Vol. 25, No. 01 (2019), pp. 62–74.
MEASURE OF NONCOMPACTNESS IN THE STUDY OF SOLUTIONS FOR A SYSTEM OF INTEGRAL
EQUATIONS
Vatan Karakaya
1, Mohammad Mursaleen2, Nour El Houda Bouzara3, and Derya Sekman 41Department of Mathematical Engineering, Yildiz Technical University, Davutpasa Campus, Esenler, 34210 Istanbul, Turkey, vkkaya@yahoo.com
2Department of Mathematics, Aligarh Muslim University, Aligarh 202002, India, mursaleenm@gmail.com
3 Faculty of Mathematics, University of Science and Technology Houari Boumedi`ene, Bab-Ezzouar, 16111 Algies, Algeria, bzr.nour@gmail.com
4 Faculty of Arts and Sciences, Department of Mathematics, Ahi Evran University, 40100 Kirsehir, Turkey, deryasekman@gmail.com
Abstract. In this work, we prove the existence of solutions for a tripled system of integral equations using some new results of fixed point theory associated with measure of noncompactness. These results extend the results in some previous works. Also, the condition under which the operator admits fixed points is more general than the others in literature.
Key words and Phrases: Measure of noncompactness, Fixed point, System of inte- gral equations.
Abstrak. Dalam paper ini, dibuktikan eksistensi solusi dari suatu sistemtripled dari persamaan integral menggunakan suatu hasil terbaru dalam teori titik tetap yang berasosiasi dengan ukurannoncompactness. Hasil-hasil ini memperluas hasil- hasil yang sudah ada sebelumnya. Lebih jauh, kondisi operator yang memenuhi syarat titik tetap di dalam paper ini juga lebih umum dibandingkan dengan yang sudah ada di dalam literatur-literatur sebelumnya.
Kata kunci: Ukurannoncompactness, titik tetap, sistem persamaan integral.
2010 Mathematics Subject Classification:47H10, 47H08, 45G15 Received: 16-07-2017, accepted: 12-04-2018.
62
1. INTRODUCTION
In recent years, measure of noncompactness which was introduced by Kura- towski [18] in 1930 and has provided powerful tools for obtaining the solutions of a large variety of integral equations and systems. One can find related references in studies involving Aghajani et al. [2], [3], [4], [5], Banas [9], Banas and Rzepka [13], Mursaleen and Mohiuddine [19], Mursaleen and Rizvi [20], Araba et al. [8], Deepmala and Pathak [16], Shaochun and Gan [21], Sikorska [22], Alotaibi et al.
[7], and many others.
In this paper, we study the solvability of the following system of integral equations
x(t) =g1(t) +f1
t, x(ξ1(t)), y(ξ1(t)), z(ξ1(t)), ψ Rq1(t)
0 h(t, s, x(η1(s)), y(η1(s)), z(η1(s)))ds y(t) =g2(t) +f2
t, x(ξ2(t)), y(ξ2(t)), z(ξ2(t)), ψRq2(t)
0 h(t, s, x(η2(s)), y(η2(s)), z(η2(s)))ds z(t) =g3(t) +f3
t, x(ξ3(t)), y(ξ3(t)), z(ξ3(t)), ψRq3(t)
0 h(t, s, x(η3(s)), y(η3(s)), z(η3(s)))ds ,
by establishing some results of existence for fixed points of condensing operators in Banach spaces.
Throughout this paper, X is assumed to be a Banach space. The family of bounded subset, closure and closed convex hull of X are denoted by BX, X and ConvX, respectively.
We now gather some well-known definitions and results from the literature which will be used throughout this paper.
Definition 1.1 ([10]). Let X be a Banach space and BX the family of bounded subset ofX.A map
µ:BX→[0,∞) which satisfies the following:
(1) µ(A) = 0⇔A is a precompact set, (2) A⊂B⇒µ(A)6µ(B),
(3) µ(A) =µ A
, ∀A∈ BX, (4) µ(ConvA) =µ(A),
(5) µ(λA+ (1−λ)B)6λµ(A) + (1−λ)µ(B),forλ∈[0,1],
(6) Let(An)be a sequence of closed sets fromBX such thatAn+1⊆An,(n>1) and lim
n→∞µ(An) = 0.Then, the intersection set A∞= ∞∩
n=1An is nonempty andA∞ is precompact.
The functionalµis called measure of noncompactness defined on the Banach spaceX.
Theorem 1.2 ([11]). Let C be a nonempty closed, bounded and convex subset of X. IfT :C→C is a continuous mapping
µ(T A)6kµ(A), k∈[0,1), thenT has a fixed point.
The following theorem is considered as a generalization of Darbo fixed point theorem.
Theorem 1.3([1]). LetC be a nonempty closed, bounded and convex subset ofX andT :C→C be a continuous mapping such that for any subsetAof C
µ(T A)6β(µ(A))µ(A),
where β:R+ →[0,1) that isβ(tn)→1 implies tn →0. Then, T has at least one fixed point.
Corollary 1.4 ([1]). Let C be a nonempty closed, bounded and convex subset of X andT :C→C be a continuous mapping such that for any subset Aof C
µ(T A)6ϕ(µ(A)),
where ϕ:R+ →R+ is a nondecreasing and upper semicontinuous functions, that is, for everyt >0, ϕ(t)< t. Then,T has at least one fixed point.
Theorem 1.5 ([6]). Let µ1, µ2,...,µn be measures of noncompactness in Banach spacesE1, E2, ...En,(respectively).
Then the function
µe(X) =F(µ1(X1), µ2(X2), ..., µn(Xn)),
defines a measure of noncompactness in E1×E2×...×En whereXi is the natural projection ofX onEi, fori= 1,2, ..., n, andF be a convex function defined by
F : [0,∞)n→[0,∞), such that,
F(x1, x2, ..., xn) = 0⇔xi= 0, fori= 1,2, ..., n.
Example 1.6 ([17]). We can notice that by taking
F(x, y, z) = max{x, y, z} for any (x, y, z)∈[0,∞)3, or
F(x, y, z) =x+y+z for any (x, y, z)∈[0,∞)3.
Then, F satisfies the conditions of Theorem 1.5. Thus, for a measure of noncom- pactnessµi (i= 1,2,3), we have that
eµ(X) = max (µ1(X1), µ2(X2), µ3(X3)), or
eµ(X) =µ1(X1) +µ2(X2) +µ3(X3),
defines a measure of noncompactness in the space E×E×E whereXi,i= 1,2,3 are the natural projections of X onEi.
2. MAIN RESULTS
Theorem 2.1. Let Abe a nonempty, bounded ,closed and convex subset of a Ba- nach space X and let ϕ :R+ → R+ be a nondecreasing and upper semicontinous function such thatϕ(t)< tfor allt >0.Then for any measure of noncompactness µ, and continuous operatorsTi:A×A×A→A(i= 1,2,3)satisfying
µ(Ti(X1×X2×X3))6ϕ(max (µ(X1), µ(X2), µ(X3))), X1, X2, X3∈A, (1) there existx∗, y∗, z∗∈A such that
T1(x∗, y∗, z∗) =x∗ T2(x∗, y∗, z∗) =y∗ T3(x∗, y∗, z∗) =z∗
. Proof. Consider the following measure of noncompactness
µe(A×A×A) = max (µ(X1), µ(X2), µ(X3)), whereX1, X2, X3∈Aand the mapping T :A×A×A→A,
T(x, y, z) = (T1(x, y, z), T2(x, y, z), T3(x, y, z)). We have,
µe(T(A×A×A)) = µe((T1(X1×X2×X3)), T2(X1×X2×X3), T3(X1×X2×X3))
= max{µ(T1(X1×X2×X3)), µ(T2(X1×X2×X3)), µ(T3(X1×X2×X3))}
6 max{ϕ(max (µ(X1), µ(X2), µ(X3))), ϕ(max (µ(X1), µ(X2), µ(X3))), ϕ(max (µ(X1), µ(X2), µ(X3)))}.
By hypothesisϕis a non-decreasing function, then
µe(T(A×A×A)) 6 ϕ[max{max (µ(X1), µ(X2), µ(X3)),max (µ(X1), µ(X2), µ(X3)) , max (µ(X1), µ(X2), µ(X3))}].
Consequently,
µe(T(A×A×A))6ϕ(µe(A×A×A)). So,
µ(T1(x, y, z), T2(x, y, z), T3(x, y, z))6ϕ(max (µ(X1), µ(X2), µ(X3))). By Corollary 1.4, we conclude that there existx∗, y∗, z∗∈Asuch that
T(x∗, y∗, z∗) = (x∗, y∗, z∗). In the other hand,
T(x∗, y∗, z∗) = (T1(x∗, y∗, z∗), T2(x∗, y∗, z∗), T3(x∗, y∗, z∗)). Hence,
T1(x∗, y∗, z∗) =x∗ T2(x∗, y∗, z∗) =y∗ T3(x∗, y∗, z∗) =z∗
.
Definition 2.2([17]). A tripled(x, y, z)of a mappingT :A×A×A→A,is called a tripled fixed point if
T(x, y, z) =x, T(y, x, z) =y and T(z, y, x) =z.
Remark 2.3. Let T : A×A×A → A be a continuous mapping. If we define T1(x, y, z) =T(x, y, z), T2(x, y, z) =T(y, x, z)andT3(x, y, z) =T(z, y, x), then main results of [17]can be considered as a result of Theorem 2.1.
It is very natural to extend the above result from three dimensions to multi- dimensional fixed point and in the same way we can prove the following theorem.
Theorem 2.4. Let Abe a nonempty, bounded ,closed and convex subset of a Ba- nach space X and let ϕ :R+ → R+ be a nondecreasing and upper semicontinous function such thatϕ(t)< tfor allt >0.Then for any measure of noncompactness µ and for continuous operatorsTi:An→Ω (i= 1, ..., n)satifying
µ(Ti(X1×...×Xn))6ϕ(max (µ(X1), ..., µ(Xn))), Xi∈A, i= 1, n, there existx∗1, ..., x∗n such that
T1(x∗1, ..., x∗n) =x∗1 ...
Tn(x∗1, ..., x∗n) =x∗n .
As a particular case we get the following corollary:
Corollary 2.5 ([3]). LetAbe a nonempty, bounded ,closed and convex subset of a Banach spaceX and letϕ:R+→R+be a nondecreasing and upper semicontinous function such thatϕ(t)< tfor allt >0.Then for any measure of noncompactness µ, the continuous operatorG:An→A satisfying
µ(G(X1×...×Xn))6kmax (µ(X1), ..., µ(Xn)), X1, ..., Xn∈A.
And for the casen= 2,we have the following result.
Corollary 2.6 ([3]). LetAbe a nonempty, bounded ,closed and convex subset of a Banach spaceX and letϕ:R+→R+be a nondecreasing and upper semicontinous function such thatϕ(t)< tfor allt >0.Then for any measure of noncompactness µ, the continuous operatorG:A×A→Asatisfying
µ(G(X1×X2))6kmax (µ(X1), µ(X2)), X1, X2∈A.
In the following we choose for the spaceX the spaceBC(R+), i.e., the space of all real functions defined, bounded and continuous on R+. Then, we get the following theorem.
Theorem 2.7. LetAbe a nonempty, bounded, closed and convex subset ofBC(R+) andTi:A×A×A→Abe a continuous operator such that for everyx, y, z, u, v, w∈ A.
kTi(x, y, z)−Ti(u, v, w)k∞6ϕ(max{kx−uk∞,ky−vk∞,kz−wk∞}), (2) whereϕ:R+→R+is a nondecreasing and upper semicontinuous function such that ϕ(t)< tfor allt >0. Then there existx∗, y∗, z∗∈Asuch that,
T1(x∗, y∗, z∗) =x∗ T2(x∗, y∗, z∗) =y∗ T3(x∗, y∗, z∗) =z∗
.
Proof. To verify that the operatorTi:A×A×A→Asatisfy the condition (1) we recall the following notions.
The measure of noncompactness onBC(R+) for a positive fixedtonBBC(R+)
is defined as follows:
µ(X) =ω0(X) + lim sup
t→∞
diamX(t),
that is,diamX(t) = sup{|x(t)−y(t)|:x, y∈X}, X(t) ={x(t) :x∈X}.and ω0(X) = lim
K→∞ωK0 (X), ω0K(X) = lim
→0ωK(X, ), ωK(X, ) = sup
ωK(x, ) :x∈X ,
ωK(x, ) = sup{|x(t)−x(s)|:t, s∈[0, K], |t−s|6}, forK >0, where ωK(x, ) for x ∈ X and > 0, is the modulus of continuity of x on the compact [0, K], whereK is a positive number.
We have
kTi(x, y, z) (t)−Ti(x, y, z) (s)k6ϕ(max{kx(t)−x(s)k,ky(t)−y(s)k,kz(t)−z(s)k}), by taking the supremum and using the fact thatϕis nondecreasing, we get
ωK(Ti(x, y, z), )6ϕ max
ωK(x, ), ωK(y, ), ωK(z, ) . Thus,
ω0(Ti(X1×X2×X3))6ϕ(max{ω0(X1), ω0(X2), ω0(X3)}). (3) Since in (2)x, yandzare arbitrary andϕis non-decreasing,
DiamTi(X1×X2×X3) (t)6ϕ(max{DiamX1(t), DiamX2(t), DiamX3(t)}). In further,X1(t), X2(t), X3(t) are subspaces ofBC(R+). Then,
lim sup
t→∞
DiamTi(X1×X2×X3) (t)6lim sup
t→∞
ϕ(max{DiamX1(t), DiamX2(t), DiamX3(t)})+Φ () 6ϕ
max
lim sup
t→∞
DiamX1(t),lim sup
t→∞
DiamX2(t),lim sup
t→∞
DiamX3(t)
. Usingϕ(t)< tfor allt >0 and from (3) and the above inequality, we get
µ(Ti(X1×X2×X3))6ϕ(max (µ(X1), µ(X2), µ(X3))), X1, X2, X3∈A.
Consequently, there existx∗, y∗, z∗∈A such that
T(x∗, y∗, z∗) = (T1(x∗, y∗, z∗), T2(x∗, y∗, z∗), T3(x∗, y∗, z∗))
= (x∗, y∗, z∗). Thus,
T1(x∗, y∗, z∗) =x∗ T2(x∗, y∗, z∗) =y∗ T3(x∗, y∗, z∗) =z∗
.
3. APPLICATION
Now, we will use the results of the previous section to resolve the following system
x(t) =g1(t) +f1
t, x(ξ1(t)), y(ξ1(t)), z(ξ1(t)), ψRq1(t)
0 h(t, s, x(η1(s)), y(η1(s)), z(η1(s)))ds y(t) =g2(t) +f2
t, x(ξ2(t)), y(ξ2(t)), z(ξ2(t)), ψRq2(t)
0 h(t, s, x(η2(s)), y(η2(s)), z(η2(s)))ds z(t) =g3(t) +f3
t, x(ξ3(t)), y(ξ3(t)), z(ξ3(t)), ψ Rq3(t)
0 h(t, s, x(η3(s)), y(η3(s)), z(η3(s)))ds .
(4) We study system (4) under the following assumptions:
(i) ξi, ηi, qi:R+→R+,(i= 1,2,3),are continuous andξi(t)→ ∞ast→ ∞.
(ii) The functionψi :R→R, (i= 1,2,3),is continuous and there exist positive δi, αi such that
|ψi(t1)−ψi(t2)|6δi|t1−t2|αi, fori= 1,2,3 and anyt1, t2∈R+.
(iii) fi : R+×R×R×R×R→R are continuous, gi : R+→R are bounded and there exists nondecreasing continuous function Φi : R+→R+ with Φi(0) = 0, i= 1,2,3,such that
|fi(t, x1, x2, x3, x4)−fi(t, y1, y2, y3, y4)|6(ϕi(max{|x1−y1|,|x2−y2|,|x3−y3|}))+Φi(|x4−y4|). (iv) The functions defined by |fi(t,0,0,0,0)|, i = 1,2,3 are bounded on R+,
i.e.,
Mi= sup{fi(t,0,0,0,0) :t∈R+}<∞. (5) (v) hi:R+×R+×R×R×R→R, are continuous functions and there exists a
positive constantDsuch that i= 1,2,3, sup
(
Z qi(t) 0
hi(t, s, x(η(s)), y(η(s)), z(η(s)))ds
: t, s∈R+, x, y, z∈BC(R+) )
< D, (6) and
t→∞lim Z qi(t)
0
[hi(t, s, x(η(s)), y(η(s)), z(η(s)))−hi(t, s, u(η(s)), v(η(s)), w(η(s)))]ds= 0, (7)
with respect to x, y, z, u, v, w∈BC(R+). Consider the following operator,
Ti(x, y, z) =gi(t)+fi t, x(ξi(t)), y(ξi(t)), z(ξi(t)), ψ
Z qi(t) 0
h(t, s, x(ηi(s)), y(ηi(s)), z(ηi(s)))ds
!!
. Solving the system (4) is equivalent to find the fixed points of the operator Ti.
Then let verify the conditions of Theorem 2.7.
First, since gi and fi (i= 1,2,3) are continuous then the operators Ti are continuous.
In further, forx, y, z∈Br (r) (forr >0) let, kTi(x, y, z) (t)k
=
gi(t) +fi t, x(ξi(t)), y(ξi(t)), z(ξi(t)), ψ
Z qi(t) 0
h(t, s, x(ηi(s)), y(ηi(s)), z(ηi(s)))ds
!!
6
fi t, x(ξi(t)), y(ξi(t)), z(ξi(t)), ψ
Z qi(t) 0
h(t, s, x(ηi(s)), y(ηi(s)), z(ηi(s)))ds
!!
−f(t,0,0,0) +f(t,0,0,0)k+kgi(t)k 6 kgi(t)k+kf(t,0,0,0)k
+ϕi(max{|x(ξi(t))|,|y(ξi(t))|,|z(ξi(t))|}) +Φi ψ
Z qi(t) 0
h(t, s, x(ηi(s)), y(ηi(s)), z(ηi(s)))ds
!!
.
Since, gi are bounded, fi are continuous functions and using hypothesis (iv)-(v), we get
kTi(x, y, z)k∞ 6 ϕi(max{kxk∞,kyk∞,kzk∞}) +G+Mi+ Φi(δiDαi) 6 ϕi(r) +G+Mi+ Φi(δiDαi),
for somer0>0,we obtainTi(Br0×Br0×Br0)⊂Br0. Moreover,
kTi(x, y, z)−Ti(u, v, w)k∞
= sup
t
gi(t) +fi t, x(ξi(t)), y(ξi(t)), z(ξi(t)), ψ
Z qi(t) 0
h(t, s, x(ηi(s)), y(ηi(s)), z(ηi(s)))ds
!!
−gi(t)−fi t, u(ξi(t)), v(ξi(t)), w(ξi(t)), ψ
Z qi(t) 0
h(t, s, u(ηi(s)), v(ηi(s)), w(ηi(s)))ds
!!
= sup
t
fi t, x(ξi(t)), y(ξi(t)), z(ξi(t)), ψ
Z qi(t) 0
h(t, s, x(ηi(s)), y(ηi(s)), z(ηi(s)))ds
!!
−fi t, u(ξi(t)), v(ξi(t)), w(ξi(t)), ψ
Z qi(t) 0
h(t, s, u(ηi(s)), v(ηi(s)), w(ηi(s)))ds
!!
6 sup
t
{ϕi(max{|x(ξi(t))−u(ξi(t))|,|y(ξi(t))−v(ξi(t))|,|z(ξi(t))−w(ξi(t))|})
+Φi
ψ Rqi(t)
0 h(t, s, x(ηi(s)), y(ηi(s)), z(ηi(s)))ds
−ψ Rqi(t)
0 h(t, s, u(ηi(s)), v(ηi(s)), w(ηi(s)))ds
6 ϕi(max{kx−uk∞,ky−vk∞,kz−wk∞})
+ sup
t
Φi δi
Z qi(t) 0
h(t, s, x(ηi(s)), y(ηi(s)), z(ηi(s)))
−h(t, s, u(ηi(s)), v(ηi(s)), w(ηi(s)))
ds
αi! .
Consider,
Z qi(t) 0
{h(t, s, x(ηi(s)), y(ηi(s)), z(ηi(s)))−h(t, s, u(ηi(s)), v(ηi(s)), w(ηi(s)))}ds . Using the condition (7),we get
Z qi(t) 0
{h(t, s, x(ηi(s)), y(ηi(s)), z(ηi(s)))−h(t, s, u(ηi(s)), v(ηi(s)), w(ηi(s)))}ds
6 and
δi
Z qi(t) 0
{h(t, s, x(ηi(s)), y(ηi(s)), z(ηi(s)))−h(t, s, u(ηi(s)), v(ηi(s)), w(ηi(s)))}ds
αi
6δiαi. Thus,
Φi δi
Z qi(t) 0
h(t, s, x(ηi(s)), y(ηi(s)), z(ηi(s)))
−h(t, s, u(ηi(s)), v(ηi(s)), w(ηi(s)))
ds
αi!
6Φi(δiαi). On the other hand Φi is continuous function and Φi(0) = 0 , is arbitrary, then for→0,we get
kTi(x, y, z)−Ti(u, v, w)k∞6ϕi(max{kx−uk∞,ky−vk∞,kz−wk∞}). Consequently by Theorem 2.7, there exist x∗, y∗, z∗such that
T1(x∗, y∗, z∗) =x∗ T2(x∗, y∗, z∗) =y∗ T3(x∗, y∗, z∗) =z∗
. Then, we had proved the following theorem.
Theorem 3.1. Under the conditions (i)−(v)the system of integral equations(4) has at least one solution in the space BC(R+)×BC(R+)×BC(R+).
Example 3.2. Let the system of integral equations
x(t) = 2+2tt24 +x(√t)+y(√t)+z(√t)
3t2+3 + arctanR
√t 0
x(s2)s|siny(s2)||cosz(s2)|
et(1+x2(s2))(1+sin2y(s2))(1+cos2z(s2))ds y(t) = 12e−t2+t2(x(t)+y(t)+z(t))
3t4+3 + sinRt 0
esy2(s)(1+cos2x(s))(1+sin2z(s))
et2(1+y2(s))(1+sin2x(s))(1+cos2z(s))ds z(t) = 2√1+t1 4 +t3(x(t)+y(t)+z(t))
3t5+3 + cosRt2 0
s2|cosz(s)|+√
es(1+z2(s))(1+sin2y(s))(1+cos2x(s)) et(1+z2(s))(1+sin2y(s))(1+cos2x(s)) ds
.
We notice that by taking
g1(t) = t2
2 + 2t4,g2(t) = 1
2e−t2, g3(t) = 1 2√
1 +t4, f1(t, x, y, z, p) = x+y+z3t2+3 +p
f2(t, x, y, z, p) = t2(x+y+z)3t4+3 +p f3(t, x, y, z, p) = t3(x+y+z)3t5+3 +p ,
h1(t, s, x, y, z) = xs|siny| |cosz|
et(1 +x2) 1 + sin2y
(1 + cos2z) h2(t, s, x, y, z) = es 1 +y2
1 + sin2x
1 + cos2z et2(1 +y2) 1 + sin2x
(1 + cos2z) h3(t, s, x, y, z) =
s2|cosz|+q
es(1 +z2) 1 + sin2y
(1 + cos2x) et(1 +z2) 1 + sin2y
(1 + cos2x) and
η1(t) = t2, η2(t) =η3(t) =t ξ1(t) = √
t, ξ2(t) =ξ3(t) =t q1(t) = √
t, q2(t) =t, q3(t) =t2
Ψ1(t) = arctant, Ψ2(t) = sint, Ψ3(t) = cost, we get the system of integral equations (4).
To solve this system we need to verify the conditions(i)−(v).
Obviously, ξi, ηi, qi : R+ → R+ are continuous and ξi → ∞ as t → ∞. In further, the functions ψi:R→Rare continuous forδi=αi = 1, we have
|ψi(t1)−ψi(t2)|6δi|t1−t2|αi, for any t1, t2∈R+. The conditions (i)and(ii)hold.
Now, let
|f1(t, x, y, z, p)−f1(t, u, v, w, ρ)| =
x+y+z 3t2+ 3 +p−
u+v+w 3t2+ 3 +ρ
6 1
3t2+ 3[|x−u|+|y−v|+|z−w|] +|p−ρ|
6 3
3t2+ 3max{|x−u|,|y−v|,|z−w|}+|p−ρ|
6 1
t2+ 1max{|x−u|,|y−v|,|z−w|}+|p−ρ|
= ϕ1(max{|x−u|,|y−v|,|z−w|}) + Φ (|p−ρ|). Similarly, we prove that
|f2(t, x, y, z, p)−f2(t, u, v, w, ρ)|6ϕ2(max{|x−u|,|y−v|,|z−w|})+Φ (|p−ρ|) and
|f3(t, x, y, z, p)−f3(t, u, v, w, ρ)|6ϕ3(max{|x−u|,|y−v|,|z−w|})+Φ (|p−ρ|). Then, (iii)also holds.
In further(iv)is valid. Indeed,
Mi = sup|{fi(t,0,0,0,0) :t∈R+}|= 0, i= 1,2,3.
Let us verify the last condition(v). First, note that
|h1(t, s, x, y, z)−h1(t, s, u, v, w)|
=
xs|siny| |cosz|
et(1 +x2) 1 + sin2y
(1 + cos2z)− us|sinv| |cosw|
et(1 +u2) 1 + sin2v
(1 + cos2w) 6
x 1 +x2
s et− u
1 +u2 s et
61
2 s et +1
2 s et
= s
et. Hence,
t→∞lim Z t
0
|h1(t, s, x(η(s)), y(η(s)), z(η(s)))−h1(t, s, u(η(s)), v(η(s)), w(η(s)))|ds 6 lim
t→∞
Z t 0
s etds= 0.
In addition,
|h2(t, s, x, y, z)−h2(t, s, u, v, w)|
=
es y2
1 + cos2x
1 + sin2z et2(1 +y2) 1 + sin2x
(1 + cos2z)− es v2
1 + cos2u
1 + sin2w et2(1 +v2) 1 + sin2u
(1 + cos2w) 6
y2 1 +y2
es
et2 − v2 1 +v2
es et2 62es
et2. Thus,
t→∞lim Z t
0
|h2(t, s, x(η(s)), y(η(s)), z(η(s)))−h2(t, s, u(η(s)), v(η(s)), w(η(s)))|ds 6 lim
t→∞
Z t 0
2es et2ds= 0.
Moreover,
|h3(t, s, x, y, z)−h3(t, s, u, v, w)|
=
s2|cosz|+q
es(1 +z2) 1 + sin2y
(1 + cos2x) et(1 +z2) 1 + sin2y
(1 + cos2x) −
s2|cosw|+q
es(1 +w2) 1 + sin2v
(1 + cos2u) et(1 +w2) 1 + sin2v
(1 + cos2u)
6
s2
et(cosz−cosw) 6 s2
et. Then,
t→∞lim Z t
0
|h2(t, s, x(η(s)), y(η(s)), z(η(s)))−h2(t, s, u(η(s)), v(η(s)), w(η(s)))|ds 6 lim
t→∞
Z t 0
s2 etds= 0.
Furthermore, for any x, y, z∈BC(R+)×BC(R+)×BC(R+), sup
Z t 0
hi(t, s, x(η(s)), y(η(s)), z(η(s)))ds
, t, s∈R+
< D.
It is easy to see that for an r0>0,we have ϕ(r0) +1
2 + Φ (D)6r0, holds and the condition(v)is valid.
Finally, the system has at least one solution inBC(R+)×BC(R+)×BC(R+).
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