R E S E A R C H
Open Access
Separated boundary value problems for
second-order impulsive
q
-integro-difference
equations
Chatthai Thaiprayoon
1, Jessada Tariboon
1*and Sotiris K Ntouyas
2*Correspondence:
1Department of Mathematics,
Faculty of Applied Science, King Mongkut’s University of Technology, North Bangkok, Bangkok, Thailand Full list of author information is available at the end of the article
Abstract
This article studies the existence and uniqueness of solutions for a boundary value problem of nonlinear second-order impulsiveqk-integro-difference equations with
separated boundary conditions. Several new results are obtained by applying a variety of fixed point theorems. Some examples are presented to illustrate the results. MSC: 26A33; 39A13; 34A37
Keywords: qk-derivative;qk-integral; impulsiveqk-difference equation; existence;
uniqueness; fixed point theorems
1 Introduction
In this paper, we study the separated boundary value problem for impulsiveqk
-integro-difference equation of the following form: ⎧
⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
D
qkx(t) =f(t,x(t), (Sqkx)(t)), t∈J:= [,T],t=tk,
x(tk) =Ik(x(tk)), k= , , . . . ,m,
Dqkx(tk+) –Dqk–x(tk) =Ik∗(x(tk)), k= , , . . . ,m,
x() +Dqx() = , x(T) +Dqmx(T) = ,
(.)
where =t<t<t<· · ·<tk<· · ·<tm<tm+=T,f :J×R→R,
(Sqkx)(t) =
t
tk
φ(t,s)x(s)dqks, t∈(tk,tk+],k= , , , . . . ,m, (.)
φ:J×J→[,∞) is a continuous function,Ik,Ik∗∈C(R,R),x(tk) =x(t+k) –x(tk) fork=
, , . . . ,m,x(tk+) =limh→x(tk+h) and <qk< fork= , , , . . . ,m.
The notions ofqk-derivative andqk-integral on finite intervals were introduced in [].
For a fixedk∈N∪ {}letJk:= [tk,tk+]⊂Rbe an interval and <qk< be a constant. We
defineqk-derivative of a functionf :Jk→Rat a pointt∈Jkas follows.
Definition . Assumef :Jk→Ris a continuous function and lett∈Jk. Then the
ex-pression
Dqkf(t) =
f(t) –f(qkt+ ( –qk)tk)
( –qk)(t–tk)
, t=tk, Dqkf(tk) = lim t→tk
Dqkf(t) (.)
is called theqk-derivative of functionf att.
We say thatf isqk-differentiable onJkprovidedDqkf(t) exists for allt∈Jk. Note that if
tk= andqk=qin (.), thenDqkf=Dqf, whereDqis the well-knownq-derivative of the
functionf(t) defined by
Dqf(t) =
f(t) –f(qt)
( –q)t . (.)
In addition, we should define the higherqk-derivative of functions.
Definition . Let f :Jk →Rbe a continuous function, we call the second-order qk
-derivativeD
qkf providedDqkf isqk-differentiable onJkwithD
qkf =Dqk(Dqkf) :Jk→R.
Similarly, we define higher orderqk-derivativeDnqk:Jk→R.
Theqk-integral is defined as follows.
Definition . Assumef :Jk→Ris a continuous function. Then theqk-integral is
de-fined by t
tk
f(s)dqks= ( –qk)(t–tk)
∞
n=
qnkfqnkt+ –qkn tk (.)
fort∈Jk. Moreover, ifa∈(tk,t) then the definiteqk-integral is defined by
t
a
f(s)dqks=
t
tk
f(s)dqks–
a
tk
f(s)dqks
= ( –qk)(t–tk)
∞
n=
qnkfqnkt+ –qnk tk
– ( –qk)(a–tk)
∞
n=
qnkfqnka+ –qnk tk .
Note that iftk= andqk=q, then (.) reduces toq-integral of a functionf(t), defined
bytf(s)dqs= ( –q)t
∞
n=qnf(qnt) fort∈[,∞).
For the basic properties ofqk-derivative andqk-integral we refer to [].
The book by Kac and Cheung [] covers many of the fundamental aspects of the quan-tum calculus. In recent years, the topic ofq-calculus has attracted the attention of several researchers and a variety of new results can be found in the papers [–] and the refer-ences cited therein.
Impulsive differential equations serve as basic models to study the dynamics of processes that are subject to sudden changes in their states. Recent development in this field has been motivated by many applied problems, such as control theory, population dynamics and medicine. For some recent works on the theory of impulsive differential equations, we refer the interested reader to the monographs [–].
value problem (.) into an equivalent integral equation. The main results are given in Section , while examples illustrating the results are presented in Section .
2 An auxiliary lemma
LetJ= [,T],J= [t,t],Jk= (tk,tk+] fork= , , . . . ,m. LetPC(J,R) ={x:J→R:x(t) is
continuous everywhere except for sometkat whichx(t+k) andx(tk–) exist andx(t–k) =x(tk),
k= , , . . . ,m}.PC(J,R) is a Banach space with the norm x PC=sup{|x(t)|;t∈J}.
We now consider the following linear case: ⎧
⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
D
qkx(t) =h(t), t∈J,t=tk,
x(tk) =Ik(x(tk)), k= , , . . . ,m,
Dqkx(tk+) –Dqk–x(tk) =Ik∗(x(tk)), k= , , . . . ,m,
x() +Dqx() = , x(T) +Dqmx(T) = ,
(.)
whereh:J→Ris a continuous function.
Lemma . The unique solution of problem(.)is given by
x(t) =
–t T
m
k=
tk
tk– s
tk–
h(r)dqk–r dqk–s+Ik
x(tk)
+
–t T
m
k=
tk
tk–
h(s)dqk–s+I ∗
k
x(tk) (T–tk+ )
+
–t T
T
tm
s
tm
h(r)dqmr dqms+
–t
T
T
tm
h(s)dqms
+
<tk<t
tk
tk– s
tk–
h(r)dqk–r dqk–s+Ik
x(tk)
+
<tk<t
tk
tk–
h(s)dqk–s+I ∗
k
x(tk) (t–tk) +
t
tk
s
tk
h(r)dqkr dqks, (.)
withbi=a(·) = for a>b.
Proof Taking theq-integral for the first equation of (.), fort∈J, we have
Dqx(t) =Dqx() + t
h(s)dqs, (.)
which leads to
Dqx(t) =Dqx() + t
h(s)dqs. (.)
Fort∈Jwe get byq-integrating (.),
x(t) =x() +Dqx()t+ t
s
h(r)dqr dqs
:=A+Bt+ t
s
In particular, fort=t, we obtain
x(t) =A+Bt+
t
s
h(r)dqr dqs. (.)
Fort∈J= (t,t],q-integrating (.), we have
Dqx(t) =Dqx
t+ + t
t
h(s)dqs.
Using the third condition of (.) with (.), it follows that
Dqx(t) =B+ t
h(s)dqs+I ∗
x(t) +
t
t
h(s)dqs. (.)
Takingq-integral to (.) fort∈J, we obtain
x(t) =xt+ +
B+ t
h(s)dqs+I ∗
x(t) (t–t)
+ t
t s
t
h(r)dqr dqs. (.)
Applying the second equation of (.) with (.) and (.), we get
x(t) =A+Bt+
t
s
h(r)dqr dqs+I
x(t)
+
B+ t
h(s)dqs+I ∗
x(t) (t–t) +
t
t s
t
h(r)dqr dqs
=A+Bt+ t
s
h(r)dqr dqs+I
x(t)
+ t
h(s)dqs+I ∗
x(t) (t–t) +
t
t s
t
h(r)dqr dqs.
Repeating the above process, fort∈Jk, we get
x(t) =A+Bt
+
<tk<t
tk
tk– s
tk–
h(r)dqk–r dqk–s+Ik
x(tk)
+
<tk<t
tk
tk–
h(s)dqk–s+I ∗
k
x(tk) (t–tk)
+ t
tk
s
tk
h(r)dqkr dqks. (.)
From the first boundary condition of (.) (i.e. x() +Dqx() = ) and (.), we have
Also, the second boundary condition of (.) (i.e. x(T) +Dqmx(T) = ) and (.), yields
Substituting constantsAandBinto (.), we obtain (.) as requested.
3 Main results
+
<tk<t
tk
tk–
fs,x(s), (Sqk–x)(s) dqk–s+I ∗
k
x(tk) (t–tk)
+ t
tk
s
tk
fr,x(r), (Sqkx)(r) dqkr dqks. (.)
It should be noticed that problem (.) has solutions if and only if the operatorAhas fixed points.
Our first result is an existence and uniqueness result for the impulsive boundary value problem (.) by using the Banach contraction mapping principle.
Letφ=max{φ(t,s) : (t,s)∈J×J}. Further, for convenience we set
ω= + T T
m+
k=
L(tk–tk–)
+qk–
+φL(tk–tk–)
+qk–+qk–
+ +T T
m+
k=
L(tk–tk–) +
φL(tk–tk–)
+qk–
+ + T T
m
k=
(T–tk)
L(tk–tk–) +
φL(tk–tk–)
+qk–
+mL( + T)
T +
mL( +T)
T +
L( + T)
T
m
k=
(T–tk), (.)
and
λ=
M( + T)
T
m+
k=
(tk–tk–)
+qk–
+M( +T) T
m+
k=
(tk–tk–)
+M( + T) T
m
k=
(T–tk)(tk–tk–) +
mM( + T)
T
+mM( +T)
T +
M( + T)
T
m
k=
(T–tk). (.)
Theorem . Assume that:
(H) The functionf :J×R→Ris continuous and there exist constantsL,L> such
that
ft,x, (Sqkx) –f
t,y, (Sqky) ≤L|x–y|+L
Sqk|x–y| ,
for eacht∈Jandx,y∈R,k= , , , . . . ,m.
(H) The functionsIk,Ik∗:R→Rare continuous and there exist constantsL,L> such
that
Ik(x) –Ik(y)≤L|x–y| and Ik∗(x) –Ik∗(y)≤L|x–y|,
If
ω≤δ< , (.)
whereωis defined by(.),andδ> ,then the boundary value problem(.)has a unique solution on J.
Proof We transform the boundary value problem (.) into a fixed point problem,x=Ax, where the operatorAis defined by (.). By using the Banach contraction mapping prin-ciple, we shall show thatAhas a fixed point which is the unique solution of the boundary value problem (.).
Let M, M, and M be nonnegative constants such that supt∈J|f(t, , )| = M,
sup{|Ik()|:k= , , . . . ,m}=M, andsup{|Ik∗()|:k= , , . . . ,m}=M. By choosing a
con-stantRas
R≥ λ –ε,
whereδ≤ε< andλdefined by (.), we will show thatABR⊂BR, where a ballBRis
defined byBR={x∈PC(J,R) : x ≤R}. Forx∈BR, we have Ax
≤sup
t∈J
–t
T
m
k=
tk
tk– s
tk–
fr,x(r), (Sqk–x)(r) dqk–r dqk–s+Ik
x(tk)
+
–t T
m
k=
tk
tk–
fs,x(s), (Sqk–x)(s) dqk–s+I ∗
k
x(tk) (T–tk+ )
+
–t T
T
tm
s
tm
fr,x(r), (Sqmx)(r) dqmr dqms
+
–t T
T
tm
fs,x(s), (Sqmx)(s) dqms
+
<tk<t
tk
tk– s
tk–
fr,x(r), (Sqk–x)(r) dqk–r dqk–s+Ik
x(tk)
+
<tk<t
tk
tk–
fs,x(s), (Sqk–x)(s) dqk–s+I ∗
k
x(tk) (t–tk)
+ t
tk
s
tk
fr,x(r), (Sqkx)(r) dqkr dqks
≤
+T
T
m
k=
tk
tk– s
tk–
fr,x(r), (Sqk–x)(r) dqk–r dqk–s+Ik
x(tk)
+
+T T
m
k=
tk
tk–
fs,x(s), (Sqk–x)(s) dqk–s+I ∗
k
x(tk)
(T–tk+ )
+
+T T
T
tm
s
tm
+ + T T
m
k=
(LR+M) +
+T T
m
k=
(LR+M) +
+ T T
m
k=
(LR+M)(T–tk)
=ωR+λ≤(δ+ –ε)R≤R,
which implies thatABR⊂BR.
For anyx,y∈PC(J,R) and for eacht∈J, we have Ax(t) –Ay(t)
≤ + T
T
m+
k=
tk
tk– s
tk–
fr,x(r), (Sqk–x)(r) –f
r,y(r), (Sqk–y)(r) dqk–r dqk–s
+ +T T
m+
k=
tk
tk–
fs,x(s), (Sqk–x)(s) –f
s,y(s), (Sqk–y)(s) dqk–s
+ + T T
m
k=
(T–tk)
tk
tk–
fs,x(s), (Sqk–x)(s) –f
s,x(s), (Sqk–x)(s) dqk–s
+ + T T
m
k=
Ik
x(tk) –Ik
y(tk) +
+T T
m
k=
Ik∗x(tk) –Ik∗
y(tk)
+ + T T
m
k=
I∗
k
x(tk) –Ik∗
y(tk) (T–tk)
≤ + T
T
m+
k=
L(tk–tk–)
+qk–
+φL(tk–tk–)
+qk–+qk–
x–y
+ +T T
m+
k=
L(tk–tk–) +
φL(tk–tk–)
+qk–
x–y
+ + T T
m
k=
(T–tk)
L(tk–tk–) +
φL(tk–tk–)
+qk–
x–y
+m( + T)L
T x–y +
m( +T)L
T x–y +
( + T)L x–y
T
m
k=
(T–tk)
=ω x–y ,
which implies that Ax–Ay ≤ω x–y . Asω< ,Ais a contraction. Therefore, by the Banach contraction mapping principle, we find thatAhas a fixed point which is the unique solution of problem (.). This completes the proof.
The second existence result is based on Schaefer’s fixed point theorem.
Theorem . Assume that:
(H) f :J×R→Ris a continuous function and there exists a constantN> such that
ft,x, (Sqkx) ≤N,
(H) The functionsIk,Ik∗:R→Rare continuous and there exist constantsN,N> such
that
Ik(x)≤N and Ik∗(x)≤N,
for allx∈R,k= , , . . . ,m.
Then the boundary value problem(.)has at least one solution on J.
Proof We will use Schaefer’s fixed point theorem to prove thatA, defined by (.), has a fixed point. We divide the proof into four steps.
Step :Continuity ofA.
Let{xn}be a sequence such thatxn→xinPC(J,R). Sincef is a continuous function on
J×RandI
k,I∗kare continuous functions onRfork= , , . . . , we have
ft,xn(t), (Sqkxn)(t) →f
t,x(t), (Sqkx)(t) ,
andIk(xn(tk))→Ik(x(tk)),Ik∗(xn(tk))→Ik∗(x(tk)) fork= , , . . . , asn→ ∞.
Then, for eacht∈J, we get (Axn)(t) – (Ax)(t)
=
–t T
m
k=
tk
tk– s
tk–
fr,xn(r), (Sqk–xn)(r)
–fr,x(r), (Sqk–x)(r) dqk–r dqk–s+Ik
xn(tk) –Ik
x(tk)
+
–t T
m
k=
tk
tk–
fs,xn(s), (Sqk–xn)(s) –f
s,x(s), (Sqk–x)(s) dqk–s
+Ik∗xn(tk) –Ik∗
x(tk)
(T–tk+ )
+
–t T
T
tm
s
tm
fr,xn(r), (Sqmxn)(r) –f
r,x(r), (Sqmx)(r) dqmr dqms
+
–t T
T
tm
fs,xn(s), (Sqmxn)(s) –f
s,x(s), (Sqmx)(s) dqms
+
<tk<t
tk
tk– s
tk–
fr,xn(r), (Sqk–xn)(r) –f
r,x(r), (Sqk–x)(r) dqk–r dqk–s
+Ik
xn(tk) –Ik
x(tk)
+
<tk<t
tk
tk– f
s,xn(s), (Sqk–xn)(s) –f
s,x(s), (Sqk–x)(s) dqk–s
+Ik∗xn(tk) –Ik∗
x(tk)
(t–tk)
+ t
tk
s
tk
fr,xn(r), (Sqkxn)(r) –f
r,x(r), (Sqkx)(r) dqkr dqks,
Step :Amaps bounded sets into bounded sets in PC(J,R).
So, let us prove that for anyr> , there exists a positive constantρsuch that for each x∈Br={x∈PC(J,R) : x ≤r}, we have Ax ≤ρ. For anyx∈Br, we have
Step :Amaps bounded sets into equicontinuous sets of PC(J,R).
+|τ–τ| T
m
k=
tk
tk–
fs,x(s), (Sqk–x)(s) dqk–s+I ∗
k
x(tk)
(T–tk+ )
+|τ–τ| T
T
tm
s
tm
fr,x(r), (Sqmx)(r) dqmr dqms
+|τ–τ| T
T
tm
fs,x(s), (Sqmx)(s) dqms
+|τ–τ| i
k=
tk
tk–
fs,x(s), (Sqk–x)(s) dqk–s+I ∗
k
x(tk)
+ τ
ti
s
ti
fr,x(r), (Sqix)(r) dqir dqis–
τ
ti
s
ti
fr,x(r), (Sqix)(r) dqir dqis
≤|τ–τ|
T
m
k=
N(tk–tk–)
+qk–
+N
+|τ–τ| T
m
k=
N(tk–tk–) +N
(T–tk+ ) +|
τ–τ|
T
N(T–tm)
+qm
+|τ–τ| T
N(T–tm)
+|τ–τ| i
k=
N(tk–tk–) +N
+|τ–τ|N +qi
(τ+τ+ ti).
Asτ→τ, the right-hand side of the above inequality (which is independent ofx) tends
to zero. As a consequence of Steps to , together with the Arzelá-Ascoli theorem, we deduce thatA:PC(J,R)→PC(J,R) is completely continuous.
Step :We show that the set
E=x∈PC(J,R) :x=θAx for some <θ<
is bounded.
Letx∈E. Thenx(t) =θ(Ax)(t) for some <θ < . Thus, for eacht∈J, by using the computations of Step , we have that
Ax ≤ρ.
This shows that the setEis bounded. As a consequence of Schaefer’s fixed point theo-rem, we conclude thatAhas a fixed point which is a solution of the impulsiveqk
-integro-difference boundary value problem (.).
The third existence result for the impulsive boundary value problem (.) is based on Krasnoselskii’s fixed point theorem.
Lemma .(Krasnoselskii’s fixed point theorem) [] Let M be a closed,bounded,convex and nonempty subset of a Banach space X.Let A,B be the operators such that
(c) Bis a contraction mapping.
Then there exists z∈M such that z=Az+Bz. For convenience we put
= + T T
m+
k=
(tk–tk–)
+qk–
+ +T T
m+
k=
(tk–tk–)
+ + T T
m
k=
(T–tk)(tk–tk–), (.)
and
λ=
m( + T)N
T +
m( +T)N
T +
( + T)N
T
m
k=
(T–tk). (.)
Theorem . Let f :J×R→Rbe a continuous function.Assume that:
(A) |f(t,x,y)| ≤μ(t),∀(t,x,y)∈J×R×Randμ∈C(J,R+).
(A) There exist constantsN,N> such that|Ik(x)| ≤N and|Ik∗(x)| ≤N,∀x∈R,for
k= , , . . . ,m.
(A) There exist constantsK,K> such that|Ik(x) –Ik(y)| ≤K|x–y|and|Ik∗(x) –Ik∗(y)| ≤
K|x–y|,∀x,y∈R,fork= , , . . . ,m.
If
m( + T)K
T +
m( +T)K
T +
( + T)K
T
m
k=
(T–tk) < , (.)
then boundary value problem(.)has at least one solution on J.
Proof We definesupt∈J|μ(t)|= μ and choose a suitable constantρas
ρ≥ μ +λ,
whereandλare defined by (.) and (.), respectively. We define the operatorsand
onBρ={x∈PC(J,R) : x ≤ρ}as
(x)(t) =
–t T
m
k=
tk
tk– s
tk–
fr,x(r), (Sqk–x)(r) dqk–r dqk–s
+
–t T
m
k=
tk
tk–
fs,x(s), (Sqk–x)(s) dqk–s
(T–tk+ )
+
–t T
T
tm
s
tm
fr,x(r), (Sqmx)(r) dqmr dqms
+
–t T
T
tm
+
<tk<t
tk
tk– s
tk–
fr,x(r), (Sqk–x)(r) dqk–r dqk–s
+
<tk<t
tk
tk–
fs,x(s), (Sqk–x)(s) dqk–s
(t–tk)
+ t
tk
s
tk
fr,x(r), (Sqkx)(r) dqkr dqks
and
(x)(t) =
–t T
m
k=
Ik
x(tk) +
–t
T
m
k=
Ik∗x(tk) (T–tk+ )
+
<tk<t
Ik
x(tk) +
<tk<t
Ik∗x(tk) (t–tk).
Forx,y∈Bρ, we have
x+y ≤ μ
+ T T
m+
k=
(tk–tk–)
+qk–
+ +T T
m+
k=
(tk–tk–)
+ + T T
m
k=
(T–tk)(tk–tk–)
+m( + T)N
T +
m( +T)N
T +
( + T)N
T
m
k=
(T–tk)
= μ +λ
≤ρ. Thus,x+y∈Bρ.
Forx,y∈PC(J,R), from (A), we have
x–y ≤ + T T
m
k=
Ik
x(tk) –Ik
y(tk) +
+T T
m
k=
Ik∗x(tk) –Ik∗
y(tk)
+ + T T
m
k=
Ik∗x(tk) –Ik∗
y(tk) (T–tk)
≤ m( + T)K x–y
T +
m( +T)K x–y
T
+( + T)K x–y T
m
k=
(T–tk),
which implies, by (.), thatis a contraction mapping.
Continuity off implies that the operatoris continuous. Also,is uniformly bounded onBρas
x ≤ μ .
We definesup(t,x)∈J×Bρ|f(t,x)|=f <∞,τ,τ∈(ti,ti+) for somei∈ {, , . . . ,m}withτ<
τand consequently we get
(x)(τ) – (x)(τ)
≤|τ–τ|
T
m
k=
tk
tk– s
tk– f
r,x(r), (Sqk–x)(r) dqk–r dqk–s
+|τ–τ| T
m
k=
tk
tk–
fs,x(s), (Sqk–x)(s) dqk–s
(T–tk+ )
+|τ–τ| T
T
tm
s
tm
fr,x(r), (Sqmx)(r) dqmr dqms
+|τ–τ| T
T
tm
fs,x(s), (Sqmx)(s) dqms
+|τ–τ| i
k=
tk
tk–
fs,x(s), (Sqk–x)(s) dqk–s
+ τ
ti
s
ti
fr,x(r), (Sqix)(r) dqir dqis–
τ
ti
s
ti
fr,x(r), (Sqix)(r) dqir dqis
≤|τ–τ|f
T
m
k=
(tk–tk–)
+qk–
+|τ–τ|f T
m
k=
(tk–tk–)(T–tk+ )
+|τ–τ|f T
(T–tm)
+qm
+|τ–τ|f
T (T–tm) +|τ–τ|f
i
k=
(tk–tk–)
+|τ–τ|f +qi
(τ+τ+ ti),
which is independent ofxand tends to zero asτ–τ→. Thus,is equicontinuous. So
is relatively compact onBρ. Hence, by the Arzelá-Ascoli theorem,is compact onBρ.
Thus all the assumptions of Lemma . are satisfied. So the boundary value problem (.) has at least one solution onJ. The proof is completed.
Our final, fourth existence result is based on the Leray-Schauder Nonlinear Alternative.
Lemma .(Nonlinear alternative for single valued maps) [] Let E be a Banach space, C a closed,convex subset of E,U an open subset of C and∈U.Suppose that F:U→C is a continuous,compact(that is,F(U)is a relatively compact subset of C)map.Then either
(i) Fhas a fixed point inU,or
(ii) there is au∈∂U(the boundary ofUinC)andθ∈(, )withu=θF(u). Theorem . Assume that:
(A) There exist a continuous nondecreasing functionψ: [,∞)→(,∞)and a continuous
functionp:J→R+such that
(A) There exist continuous nondecreasing functionsϕ,ϕ: [,∞)→(,∞)such that
Ik(x)≤ϕ
|x| and Ik∗(x)≤ϕ
|x| ,
for allx∈R,k= , , . . . ,m.
(A) There exists a constantM∗> such that
M∗
pψ(M∗)Q+φM∗Q+Q
> ,
wherep=max{p(t) :t∈J},φ=max{φ(t,s) : (t,s)∈J×J}and
Q=
+ T T
m+
k=
(tk–tk–)
+qk–
+ +T T
m+
k=
(tk–tk–)
+ + T T
m
k=
(T–tk)(tk–tk–),
Q=
+ T T
m+
k=
(tk–tk–)
+qk–+qk–
+ +T T
m+
k=
(tk–tk–)
+qk–
+ + T T
m
k=
(T–tk)
(tk–tk–)
+qk–
,
Q=
m( + T)ϕ(M∗)
T +
m( +T)ϕ(M∗)
T +
( + T)ϕ(M∗)
T
m
k=
(T–tk).
Then the impulsive boundary value problem(.)has at least one solution on J.
Proof First we show thatAmaps bounded sets (balls) into bounded sets in PC(J,R).For a positive numberρ, letBρ={x∈PC(J,R) : x ≤ρ}be a bounded ball inPC(J,R). Then
fort∈Jwe have (Ax)(t)
≤| –t|
T
m
k=
tk
tk– s
tk–
fr,x(r), (Sqk–x)(r) dqk–r dqk–s+Ik
x(tk)
+| –t| T
m
k=
tk
tk–
fs,x(s), (Sqk–x)(s) dqk–s+I ∗
k
x(tk)
(T–tk+ )
+| –t| T
T
tm
s
tm
f
r,x(r), (Sqmx)(r) dqmr dqms
+| –t| T
T
tm
fs,x(s), (Sqmx)(s) dqms
+
m
k=
tk
tk– s
tk–
fr,x(r), (Sqk–x)(r) dqk–r dqk–s+Ik
x(tk)
+
m
k=
tk
tk–
fs,x(s), (Sqk–x)(s) dqk–s+I ∗
k
x(tk)
+
Next we show thatAmaps bounded sets into equicontinuous sets of PC(J,R).
+|τ–τ|
The right-hand side of the above inequality, which is independent ofx, tends to zero as τ→τ. From all above and by the Arzelá-Ascoli theoremA:PC(J,R)→PC(J,R) is
com-pletely continuous.
+m( + T)ϕ( x )
T +
m( +T)ϕ( x )
T
+( + T)ϕ( x ) T
m
k=
(T–tk)
=pψ
x
+ T
T
m+
k=
(tk–tk–)
+qk–
+ +T T
m+
k=
(tk–tk–)
+ + T T
m
k=
(T–tk)(tk–tk–)
+φ x
+ T
T
m+
k=
(tk–tk–)
+qk–+qk–
+ +T T
m+
k=
(tk–tk–)
+qk–
+ + T T
m
k=
(T–tk)
(tk–tk–)
+qk–
+m( + T)ϕ( x ) T
+m( +T)ϕ( x )
T +
( + T)ϕ( x )
T
m
k=
(T–tk)
=pψ
x Q+φ x Q+Q.
Consequently, we have
x
pψ( x )Q+φ x Q+Q
≤.
In view of (A), there existsM∗such that x =M∗. Let us set
U=x∈PC(J,R) : x <M∗.
Note that the operatorA:U→PC(J,R) is continuous and completely continuous. From the choice ofU, there is nox∈∂U such thatx=θAxfor someθ ∈(, ). Consequently, by the nonlinear alternative of Leray-Schauder type (Lemma .), we deduce thatAhas a fixed pointx∈Uwhich is a solution of the problem (.). This completes the proof. 4 Examples
In this section, we will give examples to illustrate our main results.
Example . Consider the following boundary value problem for the nonlinear second-order impulsiveqk-integro-difference equation:
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
D k+
k+
x(t) = et(tt+) |x(t)|
|x(t)|++ t
t tk
sinπ(t–s)
x(s)dqks, t∈J,t=tk,
x(tk) = (k+)+|x(tk|)x|(tk)|, k= , , . . . , ,
Dk+
k+x(t
+ k) –D k
k+x(tk) = |x(tk)|
(k+)+|x(tk)|, k= , , . . . , ,
x() +D
x() = , x() +Dx() = .
(.)
HereJ= [, ],tk=k/,qk= (k+ )/(k+ ) fork= , , , . . . , ,m= ,T= ,f(t,x,Sqkx) =
(t|x|)/(et(t+ )(|x|+ )) + (t/)t
Ik∗(x) =|x|/((k+ ) +|x|). Since f(t,x) –f(t,y)≤
|x–y|+ (/)
Sqk|x–y| ,
Ik(x) –Ik(y)≤
|x–y| and I ∗
k(x) –Ik∗(y)≤
|x–y|,
(H) and (H) are satisfied withL= (/),L= (/),L= (/),L= (/). We can
show that
ω≈. < .
Hence, by Theorem ., boundary value problem (.) has a unique solution on [, ].
Example . Consider the following boundary value problem for the nonlinear second-order impulsiveqk-integro-difference equation:
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
D
k+ k+
x(t) = t
(+x)+ t sint
t|tkt cosπ(t–s)x(s)dqks|+, t∈J,t=tk,
x(tk) = kk+sint|x(πtt
k)|, k= , , . . . , ,
Dk+ k+x(t
+
k) –Dkk++x(tk) =
kcost
k+|x(tk)|sint, k= , , . . . , ,
x() +D
x() = , x() +Dx() = .
(.)
HereJ= [, ],tk=k/,qk= (k+)/(k+) fork= , , , . . . , ,m= ,T= ,f(t,x,Sqkx) =
((t)/(+x))+(tsint)/(t|t tk((cos
π(t–s))/)x(s)d
qks|+),Ik(x) = (ksinπt)/(k+t|x(tk)|)
andIk∗(x) = (kcost)/(k+|x(t
k)|sint). We can show that
ft,x, (Sqkx) =
( +tx) +
tsint
t|tkt cosπ(t–s)x(s)dqks|+
≤ =N,
Ik(x)=
kk+sint|x(tπtk)|≤ =N, and Ik∗(x)=
k+k|x(tcosk)|tsint≤ =N.
Hence, by Theorem ., boundary value problem (.) has at least one solution on [, ].
Example . Consider the following nonlinear second-order impulsive qk-difference
equation with separated boundary condition: ⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
Dk+ k+
x(t) =t(sint+)πt |
x|
|x|++
cosπt
|t tk et
–s t+et–sx(s)dqks|
, t∈J,t=tk,
x(tk) = (k+)+|x(tk|)x|(tk)|, k= , , . . . , ,
Dk+ k+x(t
+ k) –D k
k–x(tk) = |x(tk)|
(k+)+|x(tk)|, k= , , . . . , , x() +D
x() = , x() +Dx() = ,
(.)
where J = [, ], tk =k/, qk = (k + )/(k + ) for k = , , , . . . , , m = , T = ,
f(t,x) = (tsinπt/(t+ ))(|x|/(|x|+ )) + (cos(πt/))/(|t
tk(et–s/(t+et–s))x(s)dqks|),Ik(x) =
|x(tk)|/((k+ ) +|x(tk)|) andIk∗(x) =|x(tk)|/((k+ ) +|x(tk)|). Since
(A) is satisfied withL= (/), L= (/). It is easy to verify that|f(t,x)| ≤μ(t)≡
(tsinπt)/(t+ )+e–tcos(πt/),I
k(x)≤N= , andI∗k(x)≤N= for allt∈[, ],x∈R,
k= , . . . ,m. Thus (A) and (A) are satisfied. We can show that
m( + T)K
T +
m( +T)K
T +
( + T)K
T
m
k=
(T–tk)≈. < .
Hence, by Theorem ., boundary value problem (.) has at least one solution on [, ].
Example . Consider the following nonlinear second-order impulsive qk-difference
equation with separated boundary condition: ⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
D k+
k+
x(t) = x+xet+++
sinπt et+ –cost
t tk
(t–s)sint
et–s+ x(s)dqks, t∈J,t=tk,
x(tk) = sinππ(kx+)(tk), k= , , . . . , ,
Dk+
k+x(t
+ k) –Dk+
k+x(tk) =
x(tk)coskt
(k+) , k= , , . . . , ,
x() +D
x() = , x() +Dx() = .
(.)
Here J= [, ],tk=k/,qk= (k+ )/(k+ ) fork= , , , . . . , ,m= , T= , f(t,x) =
(x/(x + et + )) + ( +sinπt)/(et + ) –costt
tk((t–s)sint/(e
t–s+ ))x(s)d
qks,Ik(x) =
(sinπx)/(π(k+ )), andIk∗(x) = (xcoskt)/((k+ )). Clearly,
f(t,x)= x x+ et+ +
+sinπt et+ –cost
t
tk
(t–s)sint et–s+ x(s)dqks
≤
et+
| x+ |
+ tt
k
(t–s)sint et–s+ x(s)dqks
, Ik(x)=
π(ksinπx+ )≤|x|
and I ∗
k(x)=
(kxcos+ )kt≤|x|
.
Choosingp(t) =
et+,ψ(|x|) =|x+|,ϕ(|u|) =|x|, andϕ(|x|) = |x|, we obtain
M∗
.M∗+ .>
which implies thatM∗> .. Hence, by Theorem ., boundary value problem (.) has at least one solution on [, ].
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors contributed equally in this article. They read and approved the final manuscript.
Authors’ information
Sotiris K Ntouyas is a member of Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group at King Abdulaziz University, Jeddah, Saudi Arabia.
Author details
1Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology, North Bangkok,
Acknowledgements
The research of C Thaiprayoon and J Tariboon is supported by King Mongkut’s University of Technology North Bangkok, Thailand.
Received: 3 December 2013 Accepted: 28 February 2014 Published:17 Mar 2014 References
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Cite this article as:Thaiprayoon et al.:Separated boundary value problems for second-order impulsive