Hindawi Publishing Co po a ion
Ma hema ical P oblems in Enginee ing
Volume 2013, A icle ID 498781, 21 pages
h p://dx.doi.o g/10.1155/2013/498781
Resea ch A icle
Exis ence Resul s o a Coupled Sys em o Nonlinea Singula
F ac ional Di e en ial Equa ions wi h Impulse E ec s
Yuji Liu,1Juan J. Nie o,2,3 and Ósca O e o-Za aquiños2
1Depa men o Ma hema ics, Guangdong Uni e si y o Business S udies, Guangzhou 510320, China
2Depa amen o de An´
alisis Ma em´
a ico, Facul ad de Ma em´
a icas, Uni e sidad de San iago de Compos ela,
15782 San iago de Compos ela, Spain
3Depa men o Ma hema ics, Facul y o Science, King Abdulaziz Uni e si y, P.O. Box 80203, Jeddah 21589, Saudi A abia
Co espondence should be add essed o Juan J. Nie o; juanjose.nie o[email p o ec ed]
Recei ed 2 Oc obe 2012; Accep ed 15 Feb ua y 2013
Academic Edi o : Jocelyn Saba ie
Copy igh © 2013 Yuji Liu e al. This is an open access a icle dis ibu ed unde he C ea i e Commons A ibu ion License, which
pe mi s un es ic ed use, dis ibu ion, and ep oduc ion in any medium, p o ided he o iginal wo k is p ope ly ci ed.
A bounda y alue p oblem o he singula ac ional di e en ial sys em wi h impulse e ec s is p esen ed. By applying Schaude ’s
ixed poin heo em in a sui ably Banach space, we ob ain he exis ence o a leas one solu ion o his p oblem. Two examples a e
p esen ed o illus a e he main heo em.
1. In oduc ion
F ac ional di e en ial equa ions ha e ecei ed inc easing
a en ion du ing ecen yea s since he beha io o many
physical, chemical, and enginee ing p ocesses can be p ope ly
desc ibed by using ac ional di e en ial equa ions heo y;
see he books [1–3], pape s [4,5] and e e ences he ein. Fo
de ails on he geome ic and physical in e p e a ion o he
de i a i es o nonin ege o de , see, o example, [6–11]. Fo
some ecen wo ks wi h applica ions o enginee ing we e e
he eade o [12–15].
Fo an in oduc ion o he basic heo y o impulsi e
di e en ial equa ion, we e e he eade o [16]. Among
p e ious esea ch, li le is conce ned wi h di e en ial equa-
ions wi h ac ional o de wi h impulses [17]. Ahmad and
Si asunda am [18,19] ga e some exis ence esul s o wo-
poin bounda y alue p oblems in ol ing nonlinea impul-
si e hyb id di e en ial equa ions o ac ional o de 1<
𝛼≤2. Ahmad and Nie o in [20] es ablish su icien
condi ions o he exis ence o solu ions o he an ipe iodic
bounda y alue p oblem o impulsi e di e en ial equa ions
wi h he Capu o de i a i e o o de 𝑞 ∈ (1,2].Some ecen
esul s on impulsi e ini ial alue p oblems o bounda y alue
p oblems o ac ional di e en ial equa ions on a ini e
in e alcanbe oundin[21–23] and e e ences he ein. The
memo y p ope y o ac ional calculus makes s udies mo e
complica ed.
This pape is mo i a ed by [24]inwhich he ollowing
bounda y alue p oblem o he ac ional di e en ial equa-
ion 𝐷𝛼
0+𝑥(𝑡)=𝑓(𝑡,𝑦(𝑡),𝐷𝑝
0+𝑦(𝑡)), 𝑡∈(0,1),
𝐷𝛽
0+𝑦(𝑡)=𝑔(𝑡,𝑥(𝑡),𝐷𝑞
0+𝑥(𝑡)), 𝑡∈(0,1),
𝑥(0)=0, 𝑦(0)=0, 𝑥(1)−𝛾𝑥(𝜂)=0,
𝑦(1)−𝛾𝑦(𝜂)=0
(1)
was s udied, whe e 1<𝛼,𝛽<2,0<𝑝≤𝛽−1and 0<
𝑞≤𝛼−1,𝛾>0,1>𝛾𝜂
𝛼−1,1>𝛾𝜂
𝛽−1 and 𝑓,𝑔 : [0,1]×
𝑅2→𝑅a e con inuous unc ions, and 𝐷0+is he Riemann-
Liou ille ac ional de i a i e. An exis ence esul was p o ed
o BVP (1)in[24]. The g ow h assump ions imposed on 𝑓
and 𝑔a e sublinea cases (see [25, Theo em 3.1]); ha is, he e
exis unc ions 𝑎,𝑏∈𝐿1(0,1), nonnega i e cons an s 𝜖1,𝜖2>
0,𝛿1,𝛿2≥0and 𝜌1,𝜌2,𝜎1,𝜎2∈(0,1)such ha
󵄨󵄨󵄨󵄨𝑓(𝑡,𝑥,𝑦)󵄨󵄨󵄨󵄨≤𝑎(𝑡)+𝜖1|𝑥|𝜌1+𝜖2󵄨󵄨󵄨󵄨𝑦󵄨󵄨󵄨󵄨𝜌2,
󵄨󵄨󵄨󵄨𝑔(𝑡,𝑥,𝑦)󵄨󵄨󵄨󵄨≤𝑏(𝑡)+𝛿1|𝑥|𝜎1+𝛿2󵄨󵄨󵄨󵄨𝑦󵄨󵄨󵄨󵄨𝜎2.(2)
2Ma hema ical P oblems in Enginee ing
In [25], he ollowing bounda y alue p oblem o he
ac ional di e en ial equa ion
𝐷𝛼
0+𝑥(𝑡)=𝑓(𝑡,𝑦(𝑡),𝐷𝑝
0+𝑦(𝑡)), 𝑡∈(0,1),
𝐷𝛽
0+𝑦(𝑡)=𝑔(𝑡,𝑥(𝑡),𝐷𝑞
0+𝑥(𝑡)), 𝑡∈(0,1),
𝑥(0)=0, 𝑦(0)=0, 𝑥(1)=0, 𝑦(1)=0
(3)
was s udied, whe e 1<𝛼,𝛽<2,0<𝑝≤𝛽−1and 0<𝑞≤
𝛼−1,and𝑓,𝑔 : [0,1]×𝑅2→𝑅a e con inuous unc ions,
and 𝐷0+is he Riemann-Liou ille ac ional de i a i e. The
g ow h assump ions imposed on 𝑓and 𝑔a e sublinea cases
(see [25, Theo em 3.1]), ha is, he e exis unc ions 𝑎,𝑏 ∈
𝐿1(0,1), nonnega i e cons an s 𝜖1,𝜖2>0,𝛿1,𝛿2≥0,and
𝜌1,𝜌2,𝜎1,𝜎2∈(0,1]such ha
󵄨󵄨󵄨󵄨𝑓(𝑡,𝑥,𝑦)󵄨󵄨󵄨󵄨≤𝑎(𝑡)+𝜖1|𝑥|𝜌1+𝜖2󵄨󵄨󵄨󵄨𝑦󵄨󵄨󵄨󵄨𝜌2,
󵄨󵄨󵄨󵄨𝑔(𝑡,𝑥,𝑦)󵄨󵄨󵄨󵄨≤𝑏(𝑡)+𝛿1|𝑥|𝜎1+𝛿2󵄨󵄨󵄨󵄨𝑦󵄨󵄨󵄨󵄨𝜎2,(4)
o sublinea cases, ha is, he e exis nonnega i e cons an s
𝜖1,𝜖2>0,𝛿1,𝛿2≥0and 𝜌1,𝜌2,𝜎1,𝜎2∈(1,∞)such ha
󵄨󵄨󵄨󵄨𝑓(𝑡,𝑥,𝑦)󵄨󵄨󵄨󵄨≤𝜖1|𝑥|𝜌1+𝜖2󵄨󵄨󵄨󵄨𝑦󵄨󵄨󵄨󵄨𝜌2,
󵄨󵄨󵄨󵄨𝑔(𝑡,𝑥,𝑦)󵄨󵄨󵄨󵄨≤𝛿1|𝑥|𝜎1+𝛿2󵄨󵄨󵄨󵄨𝑦󵄨󵄨󵄨󵄨𝜎2.(5)
We ind ha in he supe linea cases, BVP (3)hasapai o
solu ions (𝑥,𝑦)=(0,0)wi hou needing any o he assump-
ions. Hence, hese cases a e i ial ones discussed in [25].
I is in e es ing o conside he sol abili y o BVP (1)when
he g ow h assump ions imposed on 𝑓,𝑔 a e supe linea
cases. Fu he mo e, he sol abili y o BVP (1)isno s udied
when 𝑞>𝛼−1o 𝑝>𝛽−1.
In his pape we conside he ollowing nonlinea bound-
a y alue p oblem o he singula mul i e m ac ional
di e en ial equa ion wi h impulse e ec s whose bounda y
condi ions a e o in eg al o m
𝐷𝛼
0+𝑥(𝑡)=𝜙(𝑡)𝑓(𝑡,𝑦(𝑡),𝐷𝑝
0+𝑦(𝑡)),
𝑡∈(0,1),𝑡=𝑡1,
𝐷𝛽
0+𝑦(𝑡)=𝜓(𝑡)𝑔(𝑡,𝑥(𝑡),𝐷𝑞
0+𝑥(𝑡)),
𝑡∈(0,1),𝑡=𝑡1,
lim
𝑡→0𝑡2−𝛼𝑥(𝑡)=∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠,
lim
𝑡→0𝑡2−𝛽𝑦(𝑡)=∫1
0
V(𝑠)𝐻(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠,
𝑥(1)=∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠,
(6)
𝑦(1)=∫1
0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠,
Δ𝑥(𝑡1)=lim
𝑡→𝑡+
1𝑥(𝑡)−lim
𝑡→𝑡−
1𝑥(𝑡)=𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)),
Δ𝑦(𝑡1)=lim
𝑡→𝑡+
1𝑦(𝑡)−lim
𝑡→𝑡−
1𝑦(𝑡)=𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1)),
Δ𝐷𝑞
0+𝑥(𝑡1)=lim
𝑡→𝑡+
1𝐷𝑞
0+𝑥(𝑡)−lim
𝑡→𝑡−
1𝐷𝑞
0+𝑥(𝑡)
=𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)),
Δ𝐷𝑝
0+𝑦(𝑡1)=lim
𝑡→𝑡+
1𝐷𝑝
0+𝑦(𝑡)−lim
𝑡→𝑡−
1𝐷𝑝
0+𝑦(𝑡)
=𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1)), (7)
whe e
(a) 1<𝛼,𝛽≤2,0<𝑝<𝛽and 0<𝑞<𝛼,𝐷0+is he
Riemann-Liou ille ac ional de i a i e,
(b) 𝜙,𝜓:(0,1) → 𝑅,𝑓,𝑔de ined on (0,1)×𝑅2,
(c) 𝑚,𝑛,𝑢,V: (0,1) → 𝑅wi h 𝑚,𝑛,𝑢,V∈𝐿
1(0,1),
𝐺,𝐻,𝑀,𝑁de ined on (0,1)×𝑅2,
(d) 0=𝑡0<𝑡1<𝑡2=1,
(e) 𝐼,𝐼1,𝐽,𝐽1:(0,1)×𝑅2→𝑅.
Apai o unc ions(𝑥,𝑦)de ined on (0,1)is called a
solu ion o BVP (1)andBVP(3), i 𝑥|(𝑡𝑘,𝑡𝑘+1],𝐷𝑞
0+𝑥|(𝑡𝑘,𝑡𝑘+1]and
𝑦|(𝑡𝑘,𝑡𝑘+1],𝐷𝑝
0+𝑦|(𝑡𝑘,𝑡𝑘+1](𝑘 = 0,1)a e con inuous, he e exis s
he limi s
lim
𝑡→𝑡+
𝑘𝑡2−𝛼𝑥(𝑡),lim
𝑡→𝑡+
𝑘𝑡2−𝛽𝑦(𝑡),
lim
𝑡→𝑡+
𝑘𝑡2+𝑞−𝛼𝐷𝑞
0+𝑥(𝑡),lim
𝑡→𝑡+
𝑘𝑡2+𝑝−𝛽𝐷𝑝
0+𝑦(𝑡),
𝑘=0,1,
(8)
𝐷𝛼
0+𝑥,𝐷𝛽
0+𝑦∈𝐿1(0,1)and (𝑥,𝑦)sa is ies all equa ions in (6)
and (7).
The no el y o his pape is as ollows: i s , he ac ional
di e en ial equa ions in (6) a e mul i e m ones and hei
nonlinea i ies 𝑓,𝑔depend on he lowe ac ional de i a i es;
second, bo h 𝜙and 𝜓may be singula a 𝑡=0and
𝑡=1, ha is,𝜙(𝑡)𝑓(𝑡,𝑥,𝑦)and 𝜓(𝑡)𝑔(𝑡,𝑥,𝑦)may be no
con inuous unc ions on [0,1]×𝑅2, hebounda ycondi ions
a e in eg al bounda y condi ions, and we ob ain he esul s
on he exis ence o a leas one solu ion o BVP (6)-(7); hi d,
0<𝑝<𝛽and 0<𝑞<𝛼a esupposed; heg ow h
assump ions imposed on 𝑓,𝑔,𝐺,𝐻,𝑀,𝑁 and 𝐼,𝐼1,𝐽,𝐽1a e
allowed o be sublinea cases. Finally, wo examples a e gi en
o illus a e he e iciency o he main heo em.
The emainde o his pape is as ollows: in Sec ion 2,we
p esen p elimina y esul s. In Sec ion 3, hemain heo em
and i s p oo a e gi en. In Sec ion 4, wo examples a e gi en
o illus a e he main esul s.
Ma hema ical P oblems in Enginee ing 3
2. P elimina ies
In his sec ion, we p esen some backg ound de ini ions and
p elimina y esul s.
De ini ion 1 (see [1]). The Riemann-Liou ille ac ional in e-
g al o o de 𝛼>0o a unc ion 𝑔:(0,∞)→𝑅is gi en
by
𝐼𝛼
0+𝑔(𝑡)=1
Γ(𝛼)∫𝑡
0(𝑡−𝑠)𝛼−1𝑔(𝑠)𝑑𝑠, (9)
p o ided ha he igh -hand side exis s.
De ini ion 2 (see [1]). The Riemann-Liou ille ac ional
de i a i e o o de 𝛼>0o a con inuous unc ion 𝑔:
(0,∞) → 𝑅is gi en by
𝐷𝛼
0+𝑔(𝑡)=1
Γ(𝑛−𝛼)𝑑𝑛
𝑑𝑡𝑛∫𝑡
0𝑔(𝑠)
(𝑡−𝑠)𝛼−𝑛+1 𝑑𝑠, (10)
whe e 𝑛−1≤𝛼<𝑛, p o ided ha he igh -hand side is
poin wise de ined on (0,∞).
De ini ion 3. 𝐾:(0,1)×𝑅2→𝑅is called a 𝛽-Ca a heodo y
unc ion i 𝐾sa is ies ha
(i) 𝑡→𝐾(𝑡,𝑡
𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉) is con inuous on
(𝑡𝑘,𝑡𝑘+1](𝑘=0,1) o e e y (𝑈,𝑉)∈𝑅2;
(ii) (𝑈,𝑉) → 𝐾(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)is con inuous on 𝑅2
o e e y 𝑡∈(0,1);
(iii) o each 𝑟>0 he e exis s a cons an 𝐴𝑟>0such ha
|𝐾(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)|≤𝐴𝑟,𝑡∈(0,1),|𝑈|,|𝑉|≤𝑟.
De ini ion 4. 𝑄:(0,1)×𝑅2→𝑅is called a 𝛼-Ca a heodo y
unc ion i 𝑄sa is ies ha
(i) 𝑡→𝑄(𝑡,𝑡
𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉) is con inuous on
(𝑡𝑘,𝑡𝑘+1](𝑘=0,1) o e e y (𝑈,𝑉)∈𝑅2;
(ii) (𝑈,𝑉) → 𝑄(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)is con inuous on 𝑅2
o e e y 𝑡∈(0,1);
(iii) o each 𝑟>0 he e exis s a cons an 𝐵𝑟>0such ha
|𝑄(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)|≤𝐵𝑟,𝑡∈(0,1),|𝑈|,|𝑉|≤𝑟.
Lemma 5 ( he Le ay-Schaude nonlinea al e na i e [23]).
Le 𝑋be a Banach space and 𝑇:𝑋→𝑋be a comple ely
con inuous ope a o . Suppose Ωis a nonemp y open subse o 𝑋
cen e ed a ze o. Then ei he he e exis s 𝑥∈𝜕Ωand 𝜆∈(0,1)
such ha 𝑥=𝜆𝑇𝑥o he e exis s 𝑥∈Ωsuch ha 𝑥=𝑇𝑥.
Le he gamma and be a unc ions Γ(𝛼)and B(𝑝,𝑞)be
de ined by
Γ(𝛼)=∫+∞
0𝑥𝛼−1𝑒−𝑥𝑑𝑥,
B(𝑝,𝑞)=∫1
0𝑥𝑝−1(1−𝑥)𝑞−1𝑑𝑥,
‖𝑚‖1=∫1
0|𝑚(𝑠)|𝑑𝑠 o 𝑚∈𝐿1(0,1).
(11)
Choose
𝑋
=
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
𝑥|(𝑡𝑘,𝑡𝑘+1]∈𝐶0(𝑡𝑘,𝑡𝑘+1](𝑘=0,1),
𝐷𝑞
0+𝑥|(𝑡𝑘,𝑡𝑘+1]∈𝐶0(𝑡𝑘,𝑡𝑘+1](𝑘=0,1),
𝑥:(0,1]󳨀→ 𝑅 he e exis he limi s
lim
𝑡→𝑡+
𝑘𝑡2−𝛼𝑥(𝑡),
lim
𝑡→𝑡+
𝑘𝑡2+𝑞−𝛼𝐷𝑞
0+𝑥(𝑡)
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
,
𝑌
=
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
{
𝑦|(𝑡𝑘,𝑡𝑘+1]∈𝐶0(𝑡𝑘,𝑡𝑘+1](𝑘=0,1),
𝐷𝑝
0+𝑥|(𝑡𝑘,𝑡𝑘+1]∈𝐶0(𝑡𝑘,𝑡𝑘+1](𝑘=0,1),
𝑦:(0,1]󳨀→ 𝑅 he e exis he limi s
lim
𝑡→𝑡+
𝑘𝑡2−𝛽𝑦(𝑡),
lim
𝑡→𝑡+
𝑘𝑡2+𝑝−𝛽𝐷𝑝
0+𝑦(𝑡)
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
}
.
(12)
Fo 𝑥∈𝑋, de ine he no m by
‖𝑥‖=‖𝑥‖𝑋
=max {sup
𝑡∈(0,1)𝑡2−𝛼 |𝑥(𝑡)|,sup
𝑡∈(0,1)𝑡2+𝑞−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞
0+𝑥(𝑡)󵄨󵄨󵄨󵄨󵄨}. (13)
I is easy o show ha 𝑋is a eal Banach space. Fo 𝑦∈𝑌,
de ine he no m by
󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩=󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩𝑌
=max {sup
𝑡∈(0,1)𝑡2−𝛽 󵄨󵄨󵄨󵄨𝑦(𝑡)󵄨󵄨󵄨󵄨,sup
𝑡∈(0,1)𝑡2+𝑝−𝛽 󵄨󵄨󵄨󵄨󵄨𝐷𝑝
0+𝑦(𝑡)󵄨󵄨󵄨󵄨󵄨}. (14)
I is easy o show ha 𝑌is a eal Banach space. Thus, (𝑋×
𝑌,||⋅||)is a Banach space wi h he no m de ined by ||(𝑥,𝑦)||=
max{||𝑥||𝑋,||𝑦||𝑌} o (𝑥,𝑦)∈𝑋×𝑌.
In his pape , we suppose he ollowing:
(A) 𝜙sa is ies ha he e exis cons an s 𝐿1>0,𝑘>−1,
𝛿∈(𝑞−𝛼,0]such ha 𝛼+2𝛿−𝑞>0,𝛼+𝑘+𝛿−𝑞≥0,
and |𝜙(𝑡)|≤𝐿1𝑡𝑘(1−𝑡)𝛿 o all 𝑡∈(0,1);𝜓sa is ies
ha he e exis cons an s 𝐿2>0,𝑙>−1,𝜃∈(𝑝−𝛽,0]
such ha 𝛽+2𝜃−𝑝>0,𝛽+𝑙+𝜃−𝑝≥0,and
|𝜓(𝑡)|≤𝐿2𝑡𝑙(1−𝑡)𝜃 o all 𝑡∈(0,1).
(B) 𝑓,𝐺,𝑀,𝐼,𝐼1a e 𝛽-Ca a heodo y unc ions and 𝑔,𝐻,
𝑁,𝐽,𝐽1a e 𝛼-Ca a heodo y unc ions.
Rema k 6. Suppose ha 𝑓is a 𝛽-Ca a heodo y unc ion. Fo
example, 𝛼=7/4,𝑞=1/8,choose𝑘=−1/2,𝛿=−3/4
and 𝜙(𝑡) = 𝑡𝑘(1−𝑡)𝛿, hen𝑘>−1,𝛿 ∈ (−𝛼,0]such ha
𝛼+2𝛿−𝑞>0,𝛼+𝑘+𝛿−𝑞≥0,and|𝜙(𝑡)|≤𝑡𝑘(1−𝑡)𝛿 o all
𝑡∈(0,1).I iseasy osee ha 𝜙is singula a 𝑡=0and 𝑡=1.
4Ma hema ical P oblems in Enginee ing
Lemma 7. Suppose ha 𝑦∈𝑌, and (a)–(e), (A)-(B) hold.
Then 𝑥∈𝑋is a solu ion o
𝐷𝛼
0+𝑥(𝑡)=𝜙(𝑡)𝑓(𝑡,𝑦(𝑡),𝐷𝑝
0+𝑦(𝑡)), 𝑡∈(0,1),𝑡=𝑡1,
lim
𝑡→0𝑡2−𝛼𝑥(𝑡)=∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠,
𝑥(1)=∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠,
Δ𝑥(𝑡1)=𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)),
Δ𝐷𝑞
0+𝑥(𝑡1)=𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)), (15)
i and only i 𝑥∈𝑋sa is ies he in eg al equa ion
𝑥(𝑡)=
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∫𝑡
0(𝑡−𝑠)𝛼−1
Γ(𝛼)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝
0+𝑦(𝑢))𝑑𝑢
−𝑡𝛼−1
Γ(𝛼)
×∫1
0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−2 ∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−1 ∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−1
Π
×( Γ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1−Γ(𝛼−1)
Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2
1)
×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
+𝑡𝛼−1 (𝑡𝛼−2
1−𝑡𝛼−1
1)
Π
×𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)), 𝑡∈(0,𝑡1],
∫𝑡
0(𝑡−𝑠)𝛼−1
Γ(𝛼)𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
−𝑡𝛼−1
Γ(𝛼)
×∫1
0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+(𝑡𝛼−2 −𝑡𝛼−1)
×∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−1 ∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−1 −𝑡𝛼−2
ΠΓ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1
×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
+𝑡𝛼−2 −𝑡𝛼−1
Π𝑡𝛼−1
1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)),
𝑡∈(𝑡1,1],
(16)
whe e
Π=( Γ(𝛼−1)
Γ(𝛼−𝑞−1)−Γ(𝛼)
Γ(𝛼−𝑞))𝑡2𝛼−𝑞−3
1.(17)
P oo . I 𝑦∈𝑌is a solu ion o BVP (15), hen
󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩=max {sup
𝑡∈(0,1)𝑡2−𝛽 󵄨󵄨󵄨󵄨𝑦(𝑡)󵄨󵄨󵄨󵄨,sup
𝑡∈(0,1)𝑡2+𝑝−𝛽 󵄨󵄨󵄨󵄨󵄨𝐷𝑝
0+𝑦(𝑡)󵄨󵄨󵄨󵄨󵄨}
=𝑟<+∞, (18)
and 𝑥sa is ies all equa ions in (31)F om(B),𝑓is a 𝛽-
Ca a heodo y unc ion, hen he e exis s 𝐴𝑟>0such ha
󵄨󵄨󵄨󵄨󵄨𝑓(𝑡,𝑦(𝑡),𝐷𝑝
0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨
=󵄨󵄨󵄨󵄨󵄨𝑓(𝑡,𝑡𝛽−2𝑡2−𝛽𝑦(𝑡),𝑡𝛽−𝑝−2𝑡2+𝑝−𝛽𝐷𝑝
0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟.
(19)
Simila ly we ge ha he e exis cons an s 𝐴󸀠
𝑟,𝐴󸀠󸀠
𝑟,𝐵󸀠
𝑟,𝐵󸀠󸀠
𝑟>0
such ha 󵄨󵄨󵄨󵄨󵄨𝐺(𝑡,𝑦(𝑡),𝐷𝑝
0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴󸀠
𝑟,
󵄨󵄨󵄨󵄨󵄨𝑀(𝑡,𝑦(𝑡),𝐷𝑝
0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴󸀠󸀠
𝑟,
𝑡∈(0,1),
󵄨󵄨󵄨󵄨󵄨𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))󵄨󵄨󵄨󵄨󵄨≤𝐵󸀠
𝑟,
󵄨󵄨󵄨󵄨󵄨𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))󵄨󵄨󵄨󵄨󵄨≤𝐵󸀠󸀠
𝑟.
(20)
I ollows om (15) ha , o 𝑡∈(𝑡𝑘,𝑡𝑘+1](𝑘=0,1), he e
exis cons an s 𝑐𝑘,𝑑𝑘∈𝑅such ha
𝑥(𝑡)=1
Γ(𝛼)∫𝑡
0(𝑡−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑐𝑘𝑡𝛼−1 +𝑑𝑘𝑡𝛼−2,𝑡∈(𝑡
𝑘,𝑡𝑘+1],𝑘=0,1. (21)
F om lim𝑡→0𝑡2−𝛼𝑥(𝑡) = ∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠,we
ge
𝑑0=∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠. (22)
F om 𝑥(1)=∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠,wege
1
Γ(𝛼)∫1
0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠+𝑐1+𝑑1
=∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠. (23)
F om Δ𝑥(𝑡1)=𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)),wege
(𝑐1−𝑐0)𝑡𝛼−1
1+(𝑑1−𝑑0)𝑡𝛼−2
1=𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)).
(24)
Ma hema ical P oblems in Enginee ing 5
F om Δ𝐷𝑞
0+𝑥(𝑡1)=𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)),wege
(𝑐1−𝑐0)Γ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1+(𝑑1−𝑑0)Γ(𝛼−1)
Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2
1
=𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)). (25)
I ollows ha
𝑐1−𝑐0=( Γ(𝛼−1)
Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2
1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
−𝑡𝛼−2
1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)))×(Π)−1,
𝑑1−𝑑0=(𝑡
𝛼−1
1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
−Γ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)))
×(Π)−1.(26)
Then
𝑑1=(𝑡
𝛼−1
1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
−Γ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)))×(Π)−1
+∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠. (27)
So
𝑐1=∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
−1
Γ(𝛼)∫1
0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
−(𝑡𝛼−1
1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
−Γ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)))×(Π)−1
−∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠,
𝑐0=∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
−1
Γ(𝛼)∫1
0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
−(𝑡𝛼−1
1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
−Γ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)))×(Π)−1
−∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
−( Γ(𝛼−1)
Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2
1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
−𝑡𝛼−2
1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)))×(Π)−1.
(28)
Hence, o 𝑡∈(0,𝑡1],weha e
𝑥(𝑡)=∫𝑡
0(𝑡−𝑠)𝛼−1
Γ(𝛼)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝
0+𝑦(𝑢))𝑑𝑢
−𝑡𝛼−1
Γ(𝛼)∫1
0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−2 ∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−1 ∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−1
Π(Γ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1−Γ(𝛼−1)
Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2
1)
×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
+𝑡𝛼−1 (𝑡𝛼−2
1−𝑡𝛼−1
1)
Π𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)). (29)
And o 𝑡∈(𝑡1,1],weha e
𝑥(𝑡)=∫𝑡
0(𝑡−𝑠)𝛼−1
Γ(𝛼)𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
−𝑡𝛼−1
Γ(𝛼)∫1
0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+(𝑡𝛼−2 −𝑡𝛼−1)∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−1 ∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−1 −𝑡𝛼−2
ΠΓ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1
×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
+𝑡𝛼−2 −𝑡𝛼−1
Π𝑡𝛼−1
1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)).
(30)
Hence, 𝑥∈𝑋sa is ies (16).
On he o he hand, i 𝑦∈𝑌and 𝑥∈𝑋is a solu ion o (16),
hen we can p o e ha 𝑥∈𝑋is a solu ion o BVP (6)-(7). The
p oo is comple ed.

6Ma hema ical P oblems in Enginee ing
Lemma 8. Suppose ha 𝑥∈𝑋, and (a)–(e), (A)-(B) hold.
Then 𝑦∈𝑌is a solu ion o
𝐷𝛽
0+𝑦(𝑡)=𝜓(𝑡)𝑔(𝑡,𝑥(𝑡),𝐷𝑞
0+𝑥(𝑡)), 𝑡∈(0,1),𝑡=𝑡1,
lim
𝑡→0𝑡2−𝛽𝑦(𝑡)=∫1
0
V(𝑠)𝐻(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠,
𝑦(1)=∫1
0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠,
Δ𝑦(𝑡1)=𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1)),
Δ𝐷𝑝
0+𝑦(𝑡1)=𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1)), (31)
i and only i 𝑦∈𝑌sa is ies he in eg al equa ion
𝑦(𝑡)=
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∫𝑡
0(𝑡−𝑠)𝛽−1
Γ(𝛽) 𝜓(𝑢)𝑔(𝑢,𝑥(𝑢),𝐷𝑞
0+𝑥(𝑢))𝑑𝑢
−𝑡𝛽−1
Γ(𝛽)
×∫1
0(1−𝑠)𝛽−1𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
+𝑡𝛽−2
×∫1
0
V(𝑠)𝐻(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛽−1
×∫1
0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
+𝑡𝛽−1
Ξ
×( Γ(𝛽)
Γ(𝛽−𝑝)𝑡𝛽−𝑝−1
1−Γ(𝛽−1)
Γ(𝛽−𝑝−2)𝑡𝛽−𝑝−2
1)
×𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1))
+𝑡𝛽−1 (𝑡𝛽−2
1−𝑡𝛽−1
1)
Ξ
×𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1)), 𝑡∈(0,𝑡1],
∫𝑡
0(𝑡−𝑠)𝛽−1
Γ(𝛽) 𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
−𝑡𝛽−1
Γ(𝛽)
×∫1
0(1−𝑠)𝛽−1𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
+(𝑡𝛽−2 −𝑡𝛽−1)
×∫1
0
V(𝑠)𝐻(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛽−1
×∫1
0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
+𝑡𝛽−1 −𝑡𝛽−2
ΞΓ(𝛽)
Γ(𝛽−𝑝)𝑡𝛽−𝑝−1
1
×𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1))
+𝑡𝛽−2 −𝑡𝛽−1
Ξ𝑡𝛽−1
1
×𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1)), 𝑡∈(𝑡1,1], (32)
whe e
Ξ=( Γ(𝛽−1)
Γ(𝛽−𝑝−1)−Γ(𝛽)
Γ(𝛽−𝑝))𝑡2𝛽−𝑝−3
1.(33)
P oo . The p oo is simila o ha o he p oo o Lemma 7
and is omi ed.
Now, we de ine he ope a o 𝑇on 𝑋×𝑌by 𝑇(𝑥,𝑦)(𝑡)=
((𝑇1𝑦)(𝑡),(𝑇2𝑥)(𝑡))wi h
(𝑇1𝑦)(𝑡)
=
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∫𝑡
0(𝑡−𝑠)𝛼−1
Γ(𝛼)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝
0+𝑦(𝑢))𝑑𝑢
−𝑡𝛼−1
Γ(𝛼)∫1
0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−2 ∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−1 ∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−1
Π(Γ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1−Γ(𝛼−1)
Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2
1)
×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
+𝑡𝛼−1 (𝑡𝛼−2
1−𝑡𝛼−1
1)
Π
×𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)), 𝑡∈(0,𝑡1],
∫𝑡
0(𝑡−𝑠)𝛼−1
Γ(𝛼)𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
−𝑡𝛼−1
Γ(𝛼)∫1
0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+(𝑡𝛼−2 −𝑡𝛼−1)∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−1 ∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−1 −𝑡𝛼−2
ΠΓ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
+𝑡𝛼−2 −𝑡𝛼−1
Π𝑡𝛼−1
1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)),
𝑡∈(𝑡1,1], (34)
Ma hema ical P oblems in Enginee ing 7
(𝑇2𝑥)(𝑡)
=
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∫𝑡
0(𝑡−𝑠)𝛽−1
Γ(𝛽) 𝜓(𝑢)𝑔(𝑢,𝑥(𝑢),𝐷𝑞
0+𝑥(𝑢))𝑑𝑢
−𝑡𝛽−1
Γ(𝛽)∫1
0(1−𝑠)𝛽−1𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
+𝑡𝛽−2 ∫1
0
V(𝑠)𝐻(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛽−1 ∫1
0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
+𝑡𝛽−1
Ξ(Γ(𝛽)
Γ(𝛽−𝑝)𝑡𝛽−𝑝−1
1−Γ(𝛽−1)
Γ(𝛽−𝑝−2)𝑡𝛽−𝑝−2
1)
×𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1))
+𝑡𝛽−1 (𝑡𝛽−2
1−𝑡𝛽−1
1)
Ξ𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1)),
𝑡∈(0,𝑡1],
∫𝑡
0(𝑡−𝑠)𝛽−1
Γ(𝛽) 𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
−𝑡𝛽−1
Γ(𝛽)∫1
0(1−𝑠)𝛽−1𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
+(𝑡𝛽−2 −𝑡𝛽−1)∫1
0
V(𝑠)𝐻(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛽−1 ∫1
0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
+𝑡𝛽−1 −𝑡𝛽−2
ΞΓ(𝛽)
Γ(𝛽−𝑝)𝑡𝛽−𝑝−1
1
×𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1))
+𝑡𝛽−2 −𝑡𝛽−1
Ξ𝑡𝛽−1
1𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1)),
𝑡∈(𝑡1,1]. (35)
Rema k 9. By Lemmas 7and 8,(𝑥,𝑦)∈𝑋×𝑌is a solu ion
o BVP (6)-(7)i andonlyi (𝑥,𝑦)∈𝑋×𝑌is a ixed poin o
he ope a o 𝑇.
Lemma 10. Suppose ha (a)–(e) and (A)-(B) hold. Then 𝑇:
𝑋×𝑌→𝑋×𝑌is well de ined and is comple ely con inuous.
P oo . The p oo is e y long, so we lis he s eps. Fi s , we
p o e ha 𝑇is well de ined; second, we p o e ha 𝑇is
con inuous, and, inally, we p o e ha 𝑇is compac . So 𝑇is
comple ely con inuous. Thus, he p oo is di ided in o h ee
s eps.
S ep 1. P o e ha 𝑇:𝑋×𝑌→𝑋×𝑌is well de ined.
Fo (𝑥,𝑦)∈𝑋×𝑌,weha e||(𝑥,𝑦)||=𝑟>0.Then
max {sup
𝑡∈(0,1)𝑡2−𝛼 |𝑥(𝑡)|,sup
𝑡∈(0,1)𝑡2+𝑞−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞
0+𝑥(𝑡)󵄨󵄨󵄨󵄨󵄨}≤𝑟<+∞,
max {sup
𝑡∈(0,1)𝑡2−𝛽 󵄨󵄨󵄨󵄨𝑦(𝑡)󵄨󵄨󵄨󵄨,sup
𝑡∈(0,1)𝑡2+𝑝−𝛽 󵄨󵄨󵄨󵄨󵄨𝐷𝑝
0+𝑦(𝑡)󵄨󵄨󵄨󵄨󵄨}≤𝑟<+∞.
(36)
F om (B), 𝑓,𝐺,𝑀,𝐼,𝐼1a e 𝛽-Ca a heodo y unc ions, hen
he e exis cons an s 𝐴𝑟>0such ha
󵄨󵄨󵄨󵄨󵄨𝑓(𝑡,𝑦(𝑡),𝐷𝑝
0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,𝑡∈
(0,1),
󵄨󵄨󵄨󵄨󵄨𝐺(𝑡,𝑦(𝑡),𝐷𝑝
0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,𝑡∈
(0,1),
󵄨󵄨󵄨󵄨󵄨𝑀(𝑡,𝑦(𝑡),𝐷𝑝
0+𝑦(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,𝑡∈
(0,1),
󵄨󵄨󵄨󵄨󵄨𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,𝑡∈
(0,1),
󵄨󵄨󵄨󵄨󵄨𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,𝑡∈
(0,1).
(37)
Hence,
󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨∫𝑡
0(𝑡−𝑠)𝛼−1
Γ(𝛼)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝
0+𝑦(𝑢))𝑑𝑢󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨
≤∫𝑡
0(𝑡−𝑠)𝛼−1
Γ(𝛼)󵄨󵄨󵄨󵄨󵄨𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝
0+𝑦(𝑢))󵄨󵄨󵄨󵄨󵄨𝑑𝑢
≤𝐴𝑟𝐿1
B(𝛼+𝛿,𝑘+1)
Γ(𝛼)<∞.
(38)
F om (34), (37), and (38), we see ha (𝑇1𝑦)(𝑡)is de ined on
(0,1],con inuouson(0,𝑡1]and (𝑡1,1], espec i ely.Onesees
ha
lim
𝑡→0𝑡2−𝛼 (𝑇1𝑦)(𝑡)
=lim
𝑡→0[𝑡2−𝛼 ∫𝑡
0(𝑡−𝑠)𝛼−1
Γ(𝛼)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝
0+𝑦(𝑢))𝑑𝑢
−𝑡
Γ(𝛼)∫1
0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡
Π(Γ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1−Γ(𝛼−1)
Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2
1)
×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
+𝑡(𝑡𝛼−2
1−𝑡𝛼−1
1)
Π𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))]
=∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠, (39)
and he eexi s helimi lim
𝑡→𝑡+
1(𝑇1𝑦)(𝑡).
8Ma hema ical P oblems in Enginee ing
On he o he hand, we ha e
𝐷𝑞
0+(𝑇1𝑦)(𝑡)
=
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∫𝑡
0(𝑡−𝑠)𝛼−𝑞−1
Γ(𝛼−𝑞) 𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝
0+𝑦(𝑢))𝑑𝑢
−𝑡𝛼−𝑞−1
Γ(𝛼−𝑞)
×∫1
0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−𝑞−2 Γ(𝛼−1)
Γ(𝛼−𝑞−1)
×∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−𝑞−1 Γ(𝛼)
Γ(𝛼−𝑞)
×∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−𝑞−1
ΠΓ(𝛼)
Γ(𝛼−𝑞)
×( Γ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1−Γ(𝛼−1)
Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2
1)
×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
+Γ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1 (𝑡𝛼−2
1−𝑡𝛼−1
1)
Π
×𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)), 𝑡∈(0,𝑡1],
∫𝑡
0(𝑡−𝑠)𝛼−𝑞−1
Γ(𝛼−𝑞) 𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
−𝑡𝛼−𝑞−1
Γ(𝛼−𝑞)
×∫1
0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+(𝑡𝛼−𝑞−2 Γ(𝛼−1)
Γ(𝛼−𝑞−1)−𝑡𝛼−𝑞−1 Γ(𝛼)
Γ(𝛼−𝑞))
×∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛼−𝑞−1 Γ(𝛼)
Γ(𝛼−𝑞)∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+1
Π(𝑡𝛼−𝑞−1 Γ(𝛼)
Γ(𝛼−𝑞)−𝑡𝛼−𝑞−2 Γ(𝛼−1)
Γ(𝛼−𝑞−1))
×Γ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
+1
Π(𝑡𝛼−𝑞−2 Γ(𝛼−1)
Γ(𝛼−𝑞−1)−𝑡𝛼−𝑞−1 Γ(𝛼)
Γ(𝛼−𝑞))
×𝑡𝛼−1
1𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1)), 𝑡∈(𝑡1,1],
𝐷𝑝
0+(𝑇2𝑥)(𝑡)
=
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∫𝑡
0(𝑡−𝑠)𝛽−𝑝−1
Γ(𝛽−𝑝)𝜓(𝑢)𝑔(𝑢,𝑥(𝑢),𝐷𝑞
0+𝑥(𝑢))𝑑𝑢
−𝑡𝛽−𝑝−1
Γ(𝛽−𝑝)
×∫1
0(1−𝑠)𝛽−1𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
+𝑡𝛽−𝑝−2 Γ(𝛽−1)
Γ(𝛽−𝑝−1)
×∫1
0
V(𝑠)𝐻(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛽−𝑝−1 Γ(𝛽)
Γ(𝛽−𝑝)
×∫1
0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
+𝑡𝛽−𝑝−1
ΞΓ(𝛽)
Γ(𝛽−𝑝)
×( Γ(𝛽)
Γ(𝛽−𝑝)𝑡𝛽−𝑝−1
1−Γ(𝛽−1)
Γ(𝛽−𝑝−2)𝑡𝛽−𝑝−2
1)
×𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1))
+𝑡𝛽−𝑝−1 (𝑡𝛽−2
1−𝑡𝛽−1
1)
ΞΓ(𝛽)
Γ(𝛽−𝑝)
×𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1)), 𝑡∈(0,𝑡1],
∫𝑡
0(𝑡−𝑠)𝛽−𝑝−1
Γ(𝛽−𝑝)𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
−𝑡𝛽−𝑝−1
Γ(𝛽) Γ(𝛽)
Γ(𝛽−𝑝)
×∫1
0(1−𝑠)𝛽−1𝜓(𝑠)𝑔(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
+(𝑡𝛽−𝑝−2 Γ(𝛽−1)
Γ(𝛽−𝑝−1)−𝑡𝛽−𝑝−1 Γ(𝛽)
Γ(𝛽−𝑝))
×∫1
0
V(𝑠)𝐻(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡𝛽−𝑝−1 Γ(𝛽)
Γ(𝛽−𝑝)∫1
0𝑛(𝑠)𝑁(𝑠,𝑥(𝑠),𝐷𝑞
0+𝑥(𝑠))𝑑𝑠
+1
Ξ(𝑡𝛽−𝑝−1 Γ(𝛽)
Γ(𝛽−𝑝)−𝑡𝛽−𝑝−2 Γ(𝛽−1)
Γ(𝛽−𝑝−1))
×Γ(𝛽)
Γ(𝛽−𝑝)𝑡𝛽−𝑝−1
1𝐽(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1))
+1
Ξ(𝑡𝛽−𝑝−2 Γ(𝛽−1)
Γ(𝛽−𝑝−1)−𝑡𝛽−𝑝−1 Γ(𝛽)
Γ(𝛽−𝑝))
×𝑡𝛽−1
1𝐽1(𝑡1,𝑥(𝑡1),𝐷𝑞
0+𝑥(𝑡1)), 𝑡∈(𝑡1,1]. (40)
Ma hema ical P oblems in Enginee ing 9
I is easy o see ha
󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨∫𝑡
0(𝑡−𝑠)𝛼−𝑞−1
Γ(𝛼−𝑞)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝
0+𝑦(𝑢))𝑑𝑢󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨
≤𝐴𝑟𝐿1
B(𝛼+𝛿−𝑞,𝑘+1)
Γ(𝛼−𝑞) <∞. (41)
F om (37)and(41), we see ha 𝐷𝑞
0+(𝑇1𝑦)(𝑡)is de ined on
(0,1],con inuouson(0,𝑡1]and (𝑡1,1], espec i ely.Onesees
ha
lim
𝑡→0𝑡2+𝑞−𝛼𝐷𝑞
0+(𝑇1𝑦)(𝑡)
=lim
𝑡→0[𝑡2+𝑞−𝛼
×∫𝑡
0(𝑡−𝑠)𝛼−𝑞−1
Γ(𝛼−𝑞)𝜙(𝑢)𝑓(𝑢,𝑦(𝑢),𝐷𝑝
0+𝑦(𝑢))𝑑𝑢
−𝑡
Γ(𝛼−𝑞)
×∫1
0(1−𝑠)𝛼−1𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+Γ(𝛼−1)
Γ(𝛼−𝑞−1)∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡 Γ(𝛼)
Γ(𝛼−𝑞)∫1
0𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠
+𝑡
ΠΓ(𝛼)
Γ(𝛼−𝑞)
×( Γ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1−Γ(𝛼−1)
Γ(𝛼−𝑞−2)𝑡𝛼−𝑞−2
1)
×𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))
+Γ(𝛼)
Γ(𝛼−𝑞)𝑡(𝑡𝛼−2
1−𝑡𝛼−1
1)
Π
×𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))]
=Γ(𝛼−1)
Γ(𝛼−𝑞−1)∫1
0𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))𝑑𝑠,
(42)
and he e exi s he limi lim𝑡→𝑡+
1𝐷𝑞
0+(𝑇1𝑦)(𝑡).
F om he abo e discussion, we ha e (𝑇1𝑦)∈𝑋.Simila ly,
we can show ha (𝑇2𝑥)∈𝑌.Hence,((𝑇1𝑦),(𝑇2𝑥))∈𝑋×𝑌.
Then 𝑇:𝑋×𝑌→𝑋×𝑌is well de ined.
S ep 2. We p o e ha 𝑇is con inuous. Le (𝑥𝑛,𝑦𝑛)∈𝑋×
𝑌wi h (𝑥𝑛,𝑦𝑛)→(𝑥
0,𝑦0)as 𝑛→∞.Wewillshow ha
𝑇(𝑥𝑛,𝑦𝑛)→𝑇(𝑥
0,𝑦0)as 𝑛→∞, ha is,p o e ha 𝑇1𝑦𝑛→
𝑇1𝑦0and 𝑇2𝑥𝑛→𝑇
2𝑥0as 𝑛→∞.
In ac , we ha e 𝑟>0such ha ||(𝑥𝑛,𝑦𝑛)||=𝑟>0.Then
max {sup
𝑡∈(0,1)𝑡2−𝛼 󵄨󵄨󵄨󵄨𝑥𝑛(𝑡)󵄨󵄨󵄨󵄨,sup
𝑡∈(0,1)𝑡2+𝑞−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞
0+𝑥𝑛(𝑡)󵄨󵄨󵄨󵄨󵄨}
≤𝑟<+∞, 𝑛=0,1,2,...,
max {sup
𝑡∈(0,1)𝑡2−𝛽 󵄨󵄨󵄨󵄨𝑦𝑛(𝑡)󵄨󵄨󵄨󵄨,sup
𝑡∈(0,1)𝑡2+𝑝−𝛽 󵄨󵄨󵄨󵄨󵄨𝐷𝑝
0+𝑦𝑛(𝑡)󵄨󵄨󵄨󵄨󵄨}
≤𝑟<+∞, 𝑛=0,1,2,....
(43)
F om (B), 𝑓,𝐺,𝑀,𝐼,𝐼1a e 𝛽-Ca a heodo y unc ions,
hen he e exis cons an s 𝐴𝑟>0such ha
󵄨󵄨󵄨󵄨󵄨𝑓(𝑡,𝑦𝑛(𝑡),𝐷𝑝
0+𝑦𝑛(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,
𝑡∈(0,1), 𝑛=0,1,2,...,
󵄨󵄨󵄨󵄨󵄨𝐺(𝑡,𝑦𝑛(𝑡),𝐷𝑝
0+𝑦𝑛(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,
𝑡∈(0,1), 𝑛=0,1,2,...,
󵄨󵄨󵄨󵄨󵄨𝑀(𝑡,𝑦𝑛(𝑡),𝐷𝑝
0+𝑦𝑛(𝑡))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,
𝑡∈(0,1), 𝑛=0,1,2,...,
󵄨󵄨󵄨󵄨󵄨𝐼(𝑡1,𝑦𝑛(𝑡1),𝐷𝑝
0+𝑦𝑛(𝑡1))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,
𝑡∈(0,1), 𝑛=0,1,2,...,
󵄨󵄨󵄨󵄨󵄨𝐼1(𝑡1,𝑦𝑛(𝑡1),𝐷𝑝
0+𝑦𝑛(𝑡1))󵄨󵄨󵄨󵄨󵄨≤𝐴𝑟,
𝑡∈(0,1), 𝑛=0,1,2,...,
sup
𝑡∈(0,1)𝑡2−𝛼 󵄨󵄨󵄨󵄨𝑥𝑛(𝑡)−𝑥0(𝑡)󵄨󵄨󵄨󵄨󳨀→ 0,
sup
𝑡∈(0,1)𝑡2−𝛽 󵄨󵄨󵄨󵄨𝑦𝑛(𝑡)−𝑦0(𝑡)󵄨󵄨󵄨󵄨,
sup
𝑡∈(0,1)𝑡2+𝑞−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞
0+𝑥𝑛(𝑡)−𝐷𝑞
0+𝑥0(𝑡)󵄨󵄨󵄨󵄨󵄨󳨀→ 0,
sup
𝑡∈(0,1)𝑡2+𝑝−𝛽 󵄨󵄨󵄨󵄨󵄨𝐷𝑝
0+𝑦𝑛(𝑡)−𝐷𝑝
0+𝑦0(𝑡)󵄨󵄨󵄨󵄨󵄨󳨀→ 0,
(44)
16 Ma hema ical P oblems in Enginee ing
+Γ(𝛽)
Γ(𝛽−𝑝)‖𝑛‖1[𝐵𝑁+𝐴𝑁]
+1
ΞΓ(𝛽)
Γ(𝛽−𝑝)
×󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨Γ(𝛽)
Γ(𝛽−𝑝)𝑡𝛽−𝑝−1
1−Γ(𝛽−1)
Γ(𝛽−𝑝−2)𝑡𝛽−𝑝−2
1󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨󵄨[𝐵𝐽+𝐴𝐽]
+Γ(𝛽)
Γ(𝛽−𝑝)󵄨󵄨󵄨󵄨󵄨󵄨𝑡𝛽−2
1−𝑡𝛽−1
1󵄨󵄨󵄨󵄨󵄨󵄨
Ξ[𝐵1,𝐽 +𝐴1,𝐽],
Λ1=𝐿2
B(𝛽+𝜃,𝑙+1)
Γ(𝛽) 𝐶𝑔+𝐿2B(𝛽+𝜃,𝑙+1)
Γ(𝛽) 𝐶𝑔
+‖V‖1𝐶𝐻+1
ΞΓ(𝛽)
Γ(𝛽−𝑝)𝑡𝛽−𝑝−1
1𝐶𝐽+1
Ξ𝑡𝛽−1
1𝐶1,𝐽,
Λ2=𝐿2
B(𝛽+𝜃,𝑙+1)
Γ(𝛽) [𝐵𝑔+𝐴𝑔]
+𝐿2B(𝛽+𝜃,𝑙+1)
Γ(𝛽) [𝐵𝑔+𝐴𝑔]+‖V‖1[𝐵𝐻+𝐴𝐻]
+1
ΞΓ(𝛽)
Γ(𝛽−𝑝)𝑡𝛽−𝑝−1
1[𝐵𝐽+𝐴𝐽]+1
Ξ𝑡𝛽−1
1[𝐵1,𝐽 +𝐴1,𝐽],
Λ3=𝐿2
B(𝛽+𝜃−𝑝,𝑙+1)
Γ(𝛽−𝑝) 𝐶𝑔
+𝐿2
Γ(𝛽−𝑝)B(𝛽+𝜃,𝑙+1)𝐶𝑔
+( Γ(𝛽−1)
Γ(𝛽−𝑝−1)+Γ(𝛽)
Γ(𝛽−𝑝))‖V‖1𝐶𝐻
+Γ(𝛽)
Γ(𝛽−𝑝)‖𝑛‖1𝐶𝑁
+1
Ξ(Γ(𝛽)
Γ(𝛽−𝑝)+Γ(𝛽−1)
Γ(𝛽−𝑝−1))Γ(𝛽)
Γ(𝛽−𝑝)𝑡𝛽−𝑝−1
1𝐶𝐽
+1
Ξ(Γ(𝛽−1)
Γ(𝛽−𝑝−1)+Γ(𝛽)
Γ(𝛽−𝑝))𝑡𝛽−1
1𝐶1,𝐽,
Λ4=𝐿2
B(𝛽+𝜃−𝑝,𝑙+1)
Γ(𝛽−𝑝) [𝐵𝑔+𝐴𝑔]
+𝐿2
Γ(𝛽−𝑝)B(𝛽+𝜃,𝑙+1)[𝐵𝑔+𝐴𝑔]
+( Γ(𝛽−1)
Γ(𝛽−𝑝−1)+Γ(𝛽)
Γ(𝛽−𝑝))‖V‖1[𝐵𝐻+𝐴𝐻]
+Γ(𝛽)
Γ(𝛽−𝑝)‖𝑛‖1[𝐵𝑁+𝐴𝑁]
+1
Ξ(Γ(𝛽)
Γ(𝛽−𝑝)+Γ(𝛽−1)
Γ(𝛽−𝑝−1))
×Γ(𝛽)
Γ(𝛽−𝑝)𝑡𝛽−𝑝−1
1[𝐵𝐽+𝐴𝐽]
+1
Ξ(Γ(𝛽−1)
Γ(𝛽−𝑝−1)+Γ(𝛽)
Γ(𝛽−𝑝))𝑡𝛽−1
1[𝐵1,𝐽 +𝐴1,𝐽].
(81)
P oo . To apply Lemma 5, we should de ine an open bounded
subse Ωo 𝑋×𝑌cen e ed a ze o such ha assump ions in
Lemma 5hold.
Le Ω1={(𝑥,𝑦)∈𝑋×𝑌: (𝑥,𝑦)=𝜆𝑇(𝑥,𝑦) o some
𝜆∈(0,1)}.Wep o e ha Ω1is bounded. Fo (𝑥,𝑦)∈Ω1,we
ge (𝑥,𝑦)=𝜆𝑇(𝑥,𝑦).I ollows ha 𝑥=𝜆𝑇1𝑦and 𝑦=𝜆𝑇2𝑥.
Fo 𝑡∈(0,𝑡
1],weob ain𝑡2−𝛼|𝑥(𝑡)| ≤ 𝑡2−𝛼|(𝑇1𝑦)(𝑡)| ≤
Θ1+Θ2Φ−1(||𝑦||).
Fo 𝑡∈(𝑡1,1],
𝑡2−𝛼 |𝑥(𝑡)|
≤𝑡2−𝛼 ∫𝑡
0(𝑡−𝑠)𝛼−1
Γ(𝛼)󵄨󵄨󵄨󵄨󵄨𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))󵄨󵄨󵄨󵄨󵄨𝑑𝑠
+𝑡
Γ(𝛼)∫1
0(1−𝑠)𝛼−1 󵄨󵄨󵄨󵄨󵄨𝜙(𝑠)𝑓(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))󵄨󵄨󵄨󵄨󵄨𝑑𝑠
+(1−𝑡)∫1
0󵄨󵄨󵄨󵄨󵄨𝑢(𝑠)𝐺(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))󵄨󵄨󵄨󵄨󵄨𝑑𝑠
+𝑡∫1
0󵄨󵄨󵄨󵄨󵄨𝑚(𝑠)𝑀(𝑠,𝑦(𝑠),𝐷𝑝
0+𝑦(𝑠))󵄨󵄨󵄨󵄨󵄨𝑑𝑠
+1−𝑡
ΠΓ(𝛼)
Γ(𝛼−𝑞)𝑡𝛼−𝑞−1
1󵄨󵄨󵄨󵄨󵄨𝐼(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))󵄨󵄨󵄨󵄨󵄨
+1−𝑡
Π𝑡𝛼−1
1󵄨󵄨󵄨󵄨󵄨𝐼1(𝑡1,𝑦(𝑡1),𝐷𝑝
0+𝑦(𝑡1))󵄨󵄨󵄨󵄨󵄨
≤Σ1+Σ2Φ−1 (󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩). (82)
I ollows ha
sup
𝑡∈(0,1)𝑡2−𝛼 |𝑥(𝑡)|≤max {Θ1,Σ1}+max {Θ2,Σ2}Φ−1 (󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩).
(83)
Simila ly, we ha e o 𝑡∈(0,𝑡1] ha
𝑡𝑞+2−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞
0+𝑥(𝑡)󵄨󵄨󵄨󵄨󵄨≤Θ3+Θ4Φ−1 (󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩)(84)
and o 𝑡∈(0,𝑡1]
𝑡𝑞+2−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞
0+𝑥(𝑡)󵄨󵄨󵄨󵄨󵄨≤Σ3+Σ4Φ−1 (󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩). (85)
I ollows ha
sup
𝑡∈(0,1)𝑡2+𝑞−𝛼 󵄨󵄨󵄨󵄨󵄨𝐷𝑞
0+𝑥(𝑡)󵄨󵄨󵄨󵄨󵄨
≤max {Θ3,Σ3}+max {Θ4,Σ4}Φ−1 (󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩). (86)

Ma hema ical P oblems in Enginee ing 17
Hence,
‖𝑥‖≤max {Θ1,Σ1,Θ3,Σ3}
+max {Θ2,Σ2,Θ4,Σ4}Φ−1 (󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩). (87)
Simila o heabo ediscussionwecanp o e ha
󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩≤max {Υ1,Λ1,Υ3,Λ3}+max {Υ2,Λ2,Υ4,Λ4}Φ(‖𝑥‖).
(88)
Case 1. Conside (max{Θ2,Σ2,Θ4,Σ4}](2max{Υ2,Λ2,Υ4,
Λ4})<1).
Wi hou losso gene ali y,suppose ha
‖𝑥‖≥Φ−1 (max {Υ1,Λ1,Υ3,Λ3}
max {Υ2,Λ2,Υ4,Λ4}). (89)
Then use Rema k 13, and he p e ious inequali ies o ge
‖𝑥‖≤max {Θ1,Σ1,Θ3,Σ3}
+max {Θ2,Σ2,Θ4,Σ4}](2max {Υ2,Λ2,Υ4,Λ4})‖𝑥‖.
(90)
I ollows ha he e exis s a cons an 𝑊>0such ha ||𝑥||≤
𝑊.Thus
‖𝑥‖≤max {𝑊,Φ−1 (max {Υ1,Λ1,Υ3,Λ3}
max {Υ2,Λ2,Υ4,Λ4})}. (91)
Then
󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩≤max {Υ1,Λ1,Υ3,Λ3}
+max {Υ2,Λ2,Υ4,Λ4}Φ
×(max {𝑊,Φ−1 (max {Υ1,Λ1,Υ3,Λ3}
max {Υ2,Λ2,Υ4,Λ4})}).
(92)
I ollows ha Ω1is bounded.
Case 2. Conside ((max{Υ2,Λ2,Υ4,Λ4}/𝑤((2max{Θ2,Σ2,
Θ4,Σ4})−1)) <1).
Wi hou loss o gene ali y, suppose ha
󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩≥Φ(max {Θ1,Σ1,Θ3,Σ3}
max {Θ2,Σ2,Θ4,Σ4}). (93)
Then using Rema k 12 and he p e ious inequali ies, we ge
󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩≤max {Υ1,Λ1,Υ3,Λ3}
+max {Υ2,Λ2,Υ4,Λ4}
𝑤((2max {Θ2,Σ2,Θ4,Σ4})−1)󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩.(94)
I ollows ha he e exis s a cons an 𝑊>0such ha ||𝑦||≤
𝑊.Wege
󵄩󵄩󵄩󵄩𝑦󵄩󵄩󵄩󵄩≤max {𝑊,Φ(max {Θ1,Σ1,Θ3,Σ3}
max {Θ2,Σ2,Θ4,Σ4})}. (95)
Then
‖𝑥‖≤max {Θ1,Σ1,Θ3,Σ3}
+max {Θ2,Σ2,Θ4,Σ4}Φ−1
×(max {𝑊,Φ(max {Θ1,Σ1,Θ3,Σ3}
max {Θ2,Σ2,Θ4,Σ4})}).
(96)
I ollows ha Ω1is bounded.
To apply Lemma 5,le Ωbe a nonemp y open bounded
subse o 𝑋such ha Ω⊃Ω1cen e ed a ze o.
I is easy o see om Lemma 8 ha 𝑇is a comple ely con-
inuous ope a o . One can see ha
(𝑥,𝑦) =𝜆𝑇(𝑥,𝑦) ∀(𝑥,𝑦)∈𝜕Ω,𝜆∈(0,1).(97)
Thus, om Lemma 5,(𝑥,𝑦)=𝑇(𝑥,𝑦)has a leas one solu ion
(𝑥,𝑦)∈Ω.So(𝑥,𝑦)is a pai o solu ions o BVP (3)andBVP
(6). The p oo o Theo em 14 is comple e.
4. Two Examples
To illus a e he use ulness o ou main esul , we p esen wo
examples ha Theo em 14 can eadily apply.
Example 15. Conside he ollowing impulsi e bounda y
alue p oblem:
𝐷8/5
0+𝑥(𝑡)=𝑡−1/5(1−𝑡)−1
×(𝑐+𝑏𝑡6/5[𝑦(𝑡)]3+𝑎𝑡9/5[𝐷1/5
0+𝑦(𝑡)]3),
𝑡∈(0,1),𝑡=1
2,
𝐷9/5
0+𝑦(𝑡)
=𝑡−1/5(1−𝑡)−1
×(𝑐0+𝑏0𝑡1/15[𝑥(𝑡)]1/3 +𝑎0𝑡2/15[𝐷1/5
0+𝑥(𝑡)]1/3),
𝑡∈(0,1),𝑡=𝑡1,
lim
𝑡→0𝑡2/5𝑥(𝑡)=𝐺, lim
𝑡→0𝑡1/5𝑦(𝑡)=𝐻,
𝑥(1)=𝑀, 𝑦(1)=𝑁,
Δ𝑥(1
2)=𝑐𝐼,Δ𝑦(
1
2)=𝑐𝐽,
Δ𝐷1
0+𝑥(1
2)=𝑐1,𝐼,Δ𝐷
1
0+𝑦(1
2)=𝑐1,𝐽,(98)
whe e 𝑐,𝑏,𝑎,𝑐0,𝑏0,𝑎0,𝐺0,𝐻0,𝑀0,𝑁0,𝐶𝐼,𝐶𝐽,𝐶1,𝐼,𝐶1,𝐽
a e cons an s.
Co esponding o BVP (1), we ha e
(a) 𝛼=8/5,𝛽=9/5,𝑝=𝑞=1/5,
(b) 𝜙(𝑡)=𝜓(𝑡)=𝑡−1/5(1−𝑡)−1/5,𝑓(𝑡,𝑈,𝑉)=𝑐+𝑏𝑡6/5𝑈3+
𝑎𝑡9/5𝑉3and 𝑔(𝑡,𝑈,𝑉)=𝑐0+𝑏0𝑡1/15𝑈1/3 +𝑎0𝑡2/15𝑉1/3
de ined on (0,1)×𝑅2,
18 Ma hema ical P oblems in Enginee ing
(c) 𝑢(𝑡)=V(𝑡)=𝑚(𝑡)=𝑛(𝑡)≡1,𝐺(𝑡,𝑈,𝑉)=𝐺0,𝐻(𝑡,
𝑈,𝑉)=𝐻0,𝑀(𝑡,𝑈,𝑉)=𝑀0,𝑁(𝑡,𝑈,𝑉)=𝑁0,
(d) 0=𝑡0<𝑡1=(1/2)<𝑡2=1,
(e) 𝐼(𝑡,𝑈,𝑉)=𝑐𝐼,𝐼1(𝑡,𝑈,𝑉)=𝑐1,𝐼,𝐽(𝑡,𝑈,𝑉)=𝑐𝐽,𝐽1(𝑡,
𝑈,𝑉)=𝑐1,𝐽.
I is easy o show ha
(A) 𝜙sa is ies 𝛼+2𝛿−𝑞>0,𝛼+𝑘+𝛿−𝑞≥0,and
|𝜙(𝑡)|≤𝐿1𝑡𝑘(1−𝑡)𝛿 o all 𝑡∈(0,1)wi h 𝐿1=1and
𝑘=−(1/5)=𝛿;
𝜓sa is ies 𝜂+2𝜃−𝑝>0,𝛽+𝑙+𝜃−𝑝≥0,and
|𝜓(𝑡)|≤𝐿2𝑡𝑙(1−𝑡)𝜃 o all 𝑡∈(0,1)wi h 𝐿2=1and
𝑙=−(1/5)=𝜃;
(B) 𝑓,𝐺,𝑀,𝐼,𝐼1a e 𝛽-Ca a heodo y unc ions and 𝑔,𝐻,
𝑁,𝐽,𝐽1a e 𝛼-Ca a heodo y unc ions.
Fu he mo e, we ha e Φ−1(𝑥)=𝑥3and Φ(𝑥)=𝑥1/3 wi h
𝑤(𝑥)=𝑥1/3 and ](𝑥)=𝑥3.I iseasy osee ha
(i) he inequali ies
󵄨󵄨󵄨󵄨󵄨𝑓(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝑓+𝐵𝑓Φ−1 (|𝑈|)+𝐴𝑓Φ−1 (|𝑉|),
󵄨󵄨󵄨󵄨󵄨𝐺(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝐺+𝐵𝐺Φ−1 (|𝑈|)+𝐴𝐺Φ−1 (|𝑉|),
󵄨󵄨󵄨󵄨󵄨𝑀(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)󵄨󵄨󵄨󵄨󵄨
≤𝐶𝑀+𝐵𝑀Φ−1 (|𝑈|)+𝐴𝑀Φ−1 (󵄨󵄨󵄨󵄨󵄨Φ−1 (|𝑈|)󵄨󵄨󵄨󵄨󵄨)(99)
hold o all (𝑈,𝑉)∈𝑅2,𝑡∈(0,1]wi h 𝐶𝑓=|𝑐|,𝐵𝑓=
|𝑏|,𝐴𝑓=|𝑎|,𝐶𝐺=|𝐺0|,𝐵𝐺=0,𝐴𝐺=0and 𝐶𝑀=
|𝑀0|,𝐵𝑀=0,𝐴𝑀=0;
(ii) he inequali ies
󵄨󵄨󵄨󵄨󵄨𝑔(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝑔+𝐵𝑔Φ(𝑈)+𝐴𝑔Φ(𝑉),
󵄨󵄨󵄨󵄨󵄨𝐻(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝐻+𝐵𝐻Φ(𝑈)+𝐴𝐻Φ(𝑉),
󵄨󵄨󵄨󵄨󵄨𝑁(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝑁+𝐵𝑁Φ(𝑈)+𝐴𝑁Φ(𝑉)
(100)
hold o all (𝑈,𝑉) ∈ 𝑅2,𝑡 ∈ (0,1]wi h 𝐶𝑔=|𝑐
0|,
𝐵𝑔=|𝑏
0|,𝐴𝑔=|𝑎
0|,𝐶𝐻=|𝐻
0|,𝐵𝐻=𝐴
𝐻=0,
𝐶𝑁=|𝑁0|,𝐵𝑁=𝐴𝑁=0;
(iii) he inequali ies
󵄨󵄨󵄨󵄨󵄨𝐼(𝑡1,𝑡𝛼−2
1𝑈,𝑡𝛼−𝑞−2
1𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝐼+𝐵𝐼Φ−1 (|𝑈|)+𝐴𝐼Φ−1 (|𝑉|),
󵄨󵄨󵄨󵄨󵄨𝐼1(𝑡1,𝑡𝛼−2
1𝑈,𝑡𝛼−𝑞−2
1𝑉)󵄨󵄨󵄨󵄨󵄨
≤𝐶1,𝐼 +𝐵1,𝐼Φ−1 (|𝑈|)+𝐴1,𝐼Φ−1 (|𝑉|)(101)
hold o all (𝑈,𝑉)∈𝑅2wi h 𝐶𝐼=|𝑐
𝐼|,𝐵𝐼=𝐴𝐼=0,
𝐶1,𝐼 =|𝑐1,𝐼|,𝐵1,𝐼 =𝐴1,𝐼 =0;
(i ) he inequali ies
󵄨󵄨󵄨󵄨󵄨󵄨𝐽(𝑡1,𝑡𝛽−2
1𝑈,𝑡𝛽−𝑝−2
1𝑉)󵄨󵄨󵄨󵄨󵄨󵄨≤𝐶𝐽+𝐵𝐽Φ(𝑈)+𝐴𝐽Φ(𝑉),
󵄨󵄨󵄨󵄨󵄨󵄨𝐽1(𝑡1,𝑡𝛽−2
1𝑈,𝑡𝛽−𝑝−2
1𝑉)󵄨󵄨󵄨󵄨󵄨󵄨≤𝐶1,𝐽 +𝐵1,𝐽Φ(𝑈)+𝐴1,𝐽Φ(𝑉)
(102)
hold o all (𝑈,𝑉)∈𝑅2wi h 𝐶𝐽=|𝑐
𝐽|,𝐵𝐽=𝐴𝐽=0,
𝐶1,𝐽 =|𝑐1,𝐽|,𝐵1,𝐽 =𝐴1,𝐽 =0.
By di ec compu a ion, we know ha
Θ2=2B(7/5,4/5)
Γ(8/5)[|𝑏|+|𝑎|],
Σ2=2B(7/5,4/5)
Γ(8/5)[|𝑏|+|𝑎|],
Θ4=(B(6/5,4/5)
Γ(7/5)+B(7/5,4/5)
Γ(7/5))[|𝑏|+|𝑎|],
Σ4=(B(6/5,4/5)
Γ(7/5)+B(7/5,4/5)
Γ(7/5))[|𝑏|+|𝑎|],
Υ2=2B(8/5,4/5)
Γ(9/5)[󵄨󵄨󵄨󵄨𝑏0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝑎0󵄨󵄨󵄨󵄨],
Υ4=(B(7/5,4/5)
Γ(8/5)+B(8/5,4/5)
Γ(8/5))[󵄨󵄨󵄨󵄨𝑏0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝑎0󵄨󵄨󵄨󵄨],
Λ2=2B(8/5,4/5)
Γ(9/5)[󵄨󵄨󵄨󵄨𝑏0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝑎0󵄨󵄨󵄨󵄨],
Λ4=(B(7/5,4/5)
Γ(8/5)+B(8/5,4/5)
Γ(8/5))[󵄨󵄨󵄨󵄨𝑏0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝑎0󵄨󵄨󵄨󵄨].
(103)
Then Theo em 14 implies ha he exis ence o a leas one
solu ion i
max {2B(8/5,4/5)
Γ(9/5),B(7/5,4/5)
Γ(8/5)+B(8/5,4/5)
Γ(8/5)}
×(max {2B(7/5,4/5)
Γ(8/5) ,B(6/5,4/5)
Γ(7/5) +B(7/5,4/5)
Γ(7/5) })1/3
×[󵄨󵄨󵄨󵄨𝑏0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝑎0󵄨󵄨󵄨󵄨][|𝑏|+|𝑎|]1/3 <1
3
√2.(104)
Example 16. Conside he ollowing bounda y alue p oblem
wi hou impulse e ec s:
𝐷7/4
0+𝑥(𝑡)=𝑡−1/4(1−𝑡)−1/4
×(𝐶+𝐵𝑡3/4[𝑦(𝑡)]3+𝐴𝑡15/4[𝐷1
0+𝑦(𝑡)]3),
𝑡∈(0,1),
Ma hema ical P oblems in Enginee ing 19
𝐷5/4
0+𝑦(𝑡)
=𝑡−1/8(1−𝑡)−1/8
×(𝐶0+𝐵0𝑡1/4[𝑥(𝑡)]1/3 +𝐴0𝑡7/12[𝐷1/4
0+𝑥(𝑡)]1/3),
𝑡∈(0,1),
lim
𝑡→0𝑡1/4𝑥(𝑡)=0, lim
𝑡→0𝑡3/4𝑦(𝑡)=0,
𝑥(1)=0, 𝑦(1)=0, (105)
whe e 𝐶,𝐵,𝐴,𝐶0,𝐵0,and𝐴0a e cons an s.
Co esponding o BVP (1), we ha e
(a) 𝛼=7/4,𝛽=5/4,𝑝=1and 𝑞=1/4,
(b) 𝜙(𝑡) = 𝑡−1/4(1 − 𝑡)−1/4,𝜓(𝑡) = 𝑡−1/8(1 − 𝑡)−1/8,
𝑓,𝑔 de ined on (0,1) × 𝑅2,𝑓(𝑡,𝑈,𝑉) = 𝐶 +
𝐵𝑡1/12𝑈3+𝐴𝑡5/12𝑉3and 𝑔(𝑡,𝑈,𝑉)=𝐶0+𝐵0𝑡1/4𝑈1/3+
𝐴0𝑡7/12𝑉1/3,
(c) 𝑚(𝑡)=𝑛(𝑡)=𝑢(𝑡) = V(𝑡)≡0,𝐺(𝑡,𝑈,𝑉)=𝐻(𝑡,𝑈,
𝑉)=𝑀(𝑡,𝑈,𝑉)=𝑁(𝑡,𝑈,𝑉)≡0,
(d) he e exis s no impulse poin ,
(e) 𝐼(𝑡,𝑈,𝑉)=𝐼1(𝑡,𝑈,𝑉)=𝐽(𝑡,𝑈,𝑉)=𝐽1(𝑡,𝑈,𝑉)≡0.
I is easy o show ha
(A) 𝜙sa is ies 𝛼+2𝛿−𝑞>0,𝛼+𝑘+𝛿−𝑞>0,|𝜙(𝑡)|≤
𝐿1𝑡𝑘(1−𝑡)𝛿 o all 𝑡∈(0,1)wi h 𝐿1=1,𝑘=−(1/4)=
𝛿;
𝜓sa is ies 𝛽+2𝜃−𝑝>0,𝛽+𝑙+𝜃−𝑝≥0,and
|𝜓(𝑡)| ≤ 𝐿2𝑡𝑙(1−𝑡)𝜃 o all 𝑡 ∈ (0,1)wi h 𝐿2=1,
𝑙=−(1/8)=𝜃;
(B) 𝑓,𝐺,𝑀,𝐼,𝐼1a e 𝛽-Ca a heodo y unc ions and 𝑔,𝐻,
𝑁,𝐽,𝐽1a e 𝛼-Ca a heodo y unc ions.
Fu he mo e, Φ(𝑥)=𝑥1/3 and Φ−1(𝑥)=𝑥3,weha e𝑤(𝑥)=
𝑥1/3 and ](𝑥)=𝑥3,and
(i) he inequali ies
󵄨󵄨󵄨󵄨󵄨𝑓(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝑓+𝐵𝑓Φ−1 (|𝑈|)+𝐴𝑓Φ−1 (|𝑉|),
󵄨󵄨󵄨󵄨󵄨𝐺(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝐺+𝐵𝐺Φ−1 (|𝑈|)+𝐴𝐺Φ−1 (|𝑉|),
󵄨󵄨󵄨󵄨󵄨𝑀(𝑡,𝑡𝛼−2𝑈,𝑡𝛼−𝑞−2𝑉)󵄨󵄨󵄨󵄨󵄨
≤𝐶𝑀+𝐵𝑀Φ−1 (|𝑈|)+𝐴𝑀Φ−1 (󵄨󵄨󵄨󵄨󵄨Φ−1 (|𝑈|)󵄨󵄨󵄨󵄨󵄨)(106)
hold o all (𝑈,𝑉) ∈ 𝑅2,𝑡 ∈ (0,1)wi h 𝐶𝐺=𝐵
𝐺=
𝐴𝐺=𝐶𝑀=𝐵𝑀=𝐴𝑀=0,𝐶𝑓=|𝐶|,𝐵𝑓=|𝐵|and
𝐴𝑓=|𝐴|;
(ii) he inequali ies
󵄨󵄨󵄨󵄨󵄨𝑔(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝑔+𝐵𝑔Φ(𝑈)+𝐴𝑔Φ(𝑉),
󵄨󵄨󵄨󵄨󵄨𝐻(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝐻+𝐵𝐻Φ(𝑈)+𝐴𝐻Φ(𝑉),
󵄨󵄨󵄨󵄨󵄨𝑁(𝑡,𝑡𝛽−2𝑈,𝑡𝛽−𝑝−2𝑉)󵄨󵄨󵄨󵄨󵄨≤𝐶𝑁+𝐵𝑁Φ(𝑈)+𝐴𝑁Φ(𝑉)(107)
hold o all (𝑈,𝑉)∈𝑅2,𝑡 ∈(0,1)wi h 𝐶𝐻=𝐵𝐻=
𝐴𝐻=𝐶𝑁=𝐵𝑁=𝐴𝑁=0,𝐶𝑔=|𝐶0|,𝐵𝑔=|𝐵0|and
𝐴𝑔=|𝐴0|;
(iii) he inequali ies
󵄨󵄨󵄨󵄨󵄨𝐼(𝑡1,𝑡𝛼−2
1𝑈,𝑡𝛼−𝑞−2
1𝑉)󵄨󵄨󵄨󵄨󵄨
≤𝐶𝐼+𝐵𝐼Φ−1 (|𝑈|)+𝐴𝐼Φ−1 (|𝑉|),
󵄨󵄨󵄨󵄨󵄨𝐼1(𝑡1,𝑡𝛼−2
1𝑈,𝑡𝛼−𝑞−2
1𝑉)󵄨󵄨󵄨󵄨󵄨
≤𝐶1,𝐼 +𝐵1,𝐼Φ−1 (|𝑈|)+𝐴1,𝐼Φ−1 (|𝑉|)
(108)
hold o all (𝑈,𝑉)∈𝑅2wi h 𝐶𝐼=𝐵𝐼=𝐴𝐼=𝐶1,𝐼 =
𝐵1,𝐼 =𝐴1,𝐼 =0;
(i ) he e exis he nonnega i e numbe s 𝐴𝑖,𝑘,𝐵𝑖,𝑘,
𝐶𝑖,𝑘 (𝑖=1,2)such ha
󵄨󵄨󵄨󵄨󵄨󵄨𝐽(𝑡1,𝑡𝛽−2
1𝑈,𝑡𝛽−𝑝−2
1𝑉)󵄨󵄨󵄨󵄨󵄨󵄨≤𝐶𝐽+𝐵𝐽Φ(𝑈)+𝐴𝐽Φ(𝑉),
󵄨󵄨󵄨󵄨󵄨󵄨𝐽1(𝑡1,𝑡𝛽−2
1𝑈,𝑡𝛽−𝑝−2
1𝑉)󵄨󵄨󵄨󵄨󵄨󵄨≤𝐶1,𝐽 +𝐵1,𝐽Φ(𝑈)+𝐴1,𝐽Φ(𝑉)
(109)
hold o all (𝑈,𝑉)∈𝑅2wi h 𝐶𝐽=𝐵𝐽=𝐴𝐽=𝐶1,𝐽 =
𝐵1,𝐽 =𝐴1,𝐽 =0.
By di ec compu a ion, we know ha
Θ2=2B(3/2,3/4)
Γ(7/4)[|𝐵|+|𝐴|],
Σ2=2B(3/2,3/4)
Γ(7/4)[|𝐵|+|𝐴|],
Θ4=(B(5/4,3/4)
Γ(3/2)+B(3/2,3/4)
Γ(3/2))[|𝐵|+|𝐴|],
Σ4=(B(5/4,3/4)
Γ(3/2)+B(3/2,3/4)
Γ(3/2))[|𝐵|+|𝐴|],
Υ2=2B(9/8,7/8)
Γ(5/4)[󵄨󵄨󵄨󵄨𝐵0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝐴0󵄨󵄨󵄨󵄨],
Υ4=(B(1/8,7/8)
Γ(𝛽−𝑝) +B(9/8,7/8)
Γ(1/4))[󵄨󵄨󵄨󵄨𝐵0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝐴0󵄨󵄨󵄨󵄨],
Λ2=2B(9/8,7/8)
Γ(5/4)[󵄨󵄨󵄨󵄨𝐵0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝐴0󵄨󵄨󵄨󵄨],
Λ4=(B(1/8,7/8)
Γ(1/4)+B(9/8,7/8)
Γ(1/4))[󵄨󵄨󵄨󵄨𝐵0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝐴0󵄨󵄨󵄨󵄨].
(110)
20 Ma hema ical P oblems in Enginee ing
Then Theo em 14 implies he exis ence o a leas one
solu ion i
max {2B(3/2,3/4)
Γ(7/4),B(5/4,3/4)
Γ(3/2)+B(3/2,3/4)
Γ(3/2)}
×(max {2B(9/8,7/8)
Γ(5/4) ,B(1/8,7/8)
Γ(𝛽−𝑝) +B(9/8,7/8)
Γ(1/4) })3
×[󵄨󵄨󵄨󵄨𝐵0󵄨󵄨󵄨󵄨+󵄨󵄨󵄨󵄨𝐴0󵄨󵄨󵄨󵄨]3[|𝐵|+|𝐴|]<1
8.
(111)
Rema k 17. I is easy o see ha he p e ious bounda y alue
p oblems ha e a leas one solu ion o su icien ly small
|𝐵1|,|𝐵2|and |𝐴0|,|𝐵0|,|𝑎|,|𝑏|,|𝑎0|and |𝑏0|.Theycanno be
sol ed by he heo ems in [24,25].
Acknowledgmen s
This esea chispa iallysuppo edby heNa u alScience
Founda ion o Guangdong p o ince (no. S2011010001900)
and he Guangdong Highe Educa ion Founda ion o High-
Le el Talen s. This esea ch is pa ially suppo ed by Minis-
e io de Econom´
ıa y Compe i i idad and EC und FEDER,
P ojec no. MTM2010-15314, Spain.
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