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
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