Further Properties of the Nabla Laplace Transform - Nabla Fractional Calculus - Discrete Fractional Calculus

Discrete Fractional Calculus (2015)

3. Nabla Fractional Calculus

3.13. Further Properties of the Nabla Laplace Transform

In this section we want to find the Laplace transform of a ν-th order fractional difference of a function, where 0 < ν < 1.

Theorem 3.87.

Assume $$f: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$$ is of exponential order r > 0 and 0 < ν < 1. Then

$$\displaystyle{\mathcal{L}_{a}\{\nabla _{a}^{\nu }f\}(s) = s^{\nu }\mathcal{L}_{ a}\{f\}(s)}$$

for |s − 1| < r.

Proof.

Using Theorems 3.82 and 3.83 we have that

$$\displaystyle\begin{array}{rcl} \mathcal{L}_{a}\{\nabla _{a}^{\nu }f\}(s)& =& \mathcal{L}_{ a}\{\nabla \nabla _{a}^{-(1-\nu )}f\}(s) {}\\ & =& s\mathcal{L}_{a}\{\nabla _{a}^{-(1-\nu )}f\}(s) -\nabla _{ a}^{-(1-\nu )}f(a) {}\\ & =& \frac{s} {s^{1-\nu }}s^{\nu }\mathcal{L}_{a}\{f\}(s) {}\\ & s^{\nu }& \mathcal{L}_{a}\{f\}(s) {}\\ \end{array}$$

for | s − 1 |  < 1. □ 

Next we state and prove a useful lemma (see Hein et al. [119] for n = 1 and see Ahrendt et al. [3] for general n).

Lemma 3.88 (Shifting Base Lemma).

Given $$f: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$$ and $$n \in \mathbb{N}_{1}$$ , we have that

$$\displaystyle{\mathcal{L}_{a+n}\{f\}(s) = \left ( \frac{1} {1 - s}\right )^{n}\mathcal{L}_{ a}\{f\}(s) -\sum _{k=1}^{n} \frac{f(a + k)} {(1 - s)^{n-k+1}}.}$$

Proof.

Consider

$$\displaystyle\begin{array}{rcl} \mathcal{L}_{a+n}\{f\}(s)& =& \sum _{k=1}^{\infty }(1 - s)^{k-1}f(a + n + k) {}\\ & =& \sum _{k=n+1}^{\infty }(1 - s)^{k-n-1}f(a + k) {}\\ & =& \frac{1} {(1 - s)^{n}}\mathcal{L}_{a}\{f\}(s) -\sum _{k=1}^{n} \frac{f(a + k)} {(1 - s)^{n-k+1}}, {}\\ \end{array}$$

which is what we wanted to prove. □ 

With this, we are ready to provide the general form of the Laplace transform of a ν-th order fractional difference of a function f, where 0 < ν < 1.

Theorem 3.89.

Given $$f: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$$ and 0 < ν < 1. Then we have

$$\displaystyle{\mathcal{L}_{a+1}\{\nabla _{a}^{\nu }f\}(s) = s^{\nu }\mathcal{L}_{ a+1}\{f\}(s) -\frac{1 - s^{\nu }} {1 - s}f(a + 1).}$$

Proof.

Consider

$$\displaystyle\begin{array}{rcl} & & \mathcal{L}_{a+1}\{\nabla _{a}^{\nu }f\}(s) {}\\ & & \quad = \mathcal{L}_{a+1}\{\nabla \nabla _{a}^{-(1-\nu )}f\}(s) {}\\ & & \quad = s\mathcal{L}_{a+1}\{\nabla _{a}^{-(1-\nu )}f\}(s) -\nabla _{ a}^{-(1-\nu )}f(a + 1),\quad \mbox{ by Theorem <InternalRef RefID="FPar8">3.8</InternalRef>3} {}\\ & & \quad = s\mathcal{L}_{a+1}\{\nabla _{a}^{-(1-\nu )}f\}(s) - f(a + 1),\quad \mbox{ by Exercise 3.27.} {}\\ \end{array}$$

From this and Lemma 3.88, we have that

$$\displaystyle\begin{array}{rcl} & & \mathcal{L}_{a+1}\{\nabla _{a}^{\nu }f\}(s) {}\\ & & = s\left ( \frac{1} {1 - s}\mathcal{L}_{a}\{\nabla _{a}^{-(1-\nu )}f\}(s) - \frac{1} {1 - s}\nabla _{a}^{-(1-\nu )}f(a + 1)\right ) - f(a + 1) {}\\ & & = \frac{s^{\nu }} {1 - s}\mathcal{L}_{a}\{f\}(s) - \frac{1} {1 - s}f(a + 1)\quad \text{(by Theorem <InternalRef RefID="FPar8">3.8</InternalRef>2)}. {}\\ \end{array}$$

Applying Lemma 3.88 again we obtain

$$\displaystyle\begin{array}{rcl} & & \mathcal{L}_{a+1}\{\nabla _{a}^{\nu }f\}(s) {}\\ & & = s^{\nu }\left (\mathcal{L}_{a+1}\{f\}(s) + \frac{1} {1 - s}f(a + 1)\right ) - \frac{1} {1 - s}f(a + 1), {}\\ & & {}\\ \end{array}$$

which is the desired result. □ 

The following theorem was proved by Jia Baoguo.

Theorem 3.90.

Let $$f: \mathbb{N}_{a-N+1} \rightarrow \mathbb{R}$$ and N − 1 < ν < N be given. Then we have

$$\displaystyle\begin{array}{rcl} \mathcal{L}_{a+1}\{\nabla _{a}^{\nu }f\}(s)& =& s^{\nu }\mathcal{L}_{ a+1}\{f\}(s) + \frac{s^{\nu } - s^{N-1}} {1 - s} f(a + 1) {}\\ & & -\sum _{k=2}^{N}s^{N-k}\nabla ^{k-1}f(a + 1). {}\\ \end{array}$$

Proof.

We first calculate

$$\displaystyle\begin{array}{rcl} & & \mathcal{L}_{a+1}\{\nabla _{a}^{\nu }f\}(s) {}\\ & & = \mathcal{L}_{a+1}\{\nabla ^{N}\nabla _{ a}^{-(N-\nu )}f\}(s) {}\\ & & \mathop{=}\limits^{ \text{Theorem <InternalRef RefID="FPar8">3.8</InternalRef>5}}s^{N}\mathcal{L}_{ a+1}(\nabla _{a}^{-(N-\nu )}f)(s) -\sum _{ k=1}^{N}s^{N-k}\nabla ^{k-1}\nabla _{ a}^{-(N-\nu )}f(a + 1) {}\\ & & \mathop{=}\limits^{ \text{Lemma <InternalRef RefID="FPar88">3.88</InternalRef>}} \frac{s^{N}} {1 - s}\mathcal{L}_{a}\{\nabla _{a}^{-(N-\nu )}f\}(s) - s^{N}\frac{\nabla _{a}^{-(N-\nu )}f(a + 1)} {1 - s} {}\\ & & \quad -\sum _{k=1}^{N}s^{N-k}\nabla ^{k-1}\nabla _{ a}^{-(N-\nu )}f(a + 1) {}\\ & & \mathop{=}\limits^{ \text{Theorem <InternalRef RefID="FPar8">3.8</InternalRef>2}} \frac{s^{N}} {1 - s} \cdot \frac{1} {s^{N-\nu }}\mathcal{L}_{a}\{f\}(s) - s^{N}\frac{\nabla _{a}^{-(N-\nu )}f(a + 1)} {1 - s} {}\\ & & \quad -\sum _{k=1}^{N}s^{N-k}\nabla ^{k-1}\nabla _{ a}^{-(N-\nu )}f(a + 1) {}\\ & & \mathop{=}\limits^{ \text{Theorem <InternalRef RefID="FPar8">3.8</InternalRef>8}} \frac{s^{\nu }} {1 - s}[(1 - s)\mathcal{L}_{a+1}\{f\}(s) + f(a + 1)] - s^{N}\frac{\nabla _{a}^{-(N-\nu )}f(a + 1)} {1 - s} {}\\ & & \quad -\sum _{k=1}^{N}s^{N-k}\nabla ^{k-1}\nabla _{ a}^{-(N-\nu )}f(a + 1). {}\\ \end{array}$$

Since

$$\displaystyle\begin{array}{rcl} & \nabla _{a}^{-(N-\nu )}f(a + 1) =\int _{ a}^{a+1}H_{N-\nu -1}(a + 1,a)f(s)\nabla s& {}\\ & = H_{N-\nu }(a + 1,a)f(a + 1) = f(a + 1), & {}\\ \end{array}$$

we have

$$\displaystyle\begin{array}{rcl} & & \mathcal{L}_{a+1}\{\nabla _{a}^{\nu }f\}(s) {}\\ & & = s^{\nu }\mathcal{L}_{a+1}\{f\}(s) + \frac{s^{\nu }} {1 - s}f(a + 1) -\frac{s^{N}f(a + 1)} {1 - s} {}\\ & & \quad -\sum _{k=1}^{N}s^{N-k}\nabla ^{k-1}f(a + 1) {}\\ & & = s^{\nu }\mathcal{L}_{a+1}\{f\}(s) + \frac{s^{\nu } - s^{N-1}} {1 - s} f(a + 1) -\sum _{k=2}^{N}s^{N-k}\nabla ^{k-1}f(a + 1). {}\\ \end{array}$$

 □ 

Remark 3.91.

When N = 1, Theorem 3.90 becomes the Theorem 3.89. When N = 2, we can get the following Corollary.

Corollary 3.92.

Let $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ and 1 < ν < 2 be given. Then we have

$$\displaystyle{\mathcal{L}_{a+1}\{\nabla _{a}^{\nu }f\}(s) = s^{\nu }\mathcal{L}_{ a+1}\{f\}(s) + \frac{s^{\nu } - s} {1 - s}f(a + 1) -\nabla f(a + 1)}$$

$$\displaystyle{= s^{\nu }\mathcal{L}_{a+1}\{f\}(s) + \frac{s^{\nu } - 1} {1 - s}f(a + 1) + f(a).}$$