On the Definitions of Nabla Fractional Operators

and Applied Analysis 3 Definition 2.2. Let ρ t t − 1 be the backward jump operator. Then i the nabla left fractional sum of order α > 0 starting from a is defined by ∇−α a f t 1 Γ α t ∑ s a 1 ( t − ρ s α−1f s , t ∈ Na 1 2.4 ii the nabla right fractional sum of order α > 0 ending at b is defined by b∇−αf t 1 Γ α b−1 ∑ s t ( s − ρ t α−1f s 1 Γ α b−1 ∑ s t σ s − t α−1f s , t ∈b−1 N. 2.5 We want to point out that the nabla left fractional sum operator has the following characteristics. i ∇−α a maps functions defined on Na to functions defined on Na. ii ∇−n a f t satisfies the nth order discrete initial value problem ∇ny t f t , ∇iy a 0, i 0, 1, . . . , n − 1. 2.6 iii The Cauchy function t − ρ s n−1/Γ n satisfies ∇ny t 0. In the same manner, it is worth noting that the nabla right fractional sum operator has the following characteristics. i b∇−α maps functions defined on bN to functions defined on bN. ii b∇−nf t satisfies the nth order discrete initial value problem Δy t f t , Δy b 0, i 0, 1, . . . , n − 1. 2.7 iii The Cauchy function s − ρ t n−1/Γ n satisfies Δy t 0. Definition 2.3. i The nabla left fractional difference of order α > 0 is defined by ∇af t ∇n∇− n−α a f t ∇n Γ n − α t ∑ s a 1 ( t − ρ s n−α−1f s , t ∈ Na 1. 2.8 ii The nabla right fractional difference of order α > 0 is defined by b∇ f t Δnb∇ n−α f t Δ n Γ n − α b−1 ∑ s t ( s − ρ t n−α−1f s , t ∈b−1 N. 2.9 Here and throughout the paper n α 1, where α is the greatest integer less than or equal α. 4 Abstract and Applied Analysis Regarding the domains of the fractional difference operators we observe the following. i The nabla left fractional difference ∇a maps functions defined on Na to functions defined on Na n on Na if we think f 0 before a . ii The nabla right fractional difference b∇ maps functions defined on bN to functions defined on b−nN on bN if we think f 0 after b . 3. A Relation between the Operators ∇−α a and −α a In this section we illustrate how two operators, ∇−α a and −α a are related. Lemma 3.1. The following holds: i −α a 1f t ∇−α a f t , ii −α a f t 1/Γ α t − a 1 α−1f a ∇−α a f t . Proof. The proof of i follows immediately from the above definitions 1.1 and 1.2 . For the proof of ii , we have −α a f t 1 Γ α t ∑ s a ( t − ρ s α−1f s 1 Γ α t − a 1 α−1f a 1 Γ α t ∑ s a 1 ( t − ρ s α−1f s 1 Γ α t − a 1 α−1f a ∇−α a f t . 3.1 Next three lemmas show that the above relations on the operators 1.1 and 1.2 help us to prove some identities and properties for the operator∇−α a by the use of known identities for the operator −α a . Lemma 3.2. The following holds: ∇−α a ∇f t ∇∇−α a f t − t − a α−1 Γ α f a . 3.2 Proof. It follows from Lemma 3.1 and Theorem 2.1 in 13 ∇−α a ∇f t −α a 1∇f t ∇ −α a f t − t − a 1 α−1 Γ α f a ∇ { 1 Γ α t − a 1 α−1f a ∇−α a f t } − t − a 1 α−1 Γ α f a Abstract and Applied Analysis 5 α − 1 Γ α t − a 1 α−2f a ∇∇−α a f t − t − a 1 α−1 Γ α f aand Applied Analysis 5 α − 1 Γ α t − a 1 α−2f a ∇∇−α a f t − t − a 1 α−1 Γ α f a ∇∇−α a f t − t − a α−1 Γ α f a . 3.3 Lemma 3.3. Let α > 0 and β > −1. Then for t ∈ Na, the following equality holds ∇−α a t − a μ Γ ( μ 1 ) Γ ( μ α 1 ) t − a α . 3.4 Proof. It follows from Theorem 2.1 in 13 ∇−α a t − a μ −α a 1 t − a μ Γ ( μ 1 ) Γ ( μ α 1 ) t − a α . 3.5 Lemma 3.4. Let f be a real-valued function defined on Na, and let α, β > 0. Then ∇−α a ∇−β a f t ∇− α β a f t ∇−β a ∇−α a f t . 3.6 Proof. It follows from Lemma 3.1 and Theorem 2.1 in 2 ∇−α a ∇−β a f t −α a 1 −β a 1f t − α β a 1 f t ∇ − α β a f t . 3.7 Remark 3.5. Let α > 0 and n α 1. Then, by Lemma 3.2 we have ∇∇af t ∇∇n ( ∇− n−α a f t ) ∇n ( ∇∇− n−α a f t ) 3.8 or ∇∇af t ∇n [ ∇− n−α a ∇f t t − a n−α−1 Γ n − α f a ] . 3.9 Then, using the identity ∇n t − a n−α−1 Γ n − α t − a −α−1 Γ −α 3.10 we verified that 3.2 is valid for any real α. By using Lemma 3.1, Remark 3.5, and the identity ∇ t − a α−1 α − 1 t − a α−2, we arrive inductively at the following generalization. 6 Abstract and Applied Analysis Theorem 3.6. For any real number α and any positive integer p, the following equality holds: ∇−α a p−1∇f t ∇p∇−α a p−1f t − p−1 ∑ k 0 ( t − a p − 1α−p k Γ ( α k − p 1 ∇ f ( a p − 1, 3.11 where f is defined on Na. Lemma 3.7. For any α > 0, the following equality holds: b∇−α Δf t Δ b∇−αf t − b − t α−1 Γ α f b . 3.12 Proof. By using of the following summation by parts formula Δs [( ρ s − ρ t α−1f s ] α − 1 s − ρ t α−2f s s − ρ t α−1Δf s 3.13 we have b∇−α Δf t − 1 Γ α b−1 ∑ s t ( s − ρ t α−1Δf s 1 Γ α [ − b−1 ∑ s t Δs (( ρ s − ρ t α−1f s ) α − 1 b−1 ∑ s t ( s − ρ t α−2f s ]