Solution of Fractional Impulsive Problem under Power Law

INTRODUCTION

Fractional calculus has taken a notable sector of inspection on real life problems with its tremendous application in science and engineering. It has been extended because of global character of the fractional operator, which relates the reminiscence and it authorizes to learn an adjacent look at the vital behavior and inherited things of the relevant phenomena. We referred the monographs [1-3] to know about current development in such field. The purpose of fractional derivative with power law base in the sense of Riemann Liouville was established. A modern fractional derivative is added has suggested by Caputo-Fabrizio [4] depending on the augmented kernel. To through away troubles related to Caputo-Fabrizio‘s, a new change in version of a fractional derivative with non-singular and global kernel of Mittag-Leffler function (MLF) have been introduced by Atangana and Baleanu (AB) in [5]. Generalization of MLF is noted and used as non-singular and global kernel in later on extension but it does not assure singularity. Also, the ABC derivatives comes up with a magnificent memory description [6-9]. Recently the authors [10-14] deliberated inquisitive and numerical solution for few fractional models by means of AB fractional derivative with global trouble-free kernel. We display the nature and uniqueness of solutions to impulsive fractional differential equations with initial and global conditions in this article. and Where []01,ABCaaD refers the Atangana-Baleanu-Caputo (ABC) fractional derivative of order ,:ahRR is a given non-discontinuous function. Moreover, 0,00h and also disappears at impulse points 0,1,2,....,,:,1,2,....,,,mmmmnIRRmnR satisfy Represents the right and left limits of  at m and :,fPCRR is given function. Also m if10,,0,1,2,,,,,,&0.mmm By reference from Theorem 3.11 of [24], we should have standard condition 0,00h to fix the starting data for solution. By the previous referred articles, there are only few articles on Cauchy problems for ABC impulsive fractional differential equations. The main part of this paper is to derive formula of solutions for types of impulsive fractional differential equations with ABC fractional operators. Already we proved existence and uniqueness theorem theorems by few fixed-point theorems of Banach space, Kransosekliskii, Schader and is mandatory to assure a unique solution. This paper is divided as follows: Section 1 deals with survey of literature. Section 2 comprises basic definitions, fundamental lemmas and theorems. The suggested formulae for taken problem is established in section 3. The existence of unique solution for Cauchy problem and global Cauchy problem are acquired in 4th and 5th section. Finally, examples are stated and proved to the affirmation of our results.

Consider the space The space ,PCR is a complete normed linear space with the norm max.PC Let 120,and:\,,.....,.mS Definition 2.1. [5,27] Let 0,1a and 1,.H Then the left AB Caputo and AB Riemann-Liouville fractional derivatives of order a for a function  are defined by and respectively, where Ma is a function with normalization convinces the results 011and aMME is called the MLF defined by Definition 2.2. [5,27] Let 0,1a and 1,.L. Then the left AB fractional integral of order a for a function  is denoted by Definition 2.3. [5] The ABC fractional derivative with its Laplace transformation is defined by where L is the Laplace transform initiating from α defined by Definition 2.4. [29] Let X be a complete normed linear space. Then the operator :XX is a contraction if 1212xxxx for all 12,,01.xxX Definition 2.5. A function ,PCR is a solution of (1.1) if  convinces the equation ,ABCaDh on  and conditions To prove the main concept of this paper, we need to remember the following lemmas and theorems which is proved already. Lemma 2.1. [28] Let ,,,H such that the ABC fractional derivative exists. Then we have ABCaABaDI and ABaABCaID for 01.a Also, 0ABCaD if  is a constant. Lemma 2.2. [24,28] For 0,1,a the solution of the following problem is given by Theorem 2.1. [29] Let X be a complete normed linear space and A be a non-empty closed subset of X. If :AA is a contraction, then there exists a unique fixed point of . Theorem 2.2. [29] Let X be a complete normed linear space and :XX be a non-discontinuous and compact mapping (completely continuous). Suppose :for0,1TyXyy be a bounded set. Then  has at least one fixed point in X.

space and let be two operators, such that (i) 12,oaobBabB (ii) 1o is compact and continuous (iii) 2o is a contraction mapping. Then there exist B such that 12.oo

Theorem 3.1. Let 0,1a and let :vR be continuous with 00v and also disappears at hasty points m, for 1,2,....,.mn A function ,PCR is a solution of the fractional integral equation If and only if  is a solution of the impulsive ABC-fractional FDE where Proof. By lemma 2.1, we can prove this again by assuming  satisfies (3.6)-(3.8). If 10,, then ,0.ABCaDv Using lemma (2.1), we have This implies By impulse 1111,I we get If 12,, then v vanishes at 1 implies By impulse 2222,I we get If 23,, then v vanishes at 2 implies This implies that By impulse 3333,I we get Proceeding like this, we get the general expression as given below: For 1,,mmm and hence we get the solution Hence (2.5) is satisfied. Conversely assume that  satisfies the equation (1.1). If 10,, then 000;0.v By the concept that ABCaD is the left inverse of ABaI. By using lemma 2.1, we get 01,0,.ABCaDv If 1,,1,2,.....,mmmn and by the fact that 0,ABCaD where  is a constant function, we obtain ,ABCaDv for each 1,,1,2,.....,.mmmn Hence ,1,2,.....,.mmmmImn

We need the following hypothesis to prove our results,

(A1)

,,,foreach ,,andaconstant 0.hhhhLRL

(A2) The function :mIRR are continuous and there exist a constant 0JL such that *,1,2,....,&,.mmJJJLmnR (A3) There exist,CR such that ,h, for each ,.R (A4) There exist 0N such that ,1,2,....,,.mINmnR Now let us prove main results. Theorem 4.1. Assume :hRR is continuous. If (A1) and (A2) hold with then the impulsive ABC – fractional FDE (1.1) has a solution which is unique on , where Let Let max,0.hh From (A1), we get To prove B has a fixed point, for that we need to show that .llBAA For ,lA we have Thus B maps lA into itself. Next, we have to show that B is a contraction on ,PCR. Let *,,PCR and .. Then we get

The inequality (4.9) shows that B is contraction on ,.PCR It is proved that the impulsive fractional differential equation with ABC (1.1) has a solution which is unique. Theorem 4.2. Suppose :hRR is continuous and assume (A3) and (A4) hold. Then there exists at least one solution for fractional differential equations with ABC fractional derivative and impulse condition on . Step 1: To show that :,,BPCRPCR is compact. Since h and mI are continuous, we have to verify that B is continuous. Let 11,:lPCAPCRl be a set(ball) with 1001lnN. Where 0sup and  is given by (4.10). For lA and , we have Hence 1BAl is uniformly bounded. Next, we have to prove B maps bounded sets into equicontinuous set of ,PCR. By (A3), Let us fix 10,sup,.lAhh Now As 1221,0,BB that is 1lBA relatively compact for . The operator B is compact on 1lA by Arzela Ascoli theorem. Step 2: To prove that the set ,:forsome0,1TPCRB is bounded. Let .T Then B for some 0,1. Hence for , we obtain For every , we get 00:.PCnNR Hence T is bounded. That is B has a fixed point which is a solution of ABC-fractional differential equation (1.1)

Finally, we have to prove existence of solution for the impulsive FDE with ABC fractional derivative. Let f satisfying the following condition (A5) :,fPCRR is continuous and there exists a constant 01fL such that **fffL for all *,,.PCR Theorem 5.1. Suppose :gRR is continuous and assume (A1), (A2) and (A5) hold. If where  is given by (4.10), then ABC-FDE with impulse condition has a unique solution on . Proof: Define the non-linear mapping *:,,BPCRPCR as follows Then *B has a fixed point if and only if the ABC-FDE with impulse condition (1.2) has a solution. Choose From hypothesis (A5), it is simple to verify that ***,llBAA where Next, we have to prove that *B is a contraction. Let *,,PCR and , then we have The inequality (5.12) proves that *B is contraction on ,PCR. Then the ABC-FDE with impulse condition has a unique solution. Theorem 5.2. Suppose :hRR is continuous and let (A1) to (A5) hold. If Then the global ABC-FDE (1.2) with impulse condition has a solution on given domain. Proof: Let the operator *:,,BPCRPCR defined by (5.13). Define the operator **12&BB on lA as For 120,,Al we obtain ****11221122BBBB Thus **11220BBAl. For any 12&,,PCR, we have From (5.14), *1B is a contraction mapping. Now let us show that *2B is continuous and compact. h is continuous  *2B is continuous. Also *2B is uniformly bounded on 0Al because for 0&Al we get Now the operator *2B is compactness on 0,Al since *2.BB Hence, the global ABC-FDE with impulse condition possesses a solution .

For 0,1a, consider the following impulsive ABC-FDE

Clearly 0,0.5,0.50.hh Let and ,.ThenR Also, hypothesis (A1) and (A2) hold with Also verified that the condition (4.9) holds with Thus, we get By theorem 4.1, the problem (6.15) has a one-of-a-kind approach [0,1]. Also it is noted that for each R and 0,1, we have ,0.59h and 0.1.mI Hence, condition (A3) is verified with Thus, all conditions of theorem (4.2) are satisfied. Hence theorem (4.2) implies that the given problem has at least one solution on 0,1.

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Mathematics; 5(1):9. https://doi.org/10.3390/math5010009

Assistant Professor, PG Department of Mathematics, Vellalar College for Women (Autonomous) prahalathav@gmail.com