# On the Second-Order Quantum(p,q)-Difference Equations with Separated Boundary Conditions

Authors: Chanon Promsakon; Nattapong Kamsrisuk; Sotiris K. Ntouyas; Jessada Tariboon
Journal: Advances in Mathematical Physics
 (2018)

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## Abstract

In this paper, we investigate the existence and uniqueness of solutions for a boundary value problem for second-order quantum(p,q)-difference equations with separated boundary conditions, by using classical fixed point theorems. Examples illustrating the main results are also presented.


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## Body

## 1. Introduction

Quantum calculus orq-calculus is known as the study of calculus without limits. The study of q-calculus was initiated by Euler on studying infinite series. In 1910, Jackson [1, 2] was the first one, who establised the q-derivative or q-difference operator for a function f on [0,∞) by(1)Dqft=ft-fqt1-qt,t≠0,andDqf0=limt→0Dqft,t=0, and gave its properties. The q-integral of a function f on [0,∞) is defined by(2)∫0tfsdqs=1-qt∑n=0∞qnfqnt,provided that infinite series converges. The general theory of linear quantum difference equations was published in 1912 by Carmichael [3]. Details of its basic notions, results, and methods can be found in the text [4]. For other papers on the subject, see [5–7]. In recent years, the topic has been attracting the attention of several researchers and a variety of new results can be found in [8–13] and the references therein. In addition, the classical quantum calculus was generalized to qk-calculus by Tariboon and Ntouyas [14]. For details, we refer to the recent monograph [15].There are some applications ofq-calculus and difference equations to molecular problems in physics. In 1967, Finkelstein [16] studied behaviors of hydrogen atoms by using Schrödinger equation and q-calculus. In [17], the author investigated the q-field theory. The q-Coulomb problem and q-hydrogen atom were studied by [18–21]. In addition, Yang-Mills theories and also q-Yang-Mills equation were developed by [22–24]. The theory of quantum group applied to vibration and rotation molecules with q-algebra and q-Heisemberg algebra technique was established in [25–27]. The subject of elementary particle physics and chemical physics using q-calculus was investigated in [28–31]. The string theory involving q-calculus was studied in [32].The subject of(p,q)-calculus first appeared in quantum algebras which contained two quantum numbers p and q introduced by [33]. For some recent results see [34–41] and the references cited therein. Recently in [42], the authors initiated the study on boundary value problems for (p,q)-difference equations by investigating the existence and uniqueness of solution for first-order quantum (p,q)-difference equation subject to a nonlocal condition of the form(3)Dp,qxt=ft,xpt,t∈0,Tp,(4)x0=αxT+∑i=1mβi∫0ηixsdpi,qis,where 0<q<p≤1, 0<qi<pi≤1, i=1,2,…,m, are quantum numbers, Dp,q is (p,q)-difference operator, f∈C([0,T/p]×R,R), T>0, α, βi, i=1,2,…,m, are given constants, and the points ηi∈[0,piT], i=1,2,…,m. They proved existence and uniqueness results for problem (3)-(4) by using the classical fixed point theorems, such as Banach’s fixed point theorem, Boyd and Wong fixed point theorem for nonlinear contractions, and Leray-Schauder nonlinear alternative.In this paper, we continue the study on(p,q) boundary value problems, by considering the following second-order quantum (p,q)-difference equation with separated boundary conditions(5)Dp,q2xt=ft,xp2t,t∈0,Tp2,α1x0+α2Dp,qx0=α3,β1xT+β2Dp,qxTp=β3,where 0<q<p≤1 are two given quantum numbers, Dp,q2 is the second-order (p,q)-difference operator, and f∈C([0,T/p2]×R,R), T>0, αi, βi, i=1,2,3, are given real constants. Some new existence results are established by using Schaefer fixed point theorem, Krasnoselskii’s fixed point theorem, and Lelay-Schauder’s nonlinear alternative. In addition the uniqueness of solutions is established via Banach contraction mapping principle.The paper is organized as follows: In Section2, we recall some definitions and basic facts from (p,q)-calculus. The main existence and uniqueness results are given in Section 3. Examples illustrating the obtained results are presented in Section 4.


## 2. Preliminaries

In this section, we recall some basic concepts of(p,q)-calculus (see [41]). The (p,q)-number is defined by(6)np,q=pn-qnp-q,where 0<q<p≤1. For each k,n∈N, n≥k≥0, the (p,q)-factorial and (p,q)-binomial are defined by(7)np,q!=∏k=1nkp,q,n≥1,0p,q!=1,and(8)nkp,q=np,q!n-kp,q!kp,q!,respectively.Letf:[0,T]→R. The (p,q)-derivative of function f is defined as(9)Dp,qft=fpt-fqtp-qt,t≠0,and Dp,qf(0)=limt→0Dp,qf(t). Observe that the function Dp,qf(t) is defined on [0,T/p] which is extended from [0,T] of a function f(t). In addition, we say that f is (p,q)-differentiable on [0,T/p] provided that Dp,qf(t) exists for all t∈[0,T/p].Let a functionf:[0,T]→R. Then the (p,q)-integral of f is defined by(10)∫0tfsdp,qs=p-qt∑n=0∞qnpn+1fqnpn+1t,provided that the right hand side converges. Note that the domain of function ∫0t‍f(s)dp,qs is [0,pT] which shrinks from [0,T] of a function f(t) since 0<p≤1.In the following theorems we collect the basic properties of(p,q)-differentiation and (p,q)-integration, respectively. See [37].Theorem 1 (see [37]).
Suppose thatf,g:[0,T]→R are (p,q)-differentiable on [0,T/p]. Then
(a)f+g:[0,T]→R is (p,q)-differentiable on [0,T/p], and(11)Dp,qft+gt=Dp,qft+Dp,qgt.
(b)λf:[0,T]→R is (p,q)-differentiable on [0,T/p] for any constant λ, and(12)Dp,qλft=λDp,qft.
(c)fg:[0,T]→R is (p,q)-differentiable on [0,T/p], and(13)Dp,qfgt=fptDp,qgt+gqtDp,qft.Dp,qfgt=gptDp,qft+fqtDp,qgt.
(d) Ifg(t)≠0, then f/g is (p,q)-differentiable on [0,T/p] with(14)Dp,qfgt=gqtDp,qft-fqtDp,qgtgptgqt.Dp,qfgt=gptDp,qft-fptDp,qgtgptgqt.Theorem 2 (see [37]).
Letf,g:[0,T]→R be continuous functions and constants 0<q<p≤1, a,b∈[0,T]. The following formulas hold:
(a) The(p,q)-integration by parts is given by (15)∫abfptDp,qgtdp,qt=ftgtab-∫abgqtDp,qftdp,qt.
(b)Dp,q∫0t‍f(s)dp,qs=f(t).
(c)∫0t‍Dp,qf(s)dp,qs=f(t)-f(0).
(d)∫at‍Dp,qf(s)dp,qs=f(t)-f(a) where a∈(0,t).Theorem 3 (see [42]).
Let a functionf:[0,T]→R and constants 0<q<p≤1. Then for t∈[0,p2T], we have(16)∫0t∫0sfrdp,qrdp,qs=1p∫0tt-qsf1psdp,qs.The second-order(p,q)-difference of a function f:[0,T]→R can be expressed as (17)Dp,q2ft=Dp,qDp,qft=Dp,qfpt-fqtp-qt=fp2t-fpqtpp-q2t2-fpqt-fq2tqp-q2t2=qfp2t-qfpqt-pfpqt+pfq2tpqp-q2t2=qfp2t-p+qfpqt+pfq2tpqp-q2t2,t∈0,Tp2.Now, forh:[0,T/p2]→R, we transform the linear second-order (p,q)-difference equation with separated boundary condition(18)Dp,q2xt=ht,t∈0,Tp2,α1x0+α2Dp,qx0=α3,β1xT+β2Dp,qxTp=β3, into an integral equation which will be used to define the solution of the boundary value problem (5). For convenience we put constants Λ=α1(β1T+β2)-α2β1≠0, (19)Λ1=α3β1T+β2-α2β3Λ,Λ2=α1β3-α3β1Λ.Lemma 4.
The unique solution of linear problem (18) is given by(20)xt=Λ1+Λ2t+α2-α1tβ1pΛ∫0TT-qshspdp,qs+α2-α1tβ2Λ∫0T/phsdp,qs+1p∫0tt-qshspdp,qs.Proof.
Applying(p,q)-integration to the first equation of (18), we have(21)Dp,qxt-Dp,qx0=∫0thsdp,qs,t∈0,Tp.Taking (p,q)-integration to (21) and using Theorem 3, we have(22)xt=x0+Dp,qx0t+1p∫0tt-qshspdp,qs,t∈0,T.For convenience, we set constants x(0)=A, Dp,qx(0)=B and then(23)xt=A+Bt+1p∫0tt-qshspdp,qs.In particular for t=T in (23), we have(24)xT=A+BT+1p∫0TT-qshspdp,qs.Using (21) and by replacing t=T/p, we obtain(25)Dp,qxTp=B+∫0T/phsdp,qs. From the first boundary condition of (18) and the above setting constants, we get(26)α1A+α2B=α3.The second boundary condition of (18) with (24) and (25) yields(27)β1A+BT+1p∫0TT-qshspdp,qs+β2B+∫0T/phsdp,qs=β3.Solving (26) and (27) for constants A and B, it follows that(28)A=Λ1+1Λα2β1p∫0TT-qshspdp,qs+α2β2∫0T/phsdp,qsand(29)B=Λ2-1Λα1β1p∫0TT-qshspdp,qs+α1β2∫0T/phsdp,qs.Substituting constants A and B into (23), we obtain the unique solution in (20) of linear problem (18). The proof is completed.


## 3. Main Results

LetC:=C([0,T],R) denote the Banach space of all continuous functions from [0,T] to R endowed with sup-norm x=sup{xt:t∈[0,T]}. In view of Lemma 4 we define an operator A:C→C by(30)Axt=Λ1+Λ2t+α2-α1tβ1pΛ∫0TT-qsfs,xpsdp,qs+α2-α1tβ2Λ∫0T/pfs,xp2sdp,qs+1p∫0tt-qsfs,xpsdp,qs.It should be noticed that problem (5) has solutions if and only if the operator A of equation x=Ax has fixed points. Our first result is an existence theorem for separated boundary value problem of quantum (p,q)-difference equation (5) by using Schaefer’s fixed point theorem.Theorem 5.
Assume that(H1)
f:[0,T/p2]×R→R is a continuous function and there exists a constant N>0 such that(31)ft,xp2t≤N,for eacht∈[0,T/p2] and all x∈R.
Then the separated boundary value problem (5) has at least one solution on [0,T].Proof.
Schaefer’s fixed point theorem is used to prove that the operatorA defined by (30) has at least one fixed point. So, we divide the proof into four steps.
Step 1 (continuity of A). Let {xn} be a sequence of function such that xn→x on [0,T/p2]. Given that f is continuous function on [0,T/p2], we have (32)ft,xnp2t→ft,xp2t.Therefore, we obtain (33)Axnt-Axt≤α2+α1Tβ1pΛ∫0TT-qsfs,xnps-fs,xpsdp,qs+α2+α1Tβ2Λ∫0T/pfs,xnp2s-fs,xp2sdp,qs+1p∫0tt-qsfs,xnps-fs,xpsdp,qs,which implies that Axn-Ax→0 as n→∞. Hence the operator A is continuous.
Step 2 (A maps bounded sets into bounded sets in [0,T]). Choosing r>0, we define a bounded ball Br as Br={x∈C:x≤r}. Then, for any x∈Br, we have (34)Axt≤Λ1+Λ2t+α2+α1Tβ1pΛ∫0TT-qsfs,xpsdp,qs+α2+α1Tβ2Λ∫0T/pfs,xp2sdp,qs+1p∫0tt-qsfs,xpsdp,qs≤Λ1+Λ2T+α2+α1Tβ1pΛ∫0TT-qsNdp,qs+α2+α1Tβ2Λ∫0T/pNdp,qs+1p∫0TT-qsNdp,qs=Λ1+Λ2T+α2+α1TT2β1p+qΛN+α2+α1TTβ2pΛN+T2p+qN≔ρThis means that Ax≤ρ. Therefore, the set ABr is uniformly bounded.
Step 3 (A maps bounded sets into equicontinuous sets of [0,T]). Let τ1,τ2∈[0,T] with τ1<τ2 be two points and Br be a bounded ball in C. Then for any x∈Br, we get (35)Axτ2-Axτ1≤τ2-τ1Λ2+τ2-τ1α1β1pΛ∫0TT-qsfs,xpsdp,qs+τ2-τ1α1β2Λ∫0T/pfs,xnp2sdp,qs+1p∫0τ2τ2-qsfs,xpsdp,qs-∫0τ1τ1-qsfs,xpsdp,qs≤τ2-τ1Λ2+τ2-τ1α1β1pΛ∫0TT-qsfs,xpsdp,qs+τ2-τ1α1β2Λ∫0T/pfs,xnp2sdp,qs+1p∫0τ1τ2-τ1fs,xpsdp,qs+∫τ1τ2τ2-qsfs,xpsdp,qs≤τ2-τ1Λ2+τ2-τ1α1β1pΛ∫0TT-qsfs,xpsdp,qs+τ2-τ1α1β2Λ∫0T/pfs,xnp2sdp,qs+Np∫0τ1τ2-τ1dp,qs+∫τ1τ2τ2-qsdp,qs.As τ1→τ2, the right-hand side of the above inequality (which is independent of x) tends to zero. This shows that the set ABr is equicontinuous set. From a consequence of Steps 1 to 3, together with the Arzelá-Ascoli theorem, we deduce that the operator A:C→C is completely continuous.
Step 4. Finally, we show that the set (36)E=x∈C:x=θAxforsome0<θ<1is bounded.
Letx∈E be a solution of problem (5). Then x(t)=θ(Ax)(t) for some 0<θ<1. Hence, for each t∈[0,T], by the method of computation in Step 2, we obtain (37)Ax≤ρ.This gives that the set E is bounded. By applying Schaefer’s fixed point theorem, we get that A has at least one fixed point which is a solution of the second-order quantum (p,q)-difference equation with separated boundary value problem (5) on [0,T]. The proof is completed.The second existence of a unique solution is based on Banach’s contraction mapping principle. For convenience we set a positive constant(38)Φ=α2+α1TTΛβ1Tp+q+β2p+T2p+q.Theorem 6.
Assume that the functionf:[0,T/p2]×R→R is continuous and satisfying the assumption:
(H2) There exists constant L>0 such that ft,x-ft,y≤Lx-y for each t∈[0,T/p2] and x,y∈R.
If(39)LΦ<1,where Φ is defined by (38), then the boundary value problem of quantum (p,q)-difference equation (5) has a unique solution on [0,T].Proof.
The existence of a unique fixed point of operator equationx=Ax, where the operator A is defined by (30), is proved by using Banach’s contraction mapping principle. Let us define a ball Br={x∈C:x≤r} with the value r satisfying (40)r≥Λ1+Λ2T+MΦ1-LΦ,where M=supt∈[0,T/p2]ft,0. Note that the above inequality is well defined as LΦ<1. Next, we will show that ABr⊂Br. For any x∈Br, we have (41)Axt≤supt∈0,T/p2Λ1+Λ2t+α2-α1tβ1pΛ∫0TT-qsfs,xpsdp,qs+α2-α1tβ2Λ∫0T/pfs,xnp2sdp,qs+1p∫0TT-qsfs,xpsdp,qs≤Λ1+Λ2T+α2+α1Tβ1pΛ∫0TT-qsfs,xps-fs,0+fs,0dp,qs+α2+α1Tβ2Λ∫0T/pfs,xp2s-fs,0+fs,0dp,qs+1p∫0TT-qsfs,xps-fs,0+fs,0dp,qs≤Λ1+Λ2T+Lx+Mα2+α1Tβ1pΛ∫0TT-qsdp,qs+α2+α1Tβ2Λ∫0T/pdp,qs+1p∫0TT-qsdp,qs≤Λ1+Λ2T+Lr+Mα2+α1TT2β1p+qΛ+α2+α1TTβ2pΛ+T2p+q=Λ1+Λ2T+Lr+MΦ≤r,which implies that Ax<r. Therefore, we have ABr⊂Br. Finally we shall show that A is a contraction. For any x,y∈C and for each t∈[0,T], we obtain (42)Axt-Ayt≤α2+α1Tβ1pΛ∫0TT-qsfs,xps-fs,ypsdp,qs+α2+α1Tβ2Λ∫0T/pfs,xp2s-fs,yp2sdp,qs+1p∫0TT-qsfs,xps-fs,xpsdp,qs≤x-yα2+α1TT2β1p+qΛ+α2+α1TTβ2pΛ+T2p+q=LΦx-y.Therefore, we have Ax-Ay≤LΦx-y. As LΦ<1, then the operator A is a contraction. Hence, by the Banach contraction mapping principle, it follows that the operator A has a unique fixed point which is the unique solution of problem (5) on [0,T]. This completes the proof.The third existence theorem is based on Lelay-Schauder’s nonlinear alternative.Lemma 7 (nonlinear alternative for single-value maps [43]).
LetE be a Banach space, C be a closed, convex subset of E, U be an open subset of C, and 0∈U. Suppose that A:U¯→C is continuous and compact (that is, A(U¯) is a relatively compact subset of C map). Then either (i)
A has a fixed point in U¯, or(ii)
there is ax∈∂U (the boundary of U in C) and λ∈(0,1) with x=λA(x).Theorem 8.
Assume thatf:[0,T/p2]×R→R is a continuous function. In addition we suppose that (H3)
there exist a continuous nondecreasing functionψ:[0,∞)→(0,∞) and a function z∈C([0,T/p2],R+) such that (43)ft,x≤ztψxforeacht,x∈0,Tp2×R;(H4)
there exists a constantK>0 such that (44)KΛ1+Λ2T+zψKΦ>1.Then the separated boundary value problem (5) has at least one solution on [0,T].Proof.
As in the proof of Theorem5, the operator A:C→C is completely continuous. The result will follow from the Lelay-Schauder nonlinear alternative (Lemma 7) once we have proved the boundedness of the set of all solutions to equations x=θAx for θ∈[0,1].
Letx be a solution of problem (5). Then, from (H3), we have (45)x≤Λ1+Λ2T+zψxα2+α1TT2β1p+qΛ+α2+α1TTβ2pΛ+T2p+q=Λ1+Λ2T+zψxΦ.Thus, we obtain(46)xΛ1+Λ2T+zψxΦ≤1.In view of (H4), there exists a positive constant K such that x≠K. Let us define the set(47)U=x∈C:x<K.Note that the operator A:U¯→C is continuous and completely continuous. From the choice of U, there is no x∈∂U such that x=θAx for some θ∈(0,1). Therefore, by applying the nonlinear alternative of Leray-Schauder type, we can conclude that the operator A has at least one fixed point in U¯, which is a solution of the quantum (p,q)-difference boundary value problem (5) on [0,T]. The proof is completed.The final existence theorem is established by using Krasoselskii’s fixed point theorem.Lemma 9 (Krasnoselskii’s fixed point theorem [44]).
LetM be a closed, bounded, convex, and nonemtry subset of a Banach space X. Let A,B be the operators such that (a)Ax+By∈M whenever x,y∈M; (b)A is compact and continuous; (c)B is a contraction mapping. Then there exists z∈M such that z=Az+Bz.Theorem 10.
Assume thatf:[0,T/p2]×R→R is a continuous function satisfying the assumption (H2). In addition we suppose that
(H5) ft,x≤κ(t),∀(t,x)∈[0,T/p2]×R and κ∈C([0,T/p2],R+).
If(48)T2p+qL<1,then the separated boundary value problem of quantum (p,q)-difference equation (5) has at least one solution on [0,T].Proof.
Let us definesupt∈[0,T/p2]κt=κ and choose a suitable constant r¯ as (49)r¯≥Λ1+Λ2T+κΦ,where Φ is defined by (38). Furthermore, we define the operators P and Q on Br¯={x∈C:x≤r¯} by (50)Pxt=α2-α1tβ1pΛ∫0TT-qsfs,xpsdp,qs+α2-α1tβ2Λ∫0T/pfs,xp2sdp,qs,Qyt=Λ1+Λ2t+1p∫0tt-qsfs,ypsdp,qs.Observe that Px+Qx=Ax. For x,y∈Br¯, we have (51)Px+Qy≤Λ1+Λ2T+κα2+α1TT2β1p+qΛ+α2+α1TTβ2pΛ+T2p+q=Λ1+Λ2T+κΦ≤r¯,which yields Px+Qy∈Br¯. Therefore, the condition (a) of Lemma 9 is satisfied. It follows from the assumption (H2) together with (48) that Q is a contraction mapping. Hence the condition (c) of Lemma 9 is fulfilled. Now we will show that the condition (b) of Lemma 9 is satisfied. Given that function f is continuous, we get that the operator P is continuous. It is easy to verify that (52)Px≤κα2+α1TT2β1p+qΛ+α2+α1TTβ2pΛ.Thus, the set P(Br) is uniformly bounded. Next we prove the compactness of the operator P. Give sup(t,x)∈[0,T/p2]×Br¯ft,x=f¯<∞ and let t1,t2∈[0,T] with t1<t2. Then we get (53)Pxt2-Pxt1≤α1β1t2-t1pΛ∫0TT-qsfs,xpsdp,qs+α1β2t2-t1Λ∫0T/pfs,xp2sdp,qs≤f¯α1β1T2p+qΛ+α1β2TΛt2-t1,which is independent of x and tends to zero as t1→t2. So, the set P(Br) is equicontinuous. By the Arzelá-Ascoli theorem, P is compact on Br¯. Thus all the assumptions of Lemma 9 are satisfied. Therefore the separated boundary value problem of quantum (p,q)-difference equation (5) has at least one solution on [0,T]. The proof is completed.


## 4. Examples

Example 1.
Consider the following separated boundary value problem for second-order quantum(p,q)-difference equation of the form(54)D1/2,1/52xt=ft,xt4,t∈0,4,56x0+45Dp,qx0=13,14x1+56Dp,qx2=35.
Set constantsp=1/2, q=1/5, T=1, α1=5/6, α2=4/5, α3=1/3, β1=1/4, β2=5/6, and β3=3/5. Then we can find constants as Λ=0.7027777778, Λ1=-0.1691699605, Λ2=0.5928853755, and Φ=6.132128741.
(i) Letf:[0,4]×R→R by(55)ft,xt4=5t21+x2t/4+t2sin⁡t3t+1.It is easy to see that ft,x≤5.25=N which satisfies (H1). Thus, by Theorem 5, the separated boundary value problem (54) with f given by (55) has at least one solution on [0,1].
(ii) Givenf:[0,4]×R→R with(56)ft,xt4=cos⁡t18x2t/4+2xt/41+xt/4.it follows that (57)ft,x-ft,y≤19x-y.Therefore, the condition (H2) is satisfied with L=1/9. Indeed, we have LΦ=0.6813476379<1. Hence, by Theorem 6, the separated boundary value problem of quantum (p,q)-difference equation (54) has a unique solution on [0,1].Example 2.
Consider the following separated boundary value problem for second-order quantum(p,q)-difference equation of the form(58)D1/3,1/62xt=gt,xt9,t∈0,92,73x0+56Dp,qx0=47,25x12+72Dp,qx32=65.
Set constantsp=1/3, q=1/6, T=1/2, α1=7/3, α2=5/6, α3=4/7, β1=2/5, β2=7/2, and β3=6/5. Then we can compute that Λ=8.3, Λ1=0.1342512908, Λ2=0.3098106712, and Φ=1.813253012.
(i) Letg:[0,9/2]×R→R be a nonlinear function defined by(59)gt,xt9=1t+23x4t/91+x2t/9+2.Choosing z(t)=1/(t+2)3 and ψ(x)=x2+2, we notice that the condition (H3) of Theorem 8 is satisfied. Furthermore, we can find that z=1/8. Then there exists a constant K such that K∈(0.9447899357,3.467170197). Therefore, the separated boundary value problem (58) with g given by (59) has at least one solution on [0,1/2].
(ii) Let a functiong:[0,9/2]×R→R be defined by(60)gt,xt9=3e-t2+t2·xt/9xt/9+1+t2.It follows that (61)gt,x≤3e-t2+t2+t2.Then the function g defined in (60) satisfies condition (H5) with κ(t)=((3e-t)/(2+t2))+(t/2). In addition, the condition (H2) is satisfied with L=3/2 and (62)T2p+qL=0.75<1.Hence, by applying Theorem 10, the separated boundary value problem for second-order quantum (p,q)-difference equation (58) with g defined in (60) has at least one solution on [0,1/2].Remark 3.
In Example2, Case (ii), although the function g(t,x) satisfies condition (H2), Theorem 6 cannot be applied, since LΦ=2.719879518>1.



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