Observability of fractional order differential impulsive multi control problem with fractional integral nonlocal initial condition Observability of fractional order differential impulsive multi control problem with fractional integral nonlocal condition

: In this paper ,the obsevability of fractional differential impulsive multi control abstract problem with fractional integral nonlocal initial condition have been studied as abstract Cauchy problem for using Banach fixed point which defined on space of presented problem which is piecewise continuous space and proprieties of the initial observable condition for their problem.

Observability of fractional order differential impulsive multi control problem with fractional integral nonlocal initial condition Sameer Qasim Hasan where , 0 < , is the caputo fractional derivative.
are nonnegative functions. Assume a bounded operator A(t): ( Banach space , , , , denoted the left and the right limit of at , respectively g:PC([0,T]; is a given function. y(.) is referred to as the output which is belong to Banach space Y. C: , : is a bounded linear operator.
Our aim to study and present the obsevability of fractional differential impulsive multi control abstract problem with fractional integral nonlocal initial condition (1)(2)(3)(4)(5) with necessary and sufficient conditions that which guaranty the problem initial observable.

Preliminaries:
The following definitions and results are need it later on for investigate the initial observable for problem (1)(2)(3).

Definition(2.2), [2]:
The Caputa fractional derivative of a function with order where ,and , is defined by: . (5) Where is absolutely continuous derivative up to q. If then .

Lemma (2.3), [7]:
(9) Let :Y) , now assume the operator H: as (10) As the same proving of results in [4], we can prove the following, Observability of fractional order differential impulsive multi control problem with fractional integral nonlocal initial condition Sameer Qasim Hasan Remarks( 3.1): 1. The system in (8) is initial observable if kernel={0}. 2. The system in (8) is continuously initially observable if .
3. If a system in (8) is initially observable which implies the map is injective but not surjective. 4. when system in (8) is continuous initially observable implies that :Y exists and bounded that is there exists >0 Such that for all As the same proving of result in [1], we can prove the following,:
Since the system in (8) is continuously initially observable so that the initial state of the system (8) can be obtained as follow: From (10), we have that Observability of fractional order differential impulsive multi control problem with fractional integral nonlocal initial condition Sameer Qasim Hasan (11) From (11) the equation (8) become (12) In the following formulation, we generalization of concluding remark (3.3).

4.The problem formulation :
Consider the the fractional differential impulsive multi control abstract problem with fractional integral nonlocal initial conditions(1-5) and let the output . now substitutes (7)  Observability of fractional order differential impulsive multi control problem with fractional integral nonlocal initial condition Sameer Qasim Hasan (15) From equations (10) and (15) and Substituting in (7), we get:

Remark ( 4.1):
The equation in (16) is a finite time observer which provide the mild solution x(.) to have a fixed point ,for all control functions .

Main results:
Consider the abstract control problem (1-5) and consider their mild solution (7) with hypothesis(h1-h5) and we needs the following adopted in the main result: (g1) Let : Assume that the hypotheses (h1-h5 ) and conditions g3(i),(ii) are satisfied Then the the impulsive multi control fractional differential abstract problem with fractional integral nonlocal initial condition(1-5) has a unique fixed point (.) for all control function (. Observability of fractional order differential impulsive multi control problem with fractional integral nonlocal initial condition Sameer Qasim Hasan From condition(g3)(ii).Hence , Therefore, is contraction. Thus and we had C , hence C