Periodic Solutions for Non-Linear Systems of Integral Equations

International Journal of Engineering Inventions
e-ISSN: 2278-7461, p-ISSN: 2319-6491
Volume 3, Issue 8 (April 2014) PP: 01-16
www.ijeijournal.com Page | 1
Periodic Solutions for Non-Linear Systems of Integral Equations
Prof. Dr. Raad. N. Butris1
, Baybeen S. Fars2
1, 2
Department of Mathematics, Faculty of Science, University of Zakho
Abstract: The aim of this paper is to study the existence and approximation of periodic solutions for non-linear
systems of integral equations, by using the numerical-analytic method which were introduced by Samoilenko[
10, 11]. The study of such nonlinear integral equations is more general and leads us to improve and extend the
results of Butris [2].
Keyword And Phrases: Numerical-analytic methods, existence and approximation of periodic solutions,
nonlinear system, integral equations.
I. Introduction
Integral equation has been arisen in many mathematical and engineering field, so that solving this kind
of problems are more efficient and useful in many research branches. Analytical solution of this kind of
equation is not accessible in general form of equation and we can only get an exact solution only in special
cases. But in industrial problems we have not spatial cases so that we try to solve this kind of equations
numerically in general format. Many numerical schemes are employed to give an approximate solution with
sufficient accuracy [3,4,6, ,12,13,14,15]. Many author create and develop numerical-analytic methods [1, 2,5,
7,8,9] and schemes to investigate periodic solution of integral equations describing many applications in
mathematical and engineering field.
Consider the following system of non-linear integral equations which has the form:
𝑥 𝑡, 𝑥0, 𝑦0 = 𝐹0 𝑡 + 𝑓1(𝑠, 𝑥 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑠, 𝑥0, 𝑦0 ,
𝑡
0
, 𝐺1 𝑠, 𝜏 𝑔1(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )
𝑠
−∞
𝑑𝜏,
, 𝑔1(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )𝑑𝜏
𝑏 𝑠
𝑎 𝑠
)𝑑𝑠 ⋯( I1 )
𝑦 𝑡, 𝑥0, 𝑦0 = 𝐺0 𝑡 + 𝑓2(𝑠, 𝑥 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑠, 𝑥0, 𝑦0 ,
𝑡
0
, 𝐺2 𝑠, 𝜏 𝑔2(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )
𝑠
−∞
𝑑𝜏,
, 𝑔2(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )𝑑𝜏
𝑏 𝑠
𝑎 𝑠
)𝑑𝑠 ⋯ ( I2 )
where 𝑥 ∈ 𝐷 ⊂ 𝑅 𝑛
, 𝐷 is closed and bounded domain subset of Euclidean space 𝑅 𝑛
.
Let the vector functions
𝑓1 𝑡, 𝑥, 𝑦, 𝑧, 𝑤 = 𝑓11 𝑡, 𝑥, 𝑦, 𝑧, 𝑤 , 𝑓12 𝑡, 𝑥, 𝑦, 𝑧, 𝑤 , … , 𝑓1𝑛 𝑡, 𝑥, 𝑦, 𝑧, 𝑤 ,
𝑓2 𝑡, 𝑥, 𝑦, 𝑢, 𝑣 = 𝑓21 𝑡, 𝑥, 𝑦, 𝑢, 𝑣 , 𝑓22 𝑡, 𝑥, 𝑦, 𝑢, 𝑣 , … , 𝑓2𝑛 𝑡, 𝑥, 𝑦, 𝑢, 𝑣 ,
𝑔1 𝑡, 𝑥, 𝑦 = 𝑔11 𝑡, 𝑥, 𝑦 , 𝑔12 𝑡, 𝑥, 𝑦 , … , 𝑔1𝑛 𝑡, 𝑥, 𝑦 ,
𝑔2 𝑡, 𝑥, 𝑦 = 𝑔21 𝑡, 𝑥, 𝑦 , 𝑔22 𝑡, 𝑥, 𝑦 , … , 𝑔2𝑛 𝑡, 𝑥, 𝑦 ,
𝐹0 𝑡 = (𝐹01 𝑡 , 𝐹02 𝑡 , … , 𝐹0𝑛 𝑡 ),
and
𝐺0 𝑡 = (𝐺01 𝑡 , 𝐺02 𝑡 , ⋯ , 𝐺0𝑛 𝑡 )
are defined and continuous on the domains:
𝑡, 𝑥, 𝑦, 𝑧, 𝑤 ∈ 𝑅1
× 𝐷 × 𝐷1 × 𝐷2 = −∞, ∞ × 𝐷 × 𝐷1 × 𝐷2 × 𝐷3
𝑡, 𝑥, 𝑦, 𝑢, 𝑣 ∈ 𝑅1
× 𝐷 × 𝐷1 × 𝐷2 = −∞, ∞ × 𝐷 × 𝐷1 × 𝐷2 × 𝐷3
⋯ 1.1
and periodic in t of period T, Also a(t) and b(t) are continuous and periodic in t of period 𝑇.
Suppose that the functions 𝑓1 𝑡, 𝑥, 𝑦, 𝑧, 𝑤 , 𝑓2 𝑡, 𝑥, 𝑦, 𝑢, 𝑣 , 𝑔1 𝑡, 𝑥, 𝑦
and 𝑔2 𝑡, 𝑥, 𝑦 satisfies the following inequalities:

𝑓1 𝑡, 𝑥, 𝑦, 𝑧, 𝑤 ≤ 𝑀1 , 𝑓2 𝑡, 𝑥, 𝑦, 𝑢, 𝑣 ≤ 𝑀2 , 𝑀1, 𝑀2 > 0
𝑔1 𝑡, 𝑥, 𝑦 ≤ 𝑁1 , 𝑔1 𝑡, 𝑥, 𝑦 ≤ 𝑁1 , 𝑁1, 𝑁 2 > 0
⋯ 1.2
𝑓1 𝑡, 𝑥1, 𝑦1, 𝑧1, 𝑤1 − 𝑓1 𝑡, 𝑥2, 𝑦2, 𝑧2, 𝑤2 ≤ 𝐾1 𝑥1 − 𝑥2 + 𝐾2 𝑦1−𝑦2 +
+𝐾3 𝑧1−𝑧2 + 𝐾4 𝑤1−𝑤2 ⋯ 1.3
𝑓2 𝑡, 𝑥1, 𝑦1, 𝑢1, 𝑣1 − 𝑓2 𝑡, 𝑥2, 𝑦2, 𝑢2, 𝑣2 ≤ 𝐿1 𝑥1 − 𝑥2 + 𝐿2 𝑦1−𝑦2 +
+𝐿3 𝑢1−𝑢2 + 𝐿4 𝑣1−𝑣2 , ⋯ 1.4
𝑔1 𝑡, 𝑥1, 𝑦1 − 𝑔1 𝑡, 𝑥2, 𝑦2 ≤ 𝑅1 𝑥1 − 𝑥2 + 𝑅2 𝑦1 − 𝑦2 , ⋯ 1.5
𝑔2 𝑡, 𝑥1, 𝑦1 − 𝑔2 𝑡, 𝑥2, 𝑦2 ≤ 𝐻1 𝑥1 − 𝑥2 + 𝐻2 𝑦1 − 𝑦2 , ⋯ 1.6
for all 𝑡 ∈ 𝑅1
, 𝑥, 𝑥1, 𝑥2 ∈ 𝐷 , 𝑦 , 𝑦1, 𝑦2 ∈ 𝐷1, 𝑧, 𝑧1, 𝑧2, 𝑢, 𝑢1, 𝑢2 ∈ 𝐷2 , and 𝑤, 𝑤1, 𝑤2, 𝑣, 𝑣1, 𝑣2 ∈ 𝐷3, where 𝑀1 =
𝑀11, 𝑀12, … , 𝑀1𝑛 ,
𝑀2 = 𝑀21, 𝑀22, … , 𝑀2𝑛 , 𝑁1 = (𝑁11, 𝑁12, … , 𝑁1𝑛 ) and
𝑁2 = 𝑁21, 𝑁22, … , 𝑁2𝑛 are positive constant vectors and 𝐾1 = (𝐾1 𝑖𝑗
), 𝐾2 = (𝐾2 𝑖𝑗
), 𝐾3 = (𝐾3 𝑖𝑗
), 𝐾4 = (𝐾4 𝑖𝑗
),
𝐿1 = (𝐿1 𝑖𝑗
), 𝐿2 = (𝐿2 𝑖𝑗
),
𝐿3 = ( 𝐿3 𝑖𝑗
), 𝐿4 = (𝐿4 𝑖𝑗
), 𝑅1 = (𝑅1 𝑖𝑗
), 𝑅2 = (𝑅2 𝑖𝑗
), 𝐻1 = (𝐻1 𝑖𝑗
), and
𝐻2 = (𝐻2 𝑖𝑗
) are positive constant matrices.
Also 𝐺1 𝑡, 𝑠 and 𝐺2 𝑡, 𝑠 are (n×n) continuous positive matrix and
periodic in t, s of period T in the domain 𝑅1
× 𝑅1
and satisfy the following conditions:
𝐺1 𝑡, 𝑠 ≤ 𝛾 𝑒−𝜆1 𝑡−𝑠
, 𝛾, 𝜆1 > 0 ⋯ 1.7
and
𝐺2(𝑡, 𝑠) ≤ 𝛿 𝑒−𝜆2 𝑡−𝑠
, 𝛿, 𝜆2 > 0 ⋯ 1.8
where −∞ < 0 ≤ 𝑠 ≤ 𝑡 ≤ 𝑇 < ∞ , 𝑖, 𝑗 = 1,2, ⋯ , 𝑛,
and 𝑕 = 𝑚𝑎𝑥𝑡∈ 0,𝑇 𝑏 𝑡 − 𝑎 𝑡 , . = 𝑚𝑎𝑥𝑡∈ 0,𝑇 .
Now define a non-empty sets as follows:
𝐷𝑓1
= 𝐷 −
𝑇
2
𝑀 1 ,
𝐷1𝑓2
= 𝐷1 −
𝑇
2
𝑀 2 ,
𝐷2𝑓1
= 𝐷2 −
𝑇
2
𝛾
𝜆1
𝑅1 𝑀1 + 𝑅2 𝑀2 ,
𝐷3𝑓1
= 𝐷3 −
𝑇
2
𝑕 𝑅1 𝑀1 + 𝑅2 𝑀2 ,
𝐷2𝑓2
= 𝐷2 −
𝑇
2
𝛿
𝜆2
𝐻1 𝑀1 + 𝐻2 𝑀2 ,
𝐷3𝑓2
= 𝐷3 −
𝑇
2
𝑕 𝐻1 𝑀1 + 𝐻2 𝑀2 .
⋯ 1.9
Forever, we suppose that the greatest eigen-value 𝑞 𝑚𝑎𝑥 of the following
matrix:
Q0 =
𝑇
2
[𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)]
𝑇
2
[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)]
𝑇
2
[𝐿1 + 𝐻1(𝐿3
𝛿
𝜆2
+ 𝑕𝐿4)]
𝑇
2
[𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)]
⋯ 1.10
is less than unity, i.e.
𝑞 𝑚𝑎𝑥 𝑄0 =
φ1 + 𝜑1
2 + 4(𝜑2 − 𝜑3)
2
< 1 , ⋯ 1.11
where φ1 =
𝑇
2
[𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] +
𝑇
2
[𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)],
φ2 = (
𝑇
2
[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)])(
𝑇
2
[𝐿1 + 𝐻1(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)]) and
𝜑3 = (
𝑇
2
[𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)])(
𝑇
2
[𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)]).
By using lemma 3.1[10],we can state and prove the following lemma:
Lemma 1.1. Let 𝑓1(𝑡, 𝑥, 𝑦, 𝑧, 𝑤) and 𝑓2(𝑡, 𝑥, 𝑦, 𝑢, 𝑣) be a vector functions which are defined and continuous in
the interval 0, 𝑇 , then the inequality:

𝐿1 𝑡, 𝑥0, 𝑦0
𝐿2(𝑡, 𝑥0, 𝑦0)
≤
𝑀1 𝛼 𝑡
𝑀2 𝛼 𝑡
⋯ 1.12
satisfies for 0 ≤ 𝑡 ≤ 𝑇 and 𝛼 𝑡 ≤
𝑇
2
,
where 𝛼 𝑡 = 2𝑡(1 −
𝑡
𝑇
) for all 𝑡 ∈ 0, 𝑇 ,
𝐿1 𝑡, 𝑥0, 𝑦0 = [𝑓1(𝑠, 𝑥0, 𝑦0,
𝑡
0
𝐺1 𝑠, 𝜏 𝑔1(𝜏, 𝑥0, 𝑦0)
𝑠
−∞
𝑑𝜏,
, 𝑔1(𝜏, 𝑥0, 𝑦0)𝑑𝜏
𝑏 𝑠
𝑎 𝑠
) −
1
𝑇
𝑓1(𝑠, 𝑥0, 𝑦0,
𝑇
0
, 𝐺1 𝑠, 𝜏 𝑔1(𝜏, 𝑥0, 𝑦0)
𝑠
−∞
𝑑𝜏, 𝑔1(𝜏, 𝑥0, 𝑦0)𝑑𝜏
𝑏 𝑠
𝑎 𝑠
)𝑑𝑠]𝑑𝑠
𝐿2 𝑡, 𝑥0, 𝑦0 = [𝑓2(𝑠, 𝑥0, 𝑦0,
𝑡
0
𝐺2 𝑠, 𝜏 𝑔2(𝜏, 𝑥0, 𝑦0)
𝑠
−∞
𝑑𝜏,
, 𝑔2(𝜏, 𝑥0, 𝑦0)𝑑𝜏
𝑏 𝑠
𝑎 𝑠
) −
1
𝑇
𝑓2(𝑠, 𝑥0, 𝑦0,
𝑇
0
, 𝐺2 𝑠, 𝜏 𝑔2(𝜏, 𝑥0, 𝑦0)
𝑠
−∞
𝑑𝜏, 𝑔2(𝜏, 𝑥0, 𝑦0)𝑑𝜏
𝑏 𝑠
𝑎 𝑠
)𝑑𝑠]𝑑𝑠
Proof.
𝐿1 𝑡, 𝑥0, 𝑦0 ≤ (1 −
𝑡
𝑇
) 𝑓1(𝑠, 𝑥0, 𝑦0, 𝐺1
𝑠
−∞
𝑠, 𝜏 𝑔1 𝜏, 𝑥0, 𝑦0 𝑑𝜏,
𝑡
0
, 𝑔1 𝜏, 𝑥0, 𝑦0 𝑑𝜏
𝑏(𝑠)
𝑎(𝑠)
) 𝑑𝑠 +
+
𝑡
𝑇
𝑓1(𝑠, 𝑥0, 𝑦0, 𝐺1
𝑠
−∞
𝑠, 𝜏 𝑔1 𝜏, 𝑥0, 𝑦0 𝑑𝜏,
𝑇
𝑡
𝑔1 𝜏, 𝑥0, 𝑦0 𝑑𝜏)
𝑏(𝑠)
𝑎(𝑠)
𝑑𝑠
≤ (1 −
𝑡
𝑇
) 𝑀1
𝑡
0
𝑑𝑠 +
𝑡
𝑇
𝑀1
𝑇
𝑡
𝑑𝑠
≤ 𝑀1[(1 −
𝑡
𝑇
)𝑡 +
𝑡
𝑇
(𝑇 − 𝑡)]
so that:
𝐿1 𝑡, 𝑥0, 𝑦0 ≤ 𝑀1 𝛼 𝑡 ⋯ 1.13
And similarly, we get also
𝐿2 𝑡, 𝑥0, 𝑦0 ≤ 𝑀2 𝛼 𝑡 ⋯ 1.14
From ( 1.13) and ( 1.14), then the inequality ( 1.12) satisfies for all
0 ≤ 𝑡 ≤ 𝑇 and 𝛼 𝑡 ≤
𝑇
2
. ∎
II. Approximation Solution Of (I1) And (I2)
In this section, we study the approximate periodic solution of (I1) and
(I2) by proving the following theorem:
Theorem 2.1. If the system (I1) and (I2) satisfies the inequalities ( 1.2), ( 1.3), 1.4), (1.5),(1.6) and
conditions(1.7),(1.8) has periodic solutions 𝑥 = 𝑥 𝑡, 𝑥0, 𝑦0 and 𝑦 = 𝑦(𝑡, 𝑥0, 𝑦0), then the sequence of
functions:
𝑥 𝑚+1 𝑡, 𝑥0, 𝑦0 = 𝐹0 𝑡 + [𝑓1(𝑠, 𝑥 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑚 𝑠, 𝑥0, 𝑦0 ,
𝑡
0

, 𝐺1 𝑠, 𝜏 𝑔1(𝜏, 𝑥 𝑚 𝜏, 𝑥0, 𝑦0 , 𝑦 𝑚 𝜏, 𝑥0, 𝑦0 )
𝑠
−∞
𝑑𝜏,
, 𝑔1(𝜏, 𝑥 𝑚 𝜏, 𝑥0, 𝑦0 , 𝑦 𝑚 𝜏, 𝑥0, 𝑦0 )𝑑𝜏)
𝑏 𝑠
𝑎 𝑠
−
1
𝑇
𝑓1(𝑠, 𝑥 𝑚 𝑠, 𝑥0, 𝑦0 ,
𝑇
0
, 𝑦 𝑚 𝑠, 𝑥0, 𝑦0 , 𝐺1 𝑠, 𝜏 𝑔1(𝜏, 𝑥 𝑚 𝜏, 𝑥0, 𝑦0 , 𝑦 𝑚 𝜏, 𝑥0, 𝑦0 )
𝑠
−∞
𝑑𝜏,
, 𝑔1(𝜏, 𝑥 𝑚 𝜏, 𝑥0, 𝑦0 , 𝑦 𝑚 𝜏, 𝑥0, 𝑦0 )𝑑𝜏)𝑑𝑠]𝑑𝑠
𝑏 𝑠
𝑎 𝑠
⋯ 2.1
with
𝑥0 𝑡, 𝑥0, 𝑦0 = 𝐹0 𝑡 = 𝑥0 , for all m = 0, 1,2, ⋯
𝑦 𝑚+1 𝑡, 𝑥0, 𝑦0 = 𝐺0 𝑡 + [𝑓2(𝑠, 𝑥 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑚 𝑠, 𝑥0, 𝑦0 ,
𝑡
0
, 𝐺2 𝑠, 𝜏 𝑔2(𝜏, 𝑥 𝑚 𝜏, 𝑥0, 𝑦0 , 𝑦 𝑚 𝜏, 𝑥0, 𝑦0 )
𝑠
−∞
𝑑𝜏,
, 𝑔2(𝜏, 𝑥 𝑚 𝜏, 𝑥0, 𝑦0 , 𝑦 𝑚 𝜏, 𝑥0, 𝑦0 )𝑑𝜏)
𝑏 𝑠
𝑎 𝑠
−
1
𝑇
𝑓2(𝑠, 𝑥 𝑚 𝑠, 𝑥0, 𝑦0 ,
𝑇
0
, 𝑦 𝑚 𝑠, 𝑥0, 𝑦0 , 𝐺2 𝑠, 𝜏 𝑔2(𝜏, 𝑥 𝑚 𝜏, 𝑥0, 𝑦0 , 𝑦 𝑚 𝜏, 𝑥0, 𝑦0 )
𝑠
−∞
𝑑𝜏,
, 𝑔2(𝜏, 𝑥 𝑚 𝜏, 𝑥0, 𝑦0 , 𝑦 𝑚 𝜏, 𝑥0, 𝑦0 )𝑑𝜏)𝑑𝑠]𝑑𝑠
𝑏 𝑠
𝑎 𝑠
⋯ 2.2
with
𝑦0 𝑡, 𝑥0, 𝑦0 = 𝐺0 𝑡 = 𝑦0 , for all m = 0, 1,2, ⋯
periodic in t of period T, and uniformly converges as 𝑚 → ∞ in the domain:
𝑡, 𝑥0, 𝑦0 ∈ 0, 𝑇 × 𝐷𝑓1
× 𝐷1𝑓2
⋯ 2.3
to the limit functions 𝑥0
𝑡, 𝑥0, 𝑦0 , 𝑦0
𝑡, 𝑥0, 𝑦0 defined in the domain ( 2.3) which are periodic in t of period T
and satisfying the system of integral equations:
𝑥 𝑡, 𝑥0, 𝑦0 = 𝐹0 𝑡 + [𝑓1(𝑠, 𝑥 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑠, 𝑥0, 𝑦0 ,
𝑡
0
, 𝐺1 𝑠, 𝜏 𝑔1(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )
𝑠
−∞
𝑑𝜏,
, 𝑔1(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )𝑑𝜏)
𝑏 𝑠
𝑎 𝑠
−
1
𝑇
𝑓1(𝑠, 𝑥 𝑠, 𝑥0, 𝑦0 ,
𝑇
0
, 𝑦 𝑠, 𝑥0, 𝑦0 , 𝐺1 𝑠, 𝜏 𝑔1(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )
𝑠
−∞
𝑑𝜏,
, 𝑔1(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )𝑑𝜏)𝑑𝑠]𝑑𝑠
𝑏 𝑠
𝑎 𝑠
⋯ 2.4
and
𝑦 𝑡, 𝑥0, 𝑦0 = 𝐺0 𝑡 + [𝑓2(𝑠, 𝑥 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑠, 𝑥0, 𝑦0 ,
𝑡
0

, 𝐺2 𝑠, 𝜏 𝑔2(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )
𝑠
−∞
𝑑𝜏,
, 𝑔2(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )𝑑𝜏)
𝑏 𝑠
𝑎 𝑠
−
1
𝑇
𝑓2(𝑠, 𝑥 𝑠, 𝑥0, 𝑦0 ,
𝑇
0
𝑠
−∞
𝑑𝜏,
𝑏 𝑠
𝑎 𝑠
⋯ 2.5
provided that:
𝑥0
𝑡, 𝑥0, 𝑦0 − 𝑥 𝑚 𝑡, 𝑥0, 𝑦0
𝑦0
𝑡, 𝑥0, 𝑦0 − 𝑦 𝑚 𝑡, 𝑥0, 𝑦0
≤ 𝑄0
𝑚
(𝐸 − 𝑄0)−1
𝐶0 ⋯ 2.6
for all 𝑚 ≥ 0 , where
𝐶0 =
𝑇
2
𝑀1
𝑇
2
𝑀2
, 𝐸 is identity matrix.
Proof. Consider the sequence of functions 𝑥1 𝑡, 𝑥0, 𝑦0 , 𝑥2 𝑡, 𝑥0, 𝑦0 , ⋯,
𝑥 𝑚 𝑡, 𝑥0, 𝑦0 , ⋯, and 𝑦1 𝑡, 𝑥0, 𝑦0 , 𝑦2 𝑡, 𝑥0, 𝑦0 , ⋯ , 𝑦 𝑚 𝑡, 𝑥0, 𝑦0 , ⋯, defined
by the recurrences relations ( 2.1) and ( 2.2). Each of these functions of sequence defined and continuous in the
domain ( 1.1) and periodic in 𝑡 of period 𝑇.
Now, by the lemma 1.1, and using the sequence of functions (2.1),
when 𝑚 = 0, we get:
By mathematical induction and lemma1.1 , we have the following inequality:
𝑥 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑥0 ≤
𝑇
2
𝑀1 ⋯ 2.13
i.e. 𝑥 𝑚 𝑡, 𝑥0, 𝑦0 ∈ 𝐷, for all 𝑡 ∈ 𝑅1
, 𝑥0 ∈ 𝐷𝑓1
, 𝑦0 ∈ 𝐷1𝑓2
, 𝑚 = 0,1,2, ⋯,
and
𝑦 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑦0 ≤
𝑇
2
𝑀2 ⋯ 2.14
i.e. 𝑦 𝑚 𝑡, 𝑥0, 𝑦0 ∈ 𝐷, for all 𝑡 ∈ 𝑅1
, 𝑥0 ∈ 𝐷𝑓1
, 𝑦0 ∈ 𝐷1𝑓2
, 𝑚 = 0,1,2, ⋯,
Now, from ( 2.13), we get:
𝑧 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑧0(𝑡, 𝑥0, 𝑦0) ≤
𝑇
2
(
𝛾
𝜆1
) 𝑅1 𝑀1 + 𝑅2 𝑀2 ⋯ 2.15
and
𝑤 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑤0(𝑡, 𝑥0, 𝑦0) ≤
𝑇
2
𝑕 𝑅1 𝑀1 + 𝑅2 𝑀2 ⋯ 2.16
, 𝑥0 ∈ 𝐷𝑓1
, 𝑦0 ∈ 𝐷1𝑓2
, 𝑧0 𝑡, 𝑥0, 𝑦0 ∈ 𝐷2𝑓1
and
𝑤0 𝑡, 𝑥0, 𝑦0 ∈ 𝐷3𝑓1
,
i.e. 𝑧 𝑚 𝑡, 𝑥0, 𝑦0 ∈ 𝐷2 and 𝑤 𝑚 𝑡, 𝑥0, 𝑦0 ∈ 𝐷3 , for all 𝑥0 ∈ 𝐷𝑓1
, 𝑦0 ∈ 𝐷1𝑓2
,
where
𝑧 𝑚 𝑡, 𝑥0, 𝑦0 = 𝐺1 𝑡, 𝑠 𝑔1 𝑠, 𝑥 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑚 𝑠, 𝑥0, 𝑦0 𝑑𝑠
𝑡
−∞
and
𝑤 𝑚 𝑡, 𝑥0, 𝑦0 = 𝑔1 𝑠, 𝑥 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑚 𝑠, 𝑥0, 𝑦0 𝑑𝑠
𝑏(𝑡)
𝑎(𝑡)
, 𝑚 = 0,1,2, ⋯ ,
Also from (4.2.14), we get:
𝑢 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑢0(𝑡, 𝑥0, 𝑦0) ≤
𝑇
2
(
𝛿
𝜆2
) 𝐻1 𝑀1 + 𝐻2 𝑀2 ⋯ 2.17
and
𝑉𝑚 𝑡, 𝑥0, 𝑦0 − 𝑉0(𝑡, 𝑥0, 𝑦0) ≤
𝑇
2
𝑕 𝐻1 𝑀1 + 𝐻2 𝑀2 ⋯ 2.18

, 𝑥0 ∈ 𝐷𝑓1
, 𝑦0 ∈ 𝐷1𝑓2
, 𝑢0 𝑡, 𝑥0, 𝑦0 ∈ 𝐷2𝑓2
and
𝑣0 𝑡, 𝑥0, 𝑦0 ∈ 𝐷3𝑓2
,
i.e. 𝑢 𝑚 𝑡, 𝑥0, 𝑦0 ∈ 𝐷2 and 𝑣 𝑚 𝑡, 𝑥0, 𝑦0 ∈ 𝐷3, for all 𝑥0 ∈ 𝐷𝑓1
, 𝑦0 ∈ 𝐷1𝑓2
,
where
𝑢 𝑚 𝑡, 𝑥0, 𝑦0 = 𝐺2 𝑡, 𝑠 𝑔2 𝑠, 𝑥 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑚 𝑠, 𝑥0, 𝑦0 𝑑𝑠
𝑡
−∞
and
𝑣 𝑚 𝑡, 𝑥0, 𝑦0 = 𝑔2 𝑠, 𝑥 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑚 𝑠, 𝑥0, 𝑦0 𝑑𝑠
𝑏(𝑡)
𝑎(𝑡)
, 𝑚 = 0,1,2, ⋯ ,
Next, we claim that two sequences 𝑥 𝑚 (𝑡, 𝑥0, 𝑦0)} 𝑚=0
∞
, 𝑦 𝑚 (𝑡, 𝑥0, 𝑦0)} 𝑚=0
∞
are convergent uniformly to the limit functions 𝑥, 𝑦 on the domain ( 2.3).
By mathematical induction and lemma 1.1, we find that:
𝑥 𝑚+1 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑚 𝑡, 𝑥0, 𝑦0 ≤
≤ 𝛼(𝑡)[𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑥 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑚−1 𝑡, 𝑥0, 𝑦0
+𝛼(𝑡)[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑦 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑚−1 𝑡, 𝑥0, 𝑦0
⋯ 2.19
and
𝑦 𝑚+1 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑚 𝑡, 𝑥0, 𝑦0 ≤
≤ 𝛼(𝑡)[𝐿1 + 𝐻1(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] 𝑥 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑚−1 𝑡, 𝑥0, 𝑦0 +
+𝛼(𝑡)[𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] 𝑦 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑚−1 𝑡, 𝑥0, 𝑦0
⋯ 2.20
We can write the inequalities ( 2.19) and ( 2.20) in a vector form:
𝐶 𝑚+1 𝑡, 𝑥0, 𝑦0 ≤ 𝑄 𝑡 𝐶 𝑚 𝑡, 𝑥0, 𝑦0 ⋯ 2.21
where
𝐶 𝑚+1 𝑡, 𝑥0, 𝑦0 =
𝑥 𝑚+1 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑚 𝑡, 𝑥0, 𝑦0
𝑦 𝑚+1 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑚 𝑡, 𝑥0, 𝑦0
,
and
𝑄 𝑡 =
𝛼(𝑡)[𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝛼(𝑡)[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)]
𝛼(𝑡)[𝐿1 + 𝐻1(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] 𝛼(𝑡)[𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)]
,
𝐶 𝑚 𝑡, 𝑥0, 𝑦0 =
𝑥 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑚−1 𝑡, 𝑥0, 𝑦0
𝑦 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑚−1 𝑡, 𝑥0, 𝑦0
Now, we take the maximum value to both sides of the inequality ( 2.21),
for all 0 ≤ 𝑡 ≤ 𝑇 and 𝛼 𝑡 ≤
𝑇
2
, we get:
𝐶 𝑚+1 ≤ 𝑄0 𝐶 𝑚 , ⋯ 2.22
where 𝑄0 = 𝑚𝑎𝑥𝑡∈ 0,𝑇 𝑄(𝑡) .
Q0 =
𝑇
2
[𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)]
𝑇
2
[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)]
𝑇
2
[𝐿1 + 𝐻1(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)]
𝑇
2
[𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)]
By iterating the inequality (4.2.22), we find that:
𝐶 𝑚+1 ≤ 𝑄0
𝑚
𝐶0 , ⋯ 2.23
which leads to the estimate:
𝐶𝑖 ≤ 𝑄0
𝑖−1
𝑚
𝑖=1
𝐶0
𝑚
𝑖=1
⋯ 2.24
Since the matrix 𝑄0 has maximum eigen-values of (4.1.13) and the series
(4.2.24) is uniformly convergent, i. e.

lim
𝑚→∞
𝑄0
𝑖−1
𝑚
𝑖=1
𝐶0 = 𝑄0
𝑖−1
∞
𝑖=1
𝐶0 = (𝐸 − 𝑄0)−1
𝐶0 ⋯ ( 2.25)
The limiting relation (4.2.25) signifies a uniform convergence of the sequence
𝑥 𝑚 (𝑡, 𝑥0, 𝑦0), 𝑦 𝑚 (𝑡, 𝑥0, 𝑦0) 𝑚=0
∞
in the domain (4.2.3) as 𝑚 → ∞.
Let
lim
𝑚→∞
𝑥 𝑚 𝑡, 𝑥0, 𝑦0 = 𝑥0
𝑡, 𝑥0, 𝑦0 ,
lim
𝑚→∞
𝑦 𝑚 𝑡, 𝑥0, 𝑦0 = 𝑦0
𝑡, 𝑥0, 𝑦0 .
⋯ ( 2.26)
Finally, we show that 𝑥 𝑡, 𝑥0, 𝑦0 ≡ 𝑥0
𝑡, 𝑥0, 𝑦0 ∈ 𝐷 and
𝑦 𝑡, 𝑥0, 𝑦0 ≡ 𝑦0
(𝑡, 𝑥0, 𝑦0) ∈ 𝐷1, for all 𝑥0 ∈ 𝐷𝑓1
and 𝑦0 ∈ 𝐷1𝑓2
.
By using inequalities ( 2.1) and ( 2.4) and lemma 1.1, such that:
[𝑓1(𝑠, 𝑥 𝑚 (𝑠, 𝑥0, 𝑦0), 𝑦 𝑚 (𝑠, 𝑥0, 𝑦0),
𝑡
0
𝑧 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑤 𝑚 (𝑠, 𝑥0, 𝑦0)) −
−
1
𝑇
𝑓1(𝑠, 𝑥 𝑚 (𝑠, 𝑥0, 𝑦0), 𝑦 𝑚 (𝑠, 𝑥0, 𝑦0),
𝑇
0
𝑧 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑤 𝑚 (𝑠, 𝑥0, 𝑦0))𝑑𝑠]𝑑𝑠 −
− [𝑓1(𝑠, 𝑥(𝑠, 𝑥0, 𝑦0), 𝑦(𝑠, 𝑥0, 𝑦0),
𝑡
0
𝑧 𝑠, 𝑥0, 𝑦0 , 𝑤(𝑠, 𝑥0, 𝑦0)) −
−
1
𝑇
𝑓1(𝑠, 𝑥(𝑠, 𝑥0, 𝑦0), 𝑦(𝑠, 𝑥0, 𝑦0),
𝑇
0
𝑧 𝑠, 𝑥0, 𝑦0 , 𝑤(𝑠, 𝑥0, 𝑦0))𝑑𝑠]𝑑𝑠
≤ 𝛼(𝑡) [𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑥 𝑚 𝑠, 𝑥0, 𝑦0 − 𝑥 𝑠, 𝑥0, 𝑦0 +
+[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑦 𝑚 𝑠, 𝑥0, 𝑦0 − 𝑦 𝑠, 𝑥0, 𝑦0 ≤
≤
𝑇
2
[𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑥 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑡, 𝑥0, 𝑦0 +
+[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑦 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑡, 𝑥0, 𝑦0 .
From inequality ( 2.26) and on the other hand suppose that:.
𝑥 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑡, 𝑥0, 𝑦0 ≤ 𝜖1 ,
𝑦 𝑚 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑡, 𝑥0, 𝑦0 ≤ 𝜖2 .
Thus
[𝑓1(𝑠, 𝑥 𝑚 (𝑠, 𝑥0, 𝑦0), 𝑦 𝑚 (𝑠, 𝑥0, 𝑦0),
𝑡
0
𝑧 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑤 𝑚 (𝑠, 𝑥0, 𝑦0)) −
−
1
𝑇
𝑓1(𝑠, 𝑥 𝑚 (𝑠, 𝑥0, 𝑦0), 𝑦 𝑚 (𝑠, 𝑥0, 𝑦0),
𝑇
0
𝑧 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑤 𝑚 (𝑠, 𝑥0, 𝑦0))𝑑𝑠]𝑑𝑠 −
− [𝑓1(𝑠, 𝑥(𝑠, 𝑥0, 𝑦0), 𝑦(𝑠, 𝑥0, 𝑦0),
𝑡
0
𝑧 𝑠, 𝑥0, 𝑦0 , 𝑤(𝑠, 𝑥0, 𝑦0)) −
−
1
𝑇
𝑓1(𝑠, 𝑥(𝑠, 𝑥0, 𝑦0), 𝑦(𝑠, 𝑥0, 𝑦0),
𝑇
0
𝑧 𝑠, 𝑥0, 𝑦0 , 𝑤(𝑠, 𝑥0, 𝑦0))𝑑𝑠]𝑑𝑠
≤
𝑇
2
[𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)]𝜖1 +
𝑇
2
[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)]𝜖2
Putting 𝜖1 =
𝜖3
𝑇
2
[𝐾1+𝑅1(
𝛾
𝜆1
𝐾3+𝑕 𝐾4)]
and 𝜖2 =
𝜖4
𝑇
2
[𝐾2+𝑅2(
𝛾
𝜆1
𝐾3+𝑕𝐾4)]
and substituting in the last equation, we have:
[𝑓1(𝑠, 𝑥 𝑚 (𝑠, 𝑥0, 𝑦0), 𝑦 𝑚 (𝑠, 𝑥0, 𝑦0),
𝑡
0
𝑧 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑤 𝑚 (𝑠, 𝑥0, 𝑦0)) −

−
1
𝑇
𝑓1(𝑠, 𝑥 𝑚 (𝑠, 𝑥0, 𝑦0), 𝑦 𝑚 (𝑠, 𝑥0, 𝑦0),
𝑇
0
𝑧 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑤 𝑚 (𝑠, 𝑥0, 𝑦0))𝑑𝑠]𝑑𝑠 −
− [𝑓1(𝑠, 𝑥(𝑠, 𝑥0, 𝑦0), 𝑦(𝑠, 𝑥0, 𝑦0),
𝑡
0
𝑧 𝑠, 𝑥0, 𝑦0 , 𝑤(𝑠, 𝑥0, 𝑦0)) −
−
1
𝑇
𝑓1(𝑠, 𝑥(𝑠, 𝑥0, 𝑦0), 𝑦(𝑠, 𝑥0, 𝑦0),
𝑇
0
𝑧 𝑠, 𝑥0, 𝑦0 , 𝑤(𝑠, 𝑥0, 𝑦0))𝑑𝑠]𝑑𝑠
≤
𝑇
2
[𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)]
𝜖3
𝑇
2
[𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)]
+
+
𝑇
2
[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)]
𝜖4
𝑇
2
[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)]
≤ 𝜖3 + 𝜖4,
and choosing 𝜖3 + 𝜖4 = 𝜖 , we get:
[𝑓1(𝑠, 𝑥 𝑚 (𝑠, 𝑥0, 𝑦0), 𝑦 𝑚 (𝑠, 𝑥0, 𝑦0),
𝑡
0
𝑧 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑤 𝑚 (𝑠, 𝑥0, 𝑦0)) −
−
1
𝑇
𝑓1(𝑠, 𝑥 𝑚 (𝑠, 𝑥0, 𝑦0), 𝑦 𝑚 (𝑠, 𝑥0, 𝑦0),
𝑇
0
𝑧 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑤 𝑚 (𝑠, 𝑥0, 𝑦0))𝑑𝑠]𝑑𝑠 −
− [𝑓1(𝑠, 𝑥(𝑠, 𝑥0, 𝑦0), 𝑦(𝑠, 𝑥0, 𝑦0),
𝑡
0
𝑧 𝑠, 𝑥0, 𝑦0 , 𝑤(𝑠, 𝑥0, 𝑦0)) −
−
1
𝑇
𝑓1(𝑠, 𝑥(𝑠, 𝑥0, 𝑦0), 𝑦(𝑠, 𝑥0, 𝑦0),
𝑇
0
𝑧 𝑠, 𝑥0, 𝑦0 , 𝑤(𝑠, 𝑥0, 𝑦0))𝑑𝑠]𝑑𝑠 ≤ 𝜖
for all 𝑚 ≥ 0,
i. e. lim
𝑚→∞
[𝑓1(𝑠, 𝑥 𝑚 (𝑠, 𝑥0, 𝑦0), 𝑦 𝑚 (𝑠, 𝑥0, 𝑦0),
𝑡
0
𝑧 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑤 𝑚 (𝑠, 𝑥0, 𝑦0)) −
−
1
𝑇
𝑓1(𝑠, 𝑥 𝑚 (𝑠, 𝑥0, 𝑦0), 𝑦 𝑚 (𝑠, 𝑥0, 𝑦0),
𝑇
0
𝑧 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑤 𝑚 (𝑠, 𝑥0, 𝑦0))𝑑𝑠]𝑑𝑠 =
− [𝑓1(𝑠, 𝑥(𝑠, 𝑥0, 𝑦0), 𝑦(𝑠, 𝑥0, 𝑦0),
𝑡
0
𝑧 𝑠, 𝑥0, 𝑦0 , 𝑤(𝑠, 𝑥0, 𝑦0)) −
−
1
𝑇
𝑓1(𝑠, 𝑥(𝑠, 𝑥0, 𝑦0), 𝑦(𝑠, 𝑥0, 𝑦0),
𝑇
0
𝑧 𝑠, 𝑥0, 𝑦0 , 𝑤(𝑠, 𝑥0, 𝑦0))𝑑𝑠]𝑑𝑠.
So 𝑥 𝑡, 𝑥0, 𝑦0 ∈ 𝐷, and 𝑥 𝑡, 𝑥0, 𝑦0 ≡ 𝑥0
(𝑡, 𝑥0, 𝑦0) is a periodic solution of ( I1),
Also by using the same method above we can prove 𝑦 𝑡, 𝑥0, 𝑦0 ∈ 𝐷1, and 𝑦 𝑡, 𝑥0, 𝑦0 ≡ 𝑦0
𝑡, 𝑥0, 𝑦0 is
also periodic solution of (𝐼2 ).
Theorem 2.2. With the hypotheses and all conditions of the theorem 2.1, the periodic solution of integral
equations ( 𝐼1 ) and ( 𝐼2 ) are a unique on the domain (1.3).
Proof. Suppose that 𝑥 𝑡, 𝑥0, 𝑦0 and 𝑦 𝑡, 𝑥0, 𝑦0 be another periodic solutions
for the systems (𝐼1) and ( 𝐼2 ) defined and continuous and periodic in 𝑡 of period 𝑇, this means that:
𝑥 𝑡, 𝑥0, 𝑦0 = 𝐹0 𝑡 + [𝑓1(𝑠, 𝑥 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑠, 𝑥0, 𝑦0 ,
𝑡
0
, 𝐺1 𝑠, 𝜏 𝑔1(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )
𝑠
−∞
𝑑𝜏,

, 𝑔1(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )𝑑𝜏)
𝑏 𝑠
𝑎 𝑠
−
1
𝑇
𝑓1(𝑠, 𝑥 𝑠, 𝑥0, 𝑦0 ,
𝑇
0
𝑠
−∞
𝑑𝜏,
𝑏 𝑠
𝑎 𝑠
⋯ 2.27
and
𝑦 𝑡, 𝑥0, 𝑦0 = 𝐺0 𝑡 + [𝑓2(𝑠, 𝑥 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑠, 𝑥0, 𝑦0 ,
𝑡
0
, 𝐺2 𝑠, 𝜏 𝑔2(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )
𝑠
−∞
𝑑𝜏,
, 𝑔2(𝜏, 𝑥 𝜏, 𝑥0, 𝑦0 , 𝑦 𝜏, 𝑥0, 𝑦0 )𝑑𝜏)
𝑏 𝑠
𝑎 𝑠
−
1
𝑇
𝑓2(𝑠, 𝑥 𝑠, 𝑥0, 𝑦0 ,
𝑇
0
𝑠
−∞
𝑑𝜏,
𝑏 𝑠
𝑎 𝑠
⋯ 2.28
For their difference, we should obtain the inequality:
𝑥 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑡, 𝑥0, 𝑦0 ≤
≤
𝑇
2
[𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑥 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑡, 𝑥0, 𝑦0 +
+
𝑇
2
[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑦 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑡, 𝑥0, 𝑦0 ⋯ 2.29
And also
𝑦 𝑠, 𝑥0, 𝑦0 − 𝑦 𝑠, 𝑥0, 𝑦0 ≤
≤ (1 −
𝑡
𝑇
) [𝐿1 𝑥 𝑠, 𝑥0, 𝑦0 − 𝑥 𝑠, 𝑥0, 𝑦0 + 𝐿2 𝑦 𝑠, 𝑥0, 𝑦0 − 𝑦 𝑠, 𝑥0, 𝑦0
𝑡
0
+
𝛿
𝜆2
𝐿3 𝐻1 𝑥 𝑠, 𝑥0, 𝑦0 − 𝑥 𝑠, 𝑥0, 𝑦0 + 𝐻2 𝑦 𝑠, 𝑥0, 𝑦0 − 𝑦 𝑠, 𝑥0, 𝑦0 +
+𝑕𝐿4(𝐻1 𝑥 𝑠, 𝑥0, 𝑦0 − 𝑥 𝑠, 𝑥0, 𝑦0 + 𝐻2 𝑦 𝑠, 𝑥0, 𝑦0 − 𝑦 𝑠, 𝑥0, 𝑦0 )]𝑑𝑠
+
𝑡
𝑇
𝐿1 𝑥 𝑠, 𝑥0, 𝑦0 − 𝑥 𝑠, 𝑥0, 𝑦0 + 𝐿2 𝑦 𝑠, 𝑥0, 𝑦0 − 𝑦 𝑠, 𝑥0, 𝑦0 +
𝑇
𝑡
+
𝛿
𝜆2
𝐿3 𝐻1 𝑥 𝑠, 𝑥0, 𝑦0 − 𝑥 𝑠, 𝑥0, 𝑦0 + 𝐻2 𝑦 𝑠, 𝑥0, 𝑦0 − 𝑦 𝑠, 𝑥0, 𝑦0 +
+𝑕𝐿4(𝐻1 𝑥 𝑠, 𝑥0, 𝑦0 − 𝑥 𝑠, 𝑥0, 𝑦0 + 𝐻2 𝑦 𝑠, 𝑥0, 𝑦0 − 𝑦 𝑠, 𝑥0, 𝑦0 )]𝑑𝑠
≤ 𝛼(𝑡)[𝐿1 + 𝐻1(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] 𝑥 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑡, 𝑥0, 𝑦0 +
+𝛼(𝑡)[𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] 𝑦 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑡, 𝑥0, 𝑦0
so that:
𝑦 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑡, 𝑥0, 𝑦0 ≤
≤
𝑇
2
[𝐿1 + 𝐻1(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] 𝑥 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑡, 𝑥0, 𝑦0 +
+
𝑇
2
[𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] 𝑦 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑡, 𝑥0, 𝑦0 ⋯ 2.30

and the inequalities (4.2.29) and (4.2.30) would lead to the estimate:
𝑥 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑡, 𝑥0, 𝑦0
𝑦 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑡, 𝑥0, 𝑦0
≤ 𝑄0
𝑥 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑡, 𝑥0, 𝑦0
𝑦 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑡, 𝑥0, 𝑦0
⋯ 2.31
By iterating the inequality (4.2.27), which should find:
𝑥 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑡, 𝑥0, 𝑦0
𝑦 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑡, 𝑥0, 𝑦0
≤ 𝑄0
𝑚 𝑥 𝑡, 𝑥0, 𝑦0 − 𝑥 𝑡, 𝑥0, 𝑦0
𝑦 𝑡, 𝑥0, 𝑦0 − 𝑦 𝑡, 𝑥0, 𝑦0
.
But 𝑄0
𝑚
→ 0 as 𝑚 → ∞, so that proceeding in the last inequality which is contradict the supposition It follows
immediately 𝑥 𝑡, 𝑥0, 𝑦0 = 𝑥 𝑡, 𝑥0, 𝑦0 and
𝑦 𝑡, 𝑥0, 𝑦0 = 𝑦 𝑡, 𝑥0, 𝑦0 .
III. Existence of Solution of (𝑰 𝟏) and (𝑰 𝟐)
The problem of existence of periodic solution of period T of the system
( 𝐼1) and ( 𝐼2) are uniquely connected with the existence of zero of the functions ∆1 0, 𝑥0, 𝑦0 = ∆1 and
∆2 0, 𝑥0, 𝑦0 = ∆2 which has the form:
∆1: 𝐷𝑓1
× 𝐷1𝑓2
→ 𝑅 𝑛
∆1 0, 𝑥0, 𝑦0 =
1
𝑇
𝑓1(𝑡, 𝑥0
𝑡, 𝑥0, 𝑦0 , 𝑦0
𝑡, 𝑥0, 𝑦0 ,
𝑇
0
, 𝐺1 𝑡, 𝑠 𝑔1 𝑠, 𝑥0
𝑠, 𝑥0, 𝑦0 , 𝑦0
𝑠, 𝑥0, 𝑦0
𝑡
−∞
𝑑𝑠
, 𝑔1 𝑠, 𝑥0
𝑠, 𝑥0, 𝑦0 , 𝑦0
𝑠, 𝑥0, 𝑦0 𝑑𝑠)𝑑𝑡
𝑏 𝑡
𝑎 𝑡
⋯ 3.1
∆2: 𝐷𝑓1
× 𝐷1𝑓2
→ 𝑅 𝑛
∆2 0, 𝑥0, 𝑦0 =
1
𝑇
𝑓2(𝑡, 𝑥0
𝑡, 𝑥0, 𝑦0 , 𝑦0
𝑡, 𝑥0, 𝑦0 ,
𝑇
0
, 𝐺2 𝑡, 𝑠 𝑔2 𝑠, 𝑥0
𝑠, 𝑥0, 𝑦0 , 𝑦0
𝑠, 𝑥0, 𝑦0
𝑡
−∞
𝑑𝑠
, 𝑔2 𝑠, 𝑥0
𝑠, 𝑥0, 𝑦0 , 𝑦0
𝑠, 𝑥0, 𝑦0 𝑑𝑠)𝑑𝑡
𝑏 𝑡
𝑎 𝑡
⋯ 3.2
where the function 𝑥0
𝑡, 𝑥0, 𝑦0 is the limit of the sequence of the functions 𝑥 𝑚 (𝑡, 𝑥0, 𝑦0) and the function
𝑦0
𝑡, 𝑥0, 𝑦0 is the limit of the sequence of the functions 𝑦 𝑚 𝑡, 𝑥0, 𝑦0 .
Since this two functions are approximately determined from the sequences of functions:
∆1𝑚 : 𝐷𝑓1
× 𝐷1𝑓2
→ 𝑅 𝑛
∆1𝑚 0, 𝑥0, 𝑦0 =
1
𝑇
𝑓1(𝑡, 𝑥 𝑚 𝑡, 𝑥0, 𝑦0 , 𝑦 𝑚 𝑡, 𝑥0, 𝑦0 ,
𝑇
0
, 𝐺1 𝑡, 𝑠 𝑔1 𝑠, 𝑥 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑚 𝑠, 𝑥0, 𝑦0
𝑡
−∞
𝑑𝑠
, 𝑔1 𝑠, 𝑥 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑚 𝑠, 𝑥0, 𝑦0 𝑑𝑠)𝑑𝑡
𝑏 𝑡
𝑎 𝑡
⋯ 3.3
∆2𝑚 : 𝐷𝑓1
× 𝐷1𝑓2
→ 𝑅 𝑛
∆2𝑚 0, 𝑥0, 𝑦0 =
1
𝑇
𝑓2(𝑡, 𝑥 𝑚 𝑡, 𝑥0, 𝑦0 , 𝑦 𝑚 𝑡, 𝑥0, 𝑦0 ,
𝑇
0
, 𝐺2 𝑡, 𝑠 𝑔2 𝑠, 𝑥 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑚 𝑠, 𝑥0, 𝑦0
𝑡
−∞
𝑑𝑠

, 𝑔2 𝑠, 𝑥 𝑚 𝑠, 𝑥0, 𝑦0 , 𝑦 𝑚 𝑠, 𝑥0, 𝑦0 𝑑𝑠)𝑑𝑡
𝑏 𝑡
𝑎 𝑡
⋯ 3.4
for all 𝑚 = 0,1,2, ⋯
Theorem 3.1. If the hypotheses and all conditions of the theorem 2.1 and 2.2 are satisfied, then the following
inequality satisfied:
∆1 0, 𝑥0, 𝑦0 − ∆1𝑚 0, 𝑥0, 𝑦0 ≤ 𝑑 𝑚 ⋯ 3.5
∆2 0, 𝑥0, 𝑦0 − ∆2𝑚 0, 𝑥0, 𝑦0 ≤ 𝜂 𝑚 ⋯ 3.6
satisfied for all 𝑚 ≥ 0 , 𝑥0 ∈ 𝐷𝑓1
and 𝑦0 ∈ 𝐷1𝑓2
,
where
𝑑 𝑚 = [𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] [𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] ,
, 𝑄0
𝑚+1
𝐸 − 𝑄0
−1
𝐶0
and
𝜂 𝑚 = [𝐿1 + 𝐻1(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] [𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] ,
, 𝑄0
𝑚+1
𝐸 − 𝑄0
−1
𝐶0 .
Proof. By using the relation ( 3.1) and ( 3.3), we have:
∆1 0, 𝑥0, 𝑦0 − ∆1𝑚 0, 𝑥0, 𝑦0
≤ [𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑥0
𝑡, 𝑥0, 𝑦0 − 𝑥 𝑚 𝑡, 𝑥0, 𝑦0 +
+[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑦0
𝑡, 𝑥0, 𝑦0 − 𝑦 𝑚 𝑡, 𝑥0, 𝑦0
≤ [𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] [𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] ,
, 𝑄0
𝑚+1
𝐸 − 𝑄0
−1
𝐶0 = 𝑑 𝑚
And also by using the relation ( 3.2) and ( 3.4), we get:
∆2 0, 𝑥0, 𝑦0 − ∆2𝑚 0, 𝑥0, 𝑦0 ≤
≤ [𝐿1 + 𝐻1(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] [𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] ,
, 𝑄0
𝑚+1
𝐸 − 𝑄0
−1
𝐶0 = 𝜂 𝑚
where . denotes the ordinary scalar product in the space 𝑅 𝑛
. ∎
Theorem 3.2. Let the vector functions 𝑓1 𝑡, 𝑥, 𝑦, 𝑧, 𝑤 , 𝑓2 𝑡, 𝑥, 𝑦, 𝑢, 𝑣 ,
𝑔1 𝑡, 𝑥, 𝑦 and 𝑔2 𝑡, 𝑥, 𝑦 be defined on the domain:
𝐺 = 0 ≤ 𝑠 ≤ 𝑡 ≤ 𝑇, 𝑎 ≤ 𝑥, 𝑦 ≤ 𝑏, 𝑐 ≤ 𝑧, 𝑢 ≤ 𝑑, 𝑒 ≤ 𝑤, 𝑣 ≤ 𝑓} ⊆ 𝑅1
,
and periodic in t of period T.
Assume that the sequence of functions (4.3.3) and (4.3.4) satisfies the
inequalities:
min
a+
𝑇
2
𝑀1≤x0≤b−
𝑇
2
𝑀1
c+
𝑇
2
𝑀2≤y0≤d−
𝑇
2
𝑀2
∆1m 0, x0, y0 ≤ − 𝑑m ,
max
a+
𝑇
2
𝑀1≤x0≤b−
𝑇
2
𝑀1
c+
𝑇
2
𝑀2≤y0≤d−
𝑇
2
𝑀2
∆1m 0, x0, y0 ≥ 𝑑m ,
⋯ 3.7
min
a+
𝑇
2
𝑀1≤x0≤b−
𝑇
2
𝑀1
c+
𝑇
2
𝑀2≤y0≤d−
𝑇
2
𝑀2
∆2m 0, x0, y0 ≤ − ηm
,
max
a+
𝑇
2
𝑀1≤x0≤b−
𝑇
2
𝑀1
c+
𝑇
2
𝑀2≤y0≤d−
𝑇
2
𝑀2
∆2m 0, x0, y0 ≥ ηm
,
⋯ 3.8
for all 𝑚 ≥ 0, where
𝑑 𝑚 = [𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] [𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] ,
, 𝑄0
𝑚+1
𝐸 − 𝑄0
−1
𝐶0
and

𝜂 𝑚 = [𝐿1 + 𝐻1(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] [𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] ,
, 𝑄0
𝑚+1
𝐸 − 𝑄0
−1
𝐶0
Then the system (4.1.1) and (4.1.2) has periodic solution of period T
𝑥 = 𝑥(𝑡, 𝑥0, 𝑦0) and 𝑦 = 𝑦(𝑡, 𝑥0, 𝑦0) for which 𝑥0 ∈ [𝑎 +
𝑇
2
𝑀1, 𝑏 −
𝑇
2
𝑀1] and
𝑦0 ∈ [𝑐 +
𝑇
2
𝑀2, 𝑑 −
𝑇
2
𝑀2].
Proof. Let 𝑥1, 𝑥2 be any two points in the interval [𝑎 +
𝑇
2
𝑀1, 𝑏 −
𝑇
2
𝑀1] and 𝑦1, 𝑦2 be any two points in the
interval [𝑐 +
𝑇
2
𝑀2, 𝑑 −
𝑇
2
𝑀2],
such that:
Δ1m 0, 𝑥1, 𝑦1 = min
a+
𝑇
2
𝑀1≤x0≤b−
𝑇
2
𝑀1
c+
𝑇
2
𝑀2≤y0≤d−
𝑇
2
𝑀2
∆1m 0, 𝑥0, 𝑦0 ,
Δ1m 0, 𝑥2, 𝑦2 = max
a+
𝑇
2
𝑀1≤x0≤b−
𝑇
2
𝑀1
c+
𝑇
2
𝑀2≤y0≤d−
𝑇
2
𝑀2
∆1𝑚 0, 𝑥0, 𝑦0 ,
⋯ 3.9
Δ2m 0, 𝑥1, 𝑦1 = min
a+
𝑇
2
𝑀1≤x0≤b−
𝑇
2
𝑀1
c+
𝑇
2
𝑀2≤y0≤d−
𝑇
2
𝑀2
∆2m 0, x0, y0 ,
Δ2m 0, 𝑥2, 𝑦2 = max
a+
𝑇
2
𝑀1≤x0≤b−
𝑇
2
𝑀1
c+
𝑇
2
𝑀2≤y0≤d−
𝑇
2
𝑀2
∆2m 0, x0, y0 ,
⋯ 3.10
By using the inequalities ( 3.5), ( 3.6), ( 3.7) and ( 3.8), we have:
∆1 0, 𝑥1, 𝑦1 = ∆1𝑚 0, 𝑥1, 𝑦1 + ∆1 0, 𝑥1, 𝑦1 − ∆1𝑚 (0, 𝑥1, 𝑦1) ≤ 0 ,
∆1 0, 𝑥2, 𝑦2 = ∆1𝑚 0, 𝑥2, 𝑦2 + ∆1 0, 𝑥2, 𝑦2 − ∆1𝑚 (0, 𝑥2, 𝑦2) ≤ 0 .
⋯ 3.11
∆2 0, 𝑥1, 𝑦1 = ∆2𝑚 0, 𝑥1, 𝑦1 + ∆2 0, 𝑥1, 𝑦1 − ∆2𝑚 (0, 𝑥1, 𝑦1) ≤ 0 ,
∆2 0, 𝑥2, 𝑦2 = ∆2𝑚 0, 𝑥2, 𝑦2 + ∆2 0, 𝑥2, 𝑦2 − ∆2𝑚 (0, 𝑥2, 𝑦2) ≤ 0 .
⋯ 3.12
It follows from the inequalities ( 3.11) and ( 3.12) in virtue of the
continuity of the functions ∆1(0, 𝑥0, 𝑦0) and ∆1(0, 𝑥0, 𝑦0) that there exists an isolated singular point (𝑥0
, 𝑦0
) =
(𝑥0, 𝑦0) , 𝑥0
∈ 𝑥1, 𝑥2 and 𝑦0
∈ 𝑦1, 𝑦2 ,
so that ∆1(0, 𝑥0
, 𝑦0
) = 0 and ∆2(0, 𝑥0
, 𝑦0
) = 0. This means that the system ( 3.1) and (4.3.2) has a periodic
solutions 𝑥 𝑡, 𝑥0, 𝑦0 , 𝑦 𝑡, 𝑥0, 𝑦0 for which
𝑥0 ∈ [𝑎 +
𝑇
2
𝑀1, 𝑏 −
𝑇
2
𝑀1] and 𝑦0 ∈ [𝑐 +
𝑇
2
𝑀2, 𝑑 −
𝑇
2
𝑀2]. ∎
Remark 3.1. Theorem 3.2 is proved when 𝑅 𝑛
= 𝑅1
, on the other hand as 𝑥0, 𝑦0 are a scalar singular point
which should be isolated (For this remark, see [5]).
IV. Stability Theorem Of Solution (𝑰 𝟏) And (𝑰 𝟐)
In this section, we study theorem on stability of a periodic solution for the integral equations ( 𝐼1) and ( 𝐼2).
Theorem 4.1. If the function ∆1 0, 𝑥0, 𝑦0 , Δ2(0, 𝑥0, 𝑦0) are defined by
equations ( 3.1) and ( 3.2), where the function 𝑥0
(𝑡, 𝑥0, 𝑦0) is a limit of the sequence of the functions ( 2.1) , the
function 𝑦0
(𝑡, 𝑥0, 𝑦0) is the limit of the sequence of the functions ( 2.2) , Then the following inequalities yields:
∆1 0, 𝑥0, 𝑦0 ≤ 𝑀1 ⋯ 4.1
∆2 0, 𝑥0, 𝑦0 ≤ 𝑀2 ⋯ 4.2
and
∆2 0, 𝑥0
1
, 𝑦0
1
− ∆2 0, 𝑥0
2
, 𝑦0
2
≤ 𝐹1 𝐹2 𝐸3 𝐹0
1
𝑡 − 𝐹0
2
𝑡 +
+
𝑇
2
𝐹1 𝐹2 𝐸2( 𝐸3 + 𝐸4 + 𝐹1 𝐸4(1 −
𝑇
2
𝐸1)) 𝐺0
1
𝑡 − 𝐺0
2
(𝑡)
⋯ 4.4
for all 𝑥0
, 𝑥0
1
, 𝑥0
2
∈ 𝐷𝑓1
, 𝑦0
, 𝑦0
1
, 𝑦0
2
∈ 𝐷1𝑓2
, and 𝛼 𝑡 = 2𝑡(1 −
𝑡
𝑇
) ≤
𝑇
2
,
where 𝐸1 = [𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)], 𝐸2 = [𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)],

𝐸3 = [𝐿1 + 𝐻1(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)], 𝐸4 = [𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)],
𝐹1 = [(1 −
𝑇
2
𝐸1)(1 −
𝑇
2
𝐸4)]−1
and 𝐹2 = (1 −
𝑇2
4
𝑁2 𝑁3 𝐹1)−1
Proof. From the properties of the functions 𝑥0
𝑡, 𝑥0, 𝑦0 and 𝑦0
𝑡, 𝑥0, 𝑦0 as in the theorem 4.2.1, the
functions Δ1 = ∆1 𝑥0, 𝑦0 , Δ1 = ∆1 𝑥0, 𝑦0 ,
𝑥0 ∈ 𝐷𝑓1
, 𝑦0 ∈ 𝐷1𝑓2
are continuous and bounded by 𝑀1, 𝑀2 in the domain ( 1.3).
From relation ( 3.1), we find:
∆1 (0, 𝑥0 , 𝑦0 ≤
1
𝑇
𝑓1(𝑡, 𝑥0
𝑡, 𝑥0, 𝑦0 , 𝑦0
𝑡, 𝑥0, 𝑦0 ,
𝑇
0
, 𝐺1 𝑡, 𝑠 𝑔1 𝑠, 𝑥0
𝑠, 𝑥0, 𝑦0 , 𝑦0
𝑠, 𝑥0, 𝑦0 𝑑𝑠,
𝑡
−∞
, 𝑔1 𝑠, 𝑥0
𝑠, 𝑥0, 𝑦0 , 𝑦0
𝑠, 𝑥0, 𝑦0 𝑑 𝑠)
𝑏(𝑡)
𝑎(𝑡)
𝑑𝑡
by using the Lemma 4.1.1, gives:
∆1(0, 𝑥0, 𝑦0) ≤ 𝑀1.
And from relation ( 3.2), we get:
∆2 (0, 𝑥0 , 𝑦0 ≤
1
𝑇
𝑓2(𝑡, 𝑥0
𝑡, 𝑥0, 𝑦0 , 𝑦0
𝑡, 𝑥0, 𝑦0 ,
𝑇
0
, 𝐺2 𝑡, 𝑠 𝑔2 𝑠, 𝑥0
𝑠, 𝑥0, 𝑦0 , 𝑦0
𝑠, 𝑥0, 𝑦0 𝑑𝑠,
𝑡
−∞
, 𝑔2 𝑠, 𝑥0
𝑠, 𝑥0, 𝑦0 , 𝑦0
𝑠, 𝑥0, 𝑦0 𝑑𝑠)
𝑏(𝑡)
𝑎(𝑡)
𝑑𝑡
and using Lemma 4.1.1, we have:
∆2(0, 𝑥0, 𝑦0) ≤ 𝑀2
By using equation ( 3.1) and lemma 1.1, we get:
∆1 0, 𝑥0
1
, 𝑦0
1
− ∆1 0, 𝑥0
2
, 𝑦0
2
≤
1
𝑇
[𝐾1 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
𝑇
0
+𝐾2 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
+
+
𝛾
𝜆1
𝐾3(𝑅1 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+𝑅2 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
) +
+ 𝑕𝐾4(𝑅1 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+𝑅2 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
)]𝑑𝑡
≤ [𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
so,
∆1 0, 𝑥0
1
, 𝑦0
1
− ∆1 0, 𝑥0
2
, 𝑦0
2
≤ 𝐸1 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+𝐸2 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
⋯ 4.5
And also by using equation ( 3.2) and lemma 1.1, gives:
∆2 0, 𝑥0
1
, 𝑦0
1
− ∆2 0, 𝑥0
2
, 𝑦0
2
≤
1
𝑇
[𝐿1 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
𝑇
0
+𝐿2 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
+

+
𝛿
𝜆2
𝐿3(𝐻1 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+𝐻2 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
) +
+ 𝑕𝐿4(𝐻1 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+𝐻2 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
)]𝑑𝑡
≤ [𝐿1 + 𝐻1(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+[𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
.
Therefore,
∆2 0, 𝑥0
1
, 𝑦0
1
− ∆2 0, 𝑥0
2
, 𝑦0
2
≤ 𝐸3 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+𝐸4 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
⋯ 4.6
where the functions 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
, 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
, 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
and
𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
are solutions of the equation:
𝑥 𝑡, 𝑥0
𝑘
, 𝑦0
𝑘
= 𝐹0
𝑘
𝑡 + [𝑓1(𝑠, 𝑥(𝑠, 𝑥0
𝑘
, 𝑦0
𝑘
), 𝑦(𝑠, 𝑥0
𝑘
, 𝑦0
𝑘
),
𝑡
0
, 𝐺1 𝑠, 𝜏 𝑔1(𝜏, 𝑥(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
), 𝑦(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
))
𝑠
−∞
𝑑𝜏,
, 𝑔1(𝜏, 𝑥(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
), 𝑦(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
))𝑑𝜏
𝑏 𝑠
𝑎 𝑠
) −
−
1
𝑇
𝑓1(𝑠, 𝑥(𝑠, 𝑥0
𝑘
, 𝑦0
𝑘
), 𝑦(𝑠, 𝑥0
𝑘
, 𝑦0
𝑘
)
𝑇
0
,
, 𝐺1 𝑠, 𝜏 𝑔1(𝜏, 𝑥(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
), 𝑦(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
))
𝑠
−∞
𝑑𝜏,
, 𝑔1(𝜏, 𝑥(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
), 𝑦(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
))𝑑𝜏
𝑏 𝑠
𝑎 𝑠
)𝑑𝑠]𝑑𝑠 ⋯ ( 4.7)
and
𝑦 𝑡, 𝑥0
𝑘
, 𝑦0
𝑘
= 𝐺0
𝑘
𝑡 + [𝑓2(𝑠, 𝑥(𝑠, 𝑥0
𝑘
, 𝑦0
𝑘
), 𝑦(𝑠, 𝑥0
𝑘
, 𝑦0
𝑘
),
𝑡
0
, 𝐺2 𝑠, 𝜏 𝑔2(𝜏, 𝑥(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
), 𝑦(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
))
𝑠
−∞
𝑑𝜏,
, 𝑔2(𝜏, 𝑥(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
), 𝑦(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
))𝑑𝜏
𝑏 𝑠
𝑎 𝑠
) −
−
1
𝑇
𝑓2(𝑠, 𝑥(𝑠, 𝑥0
𝑘
, 𝑦0
𝑘
), 𝑦(𝑠, 𝑥0
𝑘
, 𝑦0
𝑘
)
𝑇
0
,
, 𝐺2 𝑠, 𝜏 𝑔2(𝜏, 𝑥(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
), 𝑦(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
))
𝑠
−∞
𝑑𝜏,
, 𝑔2(𝜏, 𝑥(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
), 𝑦(𝜏, 𝑥0
𝑘
, 𝑦0
𝑘
))𝑑𝜏
𝑏 𝑠
𝑎 𝑠
)𝑑𝑠]𝑑𝑠 ⋯ ( 4.8)
where 𝑘 = 1, 2.
From (4.4.7), we get:
𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
−𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
≤ 𝐹0
1
𝑡 − 𝐹0
2
𝑡 +

+𝛼(𝑡)[𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+𝛼 𝑡 [𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
≤ 𝐹0
1
𝑡 − 𝐹0
2
𝑡 +
+
𝑇
2
[𝐾1 + 𝑅1(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+
𝑇
2
[𝐾2 + 𝑅2(
𝛾
𝜆1
𝐾3 + 𝑕𝐾4)] 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
such that:
𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
−𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
≤ 𝐹0
1
𝑡 − 𝐹0
2
𝑡 +
+
𝑇
2
𝐸1 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+
𝑇
2
𝐸2 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
therefore:
𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
−𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
≤ (1 −
𝑇
2
𝐸1)−1
𝐹0
1
𝑡 − 𝐹0
2
𝑡 +
+
𝑇
2
𝐸2(1 −
𝑇
2
𝐸1)−1
𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
⋯ 4.9
And also from relation ( 4.8), we have:
𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
−𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
≤
≤ 𝐺0
1
𝑡 − 𝐺0
2
𝑡 +
+𝛼(𝑡)[𝐿1 + 𝐻1(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+𝛼(𝑡)[𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
≤
≤ 𝐺0
1
𝑡 − 𝐺0
2
𝑡 +
+
𝑇
2
[𝐿1 + 𝐻1(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+
𝑇
2
[𝐿2 + 𝐻2(
𝛿
𝜆2
𝐿3 + 𝑕𝐿4)] 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
so that:
𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
−𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
≤ 𝐺0
1
𝑡 − 𝐺0
2
𝑡 +
+
𝑇
2
𝐸3 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
+
+
𝑇
2
𝐸4 𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
and hence:
𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
−𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
≤ (1 −
𝑇
2
𝐸4)−1
𝐺0
1
𝑡 − 𝐺0
2
𝑡 +
+
𝑇
2
𝐸3(1 −
𝑇
2
𝐸4)−1
𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
⋯ 4.10
Now, by substituting inequality ( 4.10) in ( 4.9), we get:
𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
≤ (1 −
𝑇
2
𝐸1)−1
𝐹0
1
𝑡 − 𝐹0
2
𝑡 +
+
𝑇
2
𝐸2(1 −
𝑇
2
𝐸1)−1
[(1 −
𝑇
2
𝐸4)−1
𝐺0
1
𝑡 − 𝐺0
2
𝑡 +
+
𝑇
2
𝐸3(1 −
𝑇
2
𝐸4)−1
𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
]
≤ (1 −
𝑇
2
𝐸1)−1
𝐹0
1
𝑡 − 𝐹0
2
𝑡 +
+
𝑇
2
𝐸2[(1 −
𝑇
2
𝐸1)(1 −
𝑇
2
𝐸4)]−1
𝐺0
1
𝑡 − 𝐺0
2
𝑡 +
+
𝑇2
4
𝐸2 𝐸3[(1 −
𝑇
2
𝐸1)(1 −
𝑇
2
𝐸4)]−1
𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
]
putting 𝐹1 = [(1 −
𝑇
2
𝐸1)(1 −
𝑇
2
𝐸4)]−1
,
and substituting in the last inequality, we obtain:

𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
≤ (1 −
𝑇
2
𝐸1)−1
𝐹0
1
𝑡 − 𝐹0
2
𝑡 +
+
𝑇
2
𝐸2 𝐹1 𝐺0
1
𝑡 − 𝐺0
2
𝑡 + +
𝑇2
4
𝐸2 𝐸3 𝐹1 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
]
as the 𝐹1(1 −
𝑇
2
𝐸4) = (1 −
𝑇
2
𝐸1)−1
𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
≤ 𝐹1(1 −
𝑇
2
𝐸4) 𝐹0
1
𝑡 − 𝐹0
2
𝑡 +
+
𝑇
2
𝐸2 𝐹1 𝐺0
1
𝑡 − 𝐺0
2
𝑡 +
𝑇2
4
𝐸2 𝐸3 𝐹1 𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
]
which implies that:
𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
≤≤ 𝐹1(1 −
𝑇
2
𝐸4)(1 −
𝑇2
4
𝐸2 𝐸3 𝐹1)−1
𝐹0
1
𝑡 − 𝐹0
2
𝑡 +
+
𝑇
2
𝐸2 𝐹1(1 −
𝑇2
4
𝐸2 𝐸3 𝐹1)−1
𝐺0
1
𝑡 − 𝐺0
2
𝑡
putting 𝐹2 = (1 −
𝑇2
4
𝐸2 𝐸3 𝐹1)−1
and substituting in the last inequality, we obtain:
𝑥0
𝑡, 𝑥0
1
, 𝑦0
1
− 𝑥0
𝑡, 𝑥0
2
, 𝑦0
2
≤ 𝐹1 𝐹2(1 −
𝑇
2
𝐸4) 𝐹0
1
𝑡 − 𝐹0
2
𝑡 +
+
𝑇
2
𝐸2 𝐹1 𝐹2 𝐺0
1
𝑡 − 𝐺0
2
𝑡 ⋯ 4.11
Also, substituting the inequalities ( 4.11) in ( 4.10), we find that:
𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
−𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
≤ (1 −
𝑇
2
𝐸4)−1
𝐺0
1
𝑡 − 𝐺0
2
𝑡 +
+
𝑇
2
𝐸3(1 −
𝑇
2
𝐸4)−1
[𝐹1 𝐹2(1 −
𝑇
2
𝐸4) 𝐹0
1
𝑡 − 𝐹0
2
𝑡 +
+
𝑇
2
𝐸2 𝐹1 𝐹2 𝐺0
1
𝑡 − 𝐺0
2
𝑡 ]
and hence
𝑦0
𝑡, 𝑥0
1
, 𝑦0
1
−𝑦0
𝑡, 𝑥0
2
, 𝑦0
2
≤
𝑇
2
𝐸3 𝐹1 𝐹2 𝐹0
1
𝑡 − 𝐹0
2
𝑡 +
+[𝐹1(1 −
𝑇
2
𝐸1) +
𝑇
2
𝐹1 𝐹2 𝐸2] 𝐺0
1
𝑡 − 𝐺0
2
𝑡
⋯ 4.12
so, substituting inequalities ( 4.11) and ( 4.12) in inequality ( 4.5), we get the inequality ( 4.3).
and the same, substituting inequalities ( 4.11) and ( 4.12) in inequality
( 4.6), gives the inequality ( 4.4). ∎
REFERENCES
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Education, University of Mosul.
[2] Butris, R. N. and Rafeq, A. Sh., (2011), Existence and Uniqueness, Solution for Non-linear Volterra Integral Equation, J. Duhok
Univ. Vol.No. 1, (Pure and Eng. Sciences).
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[5] Mitropolsky, Yu. A. and Martynyuk, D. I., (1979), For Periodic Solutions for the Oscillations System with Retarded Argument,
Kiev, Ukraine.
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college of Education, University of Duhok.
[8] Perestyuk, N. A., (1971), The periodic solutions for non-linear systems of differential equations, Math. and Meca. J., Univ. of Kiev,
Kiev, Ukraine,5, 136-146.
[9] Perestyuk, N. A. and Martynyuk, D. I., (1976), Periodic solutions of a certain class systems of differential equations, Math. J., Univ.
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[10] Samoilenko, A. M. and Ronto, N. I., (1976), A Numerical – Analytic Methods for Investigating of Periodic Solutions, Kiev,
Ukraine.
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[12] Shestopalov, Y. V. and Smirnov, Y. G., (2002), Integral Equations, Karlstad University.
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[14] Tarang, M., (2004), Stability of the spline collocation method for Volterra integro-differential equations, Thesis, University of
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