Volume 2009, Article ID 917614,13pages doi:10.1155/2009/917614
Research Article
Convex Solutions of a Nonlinear Integral Equation of Urysohn Type
Tiberiu Trif
Faculty of Mathematics and Computer Science, Babes¸-Bolyai University, Str. Kog˘alniceanu Nr. 1, 400084 Cluj-Napoca, Romania
Correspondence should be addressed to Tiberiu Trif,[email protected] Received 4 August 2009; Accepted 25 September 2009
Recommended by Donal O’Regan
We study the solvability of a nonlinear integral equation of Urysohn type. Using the technique of measures of noncompactness we prove that under certain assumptions this equation possesses solutions that are convex of orderpfor eachp∈ {−1,0, . . . , r}, withr≥ −1 being a given integer. A concrete application of the results obtained is presented.
Copyrightq2009 Tiberiu Trif. 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
Existence of solutions of differential and integral equations is subject of numerous investigationssee, e.g., the monographs1–3or4. Moreover, a lot of work in this domain is devoted to the existence of solutions in certain special classes of functionse.g., positive functions or monotone functions. We merely mention here the result obtained by Caballero et al. 5 concerning the existence of nondecreasing solutions to the integral equation of Urysohn type
xt at ut, xt
T
0
vt, s, xsds, t∈0, T, 1.1
whereTis a positive constant. In the special case whenut, x:x2or evenut, x:xn, the authors proved in5that ifais positive and nondecreasing,vis positive and nondecreasing in the first variable when the other two variables are kept fixed, and they satisfy some additional assumptions, then there exists at least one positive nondecreasing solution x : 0, T → Rto1.1. A similar existence result, but involving a Volterra type integral equation, has been obtained by Bana´s and Martinon6.
It should be noted that both existence results were proved with the help of a measure of noncompactness related to monotonicity introduced by Bana´s and Olszowy7. The reader is referred also to the paper by Bana´s et al.8, in which another measure of noncompactness is used to prove the solvability of an integral equation of Urysohn type on an unbounded interval.
The main purpose of the present paper is twofold. First, we generalize the result from the paper5to the framework of higher-order convexity. Namely, we show that given an integerr≥ −1, ifaandvare convex of orderpfor eachp∈ {−1,0, . . . , r}, then1.1possesses at least one solution which is also convex of orderpfor eachp ∈ {−1,0, . . . , r}. Second, we simplify the proof given in5by showing that it is not necessary to make use of the measure of noncompactness related to monotonicity introduced by Bana´s and Olszowy7.
2. Measures of Noncompactness
Measures of noncompactness are frequently used in nonlinear analysis, in branches such as the theory of differential and integral equations, the operator theory, or the approximation theory. There are several axiomatic approaches to the concept of a measure of noncompactnesssee, e.g.,9–11or12. In the present paper the definition of a measure of noncompactness given in the book by Bana´s and Goebel12is adopted.
LetEbe a real Banach space, letMEbe the family consisting of all nonempty bounded subsets ofE, and let NE be the subfamily ofME consisting of all relatively compact sets.
Given any subsetX ofE, we denote by clX and coX the closure and the convex hull ofX, respectively.
Definition 2.1see12. A functionμ:ME → 0,∞is said to be a measure of noncompactness inEif it satisfies the following conditions.
1The family kerμ:{X ∈ ME|μX 0}called the kernel ofμis nonempty and it satisfies kerμ⊆ NE.
2μX≤μYwheneverX, Y ∈ MEsatisfyX⊆Y. 3μX μclX μcoXfor allX∈ ME.
4μλX 1−λY≤λμX 1−λμYfor allλ∈0,1and allX, Y ∈ ME.
5IfXnis a sequence of closed sets fromMEsuch thatXn1 ⊆ Xnfor each positive integernand if limn→ ∞μXn 0, then the setX∞:∞
n1Xnis nonempty.
An important and very convenient measure of noncompactness is the so-called Hausdorffmeasure of noncompactnessχ:ME → 0,∞, defined by
χX:inf
ε∈0,∞|X possesses a finiteε−net inX
. 2.1
The importance of this measure of noncompactness is given by the fact that in certain Banach spaces it can be expressed by means of handy formulas. For instance, consider the Banach spaceC : Ca, bconsisting of all continuous functionsx :a, b → R, endowed with the standard maximum norm
x:max{|xt| |t∈a, b}. 2.2
GivenX∈ MC,x∈X, andε >0, let
ωx, ε:sup{|xt−xs| |t, s∈a, b,|t−s| ≤ε} 2.3
be the usual modulus of continuity ofx. Further, let
ωX, ε:sup{ωx, ε|x∈X}, 2.4
and ω0X : limε→0ωX, ε. Then it can be proved see Bana´s and Goebel 12, Theorem 7.1.2that
χX 1
2ω0X ∀X∈ MC. 2.5
For further facts concerning measures of noncompactness and their properties the reader is referred to the monographs 9, 11 or 12. We merely recall here the following fixed point theorem.
Theorem 2.2see12, Theorem 5.1. LetEbe a real Banach space, let μ : ME → 0,∞be a measure of noncompactness in E, and letQ be a nonempty bounded closed convex subset of E.
Further, letF :Q → Qbe a continuous operator such thatμFX≤ kμXfor each subsetXof Q, wherek∈0,1is a constant. ThenFhas at least one fixed point inQ.
3. Convex Functions of Higher Orders
LetI ⊆Rbe a nondegenerate interval. Given an integerp≥ −1, a functionx:I → Ris said to be convex of orderporp-convex if
t0, t1, . . . , tp1; x
≥0 3.1
for any systemt0< t1<· · ·< tp1ofp2 points inI, where t0, t1, . . . , tp1; x
: 1
t0−t1t0−t2· · ·
t0−tp1 xt0 1
t1−t0t1−t2· · ·
t1−tp1 xt1 · · · 1
tp1−t0 tp1−t1 · · ·
tp1−tp x tp1
3.2
is called the divided difference ofxat the pointst0, t1, . . . , tp1. With the help of the polynomial function defined by
ωt: t−t0t−t1· · ·
t−tp1 , 3.3
the previous divided difference can be written as
t0, t1, . . . , tp1;x
p1 k0
xtk
ωtk. 3.4
An alternative way to define the divided differencet0, t1, . . . , tp1;xis to set ti;x:xti for eachi∈
0,1, . . . , p1 , ti, ti1, . . . , tij;x
:
ti, . . . , tij−1;x
−
ti1, . . . , tij;x
ti−tij ,
3.5
wheneverj∈ {0,1, . . . , p1−i}. Finally, we mention a representation of the divided difference by means of two determinants. It can be proved that
t0, t1, . . . , tp1;x U
t0, t1, . . . , tp1;x V
t0, t1, . . . , tp1 , 3.6
where
U
t0, t1, . . . , tp1;x :
1 1 · · · 1 t0 t1 · · · tp1 t20 t21 · · · t2p1 ... ... · · · ... tp0 tp1 · · · tpp1 xt0 xt1 · · · x
tp1 ,
V
t0, t1, . . . , tp1 :
1 1 · · · 1 t0 t1 · · · tp1 t20 t21 · · · t2p1 ... ... · · · ... tp10 tp11 · · · tp1p1
.
3.7
Note that a convex function of order−1 is a nonnegative function, a convex function of order 0 is a nondecreasing function, while a convex function of order 1 is an ordinary convex function.
LetI ⊆Rbe a nondegenerate interval, letx:I → Rbe an arbitrary function, and let h∈R. The difference operatorΔhwith the spanhis defined by
Δhxt:xth−xt 3.8
for allt∈Ifor whichth∈I. The iteratesΔphp0,1,2, . . .ofΔhare defined recursively by
Δ0hx:x, Δp1h x: Δh
Δphx
forp0,1,2, . . . . 3.9
It can be provedsee, e.g.,13, page 368, Corollary 3that
Δphx t
p k0
−1p−k p
k
xtkh 3.10
for everyt∈Ifor whichtph∈I. On the other hand, the equality
t, th, t2h, . . . , tph;x
Δphx t
p!hp 3.11
holds for every nonnegative integerpand everyt∈Ifor whichtph∈I.
LetI ⊆Rbe a nondegenerate interval. Given an integerp ≥ −1, a functionx:I → R is called Jensen convex of orderpor Jensenp-convex if
Δp1h x
t≥0 3.12
for allt∈Iand allh >0 such thatt p1h∈I. Due to3.11, it is clear that every convex function of orderpis also Jensen convex of orderp. In general, the converse does not hold.
However, under the additional assumption thatxis continuous, the two notions turn out to be equivalent.
Theorem 3.1see13, page 387, Theorem 1. LetI ⊆Rbe a nondegenerate interval, letp≥ −1 be an integer, and letx:I → Rbe a continuous function. Thenxis convex of orderpif and only if it is Jensen convex of orderp.
Finally, we mention the following result concerning the difference of order p of a product of two functions:
Lemma 3.2. LetI ⊆ Rbe a nondegenerate interval, and letpbe a nonnegative integer. Given two functionsx, y:I → R, the equality
Δphxy t
p k0
p
k
Δkhx t·
Δp−kh y
tkh 3.13
holds for everyt∈Isuch thattph∈I.
4. Main Results
Throughout this section T is a positive real number. In the spaceC0, T, consisting of all continuous functionsx:0, T → R, we consider the usual maximum norm
x:max{|xt| |t∈0, T}. 4.1
Our first main result concerns the integral equation of Urysohn type1.1in whicha, u, andvare given functions, whilexis the unknown function. We assume that the functions a,u, andvsatisfy the following conditions:
C1r ≥ −1 is a given integer number;
C2a : 0, T → Ris a continuous function which is convex of order pfor eachp ∈ {−1,0, . . . , r};
C3u:0, T×R → Ris a continuous function such thatut,0 0 for allt∈0, Tand the function
t∈0, T−→ut, xt∈R 4.2
is convex of orderpfor eachp ∈ {−1,0, . . . , r}wheneverx ∈ C0, Tis convex of orderpfor eachp∈ {−1,0, . . . , r};
C4there exists a continuous function ϕ : 0,∞ × 0,∞ → 0,∞ which is nondecreasing in each variable and satisfies
ut, x−u
t, y ≤x−yϕ
x, y 4.3
for allt∈0, Tand allx, y∈0,∞;
C5v:0, T×0, T×R → Ris a continuous function such that the functionv·, s, x: 0, T → Ris convex of orderpfor eachp∈ {−1,0, . . . , r}whenevers∈0, Tand x∈0,∞;
C6there exists a continuous nondecreasing functionψ :0,∞ → 0,∞such that
|vt, s, x| ≤ψ|x| ∀t, s∈0, T, x∈R; 4.4
C7there existsr0>0 such that
aTr0ϕr0,0ψr0≤r0, Tϕr0, r0ψr0<1. 4.5
Theorem 4.1. If the conditions (C1)–( C7) are satisfied, then1.1possesses at least one solution x∈C0, Twhich is convex of orderpfor eachp∈ {−1,0, . . . , r}.
Proof. Consider the operatorF, defined onC0, Tby
Fxt:at ut, xt
T
0
vt, s, xsds, t∈0, T. 4.6
ThenFx∈C0, Twheneverx∈C0, T see5, the proof of Theorem 3.2.
We claim thatFis continuous onC0, T. To this end we fix anyx0inC0, Tand prove thatFis continuous atx0. Letc:x01, and let
M1:max{|ut, x| |t∈0, T, x∈−x0,x0}
M2:max{|vt, s, x| |t, s∈0, T, x∈−c, c}. 4.7
Further, let ε > 0. The uniform continuity of u on 0, T×−c, cas well as that of v on 0, T×0, T×−c, censures the existence of a real numberδ >0 such that
ut, x−u
t, y < ε, vt, s, x−v
t, s, y < ε 4.8
for allt, s∈0, Tand allx, y∈−c, csatisfying|x−y|< δ. Then for everyx∈C0, Tsuch thatx−x0<min{1, ε, δ}and everyt∈0, Twe have
|Fxt−Fx0t| ≤
ut, xt−ut, x0t T
0
vt, s, xsds
ut, x0t T
0
vt, s, xs−vt, s, x0sds
≤ |ut, xt−ut, x0t|
T
0
|vt, s, xs|ds |ut, x0t|
T
0
|vt, s, xs−vt, s, x0s|ds
≤εTM1M2.
4.9
Therefore, the inequalityFx−Fy ≤ εTM1M2holds for everyxinC0, Tsatisfying x−x0<min{1, ε, δ}. This proves the continuity ofFatx0.
Next, letr0be the positive real number whose existence is assured byC7, and letQ be the subset ofC0, T, consisting of all functionsxsuch thatx ≤ r0 andxis convex of orderpfor eachp∈ {−1,0, . . . , r}. Obviously,Qis a nonempty bounded closed convex subset ofC0, T. We claim thatFmapsQinto itself. To prove this, letx∈Qbe arbitrarily chosen.
For everyt∈0, Twe have
|Fxt| ≤ |at||ut, xt|
T
0
|vt, s, xs|ds. 4.10
Sincexis convex of order−1i.e., nonnegative, according toC3andC4we also have
|ut, xt||ut, xt−ut,0| ≤xtϕxt,0≤ xϕx,0. 4.11
This inequality andC6yield
|Fxt| ≤ axϕx,0 T
0
ψ|xs|ds
≤ aTxϕx,0ψx.
4.12
Taking into account thatx ≤r0, byC4,C6, andC7we conclude that
Fx ≤ aTr0ϕr0,0ψr0≤r0. 4.13
On the other hand, for everyt∈0, Twe have
Fxt at xutxvt, 4.14
wherexu, xv :0, T → Rare the functions defined by
xut:ut, xt, xvt: T
0
vt, s, xsds, 4.15
respectively. According toLemma 3.2, we have Δp1h Fx
t Δp1h a
t
p1
k0
p1 k
Δkhxv
t
Δp1−kh xu
tkh 4.16
for everyp∈ {−1,0, . . . , r}and everyt∈0, Tsuch thattph∈0, T. But Δkhxv
t k
i0
−1k−i k
i
xvtih
T
0
k i0
−1k−i k
i
vtih, s, xsds
T
0
Δkhxv,s
tds,
4.17
wherexv,st:vt, s, xs. By virtue ofC5we haveΔkhxv,st≥0, whence Δkhxv
t≥0 for eachk∈ {0,1, . . . , r1}. 4.18
This inequality together with 4.16,C2, and C3ensures that the functionFx is Jensen convex of orderpfor eachp∈ {−1,0, . . . , r}. SinceFxis continuous on0, T, byTheorem 3.1 it follows thatFxis convex of orderpfor eachp ∈ {−1,0, . . . , r}. Taking into account4.13, we conclude thatFmapsQinto itself, as claimed.
Finally, we prove that the operatorFsatisfies the Darbo condition with respect to the Hausdorffmeasure of noncompactnessχ. To this end letXbe an arbitrary nonempty subset ofQand letx∈X. Further, letε >0 and lett1, t2∈0, Tbe such that|t1−t2| ≤ε. We have
|Fxt1−Fxt2|
≤ |at1−at2|
ut1, xt1 T
0
vt1, s, xsds−ut2, xt2 T
0
vt2, s, xsds
≤ |at1−at2||ut1, xt1−ut1, xt2| T
0
|vt1, s, xs|ds
|ut1, xt2−ut2, xt2| T
0
|vt1, s, xs|ds
|ut2, xt2| T
0
|vt1, s, xs−vt2, s, xs|ds
≤ωa, ε |xt1−xt2|ϕxt1, xt2Tψx |ut1, xt2−ut2, xt2|Tψx
|xt2|ϕxt2,0 T
0
|vt1, s, xs−vt2, s, xs|ds.
4.19
Letting
ωr0u, ε:supu
t, y −u
t, y : t, t∈0, T,|t−t| ≤ε, y∈0, r0 , ωr0v, ε:supv
t, s, y −v
t, s, y : s, t, t∈0, T,|t−t| ≤ε, y∈0, r0 ,
4.20
we get
|Fxt1−Fxt2| ≤ωa, ε Tϕr0, r0ψr0ωx, ε
Tψr0ωr0u, ε Tr0ϕr0,0ωr0v, ε. 4.21
Thus
ωFx, ε≤ωa, ε Tϕr0, r0ψr0ωx, ε
Tψr0ωr0u, ε Tr0ϕr0,0ωr0v, ε, 4.22
whence
ωFX, ε≤ωa, ε Tϕr0, r0ψr0ωX, ε
Tψr0ωr0u, ε Tr0ϕr0,0ωr0v, ε. 4.23
Taking into account thatais uniformly continuous on0, T,uis uniformly continuous on 0, T×0, r0andvis uniformly continuous on0, T×0, T×0, r0, we have thatωa, ε → 0, ωr0u, ε → 0 and ωr0v, ε → 0 asε → 0. So letting ε → 0we obtain ω0FX ≤ Tϕr0, r0ψr0ω0X, that is,
χFX≤Tϕr0, r0ψr0χX 4.24
by virtue of2.5.
ByC7andTheorem 2.2we conclude the existence of at least one fixed point ofF in Q. This fixed point is obviously a solution of1.1whichin view of the definition ofQis convex of orderpfor eachp∈ {−1,0, . . . , r}.
Theorem 4.1can be further generalized as follows. Given an integer numberr ≥ −1 and a sequenceξ : ξ−1, ξ0, . . . , ξr∈ {−1,1}r2, we denote by Convr,ξ0, Tthe set consisting of all functionsx∈C0, Twith the property that for eachp∈ {−1,0, . . . , r}the functionξpx is convex of orderp. For instance, ifr 1 andξ 1,−1,1, then Convr,ξ0, Tconsists of all functions inC0, Tthat are nonnegative, nonincreasing, and convex on0, T.
Recallsee, e.g., Roberts and Varberg14, pages 233-234that a functionx:0, T → Ris called absolutely monotonicresp., completely monotonicif it possesses derivatives of all orders on0, Tand
xkt≥0
resp.,−1kxkt≥0
4.25
for each t ∈ 0, T and each integer k ≥ 0. By13, Theorem 6, page 392it follows that if x : 0, T → Ris an absolutely monotonicresp., a completely monotonicfunction, then xbelongs to every set Convr,ξ0, T with r ≥ −1 and ξk 1resp.,ξk −1k1for each k∈ {−1,0, . . . , r}.
Instead of the conditions C1, C2, C3, and C5 we consider the following conditions.
C1r ≥ −1 is a given integer number andξ : ξ−1, ξ0, . . . , ξr ∈ {−1,1}r2 is a sequence such that either
ξk1 for eachk∈ {−1,0, . . . , r} 4.26
or
ξk −1k1 for eachk∈ {−1,0, . . . , r}. 4.27
C2a:0, T → Rbelongs to Convr,ξ0, T.
C3u:0, T×R → Ris a continuous function such thatut,0 0 for allt∈0, Tand the function
t∈0, T−→ut, xt∈R 4.28
belongs to Convr,ξ0, Twheneverx∈Convr,ξ0, T.
C5v:0, T×0, T×R → Ris a continuous function such that the functionv·, s, x: 0, T → Rbelongs to Convr,ξ0, Twhenevers∈0, Tandx∈0,∞.
Theorem 4.2. If the conditions (C1)–(C3), (C4), (C5), and (C6)-(C7) are satisfied, then1.1possesses at least one solutionx∈Convr,ξ0, T.
Proof. Consider the operatorF, defined onC0, T, as in the proof ofTheorem 4.1. As we have already seen in the proof ofTheorem 4.1we haveFx∈C0, Twheneverx∈C0, TandF is continuous onC0, T.
Instead of the setQ, considered in the proof ofTheorem 4.1, we take now Q to be the subset of Convr,ξ0, T consisting of all functions x such that x ≤ r0. Then Q is a nonempty bounded closed convex subset of C0, T. We claim that F maps Q into itself.
Indeed, according to4.13we haveFx ≤ r0 wheneverx ∈ C0, Tsatisfiesx ≤ r0. On the other hand,Fxadmits the representation4.14, wherexu, xv:0, T → Rare defined by 4.15. Given anyp∈ {−1,0, . . . , r}, note that
ξpξk−1ξp−k for eachk∈
0,1, . . . , p1
, 4.29
whence
Δp1h
ξpFx t Δp1h
ξpa t p1
k0
p1 k
Δkhξk−1xv t
Δp1−kh
ξp−kxu tkh 4.30
for everyt∈0, Tsuch thattph∈0, T. By proceeding as in the proof ofTheorem 4.1one can show that
Δp1h
ξpFx t≥0 whenevert∈0, Tsatisfiestph∈0, T. 4.31
ThereforeFx∈Convr,ξ0, T.
The rest of the proof is similar to the corresponding part in the proof ofTheorem 4.1 and we omit it.
5. An Application
As an application of the results established in the previous section, in what follows we study the solvability of the integral equation
xt 1λxnt 1
0
1
ts1xsds, t∈0,1, 5.1
in whichnis a given positive integer and λis a positive real parameter. Note that5.1is similar to the Chandrasekhar equation, arising in the theory of radiative transfersee, e.g., Chandrasekhar15or Bana´s et al.16, and the references therein.
We are going to prove that if 0 < λ < 1/n11/nn1, then 5.1possesses at least one continuous nonnegative solution, which is nonincreasing and convex. To this end, we apply Theorem 4.2for r : 1 and ξ : 1,−1,1. Take T : 1, at ≡ 1, ut, x : xn and vt, s, x:λ/ts1x. It is immediately seen that all the conditionsC1–C3,C4,C5, andC6are satisfied if the functionsϕandψare defined by
ϕ
x, y :xn−1xn−2y· · ·xyn−2yn−1, ψx:λx, 5.2
respectively. It remains to show thatC7is satisfied, too. Taking into account the expressions ofϕandψ, conditionC7is equivalent to the following statement. If 0< λ <1/n11/nn1, then there exists anr0>0 such that
1λr0n1≤r0, nλr0n<1. 5.3
Clearly, such anr0must satisfyr0>1. Letf, g:1,∞ → Rbe the functions defined by
fr: r−1
rn1, gr: 1
nrn, 5.4
respectively. Since
fr n1−nr
rn2 , 5.5
one can see thatf attains a maximum at rn : n1/n, the maximum value beingλn : 1/n11/nn1. On the other hand, we have
gr−fr n−rn−1
nrn1 . 5.6
Ifn 1, then gr > frfor allr ∈ 0,∞. If n > 1 and 0 < r < n/n−1, then gr> fr, while ifn >1 andr≥n/n−1, thengr≤fr. Note that 0< rn< n/n−1.
Assume now that 0< λ < λn. Then we can select anr0sufficiently close tornsuch that λ < fr0< gr0. Obviously,r0satisfies5.3.
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