# Impulsive Multiorders Riemann-Liouville Fractional Differential Equations

Authors: Weera Yukunthorn; Sotiris K. Ntouyas; Jessada Tariboon
Journal: Discrete Dynamics in Nature and Society
 (2015)

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## Abstract

Impulsive multiorders fractional differential equations are studied.
Existence and uniqueness results are obtained for first- and second-order impulsive initial value problems by using Banach’s fixed point theorem in an appropriate weighted space. Examples illustrating the main results are presented.


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## Body

## 1. Introduction

Fractional calculus has become very useful over the last years because of its many applications in almost all applied sciences. By now, almost all fields of research in science and engineering use fractional calculus to better describe them.Fractional differential equations have been of great interest and are caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in various science such as physics, mechanics, chemistry, and engineering. For details and some recent results on the subject we refer to the papers [1–3], books [4–7], and references cited therein.Recently in [8], Wang et al. studied existence and uniqueness results for the following impulsive multipoint fractional integral boundary value problem involving multiorders fractional derivatives and deviating argument: (1)Dctk+αkut=ft,ut,uθt,1<αk≤2,Δutk=Ikutk,Δu′tk=Ik∗utk,k=1,2,…,p,u0=∑k=0pλkJtk+βkuηk,u′0=0,tk<ηk<tk+1,where Dctk+αk is theCaputo fractional derivative of order αk, Jtk+βk is fractional Riemann-Liouville integral of order βk>0,  f∈C(J×R×R,R),  Ik,Ik∗∈C(R,R), θ∈C(J,J), J=[0,T](T>0), 0=t0<t1<⋯<tk<⋯<tp<tp+1=T, Δu(tk)=u(tk+)-u(tk-), and Δu′(tk)=u′(tk+)-u′(tk-) where u(tk+), u′(tk+) and u(tk-), u′(tk-) denote the right and left hand limits of u(t) and u′(t) at t=tk(k=1,2,…,p).We notice that there are some discussions on the concept of solution for impulsive fractional differential equations for both Riemann-Liouville and Caputo fractional derivatives. We refer the interested reader to some recent papers [9–11] and the references cited therein. However, we can point out the problems caused by using the definition of Caputo fractional derivatives of order α with the lower limit 0 for a function f as(2)Dc0αft=1Γn-α∫0tt-sn-α-1fnsds.If there are impulse points tk such that tk∈(0,t) for some k∈N, then the f(n)(tk) does not exist, which leads to non-integrability of the right-hand side of (2). The key idea for solving this problem is to apply the definition of fractional derivative only on an interval (tk,tk+1) and combining all intervals by using impulsive conditions.In this paper, we study impulsive multiorders Riemann-Liouville fractional differential equations. More precisely, in Section3 we study the existence and uniqueness of solutions for the following initial value problem for impulsive multiorders Riemann-Liouville fractional differential equations of order 0<αk≤1 of the form (3)Dtkαkxt=ft,xt,t∈J,t≠tk,Δ~xtk=φkxtk,k=1,2,3,…,m,x0=0,where Dtkαk is theRiemann-Liouville fractional derivative of order 0<αk≤1 on intervals Jk, k=0,1,2,…,m, 0=t0<t1<t2<⋯<tk<⋯<tm<tm+1=T, f:J×R→R is a continuous function, and φk∈C(R,R). The notation Δ~x(tk) is defined by(4)Δ~xtk=Itk1-αkxtk+-Itk-11-αk-1xtk,k=1,2,3,…,m,where Itk1-αk is the Riemann-Liouville fractional integral of order 1  −  αk on interval Jk. It should be noticed that if αk=1 in (4), then Δ~x(tk) = Δx(tk) = x(tk+)-x(tk) for k=1,2,3,…,m.In Section4, we investigate the initial value problem of impulsive Riemann-Liouville fractional differential equations of the form(5)Dtkαkxt=ft,xt,t∈J,t≠tk,Δ~xtk=φkxtk,k=1,2,3,…,m,Δ∗xtk=φk∗xtk,k=1,2,3,…,m,x0=0,Dα0-1x0=β,where β∈R, 0=t0<t1<t2<⋯<tk<⋯<tm<tm+1=T, f:J×R→R is a continuous function, φk,φk∗∈C(R,R) for k=1,2,…,m, and Dtkαk is the Riemann-Liouville fractional derivative of order 1<αk≤2 on intervals Jk for k=0,1,2,…,m. The notation Δ~x(tk) is defined by (4) and Δ∗x(tk) is defined by(6)Δ∗xtk=Itk2-αkxtk+-Itk-12-αk-1xtk,k=1,2,…,m,where Itk2-αk is the Riemann-Liouville fractional integral of order 2-αk>0 on Jk. It should be noticed that if αk=2 in (6), then Δ~x(tk)=Dtkx(tk+)-Dtk-1x(tk) and Δ∗x(tk)=Δx(tk)=x(tk+)-x(tk) for k=1,2,…,m.By using Banach’s fixed point theorem we prove existence and uniqueness results for the problem (3) and (5) in an appropriate weighted space.The paper is organized as follows: Section2 contains some preliminary notations, definitions, and lemmas that we need in the sequel. In Section 3 we present the main results for problem (3), while in Section 4 we present the main results for problem (5). Examples illustrating the obtained results are also presented.


## 2. Preliminaries

In this section, we introduce some notations and definitions of fractional calculus and present preliminary results needed in our proofs later.Definition 1.
The Riemann-Liouville fractional derivative of orderα>0 of a continuous function f:(a,b)→R is defined by(7)Daαft=1Γn-αddtn∫att-sn-α-1fsds,n-1<α<n,t∈a,b,where n=[α]+1, [α] denotes the integer part of a real number α, provided the right-hand side is pointwise defined on (a,b), where Γ is the gamma function defined by Γ(α)=∫0∞‍e-ssα-1ds.Definition 2.
For at leastn-times differentiable function f:(a,b)→R, the Caputo derivative of fractional order α is defined as(8)Dcaαft=1Γn-α∫att-sn-α-1fnsds,n-1<α<n,t∈a,b,where n=[α]+1.Definition 3.
The Riemann-Liouville fractional integral of orderβ>0 of a continuous function f:(a,b)→R is defined by(9)Iaαft=1Γα∫att-sα-1fsds,t∈a,bprovided the right-hand side is pointwise defined on (a,b).Lemma 4 (see [5]).
Letα>0 and x∈C(a,b)∩L(a,b). Then the fractional differential equation(10)Daαxt=0has a unique solution(11)xt=k1t-aα-1+k2t-aα-2+⋯+knt-aα-n,where ki∈R, i=1,2,…,n, and n-1<q<n.Lemma 5 (see [5]).
Letα>0. Then for x∈C(a,b)∩L(a,b) it holds(12)IaαDaαxt=xt-∑j=1nIan-αxn-jaΓα-j+1t-aα-j,where n-1<α<n.


## 3. Impulsive Riemann-Liouville Fractional Differential Equations of Orders0<αk≤1

LetJ=[0,T], J0=[t0,t1], Jk=(tk,tk+1] for k=1,2,3,…,m. Let PC(J,R)={x:J→R, x(t) is continuous everywhere except for some tk at which x(tk+) and x(tk-) exist, and x(tk-)=x(tk), k=1,2,3,…,m}. For γ∈R+, we introduce the space Cγ,k(Jk,R)={x:Jk→R:(t-tk)γx(t)∈C(Jk,R)} with the norm xCγ,k=supt∈Jk|(t-tk)γx(t)| and PCγ={x:J→R: for each t∈Jk and (t-tk)γx(t)∈C(Jk,R), k=0,1,2,…,m} with the norm xPCγ=max{supt∈Jk|(t-tk)γx(t)|:k=0,1,2,…,m}. Clearly PCγ is a Banach space.In this section we study problem (3).Lemma 6.
Ifx∈PC(J,R) is a solution of (3), then, for any t∈Jk, k=0,1,2,…,m,(13)xt=t-tkαk-1Γαk∑0<tk<tItk-11ftk,xtk+φkxtk+Itkαkft,xt,with ∑0<0‍(·)=0. The converse is also true.Proof.
Fort∈J0, taking the Riemann-Liouville fractional integral of order α0 to the first equation of (3) and using Lemma 5, we have (14)xt=It0α0ft,xt+c0tα0-1Γα0,where c0=It01-α0x(0). The initial condition x(0)=0 implies c0=0. Then for x∈J0, we get(15)xt=It0α0ft,xt.Applying the Riemann-Liouville fractional integral of order 1-α0 from 0 to t1, we get(16)It01-α0xt1=It01-α0It0α0ft1,xt1=It01ft1,xt1.For t∈J1, taking the Riemann-Liouville fractional integral of order α1 to the first equation of (3) and using Lemma 5, we have(17)xt=t-t1α1-1Γα1It11-α1xt1++It1α1ft,xt.Since It11-α1x(t1+)=It01-α0x(t1)+φ1(x(t1)), it follows for t∈J1 that(18)xt=t-t1α1-1Γα1It01ft1,xt1+φ1xt1+It1α1ft,xt.Applying the Riemann-Liouville fractional integral of order 1-α1 to the above equation and substituting t=t1, one has(19)It11-α1xt=It01ft1,xt1+φ1xt1+It11ft,xt.For t∈J2, using the Riemann-Liouville fractional integral of order α2 for (3), we have(20)xt=t-t2α2-1Γα2It21-α2xt2++It2α2ft,xt=t-t2α2-1Γα2It11-α1xt2+φ2xt2+It2α2ft,xt=t-t2α2-1Γα2It01ft1,xt1+φ1xt1+It11ft2,xt2+φ2xt2+It2α2ft,xt.Repeating the above procession for each Jk, we obtain (13).
On the other hand, assume thatx is a solution of (3). Applying the Riemann-Liouville fractional derivative of order αk for (13) on Jk, k=0,1,2,…,m, and using Γ(0)=∞, it follows that (21)Dtkαkxt=ft,xt.It is easy to verify that Δ~x(tk)=φk(x(tk)), k=1,2,3,…,m, and x(0)=0. The proof is complete.Next we will prove that problem (3) has a unique solution by using Banach’s fixed point theorem.Theorem 7.
Assume that
(
H
1
)  f:J×R→R is a continuous function and satisfies(22)ft,x-ft,y≤L1x-y,L1>0,∀t∈J,x,y∈R;
(
H
2
)  φk:R→R, k=1,2,3,…,m, are continuous functions and satisfy (23)φkx-φky≤L2x-y,L2>0,∀x,y∈R.If(24)Λ1≔T∗Γ∗L1+L1T+L2m<1,where T∗=max{Tγ+αk-1,Tγ+αk}, Γ∗=min{Γ(αk),Γ(αk+1)}, and γ+αk>1 for k=0,1,2,…,m, then the initial value problem (3) has a unique solution on [0,T].Proof.
In view of Lemma6, we define the operator K:PC→PC as (25)Kxt=t-tkαk-1Γαk∑0<tk<tItk-11ftk,xtk+φkxtk+Itkαkft,xt.In addition, we define a ball Br={x∈PCγ(J,R), xPCγ≤r}. To show that Kx∈PCγ, we suppose τ∈Jk and then (26)t-tkγKxt-τ-tkγKxτ≤t-tkγ+αk-1Γαk∑j=1kItj-11ftj,xtj+φjxtj+t-tkγItkαkft,xt-τ-tkγ+αk-1Γαk∑j=1kItj-11ftj,xtj+φjxtj-τ-tkγItkαkfτ,xτ≤t-tkγ+αk-1-τ-tkγ+αk-1Γαk∑j=1kItj-11ftj,xtj+φjxtj+t-tkγItkαkft,xt-τ-tkγ·Itkαkfτ,xτ.As t→τ, we get |(t-tk)γKx(t)-(τ-tk)γKx(τ)|→0 for each k=0,1,2,…,m. Therefore, we have Kx(t)∈PCγ. Next we will show that KBr⊂Br. Suppose that supt∈J|f(t,0)|=M, max{|Ik(0)|,k=1,2,3,…,m}=N. Setting(27)Λ2=T∗Γ∗M+MT+Nmwe choose r such that(28)r≥Λ21-Λ1.For any x∈Br and for each t∈Jk, we have(29)Kxt≤t-tkαk-1Γαk∑j=1kItj-11fs,xstj+φjxtj+Itkαkfs,xst≤t-tkαk-1Γαk∑j=1kItj-11fs,xs-fs,0+fs,0·tj+φjxtj-φj0+φj0+Itkαkfs,xs-fs,0-fs,0t≤t-tkαk-1ΓαkL1r+Mtk+L2r+Nk+t-tkαkΓαk+1L1r+M.Multiplying both sides of the above inequality by (t-tk)γ for each t∈Jk, we obtain(30)t-tkγKxt≤t-tkγ+αk-1ΓαkL1r+Mtk+L2r+Nk+t-tkγ+αkΓαk+1L1r+M≤rT∗Γ∗L1+L1T+L2m+T∗Γ∗M+MT+Nm=rΛ1+Λ2≤r.This implies that KBr⊂Br.
Finally we will show thatK is a contraction mapping on Br. For x,y∈PCγ(J,R) and for each t∈J we have(31)Kxt-Kyt≤t-tkαk-1Γαk∑j=1kItj-11fs,xs-fs,ys·tj+φjxtj-φjytj+Itkαkfs,xs-fs,yst≤t-tkαk-1Γαk∑j=1kL1tj-tj-1x-y+L2x-y+t-tkαkΓαk+1L1x-y.Multiplying both sides of the above inequality by (t-tk)γ for each t∈Jk, we have (32)t-tkγKxt-t-tkγKyt≤t-tkγ+αk-1ΓαkL1tkx-y+L2kx-y+t-tkγ+αkΓαk+1L1x-y≤T∗Γ∗L1+L1T+L2mx-y.It follows that(33)Kx-Ky≤Λ1x-y.Since Λ1<1, K is a contraction mapping on Br. Therefore (3) has a unique solution on [0,T].Example 8.
Consider the following impulsive multiorders Riemann-Liouville fractional initial value problem:(34)Dtkk+1/k+2x=xtln⁡t+124t+331+2xt+etcos⁡t5-2t,t∈0,1110,t≠tk,Δ~xtk=xtk4k!+xtk,k=1,2,…,10,tk=k10,x0=0.
Hereαk=(k+1)/(k+2), k=0,1,2,…,10, m=10, T=11/10, f(t,x)=(|x|ln(t+1)2)/((4t+3)3(1+2|x|))+(etcost)/(5-2t), and φk(x)=|x|/((4k)!+|x|). Since |f(t,x)-f(t,y)|≤(2/27)|x-y| and |φk(x)-φk(y)|≤(1/24)|x-y|, for k=1,2,…,10, then (H1) and (H2) are satisfied with L1=2/27, L2=1/24.
By choosingγ=1, we find that T∗=1.200428, Γ∗=0.886227, and(35)Λ1=T∗Γ∗L1+L1T+L2m=0.775096<1.Hence, by Theorem 7, the initial value problem (34) has a unique solution on [0,11/10].


## 4. Impulsive Riemann-Liouville Fractional Differential Equations of Orders1<αk≤2

Problem (5) is studied in this section.Lemma 9.
The unique solution of problem (5) is given by(36)xt=t-tkαk-2Γαk-1βtk+∑j=1k-1tk-tjItj-11ftj,xtj+φjxtj+∑j=1kItj-12ftj,xtj+φj∗xtj+t-tkαk-1Γαkβ+∑j=1kItj-11ftj,xtj+φjxtj+Itkαkft,xt,for t∈Jk, k=0,1,2,…,m with ∑j=ab‍=0 for  b<a.Proof.
Fort∈J0, taking the Riemann-Liouville fractional integral of order α0 for the first equation of (5) and applying Lemma 5, we obtain(37)xt=It0α0ft,xt+c0tα0-1Γα0+c1tα0-2Γα0-1,where c0=It01-α0x(0) and c1=It02-α0x(0). The initial condition x(0)=0 implies c1=0 which leads to(38)xt=It0α0ft,xt+c0tα0-1Γα0.Using the Riemann-Liouville fractional derivative of order α0-1 for (38) on J0, we get (39)Dα0-1xt=It01ft,xt+c0.From the second initial condition of (5), we get (40)xt=It0α0ft,xt+βtα0-1Γα0.Taking the Riemann-Liouville fractional integral of order 1-α0 and 2-α0 for (40) and substituting t=t1, we have(41)It01-α0xt1=It01ft1,xt1+β,It02-α0xt1=It02ft1,xt1+βt1.For t∈J1, taking the Riemann-Liouville fractional integral of order α1 for (5), we have(42)xt=t-t1α1-2Γα1-1It12-α1xt1++t-t1α1-1Γα1It11-α1xt1++It1α1ft,xt.Since It11-α1x(t1+)=It01-α0x(t1)+φ1(x(t1)) and It12-α1x(t1+)=It02-α0x(t1)+φ1∗(x(t1)), it follows that, for t∈J1, (43)xt=t-t1α1-2Γα1-1βt1+It02ft1,xt1+φ1∗xt1+t-t1α1-1Γα1β+It01ft1,xt1+φ1xt1+It1α1ft,xt.The Riemann-Liouville integrating of the above equation of order 1-α1 and 2-α1 for t=t2 leads to(44)It12-α1xt2=βt1+It02ft1,xt1+φ1∗xt1+t2-t1(β+It01ft1,xt1+φ1xt1+It12ft2,xt2,It11-α1xt2=β+It01ft1,xt1+φ1xt1+It11ft2,xt2.For t∈J2, applying the Riemann-Liouville fractional integral of order α2 for (5) and substituting values It21-α2x(t2+)=It11-α1x(t2)+φ2(x(t2)) and It22-α2x(t2+)=It12-α1x(t2)+φ2∗(x(t2)), we get (45)xt=t-t2α2-2Γα2-1βt2+t2-t1It01ft1,xt1+φ1xt1+It02ft1,xt1+φ1∗xt1+It12ft2,xt2+φ2∗xt2+t-t2α2-1Γα2β+It01ft1,xt1+φ1xt1+It1ft2,xt2+φ2xt2+It2α2ft,xt.Repeating the above process, for t∈J, we obtain (36) as requested.Next, we will prove the existence and uniqueness of a solution to the initial value problem (5) by using Banach’s fixed point theorem.Theorem 10.
Assume that(H1) and (H2) hold. In addition we suppose that (H3)  φk∗:R→R, k=1,2,3,…,m, are continuous functions and satisfy(46)φk∗x-φk∗y≤L3x-y,L3>0,∀x,y∈R.If(47)Ω1≔T2∗Γ2∗L1Φ+T+1+L22m-1+L3m<1,where(48)Φ=12∑j=1mtj-tj-12tm-tj-tj-1,T2∗=max{Tγ+αk-2,Tγ+αk-1,Tγ+αk}, Γ2∗=min{Γ(αk-1),Γ(αk), and Γ(αk+1)}, γ+αk>2 for k=0,1,2,…,m, then problem (5) has a unique solution on [0,T].Proof.
We define the operatorA:PC(J,R)→PC(J,R) as follows:(49)Axt=t-tkαk-2Γαk-1βtk+∑j=1k-1tk-tjItj-11ftj,xtj+φjxtj+∑j=1kItj-12ftj,xtj+φj∗xtj+t-tkαk-1Γαkβ+∑j=1kItj-11ftj,xtj+φjxtj+Itkαkft,xt,for t∈Jk, k=0,1,2,…,m with ∑j=ab‍=0 for b<a.
It is straightforward to show thatAx∈PCγ(J,R); see Theorem 7. Next we will show that ABr⊂Br, where a ball Br is defined by Br={x∈PCγ(J,R),xPCγ≤r}. Assume that supt∈J|f(t,0)|=M, max{|φk(0)|:k=1,2,3,…,m}=N and max{|φk∗(0)|:k=1,2,3,…,m}=P. Setting(50)Ω2=T2∗Γ2∗MΦ+T+1+N2m-1+Pm+βT+1,we choose a constant r such that(51)r≥Ω21-Ω1.Let x∈Br. For any t∈Jk, k=0,1,2,…,m, we have(52)Axt≤t-tkαk-2Γαk-1βtk+∑j=1k-1tk-tjItj-11fs,xstj+φjxtj+∑j=1kItj-12fs,xstj+φj∗xtj+t-tkαk-1Γαkβ+∑j=1kItj-11fs,xstj+φjxtj+Itkαkfs,xst≤t-tkαk-2Γαk-1βtk+∑j=1k-1tk-tjItj-11fs,xs-fs,0+fs,0tj+φjxtj-φj0+φjx0+∑j=1kItj-12fs,xs-fs,0+fs,0tj+φj∗xtj-φj∗0+φj∗0+t-tkαk-1Γαkβ+∑j=1kItj-11fs,xs-fs,0+fs,0tj+φjxtj-φj0+φj0+Itkαkfs,xs-fs,0+fs,0t≤t-tkαk-2Γαk-1βtk+L1r+M∑j=1k-1tk-tjtj-tj-1+L2r+Nk-1+L1r+M∑j=1ktj-tj-122+L3r+Pk+t-tkαk-1Γαkβ+L1r+M∑j=1ktj-tj-1+L2r+Nk+L1r+Mt-tkαkΓαk+1.Multiplying both sides of the above inequality by (t-tk)γ for t∈Jk, we have(53)t-tkγAxt≤t-tkγ+αk-2Γαk-1βtk+L1r+M∑j=1k-1tk-tjtj-tj-1+L2r+Nk-1+L1r+M∑j=1ktj-tj-122+L3r+Pk+t-tkγ+αk-1Γαkβ+L1r+M∑j=1ktj-tj-1+L2r+Nk+L1r+Mt-tkγ+αkΓαk+1≤T2∗Γ2∗βT+L1r+M∑j=1m-1tm-tjtj-tj-1+L2r+Nm-1+L1r+M∑j=1mtj-tj-122+L3r+P+β+L1r+M∑j=1mtj-tj-1+L2r+Nm+L1r+M=rT2∗Γ2∗L1Φ+T+1+L22m-1+L3m+T2∗Γ2∗MΦ+T+1+N2m-1+Pm+βT+1=Ω1r+Ω2.This implies that ABr⊂Br.
Finally we will show thatA is a contraction mapping on Br. For x,y∈PCγ(J,R) and for each t∈J we have (54)Axt-Ayt≤t-tkαk-2Γαk-1∑j=1k-1tk-tjItj-11fs,xs-fs,ystj+φjxtj-φjytj+∑j=1kItj-12fs,xs-fs,ystj+φj∗xtj-φj∗ytj+t-tkαk-1Γαk∑j=1kItj-11fs,xs-fs,ystj+φjxtj-φjytj+Itkαkfs,xs-fs,ys·t.Multiplying both sides of the above inequality by (t-tk)γ, we have (55)t-tkγAxt-t-tkγAyt≤t-tkγ+αk-2Γαk-1L1x-y·∑j=1k-1tk-tjtj-tj-1+L2x-yk-1+L1x-y∑j=1ktj-tj-122+L3x-yk+t-tkγ+αk-1ΓαkL1x-y∑j=1ktj-tj-1+L2x-yk+L1x-yt-tkγ+αkΓαk+1≤Ω1x-y.It follows that(56)Ax-Ay≤Ω1x-y.Since Ω1<1, A is a contraction mapping on Br. Therefore (5) has a unique solution on [0,T].Example 11.
Consider the following impulsive multiorders Riemann-Liouville fractional initial value problem:(57)Dtk2k2+3k+5/k2+2k+3x=2txtcos⁡t7+xt2+xtsin2⁡t+12,t∈0,tan-1113π,t≠tk,Δ~xtk=xtk9k2+3k+5+xtk,k=1,2,…,10,tk=tan-1⁡k3π,Δ∗xtk=xtk371-2cos⁡kπ+xtk,k=1,2,…,10,tk=tan-1⁡k3π,x0=0Dt02/3x0=e.
Hereαk=(2k2+3k+5)/(k2+2k+3), k=0,1,2,…,10, m=10, T=tan-1(11/(3π)), f(t,x)=(2t|x|cost)/((7+|x|)2+|x|sin2t)+(1/2), φk(x)=|x|/(9(k2+3k+5)+|x|), and φk∗(x)=|x|/(37(1-2coskπ)+|x|). Since |f(t,x)-f(t,y)|≤(2/49)|x-y|, |φk(x)-φk(y)|≤(1/81)|x-y|, and |φk∗(x)-φk∗(y)|≤(3/37)|x-y|, then (H1), (H2), and (H3) are satisfied with L1=2/49, L2=1/81, and L3=3/37, respectively. By choosing γ=2, we can find that T2∗=0.781307, Γ2∗=0.902745, Φ=0.332114, and(58)Ω1=T2∗Γ2∗L1ϕ+T+1+L22m-1+L3m=0.982275<1.Hence, by Theorem 10, the initial value problem (57) has a unique solution on [0,tan-111/(3π)].



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*Source: 603893-2015-06-28.xml*