
+ ((1 − γ)α(2 + σ1 + σ2 ) − γσ2 )2 ≥ 0.
Step 2: We show that λ2 is nonnegative. We have two cases:
Case 1: If L = r1 , then
[γ(1 + σ2 ) + (1 − γ)β(2 + σ2 ) + L]2 − ζ
= 4γ 2 (1 + σ2 )2 + (1 − γ)2 β 2 (2 + σ2 )2 + 4γ(1 + σ2 )(1 − γ)β(1 + σ2 ) − ζ
= 4γ 2 (1 + σ2 )2 + 4(1 − γ)β(2 + σ2 )(r1 − r2 + γ) ≥ 0.
Case 2: If L = r2 , then
[γ(1 + σ2 ) + (1 − γ)β(2 + σ2 ) + L]2 − ζ
= [γ(2 + σ2 ) + (1 − γ)β(2 + σ2 ) + (1 − γ)α(2 + σ1 + σ2 )]2 − ζ
= γ 2 (2 + σ2 )2 + (1 − γ)2 β(2 + σ2 )(β(2 + σ2 ) + 2α(2 + σ1 + σ2 ))
+ 2γβ(1 − γ)(2 + σ2 )2 + (1 − γ)2 α2 (2 + σ1 + σ2 )2
+ 2γ(1 − γ)(2 + σ2 )α(2 + σ1 + σ2 ) − ζ
= 4γ 2 (1 + σ2 ) + 4γβ(1 − γ)(2 + σ2 ) + 4γ(1 − γ)α(2 + σ1 + σ2 )(1 + σ2 ) ≥ 0.
Which implies that λ2 ≥ 0.
We remark that λ1 > λ2 , this implies that λ1 and λ2 belong to the open unit
disc of R2 if and only if λ1 < 1, which is equivalent to
p
γ(1 + σ2 ) + (1 − γ)β(2 + σ2 ) + L + ζ < 2.
Then, by Perov’s fixed point theorem, the operator T has a unique solution x∗ =
(x∗ , y∗ ) ∈ K + (ω) × P (ω), which implies that x∗ ∈ C 1 (R), and for all t ∈ R,
h
(x∗ )0 (t) = γ(1 − σ 0 (t))(x∗ )0 (t − σ(t)) + (1 − γ) f (t, x∗ (t), y∗ (t))
i
− (1 − σ 0 (t))f (t − σ(t), x∗ (t − σ(t)), y∗ (t − σ(t))) .
Hence, by using (3.3), for all t ∈ R,
((x∗ )0 − y∗ )(t) = γ(1 − σ 0 (t))((x∗ )0 − y∗ )(t − σ(t)).
EJDE-2015/100
PERIODIC SOLUTIONS
7
Then, k(x∗ )0 − y∗ k ≤ γ(1 + σ2 )k(x∗ )0 − y∗ k. We deduce, by Condition (H3), that
(x∗ )0 = y∗ and x∗ is the unique solution of (1.3).
To illustrate this result, we have the following example.
Example 3.3. Consider (1.3) where f is ω-periodic with respect to t and σ is
5
< 1, L =
ω-periodic, γ = σ1 = σ2 = 41 , α = 16 , β = 15 , then γ(1 + σ2 ) = 16
5
9
9
max( 16 , 16 ) = 16 and
√
p
67 + 2749 ∼
γ(1 + σ2 ) + (1 − γ)β(2 + σ2 ) + L + ζ =
= 1.86 < 2.
80
Thus, by Theorem 3.2, Equation (1.3) has a unique positive ω-periodic solution.
The following proposition gives an estimation of the error between the exact
solution and the approximate solution of (1.3).
Proposition 3.4. Under the assumptions of Theorem 3.2, the solution of (1.3),
which is obtained by the successive approximations method starting from any x0 =
(x0 , y0 ) ∈ E, satisfies the estimate
m
m
1
e1 λ 1 + e2 λ m
e3 λm
m
∗
1
0
2
1 + e4 λ2
dC (x , x ) ≤
m
m × dC (x , x ),
e7 λm
µ(λ1 − λ2 ) e5 λm
1 + e6 λ 2
1 + e8 λ2
where µ = (1 − L)(1 − γ(1 + σ2 ) − β(2 + σ2 )(1 − γ)) − (1 − γ)2 αβ(2 + σ2 )(2 + σ1 + σ2 ),
xm = T (xm−1 ), xm = (xm , ym ), for all m ∈ N∗ and
e1 = (a(L − λ2 ) − c2 ), e2 = (a(λ1 − L) + c2 )
e3 = (b(L − λ2 ) − c(1 − L)), e4 = (b(λ1 − L) + c(1 − L))
a(L − λ2 )
a(L − λ1 )
− c), e6 = (L − λ2 )(c −
)
c
c
b(L − λ2 )
b(L − λ1 )
e7 = (L − λ1 )(
+ L − 1), e8 = (L − λ2 )(1 − L −
)
c
c
such that
e5 = (L − λ1 )(
(3.4)
a = 1 − γ(1 + σ2 ) − β(2 + σ2 )(1 − γ)
b = (1 − γ)β(2 + σ1 + σ2 )
c = (2 + σ2 )(1 − γ)α.
Proof. From Theorem 2.5, by the conditions of Theorem 3.2, one has
dC (xm , x∗ ) ≤ Am (I − A)−1 dC (x1 , x0 ),
∀m ∈ N∗ .
We have
A
m
m
(L − λ2 )λm
1 + (λ1 − L)λ2
1
=
λ1 − λ2
m
(L−λ1 )(L−λ2 )(λm
1 −λ2 )
(1−γ)α(2+σ2 )
!
m
(1 − γ)α(2 + σ2 )(λm
2 − λ1 )
,
m
(λ1 − L)λm
1 + (L − λ2 )λ2
and

1 − γ(1 + σ2 ) − β(2 + σ2 )(1 − γ) (1 − γ)β(2 + σ1 + σ2 )
|
{z
}
|
{z
}

1
a
b
,
= 


(2 + σ2 )(1 − γ)α
1−L
µ
|
{z
}

(I − A)−1
c
8
A. BELLOUR, E. AIT DADS
EJDE-2015/100
where µ = (1 − L)(1 − γ(1 + σ2 ) − β(2 + σ2 )(1 − γ)) − (1 − γ)2 αβ(2 + σ2 )(2 + σ1 + σ2 ).
Which implies
m
m
1
e1 λ 1 + e2 λ m
e3 λ m
2
1 + e4 λ 2
Am (I − A)−1 =
m
m ,
e7 λ m
µ(λ1 − λ2 ) e5 λm
1 + e6 λ 2
1 + e8 λ 2
where ei , i = 1, . . . , 8 are given by (3.4).
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Azzeddine Bellour
Department of Mathematics, Ecole Normale Superieure de Constantine, Constantine,
Algeria
E-mail address: [email protected]
El Hadi Ait Dads
University Cadi Ayyad, Department of Mathematics, Faculty of Sciences Semlalia B.P.
2390, Marrakech, Morocco.
´ Associe
´e au CNRST URAC 02
UMMISCO UMI 209, UPMC, IRD Bondy France. Unite
E-mail address: [email protected]