A comparative study to solve fractional initial value problems in discrete domain

Definition 2.1

(Sumudu Transform) [17]

If f : N 0 → C is a function, then its discrete Sumudu transform is defined as follows:

where S { f } ( u ) = S 0 { f } ( u ) for all values of u ≠ ‒ 1 such that the series converges. If ℵ = lim k → ∞ sup | f ( k ) | 1 k , then one of these three possibilities applies to our situation:

1. If 0 < ℵ < ∞ , and consequently u + 1 u > ℵ , then the solution to equation (1) will be convergent, while it will diverge everywhere else.

2. If ℵ is equal to zero, then equation (1) will converge for all values of u, with the possible exception of the value ‒1.

3. The equation diverges everywhere if ℵ = ∞ .

Theorem 2.2

[8]

If S { f } ( u ) = F ( u ) for u + 1 u > ℵ and assuming that u has the same values, then S { f ( k + 1 ) } ( u ) = u + 1 u F ( u ) ‒ f ( 0 ) u .

Lemma 2.3

[7]

The discrete Sumudu transforms for some elementary functions for u + 1 u > ℵ are as follows:

Definition 2.4

(Falling factorial function) [20]

Remark 2.5

[20]

Lemma 2.6

[21]

If u + 1 u > ℵ , then for the same values of u ,

and in general,

where Δ f ( k ) = f ( k + 1 ) ‒ f ( k ) and Δ 0 f ( k ) = f ( k ) .

Definition 3.1

(the sum fractional) [22]

Definition 3.2

(Riemann–Liouville difference) [3,21]

For Ƒ ( ƭ ) defined on N a and 0 < α , α ∉ N , the R ‒ L difference is defined as follows: Δ a α Ƒ ( ƭ ) ≔ Δ m Δ a ‒ ( m ‒ α ) Ƒ ( ƭ ) , ƭ ∈ N a + m ‒ α , m = [ α ] + 1 .

Remark 3.3

[23,24]

We note that for α = 1 definition equation (4) reduces to discrete sum operator Δ a ‒ 1 Ƒ ( ƭ ) = ∑ s = a ƭ ‒ 1 Ƒ ( s ) . Let a be any real number and α be any positive real number such that ϒ ‒ 1 < α < ϒ , where ϒ is an integer.

Theorem 4.1

(Composing a sum with a sum) [2]

Let ϒ : N a → R be a given and suppose α , μ > 0 . Then,

Δ a + μ ‒ α Δ a ‒ μ Ƒ ( ƭ ) = Δ a ‒ α ‒ μ Ƒ ( ƭ ) for ƭ ∈ N a + μ + α .

Lemma 4.2

[2]

Let Ƒ : N a → R be given. For any k ∈ N 0 and μ > 0 with M ‒ 1 < μ < M , we have

1. Δ k Δ a ‒ μ Ƒ ( ƭ ) = ∆ a k ‒ μ Ƒ ( ƭ ) , ƭ ∈ N a + μ ,

2. Δ k Δ a μ Ƒ ( ƭ ) = ∆ a k + μ Ƒ ( ƭ ) , ƭ ∈ N a + M ‒ μ .

Theorem 4.3

(Composing a difference with sum) [2]

Let Ƒ : N a → R be given and suppose α , μ > 0 with ϒ ‒ 1 < α ≤ ϒ . Then, we have Δ a + μ α Δ a ‒ μ Ƒ ( ƭ ) = Δ a α ‒ μ Ƒ ( ƭ ) for t ∈ N a + μ + ϒ ‒ α .

Lemma 5.1

[7]

Let m ∈ N 0 and f : N a ‒ m → R and g : N a → R are of exponential order r > 0 . Then, for all u ∈ C ∖ { ‒ 1 , 0 } such that | ( u + 1 ) / u | > r ,

Definition 5.2

[7]

For each μ ∈ R ∖ ( ‒ N ) , define the μ th-Taylor monomial to be

Theorem 5.3

[7]

Suppose f : N a → R is of exponential order r ≥ 1 and let ν > 0 with N ‒ 1 < v ≤ N . Then, for all u ∈ C ∖ { ‒ 1 , 0 } such that | ( u + 1 ) / u | > r ,

Theorem 5.4

[7]

Suppose f : N a → R is of exponential order r ≥ 1 and let v > 0 with N ‒ 1 < v ≤ N . Then, for all u ∈ C ∖ { ‒ 1 , 0 } such that | ( u + 1 ) / u | > r ,

Lemma 5.5

[9]

Let μ ∈ R ∖ ( ‒ N ) and a , b ∈ R such that b ‒ a = μ . Then, for all u ∈ C ∖ { ‒ 1 , 0 } such that | ( u + 1 ) / u | > 1 , one has

1. S b { h μ ( . , a ) } ( u ) = ( u + 1 ) μ .

2. for v , x , y ∈ R such that ( x + y ) v = ∑ k = 0 ∞ v k x k y v ‒ k .

3. v k ≔ v k ¯ k ! .

4. ‒ v k = ( ‒ 1 ) k k + v ‒ 1 v ‒ 1 .

Theorem 5.6

[6]

Let f : N a → R and ν > 0 be given with N ‒ 1 < ν ≤ N . Consider the νth-order fractional difference equation

Then,

for i ∈ { 0 , 1 , ⋯ , N ‒ 1 } .

Theorem 5.7

Let f : N 0 → R and ν > 0 be given with N ‒ 1 < ν ≤ N . Consider the ν th -order fractional difference equation (5)

Then, β i = Δ a + v ‒ N i + v ‒ N y ( a ) Γ ( i + v ‒ N + 1 ) , i ∈ { 0 , 1 , ⋯ , N ‒ 1 } , and γ i = Δ i y ( a + v ‒ N ) such that

1. β 0 = z ( 0 ) γ 0 ,

2. β 1 = z ( 1 ) γ 1 ‒ [ z ( 0 ) ‒ z ( 1 ) ] γ 0 ,

3. β 2 = 1 2 z ( 2 ) γ 2 ‒ ( z ( 1 ) ‒ z ( 2 ) ) γ 1 + 1 2 [ z ( 0 ) ‒ 2 z ( 1 ) + z ( 2 ) ] γ 0 ,

4. β 3 = 1 6 z ( 3 ) γ 3 ‒ 1 2 ( z ( 2 ) ‒ z ( 3 ) ) γ 2 + 1 2 [ z ( 1 ) ‒ 2 z ( 2 ) + z ( 3 ) ] γ 1 ,

5. β 4 = 1 24 z ( 4 ) γ 4 ‒ 1 6 ( z ( 3 ) ‒ z ( 4 ) ) γ 3 + 1 4 [ z ( 2 ) ‒ 2 z ( 3 ) + z ( 4 ) ] γ 2 ‒ 1 6 ( z ( 1 ) ‒ 3 z ( 2 ) + 3 z ( 3 ) ‒ z ( 4 ) ) γ 1 + 1 24 ( z ( 0 ) ‒ 4 z ( 1 ) + 6 z ( 2 ) ‒ 4 z ( 3 ) + z ( 4 ) ) γ 0 .

Proof

Define g : N 0 → R by ( t ) ≔ ƭ N ‒ v , we can calculate some terms of β i as follows:

1. β 0 = z ( 0 ) γ 0 , γ 0 = Δ i y ( a + v ‒ N ) , when i = 0 , we can get, ∑ p = 0 0 ∑ k = 0 0 ( ‒ 1 ) k i ! ( 0 ‒ k ) N ‒ v 0 p 0 k Δ 0 y ( a + v ‒ N ) = ( 0 ) N ‒ v γ 0 , then β 0 = z ( 0 ) γ 0 ,

2. β 1 = z ( 1 ) γ 1 ‒ [ z ( 0 ) ‒ z ( 1 ) ] γ 0 , let i = 1, ∑ p = 0 1 ∑ k = 0 1 ‒ p ( ‒ 1 ) k i ! ( 1 ‒ k ) N ‒ v 1 p 1 ‒ p k Δ 1 y ( a + v ‒ N ) = ( 1 ) N ‒ v ‒ ( 0 ) N ‒ v γ 0 + ( 1 ) N ‒ v γ 1 , then β 1 = z ( 1 ) γ 1 ‒ [ z ( 0 ) ‒ z ( 1 ) ] γ 0 ,

3. β 2 = 1 2 z ( 2 ) γ 2 ‒ ( z ( 1 ) ‒ z ( 2 ) ) γ 1 + 1 2 [ z ( 0 ) ‒ 2 z ( 1 ) + z ( 2 ) ] γ 0 , let i = 2, ∑ p = 0 2 ∑ k = 0 2 ‒ p ( ‒ 1 ) k i ! ( 2 ‒ k ) N ‒ v 2 p 2 ‒ p k Δ 2 y ( a + v ‒ N ) = 1 2 ! ( 2 ) N ‒ v ‒ 2 1 2 ! ( 1 ) N ‒ v + 1 2 ! ( 0 ) N ‒ v γ 0 + 2 1 2 ! ( 2 ) N ‒ v ‒ 2 1 2 ! ( 1 ) N ‒ v γ 1 + 1 2 ! ( 2 ) N ‒ v γ 2 , then β 2 = 1 2 z ( 2 ) γ 2 ‒ ( z ( 1 ) ‒ z ( 2 ) ) γ 1 + 1 2 [ z ( 0 ) ‒ 2 z ( 1 ) + z ( 2 ) ] γ 0 ,

4. β 3 = 1 6 z ( 3 ) γ 3 ‒ 1 2 ( z ( 2 ) ‒ z ( 3 ) ) γ 2 + 1 2 [ z ( 1 ) ‒ 2 z ( 2 ) + z ( 3 ) ] γ 1 , let i =   3, ∑ p = 0 2 ∑ k = 0 3 ‒ p ( ‒ 1 ) k i ! ( 3 ‒ k ) N ‒ v 3 p 3 ‒ p k Δ 2 y ( a + v ‒ N ) , and in the same way, we can prove that β 3 = 1 6 z ( 3 ) γ 3 ‒ 1 2 ( z ( 2 ) ‒ z ( 3 ) ) γ 2 + 1 2 [ z ( 1 ) ‒ 2 z ( 2 ) + z ( 3 ) ] γ 1 ,

5. β 4 = 1 24 z ( 4 ) γ 4 ‒ 1 6 ( z ( 3 ) ‒ z ( 4 ) ) γ 3 + 1 4 [ z ( 2 ) ‒ 2 z ( 3 ) + z ( 4 ) ] γ 2 ‒ 1 6 ( z ( 1 ) ‒ 3 z ( 2 ) + 3 z ( 3 ) ‒ z ( 4 ) ) γ 1 + 1 24 ( z ( 0 ) ‒ 4 z ( 1 ) + 6 z ( 2 ) ‒ 4 z ( 3 ) + z ( 4 ) ) γ 0 , let i = 4 , ∑ p = 0 4 ∑ k = 0 4 ‒ p ( ‒ 1 ) k i ! ( 4 ‒ k ) N ‒ v 4 p 4 ‒ p k Δ 2 y ( a + v ‒ N ) ,