mathematics Article
Nonlocal q-Symmetric Integral Boundary Value Problem for Sequential q-Symmetric Integrodifference Equations Rujira Ouncharoen 1 , Nichaphat Patanarapeelert 2, * and Thanin Sitthiwirattham 3 1 2 3
*
Center of Excellence in Mathematics and Applied Mathematics, Department of Mathematics, Faculty of Science, Chiang Mai University, Chiang Mai 50200, Thailand;
[email protected] Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok 10800, Thailand Mathematics Department, Faculty of Science and Technology, Suan Dusit University, Bangkok 10700, Thailand;
[email protected] Correspondence:
[email protected]
Received: 11 September 2018; Accepted: 22 October 2018; Published: 25 October 2018
Abstract: In this paper, we prove the sufficient conditions for the existence results of a solution of a nonlocal q-symmetric integral boundary value problem for a sequential q-symmetric integrodifference equation by using the Banach’s contraction mapping principle and Krasnoselskii’s fixed point theorem. Some examples are also presented to illustrate our results. Keywords: q-symmetric difference; equation; existence
q-symmetric integral;
q-symmetric integrodifference
(2010) Mathematics Subject Classifications: 39A05; 39A13
1. Introduction Quantum difference operators dealing with sets of nondifferentiable functions have been extensively studied as they can be used as a tool to understand complex physical systems. There are several kinds of difference operators. The q-difference operator was first introduced by Jackson [1] and was studied in intensive work especially by Carmichael [2], Mason [3], Adams [4] and Trjitzinsky [5]. The studies of quantum problems involving q-calculus have been presented. The recent works related to q-calculus theories can be found in [6–8] and the references cited therein. The q-symmetric difference operators are a useful tool in several fields, especially in quantum mechanics [9]. However, there are few research works [10–13] involving the development of q-symmetric difference operators. In 2012, A.M.C. Brito da Cruz and N. Martins [10] studied the q-deformed theory, in which the standard q-symmetric integral must be generalized to the basic integral defined. Recently, Sun, Jin and Hou [12] introduced basic concepts of fractional q-symmetric integral and derivative operator. Moreover, Sun and Hou [13] introduced basic concepts of fractional q-symmetric calculus on a time scale. In particular, the boundary value problem for q-symmetric difference equations has not been studied. The results mentioned are the motivation for this research. In this paper, we devote our attention to the estabished existence results for a nonlocal q-symmetric integral boundary value for sequential q-symmetric integrodifference equation of the form ˜ qD ˜ p u(t) = F t, u(t), (Sθ u)(t), ( Zω u)(t) , t ∈ IχT D
Mathematics 2018, 6, 218; doi:10.3390/math6110218
www.mdpi.com/journal/mathematics
Mathematics 2018, 6, 218
2 of 9
u(0) = λu( T ), Z η 0
(1)
g(s)u(s) d˜r s = 0, η ∈ IχT − {0, T }, p
q
θ1 1 where IχT := {χk T : k ∈ N} ∪ {0, T }, p, q, r, ω, θ ∈ (0, 1), p = p21 , q = q12 , r = rr12 , ω = ω ω2 , θ = θ 2 1 are the simplest form of proper fractions; g ∈ C IχT , R+ and F ∈ and χ = LCM p2 ,q2 ,r2 ,ω2 ,θ2 C IχT × R × R × R, R are given functions; and for ϕ, φ ∈ C IχT × IχT , [0, ∞) , define
( Sθ u ) ( t ) : =
Z t
ϕ(s, t)u(s)d˜θ s and ( Zω u) (t) :=
0
Z t 0
φ(s, t)u(s)d˜ω s.
This paper is organized as follows. In Section 2, we provide basic definitions, properties of q-symmetric difference operator and lemmas used in this paper. In Section 3, the existence results of problem (1) will be proved by employing Banach’s contraction mapping principle and Krasnoselskii’s fixed point theorem. In Section 4, we give some examples to illustrate our results. 2. Preliminaries We introduce some basic definitions and properties of q-symmetric difference calculus as follows. Definition 1. For q ∈ (0, 1), the q-symmetric difference of function f : R → R is defined by −1 ˜ q f (t) = f (qt) − f (q t) and D ˜ q f (0) = f 0 ( t ). D − 1 (q − q )t
The higher order q-symmetric derivatives of f is defined by ˜ qn f (t) = D ˜ q Dqn−1 f (t), n ∈ N. D ˜ q0 f (t) = f (t). We note that D Next, if f is a function defined on the interval I, q-symmetric integral is defined by Z b a
Z x
and
0
f (s)d˜q s
:=
f (s)d˜q s
:=
Z b 0
f (s)d˜q s −
x (1 − q2 )
Z a
∞
0
∑ q2k f
f (s)d˜q s
xq2k+1
k =0
where the above infinite series is convergent. We next discuss the following lemmas used to simplify our calculations. Lemma 1. Let 0 < q < 1, and f : I → R be continuous at 0. Then, Z tZ r 0
0
x (s) d˜q s d˜q r =
Z t Z qt 0
q2 s
x (qs) d˜q r d˜q s.
Proof. Using the definition of symmetric q-integral, we have Z tZ r 0
=
x (s) d˜q s d˜q r ∞
∑ q2k x
∑ q2k t(1 − q ) ∑ q2h
0 ∞
=
0
" Z t
k =0
r (1 − q2 )
k =0 ∞ 2
h =0
rq2k+1
# d˜q r =
∞
∑ q2k
k =0
"Z
t 0
r (1 − q2 ) x rq2k+1 d˜q r
tq2h+1 (1 − q2 ) x tq2h+1 q2k+1
#
Mathematics 2018, 6, 218
3 of 9
= qt2 (1 − q2 )2
∞
∞
∑ ∑ q2k+4h x
tq2k+2h+2
k =0 h =0 ∞ h 2 2 4h
∑
= qt2 (1 − q )
i q x tq2h+2 + q4h+2 x tq2h+4 + q4h+4 x tq2h+6 + ...
h =0
nh
i = qt2 (1 − q2 )2 x (tq2 ) + q2 x tq4 + q4 x tq6 + ... h i + q4 x (tq4 ) + q6 x tq6 + q8 x tq8 + ... h i o + q8 x (tq6 ) + q10 x tq8 + q12 x tq10 + ... + ... n o = qt2 (1 − q2 )2 x (tq2 ) + q2 (1 + q2 ) x tq4 + q4 (1 + q2 + q4 ) x tq6 + ... ∞ = qt2 (1 − q2 )2 ∑ q2k [k + 1]q2 x tq2k+2 k =0
∞
= q
∑ q2k t2 (1 − q2 )
1 − q2(k+1) x tq2k+2
k =0
=
Z t 0
=
qt − q2 s x (qs) d˜q s
Z t h Z qt 0
=
0
Z t Z qt 0
q2 s
x (qs) d˜q r −
Z q2 s
i x (qs) d˜q r d˜q s
0
x (qs) d˜q r d˜q s.
Lemma 2. Let 0 < q < 1. Then, Z t 0
d˜q s = t and
Z t Z qt q2 s
0
d˜q r d˜q s =
qt2 . 1 + q2
Proof. Using the definition of symmetric q-integral, we have Z t 0
Z t Z qt 0
q2 s
d˜q s
d˜q r d˜q s
= t (1 − q2 )
∞
∑
q2k = t(1 − q2 )
k =0
=
Z t 0
1 1 − q2
= t,
[qt − q2 s] d˜q s
= t (1 − q2 )
∞
∑
h i q2k qt − q2k+3 t
k =0
=
q2 1 − qt (1 − q ) 1 − q2 1 − q4 2
2
=
qt2 . 1 + q2
To study the solution of the boundary value problem (1), we first consider the solution of a linear variant of the boundary value problem (1) as follows. p q 1 are the simplest form of proper Lemma 3. Let p, q ∈ (0, 1), p = p21 , q = q12 , r = rr12 and κ = LCM p2 ,q2 ,r2 fractions; λ ∈ R; g ∈ C IκT , R+ and h ∈ C IκT , R are given functions. The solution for the problem
˜ qD ˜ p u(t) = h(t), t ∈ IκT D
(2)
u(0) = λu( T ),
(3)
Mathematics 2018, 6, 218
4 of 9
Z η 0
g(s)u(s) d˜r s = 0, η ∈ IκT − {0, T },
(4)
is in the form 1 u(t) = Λ
"Z
t + Λ
"Z
+
0 η
0
g(s) d˜q s · λ
0
0
Z TZ x 0
0
g(s) d˜q s + (1 − λ)
Rη
Z tZ x
Rη
Z TZ x
sg(s) d˜q s · λ
0
0
where Λ := λT
η
0
h(s) d˜q s d˜p x − λT
Z ηZ yZ x 0
0
h(s) d˜q s d˜p x − (1 − λ)
0
# g(y)h(s) d˜q s d˜p x d˜r y
Z ηZ yZ x 0
0
0
# g(y)h(s) d˜q s d˜p x d˜r y
(5)
h( x ) d˜q s d˜p x, 0
sg(s) d˜q s.
Proof. We first take the q-symmetric integral for (2) to obtain ˜ p u(t) = C1 + D
Z t 0
h(s) d˜q s.
(6)
Next, taking the p-symmetric integral for (6), we have Z tZ x
u(t) = C2 + C1 t +
0
0
h(s) d˜q s d˜p x.
(7)
To find C1 and C2 , we first take the r-symmetric integral for g(t)u(t). We find that Z t 0
g(s)u(s) d˜r s = C2
Z t 0
g(s) d˜r s + C1
Z t 0
sg(s) d˜r s +
Z tZ yZ x 0
0
0
g(y)h(s) d˜q s d˜p x d˜r y.
(8)
We apply condition (3) to (7). Then, we have C1 λT − (1 − λ)C2 = −λ
Z TZ x 0
0
h( x ) d˜q s d˜p x.
(9)
We next apply condition (4) to (8). Then we have C1
Z η 0
sg(s) d˜q s + C2
Z η 0
g(s) d˜q s = −
Z ηZ yZ x 0
0
0
g(y)h(s) d˜q s d˜p x d˜r y.
(10)
Constants C1 and C2 are obtained by solving the system of Equations (9) and (10) as follows "Z # Z TZ x Z ηZ yZ x η 1 g(s) d˜q s · λ h( x ) d˜q s d˜p x − (1 − λ) g(y)h(s) d˜q s d˜p x d˜r y , C1 = − Λ 0 0 0 0 0 0 "Z # Z TZ x Z ηZ yZ x η 1 ˜ ˜ ˜ ˜ ˜ ˜ C2 = sg(s) dq s · λ h( x ) dq s d p x − λT g ( y ) h ( s ) d q s d p x dr y . Λ 0 0 0 0 0 0 Employing these results in (7), we get the solution (5).
Mathematics 2018, 6, 218
5 of 9
3. Main Results To study the existence and uniqueness of solution of (1), we transform the boundary value problem (1) into a fixed point problem. Let C = C ( IχT , R) denote the Banach space of all functions u. The norm is defined by kuk = sup |u(t)|. The operator T : C → C is defined by t∈ IχT
1 (T u)(t) = Λ
"Z
η
sg(s) d˜q s · λ
0
Z ηZ yZ x
− λT t + Λ
"Z
0
η
0
g(s) d˜q s · λ
0
− (1 − λ ) +
Z tZ x 0
0
Z TZ x 0
#
g(y) F s, u(s), (Sθ u)(s), ( Zω u)(s) d˜q s d˜p x d˜r y Z TZ x 0
Z ηZ yZ x 0
0
F x, u( x ), (Sθ u)( x ), ( Zω u)( x ) d˜q s d˜p x
0
0
0
(11)
F x, u( x ), (Sθ u)( x ), ( Zω u)( x ) d˜q s d˜p x
#
g(y) F s, u(s), (Sθ u)(s), ( Zω u)(s) d˜q s d˜p x d˜r y
F x, u( x ), (Sθ u)( x ), ( Zω u)( x ) d˜q s d˜p x,
0
p1 q1 p2 , q = q2 , 1 ; LCM p2 ,q2 ,r2 ,ω2 ,θ2
where IχT := {χk T : k ∈ N} ∪ {0, T }, p, q ∈ (0, 1), p =
r =
are the simplest form of proper fractions and χ = C IχT × R × R × R, R ; and for ϕ, φ ∈ C IχT × IχT , [0, ∞) , define Sθ u ( t ) : =
Z t 0
ϕ(s, t)u(s)d˜θ s and Zω u(t) :=
Z t 0
r1 r2 ,
g ∈ C
ω1 θ1 ω2 , θ = θ 2 IχT , R+ and F ∈
ω =
φ(s, t)u(s)d˜ω s.
We note that problem (1) has solutions if and only if the operator T has fixed points. To present our results, we establish the following theorem based on Banach’s fixed point theorem. Theorem 1. Assume F ∈ C IχT × R × R × R, R , g ∈ C IχT , R+ and ϕ, φ ∈ C IχT × IχT , [0, ∞) . Let ϕ0 := sup { ϕ(t, s)} and φ0 := sup {φ(t, s)}. Assume F and g satisfy the following conditions: (t,s)∈ IχT × IχT
(t,s)∈ IχT × IχT
( H1 ) there exist positive constants L1 , L2 , L3 such that F t, u, Sθ u, Zω u − F t, v, Sθ v, Zω v ≤ L1 |u − v| + L2 Sθ u − Sθ v + L3 Zθ u − Zθ v , for all t ∈ IχT and u, v, Sθ u, Zω u ∈ R,
( H2 ) 0 < g(t) < G for all t ∈ IχT . If ( Ξ = L1 + T ( L2 ϕ0 + L3 φ0 )
pT 2 Θ+ 1 + p2
)
< 1,
where " 1 pη 2 TG Θ := ˆ 1 + p2 Λ
q r λT + λη 1 + q2 (1 + r2 )(1 + r4 )
!
(12)
Mathematics 2018, 6, 218
6 of 9
pηTG r + λT 2 + (1 − λ ) η 2 2 2 ˆ (1 + r )(1 + r4 ) Λ (1 + p )
!# ,
(13)
ˆ is given by (15), then the boundary value problem (1) has a unique solution. where Λ Proof. For any u, v ∈ C and for each t ∈ IχT , we have F t, u, Sθ u, Zω u − F t, v, Sθ v, Zω v ≤ L1 |u − v| + L2 Sθ u − Sθ v + L3 Zω u − Zω v Z t
= L1 | u − v | + L2 ϕ(s, t)|u(s) − v(s)|d˜θ s + L3 0 ≤ ku − vk L1 + T ( L2 ϕ0 + L3 φ0 ) ,
Z t 0
φ(s, t)|u(s) − v(s)|d˜ω s (14)
and
|Λ| ≥ λTg
Z η
d˜q s + |1 − λ| g
0
Z η 0
qgη 2 ˆ |1 − λ| =: Λ. 1 + q2
s d˜q s = λTgη +
(15)
From (14) and (15), we have (T u)(t) − (T v)(t) −vk ≤ k uΛ L1 + T ( L2 ϕ0 + L3 φ0 ) × ˆ # (" RηRyRx Rη RTRx d˜q s d˜p x d˜r y d˜q s d˜p x + λTG Gλ s d˜q s · 0
0
0
0
0
0
" ˜ ˜ Gλ 0 0 0 d q s d p x + (1 − λ ) G ( " 1 pη 2 TG ku − vk L1 + T ( L2 ϕ0 + L3 φ0 ) ˆ 1+ p2 Λ !#
RηRyRx
+ Λˆ pηTG λT 2 + (1+ p2 )
+ 1pT + p2 )
T +Λ ˆ
≤
Rη
d˜q s ·
RTRx
r (1 − λ ) η 2 (1+r2 )(1+r4 )
(
= ku − vk L1 + T ( L2 ϕ0 + L3 φ0 )
Θ+
pT 2 1+ p2
0
0
0
) # RTRx ˜ ˜ ˜ ˜ ˜ d q s d p x dr y + 0 0 d q s d p x !
q λT 1+ q2 2
+
(16)
r λη (1+r2 )(1+r4 )
)
.
As Ξ < 1, T is a contraction. Therefore, the proof is done based on Banach’s contraction mapping principle. Furthermore, we prove the existence of a solution to the boundary value problem (1) by using the Krasnoselskii’s fixed point theorem. Theorem 2. [14] Let K be a bounded closed convex and nonempty subset of a Banach space X. Let A, B be operators such that:
(i ) Ax + By ∈ K whenever x, y ∈ K, (ii ) A is compact and continuous, (iii ) B is a contraction mapping. Then there exists z ∈ K such that z = Az + Bz. Theorem 3. Assume that ( H1 ) and ( H2 ) hold. In addition we suppose that: ( H3 ) F t, u, Sθ u, Zω u ≤ µ(t), for all t, u, Sθ u, Zω u ∈ IχT × R × R × R, with µ ∈ C ( IχT , R+ ).
Mathematics 2018, 6, 218
7 of 9
If Θ < 1,
(17)
where Θ is given by (13), then the boundary value problem (1) has at least one solution on IχT . Proof. We let sup |µ(t)| = kµk and choose a constant t∈ IχT
(
pT 2 R ≥ kµk Θ + 1 + p2
) ,
(18)
where BR = {u ∈ C : kuk ≤ R}. Based on the results of Lemma 3, we define the operators T1 and T2 on the ball BR as
(T1 u)(t) = (T2 u)(t) =
Z tZ x 0
1 Λ
F x, u( x ), (Sθ u)( x ), ( Zω u)( x ) d˜q s d˜p x,
0
"Z
η
sg(s) d˜q s · λ
0
− λT t + Λ
"Z
η 0
Z ηZ yZ x 0
0
0
g(s) d˜q s · λ
− (1 − λ )
Z TZ x 0
0
#
g(y) F s, u(s), (Sθ u)(s), ( Zω u)(s) d˜q s d˜p x d˜r y Z TZ x 0
Z ηZ yZ x 0
0
F x, u( x ), (Sθ u)( x ), ( Zω u)( x ) d˜q s d˜p x
0
0
F x, u( x ), (Sθ u)( x ), ( Zω u)( x ) d˜q s d˜p x
#
g(y) F s, u(s), (Sθ u)(s), ( Zω u)(s) d˜q s d˜p x d˜r y .
We proceed similarly to Theorem 1 for u, v ∈ BR . Then we have (
kT1 x + T2 yk ≤
pT 2 kµk Θ + 1 + p2
)
≤ R,
which implies that T1 x + T2 y ∈ BR . Using (17), we find that T2 is a contraction mapping. From the continuity of F and the assumption ( H3 ), we see that an operator T1 is continuous and uniformly bounded on BR . For t1 , t2 ∈ IχT with t1 ≤ t2 and u ∈ BR , we have
|T1 x (t2 ) − T1 x (t1 )| =
Z Z t2 x F x, u( x ), (Sθ u)( x ), ( Zω u)( x ) d˜q s d˜p x 0 0 ˜ ˜ − F x, u( x ), (Sθ u)( x ), ( Zω u)( x ) dq s d p x 0 0 p k F k t22 − t21 . 1 + p2 Z t Z x 1
≤
We observe that the above inequality tends to zero when t2 − t1 → 0. Therefore T1 is relatively compact on BR . Hence, we can conclude by the Arzelá-Ascoli Theorem that T1 is compact on BR . We find that the assumptions of Theorem 2 are satisfied implying that the boundary value problem (1) has at least one solution on IχT . The proof is complete.
Mathematics 2018, 6, 218
8 of 9
4. Examples In this section, we provide some examples to illustrate our main results. Consider the following boundary value problem of sequential q-symetric difference equations ˜ 1D ˜ 3 u(t) = D 2
4
2 e−1−sin (πt) |u(t)| 1 1 + S 1 u (t) + Z 2 u ( t ), 2 2 2 5 3 (100 + t) (1 + |u(t)|) 100e 100π
t ∈ I 10 1 , 60
(19)
η = χ5 T =
1 21600 ,
Z 1 21600 u (0) = 1 u (10) , [100e + 20 cos2 s]u(s) d˜2 s = 0, 3 3 0 R t e −2| t − s | u ( s ) R t e−|t−s| u(s) where S 1 u (t) = 0 (t+10)2 d˜1 s and Z 2 u (t) = 0 (t+20)2 d˜2 s. 3
Set p =
3 4,
λ = 13 , ϕ(s, t) = (I)
q =
1 2,
e−|t−s| (t+10)2
r =
2 3,
5
3
ω =
and φ(s, t) =
5
2 1 5, θ = 3, e −2| t − s | , (t+20)2
T = 10, χ =
If g(t) = 100e + 20 cos2 t. We can show that ϕ0 = F t, u, S 1 u, Z 2 u − F t, v, S 1 v, Z 2 v 3
3
3
3
So, ( H1 ) is satisfied with L1 =
1 , 1002 e
L2 =
1 100e2
1 100 ,
≤
1 L.C.M (4,2,3,5,3)
φ0 =
1 400
=
1 60 ,
and
1 1 | u − v | + S 1 u − S 1 v 3 1002 e 100e2 3 1 + Z 2 u − Z 2 v . 5 100π 2 5
and L3 =
1 . 100π 2
By ( H2 ), we have g = 100e ≤ g(t) ≤ 100e + 20 = G. ˆ = 0.0419 and Θ = 1231.332. Clearly, Λ Hence, we get (
L1 + T ( L2 ϕ0 + L3 φ0 )
pT 2 Θ+ 1 + p2
)
= 0.253 < 1.
Therefore, by Theorem 1, problem (19) has a unique solution on I 10 1 .
60
(II) If g(t) = 1003 e + 30 cos2 t. We can show that F t, u, S 1 u, Z 2 u 3
3
2
≤
e−1−sin πt e−t t e−2t t + · + · 2 2 2 3 (100 + t) 100e (t + 10) 100π (t + 20)2
