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Positive solutions for fractional boundary value problems with integral boundary conditions and parameter dependence

Abbas, Hafida; Belmekki, Mohammed; Cabada Fernández, Alberto

Abstract

We study the existence of positive solutions for a Riemann fractional boundary value problem with integral boundary conditions and parameter dependence. To state our results, we use Guo-Krasnoselskii fixed point theorem. Some examples are showed to point out the applicability of the obtained results.

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Posi i e solu ions o ac ional bounda y alue p oblems wi h in eg al bounda y condi ions and pa ame e dependence Ha ida Abbas∗, Mohammed Belmekki∗∗ and Albe o Cabada∗∗∗ ∗Depa men o Ma hema ics, Uni e si y o Sa¨ıda, BP 138, 20000 Sa¨ıda, Alge ia. e-mail: a.ha ida@yahoo. ∗∗ High School o Applied Sciences, BP 165 RP. Bel Ho izon, Tlemcen, Alge ia. e-mail: m.belmekki@yahoo. ∗∗∗ Depa amen o de Es a ´ıs ica, An´alise Ma em´a ica e Op imizaci´on, Facul ade de Ma em´a icas, Uni e sidade de San iago de Compos ela, 15782 San iago de Compos ela, Spain. e-mail: alb[email p o ec ed] Abs ac We s udy he exis ence o posi i e solu ions o a Riemann ac ional bound- a y alue p oblem wi h in eg al bounda y condi ions and pa ame e dependence. To s a e ou esul s, we use Guo-K asnoselskii ixed poin heo em. Some exam- ples a e showed o poin ou he applicabili y o he ob ained esul s. Key wo ds. F ac ional di e en ial equa ion, In eg al bounda y condi ions, Posi i e solu ions, G een’s unc ion, Fixed poin heo em. AMS (MOS) subjec classi ica ion: 26A33, 34B18 1 In oduc ion F ac ional calculus appea s in many ields o enginee ing and sciences as heology, iscoelas ici y, elec ochemis y, elec omagne ism, and so o h. Many di e en books and monog aphs a e de o ed o he de elopmen o ac ional calculus. See o ins ance [5, 9, 10, 11, 14, 15, 16, 18]. The in e es o he s udy o ac ional o de di e en ial equa ions lies in he ac ha he e a e mo e deg ees o eedom in he ac ional-o de models. Fu he mo e, ac ional de i a i es p o ide an excellen ins umen o he desc ip ion o memo y and he edi a y p ope ies o a ious ma e ials and p ocesses. Recen esul s on ac ional di e en ial equa ions can be seen in [6, 7, 8, 12]. In eg al bounda y condi ions ha e a ious applica ions in applied ields such as blood low p oblems, chemical enginee ing, popula ion dynamics and so o h, o mo e de ails see [1, 4, 6, 17]. As in dynamic o popula ions, many ields o enginee ing and sciences ocus hei in e es on exis ence o posi i e solu ions. We men ion he wo ks [2, 3, 4, 13]. Mo i a ed by his and he abo e ci ed wo ks, in his pape we in es iga e he exis ence o posi i e solu ions o he ollowing ac ional di e en ial equa ion wi h in eg al bounda y condi ions. Dδu( ) + ( , u( )) = 0,0< < 1,1< δ ≤2,(1) This e sion o he a icle has been accep ed o publica ion, a e pee e iew and is subjec o Sp inge Na u e’s AM e ms o use, bu is no he Ve sion o Reco d and does no e lec pos -accep ance imp o emen s, o any co ec ions. The Ve sion o Reco d is a ailable online a : h ps://doi.o g/10.1007/s40314-021-01546-y F ac ional Bounda y Value P oblems 2 u(0) = 0, u(1) = λZ1 0h( )u( )d . (2) Whe e Dδis he Riemann-Liou ille ac ional de i a i e and is a gi en unc ion. The bounda y condi ions (2) can be hough as a mechanism pu ed a he end poin o an oscilla o , which is cha ac e ized by he weigh ed unc ion hand he pa ame e λ, ha con ols i s displacemen acco ding o he eedback om de ices measu ing he displacemen s along di e en pa s o he oscilla o . This pape is o ganized as ollows. In sec ion 2, we ecall some de ini ions conce ning he ac ional in eg al and de i a i e, and ela ed basic p ope ies which will be used in he sequel. We conside an auxilia y p oblem o de i e he G een’s unc ion. Ou main exis ence esul s a e gi en in Sec ion 3. Some examples a e in oduced in he las sec ion. 2 P elimina ies He e we p esen some basic knowledge and de ini ions o ac ional calculus which will be used in he sequel. De ini ion 2.1 ([15, 18]). The Riemann-Liou ille ac ional p imi i e o o de δ > 0 o a unc ion : (0,1] →Ris gi en by Iδ 0 ( ) = 1 Γ(δ)Z 0( −τ)δ−1 (τ)dτ, p o ided ha he igh side is poin wise de ined on (0,1], and whe e Γis he gamma unc ion. De ini ion 2.2 ([15, 18]). Fo a con inuous unc ion : (0,1] →R, The Riemann- Liou ille de i a i e o ac ional o de δ > 0is gi en by Dδ ( ) = 1 Γ(n−δ) dn d nZ 0( −τ)n−δ−1 (τ)dτ, n = [δ]+1, whe e [δ] deno es he in ege pa o he eal numbe δ. Lemma 2.1 ([15, 18]). Le δ > 0, hen he solu ions o he ac ional di e en ial equa ion Dδu( )=0 a e gi en by he ollowing exp ession u( ) = c1 δ−1+c2 δ−2+... +cn δ−n, ci∈R, i = 1, ..., n, n = [δ]+1. F om Lemma 2.1 we deduce he ollowing esul . F ac ional Bounda y Value P oblems 3 Lemma 2.2 ([15, 18]). Le δ > 0, hen IδDδu( )=u( ) + c1 δ−1+c2 δ−2+... +cn δ−n, ci∈R, i = 1, ..., n, n = [δ]+1. In o de o ge he exp ession o he G een’s unc ion o bounda y alue p oblem (1) −(2), we s a by sol ing he ollowing auxilia y p oblem: Dδu( ) + σ( ) = 0,0< <1,1< δ ≤2,(3) u(0) = 0, u(1) = λZ1 0h( )u( )d . (4) Lemma 2.3 Le 1< δ ≤2. Suppose ha 1−λR1 0h( ) δ−1d 6= 0. A unc ion u∈C[0,1] is a solu ion o he linea bounda y alue p oblem (3)-(4) i and only i i sa is ies he in eg al equa ion u( ) = Z1 0G( , s)σ(s)ds, whe e G( , s)is he G een’s unc ion gi en by G( , s) = G1( , s) + G2( , s) wi h G1( , s) =      δ−1(1−s)δ−1−( −s)δ−1 Γ(δ),0≤s≤ ≤1; δ−1(1−s)δ−1 Γ(δ),0≤ ≤s≤1. (5) and G2( , s) = λ δ−1 1−λR1 0h( ) δ−1d Z1 0h( )G1( , s)d (6) P oo . By Lemma 2.2 we ha e ha he uis a solu ion o he linea equa ion (3) i and only i i sa is ies u( ) = −Z 0 ( −s)δ−1 Γ(δ)σ(s)ds +c1 δ−1+c2 δ−2. Condi ion u(0) = 0 implies necessa ily ha c2= 0. Since u(1) = λR1 0h( )u( )d , we deduce ha c1=Z1 0 (1 −s)δ−1 Γ(δ)σ(s)ds +λc1Z1 0h(s)sδ−1ds −λ Γ(δ)Z1 0h( )Z 0( −s)δ−1σ(s)dsd . Now, since 1 −λR1 0h( ) δ−1d 6= 0, we ha e c1=1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0(1 −s)δ−1σ(s)ds −λZ1 0h( )Z 0( −s)δ−1σ(s)dsd  F ac ional Bounda y Value P oblems 4 Finally, we ha e he exp ession u( ) = −Z 0 ( −s)δ−1 Γ(δ)σ(s)ds + δ−1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0(1 −s)δ−1σ(s)ds −λ δ−1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0h( )Z 0( −s)δ−1σ(s)dsd =−Z 0 ( −s)δ−1 Γ(δ)σ(s)ds + δ−1(1 −λR1 0h( ) δ−1d +λR1 0h( ) δ−1d ) Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0(1 −s)δ−1σ(s)ds −λ δ−1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0h( )·Z 0( −s)δ−1σ(s)dsd =−Z 0 ( −s)δ−1 Γ(δ)σ(s)ds + δ−1 Γ(δ)Z1 0(1 −s)δ−1σ(s)ds +λ δ−1R1 0h( ) δ−1d Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0(1 −s)δ−1σ(s)ds −λ δ−1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0h( )·Z 0( −s)δ−1σ(s)dsd =−Z 0 ( −s)δ−1 Γ(δ)σ(s)ds + δ−1 Γ(δ)Z 0(1 −s)δ−1σ(s)ds + δ−1 Γ(δ)Z1 (1 −s)δ−1σ(s)ds +λ δ−1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0h( ) δ−1d ·Z1 0(1 −s)δ−1σ(s)ds −λ δ−1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0h( )·Z 0( −s)δ−1σ(s)dsd =−Z 0 ( −s)δ−1 Γ(δ)σ(s)ds + δ−1 Γ(δ)Z 0(1 −s)δ−1σ(s)ds + δ−1 Γ(δ)Z1 (1 −s)δ−1σ(s)ds +λ δ−1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0h( )·Z1 0 δ−1(1 −s)δ−1σ(s)dsd −λ δ−1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0h( )·Z 0( −s)δ−1σ(s)dsd F ac ional Bounda y Value P oblems 5 =−Z 0 ( −s)δ−1 Γ(δ)σ(s)ds + δ−1 Γ(δ)Z 0(1 −s)δ−1σ(s)ds + δ−1 Γ(δ)Z1 (1 −s)δ−1σ(s)ds +λ δ−1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0h( )·Z 0 δ−1(1 −s)δ−1σ(s)dsd −λ δ−1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0h( )·Z 0( −s)δ−1σ(s)dsd +λ δ−1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0h( )·Z1 δ−1(1 −s)δ−1σ(s)dsd =Z1 0G1( , s)σ(s)ds +λ δ−1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0h( )·Z1 0G1( , s)σ(s)dsd =Z1 0G1( , s)σ(s)ds +Z1 0 λ δ−1 Γ(δ)(1 −λR1 0h( ) δ−1d )Z1 0h( )G1( , s)d σ(s)ds =Z1 0G1( , s)σ(s)ds +Z1 0G2( , s)σ(s)ds. u As a di ec consequence o he p e ious esul , we deduce he ollowing p ope ies. Lemma 2.4 The unc ion G1( , s)de ined in Lemma 2.3 has he ollowing p ope ies: 1. G1( , s)∈ C ([0,1] ×[0,1]). 2. G1( , s)>0 o ( , s)∈(0,1) ×(0,1) and G1(0, s) = 0 = G1(1, s) o s∈[0,1]. 3. G1( , s) = G1(1 −s, 1− ),∀ , s ∈[0,1]. In nex esul , we deduce wo inequali ies ha , as we will see, will be undamen al o ensu e he exis ence o he solu ions o he nonlinea p oblem (1)-(2). Lemma 2.5 Le he unc ion G1( , s)be de ined in Lemma 2.3 and ix 0∈(0,1), hen G1sa is ies he ollowing inequali ies: G1( , s)≤sδ−1(1 −s)δ−1 Γ(δ),∀ ∈[0,1], s ∈[0,1] (7) and sδ−1(1 −s)δ−1k( , 0)≤G1( , s),∀ ∈[0,1], s ∈[ 0,1],(8) wi h k( , 0) :=          δ−1 Γ(δ)i 0≤ ≤ 0<1 min  δ−1 Γ(δ), δ−1(1− 0)δ−1−( − 0)δ−1 Γ(δ) δ−1 0(1− 0)δ−1i 0< 0< ≤1 . F ac ional Bounda y Value P oblems 6 P oo . Fo s> ,∂G1 ∂ ( , s) = δ−1 Γ(δ)(1 −s)δ−1 δ−2>0. Fo s < , since 1 < δ ≤2, we ha e ∂G1 ∂ ( , s) = δ−1 Γ(δ)(1 −s)δ−1 δ−2−( −s)δ−2≤δ−1 Γ(δ) δ−2−( −s)δ−2<0. As a consequence, i is ul illed ha G1( , s)≤G1(s, s) = sδ−1(1 −s)δ−1 Γ(δ)∀ , s ∈[0,1] and inequali y (7) holds. By using he hi d p ope y on Lemma 2.4, we deduce ha ∂G1 ∂s ( , s)>0 o 0 ≤s< ≤1 and ∂G1 ∂s ( , s)<0 i 0 ≤ <s≤1. Now, we in oduce he ollowing unc ion: F1( , s) = G1( , s) sδ−1(1 −s)δ−1,( , s)∈[0,1] ×(0,1), as a di ec consequence o p e ious a gumen s, we deduce ha ∂F1 ∂ ( , s)<0 o 0 ≤s< ≤1 and ∂F1 ∂ ( , s)>0 i 0 ≤ < s ≤1. As a consequence, we ha e ha G1( , s) sδ−1(1 −s)δ−1≤G1(s, s) sδ−1(1 −s)δ−1=1 Γ(δ). By he o he hand, ∂F1 ∂s ( , s) =          − δ−1 Γ(δ−1) sδ0≤ <s≤1, (δ−1)( (s2−2s + )( −s)δ−( −s)2 δ(1−s)δ) Γ(δ)( −s)2(1−s)δsδ0≤s< ≤1. F ac ional Bounda y Value P oblems 7 As a di ec consequence, we deduce ha ∂F1 ∂s ( , s)<0 o 0 ≤ <s≤1. On he o he hand, o he case 0 ≤s < ≤1 we ha e ha ∂F1 ∂s ( , s)>0 i and only i h1( , s, δ) := (1 −s)δ δ−1( −s)2−δ< s2−2s + =: h2( , s). Now, since ∂h1 ∂δ ( , s, δ) = (1 −s)δ δ−1( −s)2−δlog  − s −s, we ha e ha h1is s ic ly inc easing on he δin e al [1,2] o any 0 ≤s< ≤1 gi en. Thus, since h2( , s)−h1( , s, 2) = (1 − )s2>0, we conclude ha ∂F1 ∂s ( , s)>0 o all 0 <s< <1. So, o any 0∈(0,1) ixed, we ha e ha G1( , s) sδ−1(1 −s)δ−1≥min (lim s→1− G1( , s) s(1 −s)δ−1,G1( , 0) δ−1 0(1 − 0)δ−1) = min ( δ−1 Γ(δ),G1( , 0) δ−1 0(1 − 0)δ−1)=: k( , 0),∀ ∈[0,1], s ∈[ 0,1], and he esul is concluded. u By i ue o his lemma, we can gi e now he main esul o his sec ion. Lemma 2.6 Le 0∈(0,1) be ixed and hin oduced a he bounda y condi ions (2). Deno e by A=R1 0h( ) δ−1d ,B=R1 0h( )d and C0=R1 0k( , 0)h( )d . Assume ha h≥0on [0,1] and 1−λ A > 0. Then he G een’s unc ion G( , s)de ined in Lemma 2.3 sa is ies he inequali ies λ C0 δ−1 1−λA sδ−1(1 −s)δ−1≤G( , s)≤1 Γ(δ) 1 + λB 1−λA!sδ−1(1 −s)δ−1,∀ , s ∈[0,1]. (9) P oo . F om he de ini ion o G, he inequali y (7) and he ac ha 1 < δ ≤2, we ha e he ollowing inequali ies o all , s ∈[0,1]: G( , s)≤1 Γ(δ)sδ−1(1 −s)δ−1+λ δ−1 1−λA Z1 0 1 Γ(δ)sδ−1(1 −s)δ−1h( )d ≤1 Γ(δ) 1 + λB 1−λA!sδ−1(1 −s)δ−1. F ac ional Bounda y Value P oblems 8 On he o he hand, by Lemma 2.4 (2) and (8), we ha e o all , s ∈[0,1]: G( , s) = G1( , s) + G2( , s) ≥λ δ−1 1−λR1 0h( ) δ−1d Z1 0 h( )G1( , s)d ≥λ δ−1 1−λA C0sδ−1(1 −s)δ−1, as we wan o p o e. u As a di ec consequence, we deduce he ollowing Co olla y: Co olla y 2.1 I h≥0on [0,1] and 1−λ A > 0 hen he G een’s unc ion G( , s) de ined in Lemma 2.3 sa is ies he inequali ies λ 1−λA C0sδ−1(1−s)δ−1≤ 2−δG( , s)≤1 Γ(δ) 1 + λB 1−λA!sδ−1(1−s)δ−1,∀ , s ∈[0,1] 3 Main Resul s Now o any u: (0,1] →R, we de ine unc ion u: [0,1] →Ras ollows: u( ) = ( 2−δu( ) i ∈(0,1], lim →0+ 2−δu( ) i = 0, p o ided ha such limi exis s. Conside he Banach space E=Cδ[0,1] := {¯u: [0,1] →R,is a con inuous unc ion in [0,1]} endowed wi h he maximum no m kuk= max 0≤ ≤1|¯u( )|and de ine he cone P0⊂Eby P0={u∈E, ¯u( )≥ 2−δp( , 0)kuk, o all ∈[0,1]}, whe e p( , 0) = Γ(δ)λ δ−1 1−λA C0, 1 + λB 1−λA!, ∈[0,1], wi h 0∈(0,1) ixed, and A,Band C0in oduced in Lemma 2.6. No ice ha , p o ided ha h≥0 on [0,1] and 1 −λ A > 0, we deduce om (9) ha 0≤p( , 0)≤1 o all ∈[0,1] and 0∈(0,1). Now, we assume he ollowing hypo hesis on he nonlinea pa o he equa ion: (H1) Func ion : [0,1] ×R→[0,∞) is con inuous. F ac ional Bounda y Value P oblems 9 So, we de ine he ope a o T:P0→Eby (Tu)( ) = Z1 0G( , s) (s, ¯u(s))ds, ∈[0,1] (10) Lemma 3.1 T:P0→P0is comple ely con inuous. P oo : Le us p o e in i s ha T(P0)⊂P0. No ice om he de ini ion o Tand Co olla y 2.1 ha o u∈P0, Tu( )≥0 o all ∈[0,1] and 2−δ(Tu)( ) = Z1 0 2−δG( , s) (s, ¯u(s))ds ≥Z1 0 2−δλ δ−1 1−λA C0sδ−1(1 −s)δ−1 (s, ¯u(s))ds = 2−δΓ(δ)λ δ−1 1−λA C0 1 + λB 1−λA Z1 01 + λB 1−λA  Γ(δ)sδ−1(1 −s)δ−1 (s, ¯u(s))ds ≥ 2−δp( , 0)Z1 0max 0≤ ≤1n 2−δG( , s)o (s, ¯u(s))ds ≥ 2−δp( , 0)max 0≤ ≤1Z1 0 2−δG( , s) (s, ¯u(s))ds = 2−δp( , 0)kTuk. Thus, T(P0)⊂P0. In addi ion, since is a con inuous unc ion i ollows ha Tis a con inuous ope a o . Nex , we show ha Tis uni o mly bounded. Le D⊂Pbe a bounded se , i.e. he e exis s a cons an L > 0 such ha kuk ≤ L, o all u∈D. Se M= max 0≤s≤1,0≤u≤L{ (s, ¯u(s))}. Then, om Lemma 2.6, and o all u∈D, we ha e | 2−δTu( )|=Z1 0 2−δG( , s) (s, ¯u(s))ds ≤M Γ(δ) 1 + λB 1−λA!Z1 0sδ−1(1 −s)δ−1ds =M 1 + λB 1−λA!Γ(δ) Γ(2δ). Hence, T(D) is bounded. Finally, we show ha Tis equicon inuous, as ollows. Fo all  > 0 and o each u∈P, le 1, 2∈[0,1], be such ha 1< 2. F ac ional Bounda y Value P oblems 16 Re e ences [1] Z. 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