# The Solvability and Optimal Controls for Some Fractional Impulsive Equation

Authors: Xianghu Liu; Zhenhai Liu; Jiangfeng Han
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper is concerned with the existence and uniqueness of mild solution of some fractional impulsive equations. Firstly, we introduce the fractional calculus, Gronwall inequality, and Leray-Schauder’s fixed point theorem. Secondly with the help of them, the sufficient condition for the existence and uniqueness of solutions is presented. Finally we give an example to illustrate our main results.


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## Body

## 1. Introduction

In this paper, we study some fraction evolution with finite impulsive:(1)Dctαx(t)=Ax(t)+I1-αf(t,x(t))+B(t)U(t),iiiiiiiiiiiiiiiiiiiiiiit∈J=[0,b],t≠tk,Δx(tk)=Ik(x(tk-))k=1,2,3,…,n,x(0)=x0,
where cDtα is the standard Caputo fractional derivative of order α, b>0, 0<α<1, A:D(A)⊂X→X is a generator of a C0 semigroup {T(t),t≥0} defined on a complex Banach space X, let f:J×X→X be a given function and satisfying some assumptions that will be specified later, the function Ik:X→X is continous, and 0=t0<t1<t2<⋯<tk<⋯<tn=T,  Δx(tk)=x(tk+)-x(tk-), x(tk+) and x(tk-) denote the right and the left limits of x(t) at t=tk(k=1,2,…,n),  U is a given control function in another Banach space Y, and B is a linear operator from Y to X.The fractional calculus and fractional difference equations have attracted lots of authors during the past years, and they gave some outstanding work [1–4], because they described many phenomena in engineering, physics, science, and controllability. Delay evolution equation allows someone to think after-effect, so it is a relative important equation. There are some significant development; for example, Wang et al. [5, 6] consider the following fractional delay nonlinear integrodiffrential controlled system:
(2)Dtqx(t)+Ax(t)=f(t,xt,∫0t‍g(t,s,xs))+B(t)u(t),iiiiiiiiiiiiiiit∈I=(0,T],q∈(0,1),iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiix(t)=ϕ(t)t∈[-r,0],
and they used laplace transform and probability density functions to prove some sufficient conditions of some fractional nonlinear finite time delay evolution equations. Shu et al. [7] used the solution operator of semigroup to investigate the system given by
(3)Dαx(t)=Ax(t)+f(t),α∈(0,1),x(0)=x0,Δx|t=tk=Ik(x(tk-)).
Benchohra et al. [8] deal with existence of mild solutions of some fractional functional evolution equations with infinite delay. Balachandran et al. [9] concerned the relative controllability of fractional dynamical with delays in control. Specially, Cuevas and Lizama [10] studied some sufficient conditions for the existence and uniqueness of almost automorphic mild solutions to the following semilinear fractional differential equation fractional differential equations:
(4)Dαx(t)=Ax(t)+Dα-1f(t,x(t)),α∈(1,2),x(0)=x0.Bazhlekova [11] studied the fractional evolution equations in Banach spaces. Xue and Xiong [12] concerned the existence and uniqueness of mild solutions for abstract differential equations given by
(5)Dtqu(t)=Au(t)+Jt1-αf(t,ut)t∈I=(0,T],α∈(0,1),x(t)=ϕ(t)t∈[-r,0].Motivated by the abovementioned works, we study (1). The rest of this paper is organized as follows. In Section 2, some notation and preparation are given. In Section 3, some mainly results of (1) are obtained. At last, an example is given to demonstrate our results.


## 2. Preliminar

In this section, we will give some definitions and preliminar which will be used in the paper. The norm of the spaceX will be defined by ∥·∥X. Let C(J,X) denote the Banach space of all X value continuous functions from J=[0,T] into X, the norm ∥·∥c=sup∥·∥X. Let the another banach space PC(J,X)={x:J→X,x∈C((tk,tk+1],X),  k=0,1,2,…,n, there exist x(tk-), x(tk+), k=1,2,…,n,  x(tk-)=x(tk)},  ∥x∥PC=max{sup∥x(t+0)∥,sup∥x(t-0)∥}. We can use Lp(J,R) to denote the Banach space of all Lebesgue measurable functions from J to R with ∥f∥Lp(J,R)=(∫J‍|f(t)|pdt)1/p, and Lp(J,X) denote the Banach space of functions f:J→X which are Bochner integrable normed by ∥f∥Lp(J,X), u∈Lp(J,R).Let us recall some known definitions; for more details, see [2–4].Letα,β>0 that n-1<α<n,  n-1<β<n, and f is a suitable function.Definition 1 (Riemann-Liouville fractional integral and derivative operators).
The integral operatorIaα is defined on L1[a,b] by
(6)Iaαf(x)=1Γ(α)∫ax(x-t)α-1f(t)dt,(a≤x≤b).
The derivative operators are defined as Daαf(x)=Dan(Ian-α)f(x), where Dan=d^n/dt^n and
(7)IaαIaβf(x)=Iaα+βf(x).Definition 2.
Caputo fractional derivative off(x) of order α is defined as
(8)cDaαf(x)=1Γ(n-α)∫ax(x-t)n-α-1f(n)(t)dt.
If a=0, we can write the Caputo derivative of the function f(t)∈Cn[0,∞),  f:[0,∞)→R via the above Riemann-Liouville fractional derivative as
(9)cD0αf(x)=DLα[f(x)-∑k=0n-1xkk!f(k)(0)].
Let us recollect the generalized Gronwall inequality which can be found in [13] and will be used in our main result.Lemma 3.
Supposeβ>0, a(t) is a nonnegative function locally integrable on [0,T], and b(t) is a nonnegative, nondecreasing continuous function defined on [0,T], b(t)≤M (constant), and y(t) is nonnegative and locally integrable on [0,T] with
(10)y(t)≤a(t)+b(t)∫0t(t-s)β-1y(s)ds,t∈[0,T].
Then
(11)y(t)≤a(t)+∫0t[∑n=1∞‍[b(t)Γ(β)]nΓ(nβ)(t-s)nβ-1a(s)]ds,iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiit∈[0,T].Remark 4.
Under the hypothesis of Lemma3, let a(t) be a nondecreasing function on [0,T]. Then
(12)y(t)≤a(t)Eβ(b(t)Γ(β)tβ),
where Eβ is the Mittag-Leffler function defined by
(13)Eβ(z)=∑k=0∞‍zkΓ(kβ+1).By Lemma3 and Remark 4, we can establish a useful nonlinear impulsive Gronwall inequality which will be used in calculating.Lemma 5 (see [14]).
Letx∈PC(J,X) satisfy the following inequality:
(14)∥x(t)∥≤c1+c2∫0t(t-s)β-1∥x(s)∥ds+∑0<tk<thk∥x(tk-)∥,
where c1,c2,hk≥0 are constants. Then
(15)∥x(t)∥≤c1(1+H*Eβ(c2Γ(β)tβ))k×Eβ(c2Γ(β)tβ)fort∈(tk,tk+1],
where H*=max{hk:k=1,2,…,m}.Specially, ifβ=1,
(16)∥x(t)∥≤c1(1+H*E1(c2t)k)E1(c2t)fort∈(tk,tk+1].We also introduce the following theorem that will be used in our mainly result.Theorem 6 (Hölder’s inequality).
Assume thatp>0, q>0, and 1/p+1/q=1;  if f∈Lp(Ω) and g∈Lq(Ω) then f·g∈L1(Ω) and ∥fg∥L1(Ω)≤∥f∥Lp(Ω)∥g∥Lq(Ω).Theorem 7 (Arzela-Ascoli theorem).
If a sequence(fn) in C(x) is bounded and equicontinuous, then it has a uniformly convergent subsequence.Theorem 8 (Leray-Schauder’s fixed point theorem).
IfC is a closed bounded and convex subset of Banach space X and F:C→C is completely continuous, then the F has a fixed point in C.


## 3. Existence and Uniqueness of Mild Solution

In this section, we will investigate the existence and uniqueness for impulsive fractional differential equations with the help of the Leray-Schauder’s fixed point theorem and someone else. Without loss of generality, lett∈(tk,tk+1], 1≤k≤n-1.Firstly, we will make the following assumptions be satisfied on the data of our problem.H(1):
{T(t),t>0} is a compact semigroup, and there exists a constant M>0, such that M=supt∈[0,∞)∥T(t)∥Lb(X)<∞.H(2):
The functionf:J×X→X satisfies the following:(i)
f is measurable for all t∈J;(ii)
there exists a constantLf>0 such that ∥f(t,x)-f(t,y)∥≤Lf∥x-y∥, for all x,y∈X;(iii)
there exists a real functionϕ(t)∈L1/γ(J,R+), γ∈(0,α), and a constant θ>0, such that ∥f(t,x)∥≤ϕ(t)+θ∥x∥,  for a.e. t>0 and all x∈X.H(3):
Ii:X→X (i=1,2,…,n) satisfies the following:(i)
Ii maps a bounded set to a bounded set;(ii)
there exist constantshi>0 (i=1,2,…,n) such that
(17)∥Ii(x)-Ii(y)∥≤hi∥x-y∥,x,y∈X;(iii)
∥I(0)∥=max(∥I1(0)∥,∥I2(0)∥,…,∥In(0)∥).H(4):
LetY be a separable reflexive Banach space. Operator B∈L∞(J,L(Y,X)), ∥B∥∞, stands for the norm of operator B on Banach space L∞(J,L(Y,X)).H(5):
The multivalued mapsU:J→Pf(Y) (where Pf(Y) is a class of nonempty closed and convex subsets of Y) are measurable and U(·)⊆Ω where Ω are a bounded set of Y.Set the admissible control set:(18)Uad=SUp={u∈Lp(Ω):u(t)∈U(t)a.e.},1<p<∞.
Then, Uad≠∅ (see Proposition  2.1.7 and Lemma  2.3.2 of [15]). And it is obvious that Bu∈Lp(J,X) for all u∈Uad.According to Definitions1 and 2 and by comparison with the fractional differential equations given in [5, 16, 17], then we shall define the concept of mild solution for problem (1) as follows.Definition 9.
A functionx∈PC(J,X) is said to be a solution (mild solution) of the problem (1) if x(0)=x0 such that
(19)x(t)=Sα(t)x0+∑i=1k‍Sα(t-ti)Ii(x(ti-))+∫0tSα(t-s)f(s,x(s))ds+∫0t(t-s)α-1Tα(t-s)B(s)u(s)ds,
where
(20)Sα(t)=∫0∞ξα(θ)T(tαθ)dθ,Tα(t)=α∫0∞θξα(θ)T(tαθ)dθ,ξα(θ)=1αθ-1-1/αϖα(θ-1/α)≥0,ϖα(θ)=1π∑n=1∞(-1)n-1θ-nα-1Γ(nα+1)n!sin(nπα),iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiθ∈(0,∞),
where ξα is a probability density function defined on (0,∞), that is
(21)ξα(θ)≥0,θ∈(0,∞),∫0∞ξα(θ)=1.Lemma 10 (see [17]).
The operatorsSα(t) and Tα(t) have the following properties and there exists M as described in H(1).(i)
For any fixedt≥0, Sα(t) and Tα(t) are linear and bounded operators; that is,  for any x∈X,
(22)∥Sα(t)x∥≤M∥x∥,∥Tα(t)x∥≤MΓ(α)∥x∥.(ii)
{Sα(t),t≥0} and {Tα(t),t≥0} are strongly continuous.(iii)
For anyt≥0, Sα(t) and Tα(t) are also compact operators if  T(t) is compact.Lemma 11.
If the assumptionsH(1)–H(4) are satisfied and (1) is mildly solvable on [0,b], then there exists a constant ω>0 such that ∥x(t)∥≤ω.Proof.
If (1) can be solvable on [0,b], we may suppose x(t) is the mild solution of it, so x(t) must satisfy (19) as follows:
(23)x(t)=Sα(t)x0+∑i=1kSα(t-ti)Ii(x(ti-))+∫0tSα(t-s)f(s,x(s))ds+∫0t(t-s)α-1Tα(t-s)B(s)u(s)ds.
For t∈(tk,tk+1], 1≤k≤n-1, through calculating, we can get that
(24)∥x(t)∥≤∥Sα(t)x0∥+∥∑i=1kSα(t-ti)Ii(x(ti-))∥+∫0t∥Sα(t-s)f(s,x(s))∥ds+∫0t(t-s)α-1∥Tα(t-s)B(s)u(s)∥ds≤M∥x0∥+M∑i=1nhi∥x(ti-)∥+Mn∥I(0)∥+M∫0t[ϕ(s)+θ∥x(s)∥]ds+M∥B∥∞Γ(α)∫0t‍(t-s)α-1∥u(s)∥ds≤M∥x0∥+M∑i=1nhi∥x(ti-)∥+Mn∥I(0)∥+Mb1-r∥ϕ∥L1/γ+Mθ∫0t‍∥x(s)∥ds+M∥B∥∞Γ(α)(p-1pα-1)(p-1)/pbα-1/p∥u∥Lp.
Let ρ=M∥x0∥+Mn∥I(0)∥+Mb1-r∥ϕ∥L1/γ+(M∥B∥∞/Γ(α))(p-1/pα-1)(p-1)/pbα-1/p∥u∥Lp, then
(25)∥x(t)∥≤ρ+M∑i=1nhi∥x(ti-)∥+Mθ∫0t∥x(s)∥ds,
so it follows from Lemma 5,
(26)∥x(t)∥≤ρ(1+H*E1(Mθb))kE1(Mθb)=ω,
where
(27)H*=max{Mhi:i=1,2,…,n}.
The proof is completed.Theorem 12.
Assume that the hypothesesH(1)–H(4) are satisfied, and then the problem (1) has an unique mild solution on J provided that
(28)(∑i=1n‍hi+θb)M<1.Proof.
Transform the problem (1) into a fixed point theorem. Consider the operator F:PC(J,X)→PC(J,X) defined by
(29)(Fx)(t)=Sα(t)x0+∑i=1kSα(t-ti)Ii(x(ti-))+∫0tSα(t-s)f(s,x(s))ds+∫0t(t-s)α-1Tα(t-s)B(s)u(s)ds.
Clearly, the problem of finding mild solutions of (1) is reduced to find the fixed points of the F, the proof base on Theorem 8. Now we prove that the operator F satisfies all the conditions of the Theorem 8.
Firstly, choose(30)M[∥B∥∞Γ(α)(p-1pα-1)(p-1)/p∥x0∥+n∥I(0)∥+b1-r∥ϕ∥L1/γ000+∥B∥∞Γ(α)(p-1pα-1)(p-1)/pbα-1/p∥u∥Lp]×(1-M∑i=1nhi-Mbθ)-1≤r,
and consider the bounded set Br={x∈PC:∥x∥≤r}.
Next, for the sake of convenient, we divide the proof into several steps.
Step  1. We prove that FBr⊆Br.
In fact, for eachx∈Br, t∈(tk,tk+1], 1≤k≤n-1, we have
(31)∥(Fx)(t)∥≤∥Sα(t)x0∥+∥∑i=1kSα(t-ti)Ii(x(ti-))∥+∫0t∥Sα(t-s)f(s,x(s))∥ds+∫0t(t-s)α-1∥Tα(t-s)B(s)u(s)∥ds≤M∥x0∥+M∑i=1nhi∥x(ti-)∥+Mn∥I(0)∥+M∫0t[ϕ(s)+θ∥x(s)∥]ds+M∥B∥∞Γ(α)∫0t(t-s)α-1∥u(s)∥ds≤M∥x0∥+M∑i=1nhi∥x(ti-)∥+Mn∥I(0)∥+Mb1-γ∥ϕ∥L1/γ+Mθ∫0t∥x(s)∥ds+M∥B∥∞Γ(α)(p-1pα-1)(p-1)/pbα-(1/p)∥u∥Lp≤M∥x0∥+Mn∥I(0)∥+Mb1-γ∥ϕ∥L1/γ+M∥B∥∞Γ(α)(p-1pα-1)(p-1)/p×bα-(1/p)∥u∥Lp+(M∑i=1nhi+Mbθ)r≤r.
Hence, we can deduce that FBr⊆Br.
Step  2. We show that F is continuous.
Let{xn} be a sequence such that xn→x in PC(J,X) as n→∞. Then, for each t∈(tk,tk+1], 1≤k≤n-1, we obtain(32)∥(Fxn)(t)-(Fx)(t)∥≤∥∑i=1kSα(t-ti)[Ii(xn(ti-))-Ii(x(ti-))]∥+∫0t∥Sα(s)[f(s,xn(s))-f(s,x(s))]∥ds≤M∑i=1nhi∥xn-x∥+MLf∫0t∥xn(s)-x(s)∥ds≤[M∑i=1nhi+MLfb]∥xn-x∥,
as xn→x, and it is easy to see that
(33)∥Fxn-Fx∥⟶asn⟶∞;
that is, F is continuous.
Step  3.  F is equicontinuous on Br.
Let0≤τ1<τ2≤b; then, for each x∈Br, we obtain
(34)∥(Fx)(τ2)-(Fx)(τ1)∥≤∥[Sα(τ2)-Sα(τ1)]x0∥+∥[∑i=1kSα(τ2-ti)-∑i=1kSα(τ1-ti)]Ii(x(ti-))∥+∥∫0τ2Sα(τ2-s)f(s,x(s))dsiiiiiiiiiiiii-∫0τ1Sα(τ1-s)f(s,x(s))ds∥+∥∫0τ2(τ2-s)α-1Tα(τ2-s)B(s)u(s)dsiiiiiiiiiiiii-∫0τ1(τ1-s)α-1Tα(τ1-s)B(s)u(s)ds∥≤∥Sα(τ2)-Sα(τ1)∥∥x0∥+∑i=1k∥Sα(τ2-ti)-Sα(τ1-ti)∥×(hi∥x(ti-)∥+∥Ii(0)∥)+∥∫0τ1(Sα(τ2-s)-Sα(τ1-s))f(s,x(s))ds∥+∥∫τ1τ2Sα(τ2-s)f(s,x(s))ds∥denotedbyQ1+∥∫τ1τ2(τ2-s)α-1Tα(τ2-s)B(s)u(s)ds∥00000000000000000000000000000000denotedbyQ2+∥∫0τ1‍[(τ2-s)α-1-(τ1-s)α-1]00000000×Tα(τ2-s)B(s)u(s)ds∫0τ1‍∥00denotedbyQ3+∥∫0τ1(τ1-s)α-1[Tα(τ2-s)-Tα(τ1-s)]×B(s)u(s)ds∫0τ1(τ1-s)α-1[Tα(τ2-s)-Tα(τ1-s)]∥0000000000denotedbyQ4.
Let(35)Λ=∥Sα(τ2)-Sα(τ1)∥(∥x0∥+N)+∑i=1k∥Sα(τ2-ti)-Sα(τ1-ti)∥(hi∥x(ti-)∥+∥Ii(0)∥)+∥∫0τ1‍(Sα(τ2-s)-Sα(τ1-s))f(s,x(s))ds∥.
By (ii) of Lemma 10, we have
(36)limτ2→τ1Λ=0.
By the assumption H(2), we obtain
(37)Q1≤M(∥ϕ∥L1/γ+θbγr)(τ2-τ1)1-γ,
and we get
(38)Q2≤M∥B∥∞Γ(α)(p-1pα-1)(p-1)/p∥u∥Lp(τ2-τ1)α-(1/p),Q3≤2M∥B∥∞Γ(α)(p-1pα-1)(p-1)/p∥u∥Lp(τ2-τ1)α-(1/p),Q4≤sups∈[0,τ1-ε]∥Tα(τ2-s)-Tα(τ1-s)∥×(p-1pα-1)(p-1)/p(τ1(pα-1)/(p-1)-ε(pα-1)/(p-1))(p-1)/p×∥B∥∞∥u∥Lp+2M∥B∥∞Γ(α)(p-1pα-1)(p-1)/p∥u∥Lpεα-(1/p).
Combining the estimations forΛ, Qi(i=1,…,4), let τ2→τ1 and ε→0, and we know that ∥(Fx)(τ2)-(Fx)(τ1)∥→0, which implies that F is equicontinuous.
Step  4. Now we show that F is compact.
Lett∈(tk,tk+1], 1≤k≤n-1 be fixed, and we show that the set Π(t)={(Fx)(t):x∈Br} is relatively compact in X.
Clearly,Π(0)={x0-g(x)} is compact, so it is only necessary to consider t>0. For each ϵ∈(0,t), t∈(0,b], x∈Br and any δ>0, we define
(39)Πϵ,δ(t)={Fϵ,δ(x)(t):x∈Br},
where
(40)Fϵ,δ(x)(t)=Sα(t)x0+∑i=1kSα(t-ti)Ii(x(ti-))+∫0t-ϵ‍∫δ∞ξα(θ)T((t-s)αθ)f(s)dθds+α∫0t-ϵ(t-s)α-1×∫δ∞θξα(θ)T((t-s)αθ)B(s)u(s)dθds=Sα(t)x0+∑i=1kSα(t-ti)Ii(x(ti-))+T(ϵαδ)×∫0t-ϵ‍∫δ∞ξα(θ)T((t-s)αθ-ϵαδ)f(s)dθds+αT(ϵαδ)∫0t-ϵ‍∫δ∞θ(t-s)α-1ξα(θ)T0000000000000000×((t-s)αθ-ϵαδ)0000000000000000×B(s)u(s)dθds.
From the compactness of  T(ϵαδ)(ϵδ>0), we obtain that the set Πϵ,δ(t)={Fϵ,δ(x)(t):x∈Br} is relatively compact set in X for each ϵ∈(0,t) and δ>0. Moreover, we have
(41)∥F(x)(t)-Fϵ,δ(x)(t)∥=∥∫0t‍∫0∞‍ξα(θ)T((t-s)αθ)f(s)dθds0-∫0t-ϵ‍∫δ∞‍ξα(θ)T((t-s)αθ)f(s)dθds∥+∥α∫0t‍∫0∞‍θ(t-s)α-1ξα(θ)000000000×T((t-s)αθ)B(s)u(s)dθds00-α∫0t-ϵ‍∫δ∞‍θ(t-s)α-1ξα(θ)T((t-s)αθ)000000000000000×B(s)u(s)dθds∫0t‍∥≤∥∫0t‍∫0δ‍ξα(θ)T(sαθ)f(s)dθds∥+∥∫t-ϵt‍∫δ∞‍ξα(θ)T(sαθ)f(s)dθds∥+α∥∫0t‍∫0δ‍θ(t-s)α-1ξα(θ)00000000×T((t-s)αθ)B(s)u(s)dθds∫0t‍∥+α∥∫t-ϵt‍∫δ∞‍θ(t-s)α-1ξα(θ)0000000000×T((t-s)αθ)B(s)u(s)dθds∫0t‍∥≤M(∥b1-γϕ∥L(1/γ)+θrb)∫0δ‍ξα(θ)dθ+M(ϵ1-γ∥ϕ∥L(1/γ)+θrϵ)∫δ∞‍ξα(θ)dθ+αM∥B∥∞Γ(α)(p-1pα-1)(p-1)/p∥u∥Lpbα-1/p×∫0δθξα(θ)dθ+αM∥B∥∞Γ(α)(p-1pα-1)(p-1)/p∥u∥Lpϵα-1/p×∫δ∞θξα(θ)dθ,
when ϵ→0 and δ→0, we can easily find (1)→0, (2)→0, (3)→0, (4)→0. Therefore, there are relatively compact sets arbitrarily close to the set Π(t),t>0. Hence the set Π(t),t>0 is also relatively compact in X.
As a result, by the conclusion of Theorem8, we obtain that F has a fixed point x on Br. So system (1) has a unique mild solution on J. The proof is completed.


## 4. Optimal Control Results

In the following, we will consider the Lagrange problem (P).Find a control pair(x0,u0)∈PC(J,X)×Uad such that
(42)𝒥(x0,u0)≤J(xu,u),∀(x,u)∈PC(J,X)×Uad,
where
(43)𝒥(xu,u):=∫0bℒ(t,xu(t),u(t))dt,
and xu denotes the mild solution of system (1) corresponding to the control u∈Uad.For the existence of solution for problem (P), we shall introduce the following assumption.H(6):
The functionℒ:J×X×Y→R∪{∞} satisfies the following.(i)
The functionℒ:J×X×Y→R∪{∞} is Borel measurable;(ii)
ℒ(t,·,·) is sequentially lower semicontinuous on X×Y for almost all t∈J;(iii)
ℒ(t,x,·) is convex on Y for each x∈X and almost all t∈J;(iv)
there exist constantsc≥0, d>0, φ is nonnegative, and φ∈L1(J,R) such that
(44)ℒ(t,x,u)≥φ(t)+c∥x∥X+d∥u∥Yp.Next, we can give the following result on existence of optimal controls for problem (P).Theorem 13.
Let the assumptions of Theorem12 and H(6) hold. Suppose that B is a strongly continuous operator. Then Lagrange problem (P) admits at least one optimal pair; that is, there exists an admissible control pair (x0,u0)∈PC(J,X)×Uad such that
(45)𝒥(x0,u0)=∫0bℒ(t,x0(t),u0(t))dt≤𝒥(xu,u),00000000∀(xu,u)∈PC(J,X)×Uad.Proof.
Ifinf{𝒥(xu,u):(xu,u)∈PC(J,X)×Uad}=+∞, there is nothing to prove.
Without loss of generality, we assume thatinf{J(xu,u):(xu,u)∈PC(J,X)×Uad}=ρ<+∞. Using H(6), we have ρ>-∞. By definition of infimum, there exists a minimizing sequence feasible pair {(xm,um)}⊂𝒫ad≡{(x,u):x is a mild solution of system (31) corresponding to u∈Uad}, such that J(xm,um)→ρ as m→+∞. Since {um}⊆Uad,m=1,2,…,{um} is a bounded subset of the separable reflexive Banach space Lp(J,Y), there exists a subsequence, relabeled as {um}, and u0∈Lp(J,Y) such that
(46)um⟶wu0inLp(J,Y).
SinceUad is closed and convex, due to Marzur lemma, u0∈Uad. Let {xm} denote the sequence of solutions of the system (1) corresponding to {um}, x0 is the mild solution of the system (1) corresponding to u0.  xm and x0 satisfy the following integral equation, respectively:
(47)xm(t)=Sα(t)x0+∑i=1kSα(t-ti)Ii(xm(ti-))+∫0tSα(t-s)f(s,xm(s))ds+∫0t(t-s)α-1Tα(t-s)B(s)um(s)ds,x0(t)=Sα(t)x0+∑i=1kSα(t-ti)Ii(x0(ti-))+∫0tSα(t-s)f(s,x0(s))ds+∫0t(t-s)α-1Tα(t-s)B(s)u0(s)ds.
It follows the boundedness of{um}, {u0} and Lemma 11, one can check that there exists a positive number ω such that ∥xm∥≤ω, ∥x0∥≤ω.
Fort∈J, we obtain
(48)∥xm(t)-x0(t)∥=∥∑i=1kSα(t-ti)[Ii(xm(ti-))-Ii(x0(ti-))]∥0000000000000000000000000000denotedbyη1(t)+∫0t∥Sα(t-s)[[f(s,xm(s))-f(s,x0(s))]∥ds0000000000000000000000000000denotedbyη2(t)+∫0t(t-s)α-100000×∥Tα(t-s)[B(s)um(s)-B(s)u0(s)]∥ds,00000000000000000000000000denotebyη3(t).
By H(3)(ii), we have
(49)η1(t)=∥∑i=1k‍Sα(t-ti)[Ii(xm(ti-))-Ii(x0(ti-))]∥≤M∑i=1n‍hi∥xm-x0∥⟶0,asm⟶∞.
Using Lemma10(i) and by H(2)(ii), one can obtain
(50)η2(t)=∫0t∥Sα(t-s)[f(s,xm(s))-f(s,x0(s))]∥ds≤MLf∫0t∥xm(s)-x0(s)∥ds.
Similarly, one has(51)η3(t)=∫0t‍(t-s)α-1000×∥Tα(t-s)[B(s)um(s)-B(s)u0(s)]∥ds≤MΓ(α)(p-1pα-1)(p-1)/ptα-(1/p)×(∫0t‍∥B(s)um(s)-B(s)u0(s)∥pds)1/p≤MΓ(α)(p-1pα-1)(p-1)/pbα-(1/p)∥Bum-Bu0∥Lp(J,Y).
SinceB is strongly continuous, we have
(52)∥Bum-Bu0∥Lp(J,Y)⟶s0asm⟶∞,
which implies
(53)η3(t)⟶0asm⟶∞.
Thus
(54)∥xm(t)-x0(t)∥≤η1(t)+η3(t)+MLf∫0t‍∥xm(s)-x0(s)∥ds;
by virtue of singular version Gronwall inequality (i.e., Lemma 5), we obtain
(55)∥xm(t)-x0(t)∥≤[η1(t)+η3(t)]E1(MLfb).
This yields that
(56)xm⟶Sx0inPC(J,X)asm⟶∞.
Note thatH(6) implies all of the assumptions of Balder (see [18], Theorem  2.1) are satisfied. Hence, by Balders theorem, we can conclude that (x,u)→∫0b‍ℒ(t,x(t),u(t))dt is sequentially lower semicontinuous in the strong topology of L1(J,X). Since Lp(J,Y)⊂L1(J,Y), 𝒥 is weakly lower semicontinuous on Lp(J,Y), and since, by H(6)(iv), 𝒥>-∞,𝒥 attains its infimum at u0∈Uad; that is,
(57)ρ=limm→∞∫0bℒ(t,xm(t),um(t))dt≥∫0bℒ(t,x0(t),u0(t))dt=J(x0,u0)≥ρ.
The proof is completed.


## 5. An Example

Consider the following initial-boundary value problem of fractional impulsive parabolic control system(58)∂α∂tαx(t,y)=∂2∂y2x(t,y)+I1-α(e-t+1(t+10)x(t,y))+∫01q(y,τ)u(τ,t)dτ,t∈J′=[0,1]{1/2},y∈[0,π],Δx(12,y)=|x(y)|5+|x(y)|,y∈[0,π],x(t,0)=x(t,π)=0,t∈J=[0,1],x(0,y)=x0(y),y∈[0,π]
with the cost function
(59)𝒥(x,u)=∫01‍∫0π|x(t,y)|2dydt+∫01‍∫0π|u(t,y)|2dydt,
where α=1/2, q:[0,1]×[0,1]→R is continuous, u∈L2(J,[0,1]), bi∈L2(J).TakeX=Y=L2[0,π] and the operator A:D(A)⊂X→X is defined by
(60)Aω=ω′′,
where the domain D(A) is given by
(61){ω∈X:ω,ω′areabsolutelycontinuous,0ω′′∈X,ω(0)=ω(π)=0}.
Then A can be written as
(62)Aω=∑n=1∞‍n2(ω,ωn)ωn,ω∈D(A),
where ωn(x)=2/πsinnx(n=1,2,…) is an orthonormal basis of X. It is well known that A is the infinitesimal generator of a compact semigroup T(t)(t>0) in X given by
(63)T(t)x=∑n=1∞‍exp-n2t(x,xn)xn,x∈X,∥T(t)∥≤e-1≤1=M,f(t,x(t,y))=e-t+1(t+10)x(t,y),Ik(x(t,y))=|x(y)|5+|x(y)|,B(t,y)=[∫01‍q(y,τ)u(τ,t)dτ].It is easy to see that(64)∥f(t,x(t))∥≤πe-t+110∥x(t)∥,∥Ik(x(t))∥≤∥x(t)∥5,∥f(t,x(t))-f(t,y(t))∥≤110∥x-y∥,∥I1(x(t))-I1(y(t))∥≤15∥x-y∥,
and then
(65)(∑i=11‍hi+θb)M=(15+110×1)×1<1.Hence, all the conditions of Theorem12 are satisfied, and system (58) has a unique optimal solution.



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*Source: 914592-2013-10-21.xml*