American Journal of Computational Mathematics, 2016, 6, 166-176 Published Online June 2016 in SciRes. http://www.scirp.org/journal/ajcm http://dx.doi.org/10.4236/ajcm.2016.62018
On Existence of Solutions of q-Perturbed Quadratic Integral Equations
Maryam Al-Yami Department of Mathematics, Al Faisaliah Campus, Sciences Faculty, King Abdulaziz University, Jeddah, Saudi Arabia
Received 18 April 2016; accepted 26 June 2016; published 29 June 2016
Copyright © 2016 by authors and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/
Abstract We investigate a q-fractional integral equation with supremum and prove an existence theorem for it. We will prove that our q-integral equation has a solution in C [0,1] which is monotonic on [0,1] . The monotonicity measures of noncompactness due to Banaś and Olszowy and Darbo’s theorem are the main tools used in the proof of our main result.
Keywords q-Fractional, Integral Equation, Monotonic Solutions, Darbo Theorem, Monotonicity Measure of Noncompactness
1. Introduction Jackson in [1] introduced the concept of quantum calculus (q-calculus). This area of research has rich history and several applications, see [2]-[4] and references therein. There are several developments and applications of the q-calculus in mathematical physics, especially concerning quantum mechanics, the theory of relativity and special functions [5] [4]. Recently, several researchers attracted their attention by the concept of q-calculus, and we could find several new results in [6] [7] and the references therein. In several papers among them [8]-[11], integral equations with nonsigular kernels have been studied. In [12]- [14] Darwish et al. introduced and studied the quadratic Volterra equations with supremum. Also, Banaś et al. and Darwish [13] [15]-[17] studied quadratic integral equations of arbitrary orders with singular kernels. In [18], Darwish generalized and extended Banaś et al. [15] results to the perturbed quadratic integral equations of arbi- trary orders with singular kernels. In this paper, we will study the q-perturbed quadratic integral equation with supremum
How to cite this paper: Al-Yami, M. (2016) On Existence of Solutions of q-Perturbed Quadratic Integral Equations. American Journal of Computational Mathematics, 6, 166-176. http://dx.doi.org/10.4236/ajcm.2016.62018 M. Al-Yami
(yt)() t (β −1) yt( ) = f tyt, ( ) + ktstqs( , )( − ) ( ys)( )d stI ,∈=[ 0,1] , (1) ( ) ∫0 q Γq (β ) where 0<β ,q ∈( 0,1) , fI : ×→, ,:CI( ) → CI( ) , and kI: ×→ I . By using Darbo fixed point theorem and the monotonicity measure of noncompactness due to Banaś and Ols- zowy [19], we prove the existence of monotonic solution to Equation (1) in C [0,1] . 2. q-Calculus and Measure of Noncompactness First, we collect basic definitions and results of the q-fractional integrals and q-derivatives, for more details, see [5] [6] [20] [21] and references therein. ∈ First, for a real parameter q (0,1) , we define a q-real number [a]q by 1− qa [aa] =, ∈ , (2) q 1− q and a q-analog of the Pochhammer symbol (q-shifted factorial) is defined by
1, n = 0, (aq; ) = n−1 (3) n 1,−∈aqk n . ∏k =1( ) n Also, the q-analog of the power (ab− ) is given by 1, n = 0, n (ab− ) = n−1 (4) a−∈ bqk , n ;, a b ∈ . ∏k =1( ) Moreover, −=(n) n ≠ (a b) a( baq;)n , a 0. (5)
Notice that, limn→∞ (aq ; )n exists and we will denote it by (aq; )∞ . β −n More generally, for β ∈,,aq ≠∈ q( n ) we define (aq; ) (aq; ) = ∞ (6) β aqβ ; q ( )∞ and
(β ) (baq; ) (ab−=) aβ ∞ (7) qbaqβ ; ( )∞ (β ) β (β ) β Notice that (a−= b) a( baq; )β . Therefore, if b = 0 , then aa= . The q-gamma function is defined by Gq( x ) = ∈ −− Γq ( xx) x−1 , \{ 0, 1, 2, } , (8) (1− q) Gq( )
( x−1) 1 (1− q) x = = += where Gq( ) . Or, equivalently, Γq ( x) x−1 and satisfies Γqq( x1) [ xx] Γ ( ). qqx ; 1− q q ( )∞ ( ) Next, the q-derivative of a function f is given by f( t) − f( qt) (Dfq)( t) = ,(Dfq)( 0) = lim tq→0 (Df)( t), (9) t− qt and the q-derivative of higher order of a function f is defined by
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=ft( ), n 0, (Dfq )( t) = (10) n−1 ∈ DDqq( f)( t),. n The q-integral of a function f defined on the interval [0,b] is defined by ∞ t I f( t) == f( s)d s t( 1−∈ q) qnn f tq, t[ 0, b] . (11) ( qq) ∫a ∑ ( ) n=0 If f is given on the interval [0,b] and ab∈[0, ] then b ba fs( )d s= fs( ) d s − fs( ) d. s (12) ∫∫∫a q00 qq n The operator Iq is defined by =ft( ), n 0. n (Ifq )( t) = (13) n−1 ∈ IIqq( ft)( ),, n
The fundamental theorem of calculus satisfies for Dq and Iq , i.e., (DIqq f)( t) = f( t) , and if f is continuous at t = 0 , then (IDfqq)( t) = f( t) − f(0) . The following four formulas will be used later in this paper
(β ) β (β ) ats( −=−) ats( ) −=(ββ) β −( −1) tqDts( ) [ ]q ( ts) (14) −=−−(ββ) β ( −1) sqD( t s) [ ]q (t qs) and t t D f( ts,d) s= Df( ts ,d) s+ f( qtt ,,) (15) tq∫∫00qt q q where D denotes the q-derivative with respect to variable t. tq (ββ) ( ) Notice that, if β > 0 and abt≤≤, then (tb−) ≤−( ta) . Definition 1. [2] Let f be a function defined on [0,1] . The fractional q-integral of the Riemann-Liouville type of order β ≥ 0 is given by =ft( ), β 0, β β ∞ Ifq ( t) = qq; (16) ( ) 1 t (ββ−1) β nn( )n (tqsfsstqqftqt−) ( )dq =( 1 −) ∑ ( ),β >∈ 0,[ 0,1] . β ∫0 Γq ( ) n=0 (qq; )n Notice that, for β = 1, the above q-integral reduces to (11). Definition 2. [2] The fractional q-derivative of the Riemann-Liouville type of order β ≥ is given by 0 ft( ), β = 0, β (Dfq )( t) = β ββ− (17) DI[ ] [ ] f t,β > 0, ( qq )( ) where [β ] denotes the smallest integer greater than or equal to β . In q-calculus, the derivative rule for the product of two functions and integration by parts formulas are
(Dfgtqq)( ) = ( Df)( tgt) ( ) + f( qtDg)( q)( t), ttt (18) f( s) Dg( s)d s= f( sgs) ( ) − Df( sgqs) ( )d. s ∫∫00( qq) 0 ( q) q Lemma 1. Let γβ,0≥ and f be a function defined on [0,1] . Then the following formulas are verified: γ β γβ+ 1) (IIftqq )( ) = ( Iq ft)( ), (19) ββ 2) (DIqq f)( t) = f( t).
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Lemma 2. [21] For β > 0 , using q-integration by parts, we have
(β ) β t (Itq 1)( ) = (20) Γq (β +1) or
(β ) t (β −1) t (t−= qs) d. s (21) ∫0 q β [ ]q
Second, we recall the basic concepts which we need throughout the paper about measure of noncompactness. We assume that E, ⋅ is a real Banach space with zero element θ and we denote by B( xr, ) the closed ( ) ball with radius r and centered x, where BBrr ≡ (θ, ) . Now, let XE⊂ and denote by X and Conv X the closure and convex closure of X, respectively. Also, the symbols XY+ and λ X stands for the usual algebraic operators on sets. Moreover, the families M and N are defined by M ={A ⊂ EA: ≠∅ , A is bounded} and E E E NMEE={BB ⊂ : is relatively compact} , respectively. Definition 3. [22] Let µ :.M E → + If the following conditions
1) ∅≠{XX ∈MNEE:µµ( ) = 0} = ker ⊂ . 2) XY⊂≤, then µµ( X) ( Y). 3) µµµ( XX) =( ) = (Conv X) . 4) µ( λX+−(1 λ) YX) ≤ λµ( ) +−(1 λ) µ( Y) ,0 ≤≤ λ 1 and
5) if ( X n ) is a sequence of closed subsets of M E with Xnn+1 ⊂= Xn, 1, 2, 3, , and limnn→∞ ( X ) = 0 ∞ then XX= ≠∅ hold. Then, the mapping µ is said to be a measure of noncompactness in E. ∞ ∩n=1 n Here, kerµ is the kernel of the measure of noncompactness µ . Our result will establish in C(I) the Banach space of all defined, continuous and real functions on I ≡ [0,1] with y= maxtI∈ yt( ) . Next, we defined the measure of noncompactness related to monotonicity in CI( ) , see [19] [22]. We fix a bounded subset Y ≠∅ of CI( ) . For ε ≥ 0 and yY∈ ,,ωε( y) denotes the modulus of conti- nuity of the function y given by ωε( y,) = sup{ yt( ) − ys ( ) : ts , ∈ I , t −≤ s ε} . (22)
Moreover, we let ωε(Y,) = sup{ ωε( y ,) : yY∈ } (23) and
ω00(YY) = limε → ωε( ,) . (24)
Define
d( y) =supst,,∈≤ I s t( yt( ) −− ys( ) yt( ) − ys( ) ) (25) and
dY( ) = supyY∈ d( y) . (26)
Notice that, all functions in Y are nondecreasing on I if and only if dY( ) = 0 .
Now, we define the map µ on MCI( ) as
µω(Y) = dY( ) + 0 ( Y). (27)
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Clearly, μ verifies all conditions in Definition 3 and, therefore it is a measure of noncompactness in CI( ) [19]. Definition 4.Let ∅≠Ω ⊂E. Let : Ω → E be a continuous operator. Suppose that maps bounded sets onto bounded ones. If there exists a bounded Y ⊂ Ω with µ(YY) ≤≥ γµ( ),0 γ , then is said to be satisfies the Darbo condition with respect to a measure of noncompactness µ . If γ < 1 , then is called a contraction operator with respect to µ . Theorem 1. [23] Let Q ≠∅ be a bounded, convex and closed subset of E. If : QQ→ is a Contraction operator with respect to µ . Then has at least one fixed point belongs to Q.
3. Existence Theorem Let us consider the following suggestions: a1) fI: ×→ is continuous and
∃0 ≤c < 1 stfty . .( ,) − ftx( , ) ≤ cyx − ∀∈ t Iand xy , ∈
* Moreover, fI: ×→+ and f= maxtI∈ ft( ,0) . a2) The superposition operator F generated by the function f satisfies for any nonnegative function y the con- dition d( Fy) ≤ cd( y) , where c is the same constant as in a1). a3) : CI( ) → CI( ) is a continuous operator which satisfies the Darbo condition for the measure of non- compactness µ with a constant η . Also, y ≥ 0 if y ≥ 0 .
a4) ∃a, b ≥ 0, st . . ( y)( t) ≤ a + y ∀∈ y C( I), t ∈ I.
a5) The function kI: ×→ I + is continuous on II× and nondecreasing ∀t and separately. Moreo- ver, k* = supkts( ,) . (ts, )∈× I I 𝑠𝑠 a6) : CI( ) → CI( ) is a continuous operator and there is a nondecreasing function φ : ++→ such that yy≤ φ ( ) for any y∈ CI( ) . Moreover, for every function y∈ CI( ) which is nonnegative on I, the function y is nonnegative and nondecreasing on I.
a7) ∃>r0 0 such that (a+ br) k*φ ( r) f* ++ cr ≤r (28) Γq (β +1)
ηφkr* ( ) and c +<0 1. Γq (β +1) Before, we state and prove our main theorem, we define the two operators and on CI( ) as fol- lows
1 t (β −1) (y)( t) = k( t,d s)( t− qs) ( y)( s) s (29) ∫0 Γq (β )
and (yt)( ) = ftyt( , ( )) + ( yt)( )( yt)( ) (30)
respectively. Finding a fixed point of the operator defined on the space CI( ) is equivalent to solving Eq- uation (1). Theorem 2. Assume the suggestions (a1)-(a7) be verified, then Equation (1) has at least one solution y∈ CI( ) which is nondecreasing on I. Proof. We divide the proof into seven steps for better readability. Step 1: We will show that the operator maps CI( ) into itself. For this, it is sufficient to show that y∈ CI( ) if y∈ CI( ) . Fix ε > 0 and let y∈ CI( ) and
tt12, ∈≤ It( 1 t 2) with tt21−≤ε . We have
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(yt)( 21) − ( yt)( )
1 tt(ββ−−11) ( ) =21k t, s t − qs y sd, s −− k t s t qs y sd s ∫∫00( 2)( 2) ( )( ) qq( 11)( ) ( )( ) Γq (β )
1 tt(ββ−−11) ( ) ≤22k t,d s t − qs y s s −− k t,d s t qs y s s ∫∫00( 22)( ) ( )( ) qq( 12)( ) ( )( ) Γq (β )
1 tt(ββ−−11) ( ) +21k t,d s t − qs y s s −− k t,d s t qs y s s ∫∫00( 12)( ) ( )( ) qq( 12)( ) ( )( ) Γq (β )
1 tt(β −1) (β −1) +11kt,d s t −− qs y s s kt,ds t− qs y s s ∫∫00( 12)( ) ( )( ) q ( 11)( ) ( )( ) q Γq (β )
1 t (β −1) ≤2 kt,, s −− kt s t qs y sd s ∫0 ( 2) ( 12) ( ) ( )( ) q Γq (β )
1 t2 (β −1) +−k( t12,d s) ( t qs) ( y)( s) q s ∫t1 Γq (β ) (31) 1 t (ββ−−11) ( ) + 1 k( t,d s) ( t− qs) −−( t qs) ( y)( s) s ∫0 11 2 q Γq (β )
φ( y ) ωε( ,.) t (β −1) ≤−k 2 t qsd s ∫0 ( 2 ) q Γq ()β * kyφ ( ) tt(ββ−−11) ( ) (β−1) + 12(t− qs) −−( t qs) dd s +( t − qs) s {∫∫0 12q t 2q } 1 Γq (β ) * φ( y) ωε( ,.) ky φ( ) (β ) = k t()β +t(ββ) −+ t( ) 2 tt − 2 1 2( 21) Γqq(ββ++1) Γ ( 1) * φ( y) ωεk ( ,.) β 2 ky φ( ) (β ) ≤t2 +−( tt21) Γqq(ββ++1) Γ ( 1) φ( y) ωε( ,.) 2 ky* φ( ) ≤+k ε β . Γqq(ββ++1) Γ ( 1) Notice that, we have used
ωεk ( ,.) = sup{ kts( ,) − k(τ , s) : ts , , τ ∈ I and t −≤ τ ε} . (32)
Notice that, since the function k is uniformly continuous on II× , then when ε → 0 we have that ωεk ( ,.) → 0. Thus y∈ CI( ) , and therefore, y∈ CI( ). Step 2: applies B into itself. r0 Now, ∀∈tI, we have
(yt)( ) t (β −1) (y)( t) ≤+ f t,, y( t) k( t s)( t − qs) ( y)( s) d s ( ) ∫0 q Γq (β )
(yt)( ) t (β −1) ≤ftyt,( ) −+ ft( ,0) ft( ,0) + kts( , ) ( t− qs) ( y)( s) d s ( ) ∫0 q Γq (β ) (33) * (a+ by) kφ ( y) t (β −1) ≤c y ++ f * (t− qs) d s ∫0 q Γq (β ) (a+ by) k*φ ( y) =cy ++ f* . Γq (β +1)
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Hence (a+ by) k*φ ( y) y≤ cy ++ f* . (34) Γ+q (β 1)
Therefore, if yr≤ 0 we get from assumption a7) the following * * (a+ br00) kφ ( r ) y≤++ cr00 f ≤r . (35) Γq (β +1) Therefore, maps B into itself. r0 We define the subset B+ of B by r0 r0 B+ =∈≥∈ y B: yt 0,for t I (36) rr00{ ( ) } It is clear that B+ ≠∅ is closed, convex and bounded. r0 Step 3: applies the set B+ into itself. r0 By this facts and suggestions a ), a ) and a ), we obtain transforms B+ into itself. 1 4 6 r0 Step 4: The operator is continuous on B+ . r0 To prove this, we fix to be a sequence in B+ with → . We will show that → . ( yn ) r0 yyn yyn Thus, we have ∀∈tI,
(ynn)( t) −≤( y)( t) f( ty,,( t)) − f( tyt( ))
(yt)( ) tt(ββ−−11) (yt)( ) ( ) +n k( t, s)( t − qs) ( y)( s) d, s −−k( t s)( t qs) ( y)( s) d s ∫∫00nq q Γqq()ββΓ ()
(yt)( ) tt(ββ−−11) (yt)( ) ( ) ≤c y( t) −+ y( t) n k( t,d,d s)( t− qs) ( y)( s) s − k( t s)( t− qs) ( y)( s) s n ∫∫00nq nq Γqq(ββ) Γ () (37) (yt)( ) t (ββ−−11) ( yt)( ) t ( ) +k( t, s)( t − qs) ( y)( s) d, s −−k( t s)( t qs) ( y)(ss)d ∫0 nq ∫0 q Γqq()ββΓ ()
(yn )( t) − ( yt)( ) t (β −1) ≤c y( t) −+ y( t) k( t,d s) ( t− qs) ( y)( s) s n ∫0 nq Γq (β )
(yt)( ) t (β −1) +k( t, s) ( t −− qs) ( y)( s) ( y)( s) d. s ∫0 nq Γq (β ) Consequently, ** kφ ( r00) ynn−+ y( a br) k y − y ynn− y ≤ cy −+ y + . (38) Γqq(ββ++1) Γ ( 1)
As and are continuous operators, ∃∈n1 such that εβΓ ( +1) − ≤q ∀≥ yn y * ,.nn1 (39) 3krφ ( 0 )
Also, ∃∈n2 such that εβΓ ( +1) − ≤ q ∀≥ yn y * , nn2 . (40) 3k( a+ br0 )
Furthermore, ∃∈n3 such that ε y− y ≤,. ∀≥ nn (41) n 3c 3
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Now, take max{nnn123 , , } ≤ n, then (38) gives us that
yyn −≤ε . (42)
This shows that is continuous in B+ . r0 Step 5: In recognition of with respect to the quantity ω0 . Now, we take ∅≠YB ⊂ + . Let us fix an arbitrarily number ε > 0 and choose ∈ and ∈ with r0 yY tt12, I tt21−≤ε . We will be supposed that tt12≤ because no generality will be loss. Then, by using our suggestions and inequality (31), we get
(yt)( 21) − ( yt)( )
≤ftyt( 2,,( 2)) −+ ftyt( 11( )) ( yt)( 2)( yt)( 2) −( yt)( 2)( yt)( 1)
+−(yt)( 21)( yt)( ) ( yt)( 11)( yt)( )
≤ftyt( 22,,( )) −+− ftyt( 12( )) ftyt( 12 ,,( )) ftyt( 11( ))
+(yt)( 2) ( yt)( 2) −+( yt)( 1) ( yt)( 2) −( yt)( 11) ( yt)( ) ≤γf, ε ++ c ωε (,) y yt yt − yt r0 ( ) ( )( 22) ( )( ) ( )( 1)
+−(yt)( 2) ( yt)( 11) ( yt)( ) (43) (a+ by)φ ( y) ≤γf,, ε ++ cy ω ε ωε,.+ 2k* εβ rk0 ( ) ( ) ( ) Γq (β +1) ωε(y, ) + ky*φ ( ) Γq (β +1) (a+ br)φ ( r ) ≤γf,, ε ++ cy ω ε 00ωε,.+ 2k* εβ rk0 ( ) ( ) ( ) Γq (β +1)
ωε(y, ) * + krφ ( 0 ). Γq (β +1)
The last estimate implies ωy, εγ≤+ f ,, ε cy ωε ( ) r0 ( ) ( ) (a+ br)φ ( r ) ωε(y, ) (44) +00ωε++** εβ φ k ( ,.) 2k kr( 0 ) Γqq(ββ++1) Γ ( 1) and, consequently, ωY, εγ≤+ f ,, ε cY ωε ( ) r0 ( ) ( ) (a+ br)φ ( r ) ωε(Y, ) (45) +00ωε++** εβ φ k ( ,.) 2k kr( 0 ). Γqq(ββ++1) Γ ( 1)
Since the function k is uniformly continuous on II× and the function f is continuous on Ir×[0, 0 ] , then the last inequality gives us that
* krφ ( 0 ) ωω00(YcY) ≤+( ) ω 0( Y). (46) Γq (β +1)
Step 6: In recognition of with respect to the quantity d.
Here, we fix an arbitrary yY∈ and tt12, ∈ I with tt21> . Then, by our assumption, we obtain our sugges- tions, we have
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(yt)( 21) −−( yt)( ) ( yt)( 21) −( yt)( )
(yt)( ) t (β −1) =+−f t, y t 2 2 k t ,d s t qs y s s ( 22( )) ∫0 ( 2)( 2 ) ( )( ) q Γq (β )
(yt)( ) t (β −1) −−f t, y t 1 1 k t ,d s t − qs y s s ( 11( )) ∫0 ( 1)( 1 ) ( )( ) q Γq (β )
(yt)( ) t (β −1) −+ f( t, y( t )) 2 2 k( t ,d s)( t − qs) ( y)( s) s 22 β ∫0 2 2 q Γq ( )
(yt)( ) t (β −1) −−f( t, y( t )) 1 1 k( t ,d s)( t − qs) ( y)( s) s 11 β ∫0 1 1 q Γq ( ) ≤−−− { ft( 22,, yt( )) ft( 1yt( 1)) f( t 2,, yt( 2)) f( t 11 yt( )) }
(y)( t ) tt(ββ−−11) (Tx)() t ( ) +2122k t,d,d s t −− qs y s s k t s t − qs y s s ∫∫00( 22)( ) ( )( ) qq( 22)( ) ( )( ) Γqq(ββ) Γ ( ) (47) (y)( t ) tt(ββ−−11) (Tx)() t ( ) +1121k t, s t −− qs y s d, s k t s t − qs y s d s ∫∫00( 2)( 2) ( )( ) qq( 11)( ) ( )( ) Γqq(ββ) Γ ( ) (y)( t ) tt(β −1) (Tx)() t (β −1) −2122kt( , s)( t −− qs) ( y)( s) d, s kt( s)( t− qs) ( y)( s)d s ββ∫∫0022 q 22 q Γqq( ) Γ ( ) (y)( t ) tt(ββ−−11) (Tx)() t ( ) +1121k( t, s)( t −− qs) ( y)( s) d, s k( t s)( t − qs) ( y)( s) d s ββ∫∫002 2 qq11 Γqq( ) Γ ( ) ≤ −− − { f( t2,, yt( 2)) f( t 11 yt( )) f( t 2 ,, yt( 2)) f( t 11 yt( )) }
1 t (β −1) +yt −− yt yt − yt2 ktst,d− qs ys s {( )( 21) ( )( ) ( )( 21) ( )( ) } ∫0 ( 22)( ) ( )( ) q Γq (β )
(yt)( ) tt(ββ−−11) ( ) + 1 21k( t, s)( t− qs) ( y)( s)d, s −− k( t s)( t qs) ( y)( s) d s {∫∫002 2 qq11 Γq (β )
tt(ββ−−11) ( ) −21k( t, s)( t − qs) ( y)( s)d s −− k( t , s)( t qs) ( y)( s)d. s ∫∫002 2 qq11 }
Now, we will prove that
tt2 (ββ−−11) 1 ( ) k( t, s)( t− qs) ( y)( s)d s −− k( t , s)( t qs) ( y)( s)d s ≥ 0. (48) ∫∫002 2 qq11 We find that
tt21(ββ−−11) ( ) k( t, s)( t− qs) ( y)( s)d, s −− k( t s)( t qs) ( y)( s) d s ∫∫002 2 qq11
tt22(ββ−−11) ( ) =k( t,d,d s)( t − qs) ( y)( s) s −− k( t s)( t qs) ( y)( s) s ∫∫0022 qq12
tt21(ββ−−11) ( ) +−k( t,d s)( t qs) ( y)( s) s −− k( t,d s)( t qs) ( y)( s) s ∫∫0012 qq12 (49) tt11(ββ−−11) ( ) +−k( t, s)( t qs) ( y)( s)d, s −− k( t s)( t qs) ( y)( s) d s ∫∫0012 qq11
tt22(ββ−−11) ( ) =(kt( 2 , s) − kt( 1,s))( t 2− qs) ( y)( s)d,qq s +− k( t12 s)( t qs) ( y)( s) d s ∫∫0 t1
t1 (ββ−−11) ( ) +k( t, s) ( t − qs) −−( t qs) ( y)( s)d. s ∫0 12 1 q
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But, kt( 12,, s) ≤ kt( s) because kts( , ) is increasing with respect to t, then
t2 (β −1) kt( ,, s) −− kt( s) ( t qs) ( y)( s)d s ≥ 0, (50) ∫0 ( 2 12) q (ββ−−11) ( ) and, since (t21− qs) −−( t qs) is negative for st∈[0, 1 ) then tt12(ββ−−11) ( ) (β−1) k( t1, s) ( t 2− qs) −−( t1 qs) ( y)( s)d,qq s+ k( t12 s)( t − qs) ( y)( s) d s ∫∫0 t1 tt12(ββ−−11) ( ) (β−1) ≥k( t11, t) ( t 2 − qs) −−( t1 qs) ( y)( t1)d,qq s+ k( t11 t)( t 2 − qs) ( y)( t1)d s ∫∫0 t1
tt21(ββ−−11) ( ) (51) =k( t, t)( y)( t) ( t− qs) dd s −−( t qs) s 11 1∫∫00 2 qq1 tt(ββ) − ( ) = ktt( , ) 21(y)( t) ≥ 0. 11 β 1 [ ]q Inequalities (50) and (51) imply that
tt21(ββ−−11) ( ) k( t, s)( t− qs) ( y)( s)d s −− k( t , s)( t qs) ( y)( s)d s ≥ 0. ∫∫002 2 qq11 This inequality and (47) gives us
(yt)( 21) −−( yt)( ) ( yt)( 21) −( yt)( ) ≤ −− − { ftyt( 2,,( 2)) ftyt( 11( )) ftyt( 2 ,,( 2)) ftyt( 11( )) }
+{ (yt)( 21) −−( yt)( ) ( yt)( 21) −( yt)( ) } (52)
1 t2 (β −1) ×−k( t,d s)( t qs) ( y)( s) s ∫0 22 q Γq (β ) kr*φ ( ) ≤+d( Fy) 0 d( y). Γq (β +1) The above estimate implies that kr*φ ( ) d( y) ≤+ cd( y) 0 d( y). (53) Γq (β +1) Therefore, kr*φ ( ) d( Y) ≤+ cd( Y ) 0 d( Y ). (54) Γq (β +1) Step 7: is contraction with respect to the measure of noncompactness µ . Inequalities (46) and (54) give us that * krφ ( 0 ) ωω00(YdYc) +( ) ≤( ( YdY) ++( )) ( ω 0(YdY) +( )) (55) Γq (β +1) or kr**φ( ) ηφkr( ) µµ(Y) ≤ cY( ) + 00 µ( Y) ≤+ c µ(Y). (56) ββ++ Γqq( 1) Γ ( 1) ηφkr* ( ) But c +<0 1, then Γq (β +1) µµ(YY) ≤ ( ). (57)
175 M. Al-Yami
Inequality (57) enables us to use Theorem 1, then there are solutions to Equation (1) in CI( ) . This finishes our proof.
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