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# The general solution of impulsive systems with Riemann-Liouville fractional derivatives

Xianmin Zhang
• Corresponding author
• School of Electronic Engineering, Jiujiang University, Jiujiang, Jiangxi 332005, China
• Email:
/ Wenbin Ding
• School of Electronic Engineering, Jiujiang University, Jiujiang, Jiangxi 332005, China
/ Hui Peng
• School of Electronic Engineering, Jiujiang University, Jiujiang, Jiangxi 332005, China
/ Zuohua Liu
• School of Chemistry and Chemical Engineering, Chongqing University, Chongqing 400044, China
/ Tong Shu
• School of Electronic Engineering, Jiujiang University, Jiujiang, Jiangxi 332005, China
Published Online: 2016-12-30 | DOI: https://doi.org/10.1515/math-2016-0096

## Abstract

In this paper, we study a kind of fractional differential system with impulsive effect and find the formula of general solution for the impulsive fractional-order system by analysis of the limit case (as impulse tends to zero). The obtained result shows that the deviation caused by impulses for fractional-order system is undetermined. An example is also provided to illustrate the result.

MSC 2010: 34A08; 34A37

## 1 Introduction

Fractional calculus was utilized as a powerful tool to reveal the hidden aspects of the dynamics of the complex or hypercomplex systems [1-3], and the subject of fractional differential equations is gaining much attention. For details see [4-14] and the references therein.

Impulsive effects exist widely in many processes in which their states can be described by impulsive differential equations. Moreover, in case of impulsive differential equations with Caputo fractional derivative there have been numerous works about the subject [15-23], and impulsive fractional partial differential equations are widely considered in [24-29].

Motivated by the above-mentioned works, we will study the following impulsive Cauchy problem with Riemann-Liouville fractional derivative: $Da+qu(t)=f(t,u(t)),t∈(a,T]andt≠ti(i=1,2,…,m),Δ(Ja+1−qu)|t=ti=Ja+1−qu(ti+)−Ja+1−qu(ti−)=Δi(u(ti−)),i=1,2,…,m,Ja+1−qu(a)=ua,ua∈C,$(1) where q ∊ ℂ and ℜ(q) ∊ (0, 1), ${D}_{a+}^{q}$ denotes left-sided Riemann-Liouville fractional derivative of order q and ${\mathcal{J}}_{a+}^{1-q}$ denotes left-sided Riemann-Liouville fractional integral of order 1 − q. f : J × ℂ → ℂ is an appropriate continuous function, and a = t0 < t1 < ... <tm < tm+1 = T. Here ${\mathcal{J}}_{a+}^{1-q}u\left({t}_{i}^{+}\right)\phantom{\rule{thickmathspace}{0ex}}=\underset{\epsilon \to {0}^{+}}{lim}{\mathcal{J}}_{a+}^{1-q}u\left({t}_{i}+\epsilon \right)$ and ${\mathcal{J}}_{a+}^{1-q}u\left({t}_{i}^{-}\right)\phantom{\rule{thickmathspace}{0ex}}=\underset{\epsilon \to {0}^{-}}{lim}{\mathcal{J}}_{a+}^{1-q}u\left({t}_{i}+\epsilon \right)$ represent the right and left limits of ${\mathcal{J}}_{a+}^{1-q}u\left(t\right)\phantom{\rule{thickmathspace}{0ex}}\text{at}\phantom{\rule{thickmathspace}{0ex}}t={t}_{i},$ respectively.

For impulsive system (1) we have $limΔ1→0,Δ2→0,⋯,Δm→0impulsivesystem1=Da+qu(t)=f(t,u(t)),q∈0,1,t∈a,T,Ja+1−qu(a)=ua,ua∈C,$(2)

Therefore, it means that there exists a hidden condition $limΔ1→0,Δ2→0,⋯,Δm→0thesolutionof impulsivesystem1=thesolutionof system2$(3)

Therefore, the definition of solution for impulsive system (1) is provided as follows:

Definition 1.1: A function z(t) : [a,T] → ℂ is said to be a solution of the fractional Cauchy problem (1) if ${\mathcal{J}}_{a+}^{1-q}z\left(a\right)={u}_{a},$ the equation condition ${D}_{a+}^{q}z\left(t\right)=f\left(t,z\left(t\right)\right)$ for each t ∊ (a, T] is verified, the impulsive conditions $\mathrm{\Delta }\left({\mathcal{J}}_{a+}^{1-q}z\right){|}_{t={t}_{i}}={\mathrm{\Delta }}_{i}\left(z\left({t}_{i}^{-}\right)\right)$ (here i = 1,2,..., m)are satisfied, the restriction of z(t) to the interval (tk, tk+1] (here k = 0, 1, 2,...,m) is continuous, and the condition (3) holds.

Therefore, we will consider impulsive system (1) and seek some solutions of impulsive system (1) according to Definition 1.1.

The rest of this paper is organized as follows. In Section 2, some preliminaries are presented. In Section 3, we give the formula of general solution for impulsive differential equations with Riemann-Liouville fractional derivatives. In Section 4, an example is provided to expound the main result.

## 2 Preliminaries

Firstly, we recall some concepts of fractional calculus [2] and a property for nonlinear fractional differential equations.

Definition 2.1: The left-sided Riemann-Liouville fractional integral of order α ∊ ℂ (ℜ(α) > 0) of function x(t) is defined by $Jaα+xt=1Γq∫att−sα−1xsds,t>a,$ where Γ is the gamma function.

Definition 2.2: The left-sided Riemann-Liouville fractional derivative of order q ∊ ℂ (ℜ(q)0) of function x(t) is defined by $Daq+xt=1Γn−qddtn∫att−sn−q−1xsds,n=ℜq+1,t>a.$By Lemma 2.2 in [11], the initial value problem $Daq+ut=ft,ut,q∈Candℜq∈0,1,t∈a,T,Ja1−q+ua=ua,ua∈C,$(4) is equivalent to the following nonlinear Volterra integral equation of the second kind, $ut=uaΓqt−aq−1+1Γq∫att−sn−q−1fs,usds.$(5)

## 3 Main results

Define a piecewise function $u~t=1ΓqJa+1−qutk+t−tkq−1+1Γq∫tktt−sq−1fs,usdsfort∈tk,tk+1wherek=0,1,2,…,m$ with ${\mathcal{J}}_{a+}^{1-q}u\left({t}_{k}^{+}\right)={\mathcal{J}}_{a+}^{1-q}u\left({t}_{k}^{-}\right)+{\mathrm{\Delta }}_{k}\left(u\left({t}_{k}^{-}\right)\right).$ By Definition 2.2, ..., we have $Daq+u~t=1Γn−qΓqddt∫att−η1−q−1Ja+1−qutk+η−tkq−1+∫tknη−sq−1fs,usdsdη=1Γn−qΓqddt∫tktt−η1−q−1Ja+1−qutk+η−tkq−1+∫tknη−sq−1fs,usdsdη=ft,ut|t∈tk,tk+1$ So, ũ(t) satisfies the condition of fractional derivative of (1), and it doesn’t satisfy the condition (3). Thus, we assume that ũ(t) is an approximate solution to seek the exact solution of impulsive system (1).

Theorem 3.1: Let ξ be a constant. A function u(t) is a general solution of system (1) if and only if u(t) satisfies the fractional integral equation $ut=uaΓqt−aq−1+1Γq∫att−sq−1fs,usdsfort∈a,t1,uaΓqt−aq−1+1Γq∫att−sq−1fs,usds+∑i=1kΔiuti−Γqt−tiq−1−∑i=1kξΔiuti−Γquat−aq−1+∫att−sq−1+fs,usds−ua+∫atifs,usdst−tiq−1−∫titt−sq−1+fs,usdsfort∈tk,tk+1,$(6) provided that the integral in (6) exists.

Proof. “Necessity”. First we can easily verify that Eq. (6) satisfies the hidden condition (3).

Next, Taking Riemann-Liouville fractional derivative to Eq. (6) for each t ∊ (tk, tk+1] (where k = 0, 1, 2, ...,m), we have $Daq+ut=Daq+uaΓqt−aq−1+1Γq∫att−sq−1+fs,usds+∑i=1kΔiuti−Γqt−tiq−1−∑i=1kξΔiuti−Γquat−aq−1+∫att−sq−1+fs,usds−ua+∫atifs,usdst−tiq−1−∫titt−sq−1+fs,usds=ft,utt≥a−ξ∑i=1kΔiuti−ft,utt≥a−ft,utt≥tit∈tk,tk+1=ft,ut|t∈tk,tk+1.$

So, Eq. (6) satisfies Riemann-Liouville fractional derivative of system (1). Using (6) for each tk (here k = 1,2, ...,m), we get $Ja+1−qutk+−Ja+1−qutk−=1Γ1−q∫att−η1−q−1uηdηt→tk+−1Γ1−q∫att−η1−q−1uηdηt=tk=Δkutk−−ξΔkutk−ua+∫atfs,usds−ua+∫atkfs,usds−∫tktfs,usdst→tk=Δkutk−.$

Therefore, Eq. (6) satisfies impulsive conditions of (1). Then, Eq. (6) satisfies the conditions of system (1).

“Sufficiency”. We prove that the solutions of system (1) satisfy Eq. (6) by mathematical induction. By Definition 2.1, the solution of (1) satisfies $ut=uaΓqt−aq−1+1Γq∫att−sq−1fs,usdsfort∈a,t1.$(7)

By (7), we have ${\mathcal{J}}_{a+}^{1-q}u\left({t}_{1}^{+}\right)={\mathcal{J}}_{a+}^{1-q}u\left({t}_{1}^{-}\right)+{\mathrm{\Delta }}_{1}\left(u\left({t}_{1}^{-}\right)\right)={u}_{a}+{\mathrm{\Delta }}_{1}\left(u\left({t}_{1}^{-}\right)\right)+{\int }_{a}^{{t}_{1}}f\left(s,u\left(s\right)\right)ds,$ and the approximate solution ũ(t) (for t ∊ (t1, t2]) is given by $u~t=1ΓqJa+1−qut1+t−t1q−1+1Γq∫t1tt−sq−1fs,usds=ua+Δ1ut1−+∫at1fs,usds,Γqt−t1q−1+1Γq∫t1tt−sq−1fs,usdsfort∈t1,t2,$(8) with e1(t) = u(t) – ũ(t) for t ∊ (t1, t2]. By $limΔ1ut1−→0⁡ut=uaΓqt−aq−1+1Γq∫att−sq−1fs,usdsfort∈t1,t2,$ we get $limΔ1ut1−→0e1t=limΔ1ut1−→0ut−u~t=uaΓqt−aq−1+1Γq∫att−sq−1fs,usds−ua+∫at1fs,usdsΓqt−t1q−1−1Γq∫t1tt−sq−1fs,usds.$

Then, we assume $e1t=σΔ1ut1−limΔ1ut1−→0e1t=σΔ1ut1−Γquat−aq−1+∫att−sq−1fs,usds−ua+∫at1fs,usdst−t1q−1−∫t1tt−sq−1fs,usds.$ where function σ(.) is an undetermined function with σ(0) = 1. Thus, $ut=u~t+e1t=1ΓqσΔ1ut1−uat−aq−1+∫att−sq−1fs,usds+Δ1ut1−t−t1q−1+1−σΔ1ut1−ua+∫at1fs,usdst−t1q−1+∫t1tt−sq−1fs,usdsfort∈t1,t2.$(9)

Using (9), we get ${\mathcal{J}}_{a+}^{1-q}u\left({t}_{2}^{+}\right)={\mathcal{J}}_{a+}^{1-q}u\left({t}_{2}^{-}\right)+{\mathrm{\Delta }}_{2}\left(u\left({t}_{2}^{-}\right)\right)={u}_{a}+{\mathrm{\Delta }}_{1}\left(u\left({t}_{1}^{-}\right)\right)+{\mathrm{\Delta }}_{2}u\left({t}_{2}^{-}\right)+{\int }_{a}^{{t}_{2}}f\left(s,u\left(s\right)\right)ds.$ Therefore, the approximate solution ũ(t) (for t ∊ (t2, t3])is given by $u~t=1ΓqJa+1−qut2+t−t2q−1+1Γq∫t2tt−sq−1fs,usds=ua+Δ2ut1−+Δ2ut2−∫at2fs,usds,Γqt−t2q−1+1Γq∫t2tt−sq−1fs,usdsfort∈t2,t3$(10) with e2(t) = u(t) – ũ(t) for t ∊ (t2, t3]. Moreover, by (9), the exact solution u(t) of (1) satisfies $limΔ1ut1−→0,Δ2ut2−→0⁡ut=uaΓqt−aq−1+1Γq∫att−sq−1fs,usds,fort∈t2,t3,limΔ1ut1−→0⁡ut=1ΓqσΔ2ut2−uat−aq−1+∫att−sq−1fs,usds+Δ2ut2−t−t2q−1+1−σΔ2ut2−ua+∫at2fs,usdst−t2q−1+∫t2tt−sq−1fs,usdsfort∈t2,t3,$

$limΔ2ut2−→0⁡ut=1ΓqσΔ1ut1−uat−aq−1+∫att−sq−1fs,usds+Δ1ut1−t−t1q−1+1−σΔ1ut1−ua+∫at1fs,usdst−t1q−1+∫t1tt−sq−1fs,usdsfort∈t2,t3.$

Therefore, $limΔ1ut1−→0,Δ2ut2−→0e2t=limΔ1ut1−→0,Δ2ut2−→0ut−u~t=1Γquat−aq−1+∫att−sq−1fs,usds−ua+∫at2fs,usdst−t2q−1−∫t2tt−sq−1fs,usds,$(11)

$limΔ1ut1−→0e2t=limΔ1ut1−→0ut−u~t=σΔ2ut2−Γquat−aq−1+∫att−sq−1fs,usds−ua+∫at2fs,usdst−t2q−1−∫t2tt−sq−1fs,usds,$(12)

$limΔ2ut2−→0e2t=limΔ2ut2−→0ut−u~t=1ΓqσΔ1ut1−uat−aq−1+∫att−sq−1fs,usds+Δ1ut1−t−t1q−1−Δ1ut1−t−t2q−1+1−σΔ1ut1−ua+∫at1fs,usdst−t1q−1−∫t1tt−sq−1fs,usds−ua+Δ1ut1−+∫at2fs,usdst−t2q−1−∫t2tt−sq−1fs,usds.$(13)

Then, by (11)(13), we obtain $e2t=1Γq[σ(Δ1(u(t1−)))+σ(Δ2(u(t2−)))−1]uat−aq−1+∫att−sq−1fs,usds+Δ1(u(t1−))(t−t1)q−1−Δ1(u(t1−))(t−t2)q−1+1−σut1−ua+∫at1fs,usdst−t1q−1+∫t1tt−sq−1fs,usds−σ(Δ2(u(t2−)))ua+∫at2fs,usdst−t2q−1+∫t2tt−sq−1fs,usds.$(14)

Thus, $ut=u~t+e2t=1Γq[σ(Δ1(u(t1−)))+σ(Δ2(u(t2−)))−1]uat−aq−1+∫att−sq−1fs,usds+Δ1(u(t1−))(t−t1)q−1−Δ2(u(t2−))(t−t2)q−1+1−σut1−ua+∫at1fs,usdst−t1q−1+∫t1tt−sq−1fs,usds+1−σ(Δ2(u(t2−)))ua+∫at2fs,usdst−t2q−1+∫t2tt−sq−1fs,usdsfort∈(t2,t3].$(15)

Moreover, letting t2 → t1, we have $limt2→t1Daq+ut=ft,ut,q∈C,andℜq∈0,1,t∈(a,t3]andt≠t1andt≠t2,ΔJa+1−qut=tk=Ja+1−qutk+−Ja+1−qutk−=Δkutk−,k=1,2,Ja+1−qua=ua,ua∈C,$(16) $=Daq+ut=ft,ut,q∈C,andℜq∈0,1,t∈(a,t3]andt≠t1,ΔJa+1−qut=t1=Ja+1−qut1+−Ja+1−qut1−+Ja+1−qut2+−Ja+1−qut2−=Δ1ut1−+Δ2ut2−,Ja+1−qua=ua,ua∈C,$(17)

Using (9) and (15), we have 1 – σ1 + Δ2) = 1 – σ1) + 1 – σ2). Letting ρ(z) = 1 – σ(z), we get ρ(z + w) = ρ(z) + ρ(w) for ∀z, w ∊ ℂ. So, ρ(z)ξz, here ξ is a constant. Thus, $ut=uaΓqt−aq−1+1Γq∫att−sq−1fs,usds+Δ1ut1−Γqt−t1q−1=ξΔ1ut1−Γquat−aq−1+∫att−sq−1fs,usds−ua+∫at1fs,usdst−aq−1−∫t1tt−sq−1fs,usdsfort∈t1,t2.$(18) and $ut=uaΓqt−aq−1+1Γq∫att−sq−1fs,usds+Δ1ut1−Γqt−t1q−1+Δ2ut2−Γqt−t2q−1−ξΔ1ut1−Γquat−aq−1+∫att−sq−1fs,usds−ua+∫at1fs,usdst−t1q−1+∫titt−sq−1fs,usds−ξΔ2ut2−Γquat−aq−1+∫att−sq−1fs,usds−ua+∫at2fs,usdst−t2q−1+∫t2tt−sq−1fs,usdsfort∈t2,t3.$(19) Next, for t ∊ (tn, tn+1], suppose $ut=uaΓqt−aq−1+1Γq∫att−sq−1fs,usds+∑i=1nΔiuti−Γqt−tiq−1−∑i=1nξΔiuti−Γquat−aq−1+∫att−sq−1fs,usds−ua+∫atifs,usdst−tiq−1−∫att−sq−1fs,usdsfort∈tn,tn+1.$(20) Using (20), we have $Ja+1−qutn+1+=Ja+1−qutn+1−+Δn+1utn+1−=ua+∑i=1n+1Δiuti−+∫atn+1fs,usds$ Thus, the approximate solution ũ(t) for t ∊ (tn+1, tn+2] is given by $u~t=1ΓqJa+1−qutn+1++t−tn+1q−1+1Γq∫tn+1tt−sq−1fs,usds=ua+∑i=1n+1Δiuti−+∫atn+1fs,usds1Γqt−tn+1q−1+1Γq∫tn+1tt−sq−1fs,usdsfort∈tn+1,tn+2$(21) with en+1(t) = u(t) – ũ(t) for t ∊ (tn+1, tn+2]. By (20), the exact solution u(t) of (1) satisfies $limΔ1ut1−→0,ut=uaΓqt−aq−1+1Γq∫att−sq−1fs,usdsfort∈tn+1,tn+2,⋮Δn+1utn+1−→0$ $limΔjutj−→0,1≤j≤n+1⁡ut=uaΓqt−aq−1+1Γq∫att−sq−1fs,usds+∑1≤i≤n+1,andi≠jΔiuti−Γqt−tiq−1−∑1≤i≤n+1,andi≠jξΔiuti−Γquat−aq−1+∫att−sq−1fs,usds−ua+∫atifs,usdst−tiq−1−∫titt−sq−1fs,usdsfor∈tn+1,tn+2.$ Therefore, $limΔ1ut1−→0,en+1t=limΔ1ut1−→0,ut−u~t⋮⋮Δn+1utn+1−→0Δn+1utn+1−→0=1Γquat−aq−1+∫att−sq−1fs,usds−ua+∫atn+1fs,usdst−tn+1q−1−∫tn+1tt−sq−1fs,usds,$(22) $limΔjutj−→0,1≤j≤n+1en+1t=limΔjutj−→0,1≤j≤n+1ut−u~t=1Γq1−∑1≤j≤n+1andi≠jξΔiuti−uat−aq−1+∫att−sq−1fs,usds+∑1≤j≤n+1andi≠jΔiuti−t−tiq−1+∑1≤j≤n+1andi≠jξΔiuti−ua+∫atifs,usdst−tiq−1+∫titt−sq−1fs,usds−ua+∑1≤j≤n+1andi≠jΔiuti−+∫atn+1fs,usdst−tn+1q−1−∫tn+1tt−sq−1fs,usds.$(23)

By (22) and (23), we obtain $en+1t=1Γq1−∑1≤i≤n+1ξΔiuti−uat−aq−1+∫att−sq−1fs,usds+∑1≤i≤n+1Δiuti−t−tiq−1−∑1≤i≤n+1Δiuti−t−tiq−1+∑1≤i≤n+1ξΔiuti−ua+∫atifs,usdst−tiq−1+∫titt−sq−1fs,usds−ua+∫atn+1fs,usdst−tn+1q−1+∫titt−sq−1fs,usds.$(24) Thus, $ut=u~t+en+1t=uaΓqt−aq−1+1Γq∫att−sq−1fs,usds+∑i=1n+1Δiuti−Γqt−tiq−1−∑i=1n+1ξΔiuti−Γquat−aq−1+∫att−sq−1fs,usds−ua+∫atifs,usdst−tiq−1+∫att−sq−1fs,usdsfort∈tn+1,tn+2.$ So, the solution of system (1) satisfies Eq. (6). So, impulsive system (1) is equivalent to the integral equation (6). The proof is now completed. ⏱

## 4 Example

For system (1) it is difficult to get the analytical solution when ƒ is a nonlinear function in (1). So, a linear example is given to illustrate the obtained result.

Example 4.1: Let us consider the general solution of the impulsive fractional system $D1+12ut=t,t∈1,3andt≠2,ΔJ1+1−12ut=2=J1+1−12u2+−J1+1−12u2−=δ∈R,J1+1−12u1=u0∈R,$(25)By Theorem 3.1, the general solution of impulsive system (25) is obtained as follows: $ut=u0Γ(12)t−1−12+1Γ(12)∫1tt−s−12sds,fort∈1,2,u0Γ(12)t−1−12+1Γ(12)∫1tt−s−12sds+δΓ(12)t−2−12−ξδΓ(12)u0t−1112+∫1tt−s−12sds−u0+∫12sdst−2−12∫2tt−s−12sdsfort∈2,3.$(26) Next, it is verified that Eq. (26) satisfies the condition of system (25). Taking Riemann-Liouville fractional derivative to the both sides of Eq. (26), we have(i) for t ∊ (1,2] $D1+12ut=1Γ(12)ddt∫1tt−η1−12−1u0Γ(12)η−112−1+1Γ(12)∫1nη−s12−1sdsdηt∈1,2=t|t∈1,2,$(ii)for t ∊ (2, 3] $D1+12ut=1Γ(12)ddt∫1tt−η12−1u0Γ(12)η−1−12+1Γ(12)∫1nη−s−12sds+δΓ(12)η−2−12−ξδΓ(12)×u0η−1112+∫1nη−s−12sds−u0+∫12sdsη−2−12−∫2ηη−s−12sdsdηt∈2,3={t|t⩾1−ξδ×t|t⩾1−1Γ(12)Γ(12)ddt∫12t−η12−1u0+∫12sdsη−2−12−∫2ηη−s−12sdsdηt∈2,3={t|t⩾1−ξδt|t⩾1−t|t⩾2}t∈2,3=t|t∈2,3$So, Eq. (26) satisfies Riemann-Liouville fractional derivative condition of system (25). By Definition 2.1, we obtain $J1+1−12u2+−J1+1−12u2−=1Γq∫1tt−η12−1uηdηt→2+−1Γq∫1tt−η12−1uηdηt=2−=δΓ(12)Γ(12)∫2nt−η12−1η−2dηt→2+−ξδ1Γ(12)Γ(12)∫1tt−η12−1u0η−1−12+∫1ηη−s−12sds−u0+∫12sdsη−s−12−∫2nη−s−12sdsdηt→2+=δ−ξδu0+∫1tsds−u0+∫12sds−∫2tsdst→2+=δ.$ That is, Eq. (26) satisfies impulsive condition in system (25).Finally, it is obvious that the Eq. (26) satisfies the following limit case $limδ→0D1+12ut=t,t∈1,3andt≠2ΔJ1+1−12ut=2=J1+1−12u2+−J1+1−12u2+=δ∈R,J1+1−12u1=u0∈R,=D1+12ut=t,t∈1,3,J1+1−12u1=u0∈R,$(27) So, Eq. (26) is the general solution of system (25).

## Acknowledgement

The authors are deeply grateful to the anonymous referees for their kind comments, correcting errors and improving written language, which have been very useful for improving the quality of this paper.

The work described in this paper is financially supported by the National Natural Science Foundation of China (Grant No. 21576033, 21636004, 61563023, 61261046) and State Key Development Program for Basic Research of Health and Family planning Commission of Jiangxi Province China (Grant No. 20143246), the Natural Science Foundation of Jiangxi Province (Grant No. 20151BAB207013) and the Jiujiang University Research Foundation (Grant No. 8400183).

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## About the article

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Accepted: 2016-11-15

Published Online: 2016-12-30

Published in Print: 2016-01-01

Citation Information: Open Mathematics, ISSN (Online) 2391-5455, Export Citation