# Existence and Uniqueness of Positive Solutions for a Fractional Switched System

Authors: Zhi-Wei Lv; Bao-Feng Chen
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We discuss the existence and uniqueness of positive solutions for the following fractional switched system: (Dc0+αu(t)+fσ(t)(t,u(t))+gσ(t)(t,u(t))=0,t∈J=[0,1]); (u(0)=u′′(0)=0,u(1)=∫01u(s)ds), where Dc0+α is the Caputo fractional derivative with 2<α≤3, σ(t):J→{1,2,…,N} is a piecewise constant function depending on t, and ℝ+=[0,+∞),fi,gi∈C[J×ℝ+,ℝ+], i=1,2,…,N. Our results are based on a fixed point theorem of a sum operator and contraction mapping principle. Furthermore, two examples are also given to illustrate the results.


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## Body

## 1. Introduction

Fractional differential equations arise in various areas of science and engineering. Due to their applications, fractional differential equations have gained considerable attention (cf., e.g., [1–15] and references therein). Moreover, the theory of boundary value problems with integral boundary conditions has various applications in applied fields. For example, heat conduction, chemical engineering, underground water flow, thermoelasticity, and population dynamics can be reduced to the nonlocal problems with integral boundary conditions. In [2], Cabada and Wang considered the followingm-point boundary value problem for fractional differential equation
(1)Dc0+αu(t)+f(t,u(t))=0,0<t<1,u(0)=u′′(0)=0,u(1)=λ∫01‍u(s)ds,
where 2<α≤3, cD0+α is the Caputo fractional derivative, and f:[0,1]×[0,∞)→[0,∞) is a continuous function.On the other hand, a switched system consists of a family of subsystems described by differential or difference equations, which has many applications in traffic control, switching power converters, network control, multiagent consensus, and so forth (see [16–18]). When we consider a switched system, we always suppose that the solution is unique. So it is important to study the uniqueness of solution for a switched system. In [1], Li and Liu investigated the uniqueness of positive solution for the following switched system:
(2)x′′(t)+fσ(t)(t,x(t))=0,t∈J=[0,1],x(0)=0,x(1)=∫01‍a(s)x(s)ds,
where σ(t):J→{1,2,…,N} is a piecewise constant function depending on t, and ℝ+=[0,+∞), fi∈C[J×ℝ+,ℝ+], i=1,2,…,N.In this paper, we discuss the existence and uniqueness of positive solutions for the following fractional switched system:(3)Dc0+αu(t)+fσ(t)(t,u(t))+gσ(t)(t,u(t))=0,hhhhhhhhhhhhhhhhhhhhhihhht∈J=[0,1],u(0)=u′′(0)=0,u(1)=∫01‍u(s)ds,
where cD0+α is the Caputo fractional derivative with 2<α≤3, σ(t):J→{1,2,…,N} is a piecewise constant function depending on t, and ℝ+=[0,+∞), fi,gi∈C[J×ℝ+,ℝ+], i=1,2,…,N.The paper is organized as follows. In Section2, we present some background materials and preliminaries. Section 3 deals with some existence results. In Section 4, two examples are given to illustrate the results.


## 2. Background Materials and Preliminaries

Definition 1 (see [3, 4]).
The fractional integral of orderα with the lower limit t0 for a function f is defined as
(4)Iαf(t)=1Γ(α)∫t0t‍(t-s)α-1f(s)ds,t>t0,α>0,
where Γ is the gamma function.Definition 2 (see [3, 4]).
For a functionf:[0,∞)→ℝ, the Caputo derivative of fractional order is defined as
(5)cD0+αf(t)=1Γ(n-α)∫0t‍(t-s)n-α-1f(n)(s)ds,α>0,n=[α]+1.In the following, let us recall some basic information on cone (see more from [19, 20]). Let E be a real Banach space and let P be a cone in E which defined a partial ordering in E by x≤y if and only if y-x∈P. P is said to be normal if there exists a positive constant N such that θ≤x≤y implies ∥x∥≤N∥y∥. P is called solid if its interior P° is nonempty. If x≤y and x≠y, we write x<y. We say that an operator A is increasing if x≤y implies Ax≤Ay.For allx,y∈E, the notation x~y means that there exist λ>0 and μ>0 such that λx≤y≤μx. Clearly, ~ is an equivalence relation. Given h>θ (i.e., h≥θ and h≠θ), we denote by Ph the set Ph={x∈E∣x~h}. It is easy to see that Ph⊂P.Definition 3.
LetD=P or D=P° and let γ be a real number with 0≤γ<1. An operator A:P→P is said to be γ-concave if it satisfies
(6)A(tx)≥tγAx,∀t∈(0,1),x∈D.Definition 4.
An operatorA:E→E is said to be homogeneous if it satisfies
(7)A(tx)=tAx,∀t>0,x∈E.
An operator A:P→P is said to be subhomogeneous if it satisfies
(8)A(tx)≥tAx,∀t∈(0,1),x∈P.From [2], we have the following result.Lemma 5.
Assume that2<α≤3 and fi,gi∈C[J×ℝ+,ℝ+], i=1,2,…,N. Then the problem (3) has a solution if and only if u is a solution of the integral equation
(9)u(t)=∫01‍G(t,s)(fσ(s)(s,u(s))+gσ(s)(s,u(s)))ds,
where
(10)G(t,s)={2t(1-s)α-1(α-1+s)-α(t-s)α-1Γ(α+1),0≤s≤t≤1,2t(1-s)α-1(α-1+s)Γ(α+1),0≤t≤s≤1.Lemma 6.
G
(
t
,
s
) in Lemma 5 has the following property:(i)
G(t,s)>0forallt,s∈(0,1).(ii)
(1/Γ(α+1))h(t)(1-s)α-1(α-2+2s)≤G(t,s)≤(2/Γ(α+1))h(t)(1-s)α-1(α-1+s),t,s∈[0,1],2<α≤3,h(t)=t.Proof.
From [2], we know that (i)is obvious. For 0≤s≤t≤1, 2<α≤3, we have
(11)2t(1-s)α-1(α-1+s)-α(t-s)α-1=2t(1-s)α-1(α-1+s)-αtα-1(1-st)α-1≥2t(1-s)α-1(α-1+s)-αt(1-s)α-1=t(1-s)α-1(α-2+2s).
This means that (ii)holds.Theorem 7 (see [19]).
LetP be a normal cone in a real Banach space E, A:P→P an increasing γ-concave operator, and B:P→P an increasing subhomogeneous operator. Assume that (i)
there ish>θ such that Ah∈Ph and Bh∈Ph;(ii)
there exists a constantδ0>0 such that Ax≥δ0Bx, ∀x∈P.Then the operator equation Ax+Bx=x has a unique solution x* in Ph. Moreover, constructing successively the sequence yn=Ayn-1+Byn-1, n=1,2,…, for any initial value y0∈Ph, we have yn→x* as n→∞.


## 3. Main Results

In this section, we will deal with the existence and uniqueness of positive solutions for problem (3). Let
(12)G1(s,s)=2Γ(α+1)(1-s)α-1(α-1+s).
It is obvious that
(13)G(t,s)≤G1(s,s),t,s∈[0,1].
We consider the Banach space E=C[[0,1],ℝ] endowed with the norm defined by ∥u∥=sup⁡0≤t≤1|u(t)|. Letting P={u∈E∣u(t)≥0}, then P is a cone in E. Define an operator Ϝ:E→E as
(14)(Ϝu)(t)=∫01‍G(t,s)(fσ(s)(s,u(s))+gσ(s)(s,u(s)))ds.
Then Ϝ has a solution if and only if the operator Ϝ has a fixed point.Theorem 8.
Letfi,gi∈C[J×ℝ+,ℝ+], i=1,2,…,N. Suppose that the following conditions are satisfied:
(15)|fi(t,u(t))-fi(t,v(t))|≤li(t)|u(t)-v(t)|,|gi(t,u(t))-gi(t,v(t))|≤l¯i(t)|u(t)-v(t)|,0<∫01‍G1(s,s)(li(s)+l¯i(s))ds<1,
where
(16)li,l¯i∈C[J,ℝ+],i=1,2,…,N.
Then the problem (3) has a unique solution on [0,1].Proof.
It follows from Lemma6 that Ϝ:P→P. For t∈J, i=1,2,…,N, we set max⁡i=1,2,…,Nsup⁡t∈J|fi(t,0)|=M, max⁡i=1,2,…,Nsup⁡t∈J|gi(t,0)|=M¯, and Br={u∈C[J,ℝ+]:∥u∥≤r}, where
(17)r≥(M+M¯)∫01‍G1(s,s)ds1-max⁡i=1,2,…,N∫01‍G1(s,s)(li(s)+l¯i(s))ds.
Step  1. We show that Ϝ(Br)⊂Br.
Foru∈Br and t∈J, i=1,2,…,N,
(18)∫01G(t,s)|fi(s,u(s))+gi(s,u(s))|ds≤∫01G1(s,s)(|fi(s,u(s))-fi(s,0)|+|fi(s,0)|)ds+∫01G1(s,s)(|gi(s,u(s))-gi(s,0)|+|gi(s,0)|)ds≤rmax⁡i=1,2,…,N⁡∫01G1(s,s)(li(s)+l¯i(s))ds+(M+M¯)∫01G1(s,s)ds≤r,
which implies that |(Ϝu)(t)|≤r. Thus, ∥Ϝu∥≤r. Therefore,
(19)Ϝ(Br)⊂Br.
Step  2. We show that Ϝ is a contraction mapping.
Foru,v∈Br and for each t∈J, i=1,2,…,N, we have
(20)∫01G(t,s)|fi(s,u(s))-fi(s,v(s))|ds+∫01G(t,s)|gi(s,u(s))-gi(s,v(s))|ds≤∫01G1(s,s)li(s)|u(s)-v(s)|ds+∫01G1(s,s)l¯i(s)|u(s)-v(s)|ds≤∫01G1(s,s)(li(s)+l¯i(s))ds∥u-v∥.
This, together with 0<∫01‍G1(s,s)(li(s)+l¯i(s))ds<1, i=1,2,…,N, yields that
(21)|(Ϝu)(t)-(Ϝv)(t)|≤k∥u-v∥,
where
(22)0<k=max⁡i=1,2,…,N∫01‍G1(s,s)(li(s)+l¯i(s))ds<1.
Thus,
(23)∥Ϝu-Ϝv∥≤k∥u-v∥.
This means that Ϝ is a contraction mapping.
It follows from Banach’s contraction mapping thatϜ has a unique fixed point in Br. Therefore, the problem (3) has a unique solution.Corollary 9.
Letfi∈C[J×ℝ+,ℝ+], i=1,2,…,N. Suppose that the following conditions are satisfied:
(24)|fi(t,u(t))-fi(t,v(t))|≤li(t)|u(t)-v(t)|,0<∫01‍G1(s,s)li(s)ds<1,
where
(25)li∈C[J,ℝ+],i=1,2,…,N.
Then the following fractional switched system
(26)D0+αcu(t)+fσ(t)(t,u(t))=0,t∈J=[0,1],u(0)=u′′(0)=0,u(1)=∫01‍u(s)ds,
has a unique solution on [0,1].Theorem 10.
Assume that;(H1)
fi,gi∈C[J×ℝ+,ℝ+] and fi(t,x), gi(t,x) are increasing in x for x∈ℝ+, gi(t,0)≠0, i=1,2,…,N;(H2)
gi(t,λx)≥λgi(t,x) for λ∈(0,1),t∈J,x∈ℝ+, and there exists a constant γ∈(0,1) such that fi(t,λx)≥λγfi(t,x),∀t∈J,λ∈(0,1),x∈ℝ+, i=1,2,…,N;(H3)
there exists a constantδ0>0 such that fi(t,x)≥δ0gi(t,x), t∈J,x∈ℝ+, i=1,2,…,N.Then problem (3) has a unique solution u* in Ph, where h(t)=t,t∈J. Moreover, for any initial value u0∈Ph, constructing successively the sequence
(27)un+1(t)=∫01‍G(t,s)(fσ(s)(s,un(s))+gσ(s)(s,un(s)))ds,jjjjjjjjjjjjjjjjjjjjjjjjjjjjijjjjjjjn=0,1,2,…,
we have un(t)→u*(t) as n→∞.Proof.
Define the two operators(28)Au(t)=∫01‍G(t,s)fσ(s)(s,u(s))ds,Bu(t)=∫01‍G(t,s)gσ(s)(s,u(s))ds.
From Lemma 6, we have A:P→P and B:P→P. It is obvious that u is the solution of problem (3) if and only if u=Au+Bu. It follows from (H1) that A and B are two increasing operators. Thus, for u,v∈P,u≥v, we have Au≥Av and Bu≥Bv.
Step
1. We show that A is a γ-concaveoperator and B is a subhomogeneous operator.
In fact, forλ∈(0,1),u∈P,t∈J,i=1,2,…,N, from (H2), we have
(29)∫01‍G(t,s)fi(s,λu(s))ds≥λγ∫01‍G(t,s)fi(s,u(s))ds,
which yields that
(30)A(λu)(t)≥λγAu(t).
Thus, A is a γ-concaveoperator. By a closely similar way, we can see that B is a subhomogeneous operator.
Step
2. We show that Ah∈Ph and Bh∈Ph.
From Lemma6 and (H1), we have, for t∈J, i=1,2,…,N,
(31)∫01‍G(t,s)fi(s,h(s))ds≤2Γ(α+1)h(t)×∫01‍(1-s)α-1(α-1+s)fi(s,1)ds,∫01‍G(t,s)fi(s,h(s))ds≥1Γ(α+1)h(t)×∫01‍(1-s)α-1(α-2+2s)fi(s,0)ds.
For i=1,2,…,N, let
(32)mi=1Γ(α+1)∫01‍(1-s)α-1(α-2+2s)fi(s,0)ds,m¯i=2Γ(α+1)∫01‍1-sα-1(α-1+s)fi(s,1)ds.
It follows from g(t,0)≠0 that ∫01‍fi(s,1)ds≥∫01‍fi(s,0)ds≥δ0∫01‍gi(s,0)ds>0.
Thus,(33)mi>0,m¯i>0,i=1,2,…,N.
Letting m~=min⁡{mi,i=1,2,…,N} and m^=max⁡{m¯i,i=1,2,…,N}, then m~>0 and m^>0. Therefore,
(34)m~h(t)≤Ah(t)≤m^h(t),
which implies that
(35)Ah∈Ph.
Similarly, we have Bh∈Ph.
Step
3. There exists a constant δ0>0 such that Au≥δ0Bu, ∀u∈P.
Foru∈P and t∈J, i=1,2,…,N, by (H3), we have
(36)∫01‍G(t,s)fi(s,u(s))ds≥δ0∫01‍G(t,s)gi(s,u(s))ds.
This means that
(37)Au≥δ0Bu,u∈P.
Therefore, the conditions of Theorem 7 are satisfied. By means of Theorem 7, we obtain that the operator equation Au+Bu=u has a unique solution u* in Ph. Moreover, for any initial value u0∈Ph, constructing successively the sequence
(38)un+1(t)=∫01‍G(t,s)(fσ(s)(s,un(s))+gσ(s)(s,un(s)))ds,hhhhhhhhhhhhhhhhhhhhhhhhhhn=0,1,2,…,
we have un(t)→u*(t) as n→∞.In Theorem10, if we let B be a null operator, we have the following conclusion.Corollary 11.
Assume that;(H4)
fi∈C[J×ℝ+,ℝ+] and fi(t,x) is increasing in x for x∈ℝ+, fi(t,0)≠0, i=1,2,…,N;(H5)
there exists a constantγ∈(0,1) such that fi(t,λx)≥λγfi(t,x),∀t∈J,λ∈(0,1),x∈ℝ+, i=1,2,…,N.
Then the following fractional switched system(39)D0+αcu(t)+fσ(t)(t,u(t))=0,t∈J=[0,1],u(0)=u′′(0)=0,u(1)=∫01‍u(s)ds,
has a unique solution u* in Ph, where h(t)=t,t∈J. Moreover, for any initial value u0∈Ph, constructing successively the sequence
(40)un+1(t)=∫01‍G(t,s)fσ(s)(s,un(s))ds,n=0,1,2,…,
we have un(t)→u*(t) as n→∞.


## 4. Examples

Example 1.
Consider the following boundary value problem:(41)D0+αcu(t)+fσ(t)(t,u(t))+gσ(t)(t,u(t))=0,hhhhhhhhhhhhhhhhiihhhhhhht∈J=[0,1],u(0)=u′′(0)=0,u(1)=∫01‍u(s)ds,
where α=5/2, σ(t):J→M={1,2} is a finite switching signal,
(42)f1(t,x)=14(t+2)2x1+x+1,g1(t,x)=116sin2x+t2,f2(t,x)=18(t+2)2x1+x+1,g2(t,x)=132sin2x+t2.
Thus,
(43)fi,gi∈C[J×ℝ+,ℝ+],i=1,2.
By computation, we deduce that
(44)|f1(t,x1)-f1(t,x2)|≤116|x2-x1|,|g1(t,x1)-g1(t,x2)|≤116|x2-x1|,|f2(t,x1)-f2(t,x2)|≤132|x2-x1|,|g2(t,x1)-g2(t,x2)|≤132|x2-x1|.
On the other hand,
(45)∫01‍G1(s,s)(l1(s)+l¯1(s))ds=∫01‍2(1-s)(5/2)-1((5/2)-1+s)Γ((5/2)+1)(116+116)ds=14Γ(7/2)∫01‍(1-s)3/2(32+s)ds≤14Γ(7/2)∫01‍(1-s)3/2(32+1)ds=13π×25<1,∫01‍G1(s,s)(l1(s)+l¯1(s))ds=∫01‍2(1-s)(5/2)-1((5/2)-1+s)Γ((5/2)+1)(132+132)ds=13π×15<1.
Hence, by Theorem 8, BVP (41) has a unique positive solution on [0,1].Example 2.
Consider the following boundary value problem:(46)cD0+αu(t)+fσ(t)(t,u(t))+gσ(t)(t,u(t))=0,hhhhhhhhhhhhhhhhhihhhhhht∈J=[0,1],u(0)=u′′(0)=0,u(1)=∫01‍u(s)ds,
where α=5/2, σ(t):J→{1,2,3} is a finite switching signal,
(47)f1(t,x)=x1/3+t2+c,g1(t,x)=x(1+t2)(1+x)+b-c,f2(t,x)=2x1/3+t2+2c,g2(t,x)=2x(1+t2)(1+x)+2(b-c),f3(t,x)=3x1/3+t2+3c,g3(t,x)=3x(1+t2)(1+x)+3(b-c).
Let γ=1/3 and 0<c<b. It is obvious that fi,gi∈C[J×ℝ+,ℝ+] and are increasing with respect to the second argument, gi(t,0)=b-c>0,i=1,2,3. On the other hand, for λ∈(0,1),t∈J,x∈[0,+∞),i=1,2,3, we have
(48)gi(t,λx)=iλx(1+t2)(1+λx)+i(b-c)≥iλx(1+t2)(1+λx)+iλ(b-c)=λgi(t,x),fi(t,λx)=iλ1/3x1/3+t2+ic≥λ1/3(ix1/3+t2+ic)=λγfi(t,x).
Moreover, for t∈J,x∈ℝ+,i=1,2,3, we have
(49)fi(t,x)=ix1/3+t2+ic≥ic≥c3+(b-c)(i+i(b-c))≥c3+(b-c)(ix(1+t2)(1+x)+i(b-c))=δ0gi(t,x),
where
(50)δ0=c3+(b-c).
Hence all the conditions of Theorem 10 are satisfied. Thus, BVP (46) has a unique positive solution in Ph, where h(t)=t, t∈[0,1].



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