Impulsive Caputo-Fabrizio fractional differential equations in b-metric spaces

Definition 1.1

For a function φ ∈ L 1 ( I ) , the Caputo-Fabrizio fractional integral of order 0 < r < 1 is

where M ( r ) is a normalization constant depending on r .

Example 1.2

[7]

1. For φ ( ϑ ) = ϑ and 0 < r ≤ 1 , we have

   ( D r Cℱ φ ) ( ϑ ) = M ( r ) r 1 − exp − r 1 − r ϑ .

2. For φ ( t ) = e ℓ ϑ , ℓ ≥ 0 and 0 < r ≤ 1 , we have

   ( D r Cℱ φ ) ( ϑ ) = ℓ M ( r ) ℓ + ( 1 − ℓ ) r e ℓ ϑ 1 − exp − ℓ − r 1 − r ϑ .

Lemma 1.3

For ψ ∈ L 1 ( I ) , the given linear problem

admits the following solution:

where

Proof

Let ω satisfy (2). On account of Proposition 1 in [38]; the equation

implies that

Taking the initial condition ω ( 0 ) = ω 0 into account, we find that

Hence, we get (3).□

Definition 1.4

[22,23] For a non empty set M , and c ≥ 1 , a distance d : M × M → R 0 + is called b -metric if

1. d ( μ , ν ) = 0 if and only if μ = ν ;

2. d ( μ , ν ) = d ( ν , μ ) ;

3. d ( μ , ξ ) ≤ c [ d ( μ , ν ) + d ( ν , ξ ) ] ;

Example 1.5

[22,23] Let d : C ( I ) × C ( I ) → R 0 + be defined by

It is clear that d is a b -metric with c = 2 .

Example 1.6

[22,23] Let X = [ 0 , 1 ] and d : X × X → R 0 + be defined by

Clearly, d is not a metric, but is a b -metric space with r ≥ 2 .

Definition 1.7

[22,23] For a b -metric space ( M , d , c ) , an operator F : M → M is called a generalized α - ϕ -Geraghty contraction-type mapping whenever there exist α : M × M → R 0 + and some L ≥ 0 such that

for all μ , ν ∈ M , where λ ∈ ℱ , ϕ ψ ∈ Φ , where

and

Remark 1.8

In the case when L = 0 in Definition 1.7, and the fact that

for all x , y ∈ M , inequality (4) becomes

Definition 1.9

[22,23,25] Let M be a non-empty set, F : M → M and α : M × M → R 0 + be given mappings. We say that F is α -admissible if for all μ , ν ∈ M , we have

Definition 1.10

[22,23] Let α : M × M → R 0 + , where ( M , d , c ) is a b -metric space. We say that M is an α -regular if for each sequence { x n } n ∈ N in M with α ( x n , x n + 1 ) ≥ 1 for all n and x n → x as n → ∞ , there exists a subsequence { x n ( k ) } k ∈ N of { x n } n with α ( x n ( k ) , x ) ≥ 1 for all k .

Theorem 1.11

[22,23,25] Suppose a self-mapping F on complete b -metric space ( M , d , c ) forms a generalized α - ϕ -Geraghty contraction-type mapping with the following additional assumptions:

1. F is α -admissible;

2. there exists μ 0 ∈ M such that α ( μ 0 , F ( μ 0 ) ) ≥ 1 ;

3. either M is α -regular or F is continuous.

1. for all fixed points μ , ν of F , either α ( μ , ν ) ≥ 1 or α ( ν , μ ) ≥ 1 ,

Definition 2.1

By a solution of problem (1) we mean a function ω ∈ PC that satisfies ω ( ϑ k + ) = ω ( ϑ k − ) + L k ( ω ( ϑ k − ) ) , k = 1 , … , m , the equation ( D ϑ k r Cℱ ω ) ( ϑ ) = f ( ϑ , ω ( ϑ ) ) , ϑ ∈ I k , k = 0 , … , m , on I, and the condition ω ( 0 ) = ω 0 .

Lemma 2.2

Let h : I → R be a continuous function. A function ω ∈ PC is a solution of the fractional integral equation:

if and only if ω is a solution of the problem

Proof

Assume u satisfies (7). If ϑ ∈ I 0 , then

Lemma 1.3 implies that

If ϑ ∈ I 1 , then

Lemma 1.3 implies that

Thus,

If ϑ ∈ I 2 , then

Then,

If ϑ ∈ I k , we get (6).

Conversely, assume that ω satisfies (6). If ϑ ∈ I 0 , then

Thus, ω ( 0 ) = ω 0 and since D t k r Cℱ is the left inverse of I 0 r Cℱ we get ( D 0 r Cℱ ω ) ( ϑ ) = h ( ϑ ) .

Now, if ϑ ∈ I k , k = 1 , … , m , we get ( D ϑ k r Cℱ ω ) ( ϑ ) = h ( ϑ ) . Also,

Hence, if ω satisfies (6) then we get (7).□

Lemma 2.3

A function ω is a solution of problem (1), if and only if ω satisfies the following integral equation:

where c = ω 0 − a r f ( 0 , ω 0 ) .

Theorem 2.4

Under assumptions ( A x 1 )–( A x 4 ), problem (1) has at least one solution defined on I . Moreover, if ( A x 5 ) holds, then (1) has a unique solution.

Proof

Let the operator ϒ : PC → PC be defined by

Let α : PC × PC → R 0 + be the function defined by:

Our following result is based on Theorem 1.11.

First, we prove that ϒ is a generalized α - ϕ -Geraghty contraction operator:

Let ω , ϖ ∈ PC . For each ϑ ∈ I 0 , we have

Thus, from ( A x 1 ) we get

This gives

where λ ∈ ℱ , ϕ , ψ ∈ Φ with λ ( ϑ ) = 1 8 ϑ , L = 0 and ϕ ( ϑ ) = ψ ( ϑ ) = ϑ .

On the other hand, for each ϑ ∈ I k : k = 1 , … , m , we have

Hence, we obtain (10). So, ϒ is generalized α - ϕ -Geraghty contraction.

Next, we verify that ϒ is α -admissible:

Let ω , ϖ ∈ PC such that α ( ω , ϖ ) ≥ 1 . For each ϑ ∈ I 0 , we have

This implies from ( A x 3 ) that δ ( ϒ ω ( ϑ ) , ϒ ϖ ( ϑ ) ) ≥ 0 , which gives α ( ϒ ( ω ) , ϒ ( ϖ ) ) ≥ 1 . Hence, ϒ is α -admissible.

Now, from ( A x 2 ) , there exists μ 0 ∈ C ( I ) such that α ( μ 0 , ϒ ( μ 0 ) ) ≥ 1 .

Finally, from ( A x 4 ) , PC is α -regular. Indeed, for a given sequence { μ n } n ∈ N ⊂ PC with μ n → μ as n → ∞ , there exists a subsequence { μ n ( k ) } k ∈ N of { μ n } n with δ ( μ n ( k ) , μ ) ≥ 1 for all k . This gives α ( μ n ( k ) , μ ) ≥ 1 for all k .