Fractional Sums and Differences - Nabla Fractional Calculus - Discrete Fractional Calculus

Discrete Fractional Calculus (2015)

3. Nabla Fractional Calculus

3.9. Fractional Sums and Differences

With the relevant preliminaries established, we are now ready to develop what we mean by fractional nabla differences and fractional nabla sums. We first give the motivation for how we define nabla integral sums.

In the previous section (see Corollary 3.52) we saw that

 $$\displaystyle{y(t) =\int _{ a}^{t}H_{ n-1}(t,\rho (s))f(s)\nabla s}$$

is the unique solution of the nabla difference equation  $$\nabla ^{n}y(t) = f(t),t \in \mathbb{N}_{a+1}$$ satisfying the initial conditions ∇ i y(a) = 0, 0 ≤ i ≤ n − 1, for  $$t \in \mathbb{N}_{a}$$ . Integrating n times both sides of ∇ n y(t) = f(t) and using the initial conditions ∇ iy(a) = 0, 0 ≤ i ≤ n − 1, we get by uniqueness

 $$\displaystyle\begin{array}{rcl} & & \int _{a}^{t}\int _{ a}^{\tau _{1} }\cdots \int _{a}^{\tau _{n-1} }f(\tau _{n})\nabla \tau _{n}\cdots \nabla \tau _{2}\nabla \tau _{1} \\ & & \qquad \qquad \qquad \qquad \qquad \qquad \quad =\int _{ a}^{t}H_{ n-1}(t,\rho (s))f(s)\nabla s.{}\end{array}$$

(3.31)

The formula (3.31) can also be easily proved by repeated integration by parts. Motivated by this we define the nabla integral order sum as in the following definition.

Definition 3.54 (Integral Order Sum).

Let  $$f: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$$ be given and  $$n \in \mathbb{N}_{1}$$ . Then

 $$\displaystyle{ \nabla _{a}^{-n}f(t):=\int _{ a}^{t}H_{ n-1}(t,\rho (s))f(s)\nabla s,\quad t \in \mathbb{N}_{a}. }$$

Also, we define  $$\nabla ^{-0}f(t):= f(t)$$ .

Note that the function ∇ a n f depends on the values of f at all the points a + 1 ≤ s ≤ t, unlike the positive integer nabla difference ∇ n f(t), which just depends on the values of f at the n + 1 points tn ≤ s ≤ t. Another interesting observation is that we could think of  $$\nabla _{a}^{-n}f(t)$$ as defined on  $$\mathbb{N}_{a-n+1}$$ , from which we obtain that  $$f(t) = 0,\quad a + n - 1 \leq t \leq a$$ by our convention that the nabla integral from a point to a smaller point is zero (see Definition 3.31). The following example appears in Hein et al. [119].

Example 3.55.

Use the definition (Definition 3.54) of the fractional sum to find  $$\nabla _{a}^{-2}E_{p}(t,a),$$ where p ≠ 0, 1 is a constant. By definition we obtain, using the second integration by parts formula (3.20),

 $$\displaystyle\begin{array}{rcl} \nabla _{a}^{-2}E_{ p}(t,a)& =& \int _{a}^{t}H_{ 1}(t,\rho (s))E_{p}(s,a)\nabla s {}\\ & =& \frac{1} {p}E_{p}(s,a)H_{1}(t,s)\big\vert _{s=a}^{t} + \frac{1} {p}\int _{a}^{t}E_{ p}(s,a)\nabla s {}\\ & =& -\frac{1} {p}H_{1}(t,a) + \frac{1} {p^{2}}E_{p}(s,a)\big\vert _{s=a}^{t} {}\\ & =& -\frac{1} {p}H_{1}(t,a) + \frac{1} {p^{2}}E_{p}(t,a) - \frac{1} {p^{2}} {}\\ & =& -\frac{1} {p}(t - a) + \frac{1} {p^{2}}(1 - p)^{a-t} - \frac{1} {p^{2}}. {}\\ \end{array}$$

Note that if  $$n \in \mathbb{N}_{1}$$ , then

 $$\displaystyle{H_{n}(t,a) = \frac{(t - a)^{\overline{n}}} {n!} = \frac{(t - a)^{\overline{n}}} {\Gamma (n + 1)}.}$$

Motivated by this we define the fractional μ-th order nabla Taylor monomial as follows.

Definition 3.56.

Let  $$\mu \neq - 1,-2,-3,\cdots $$ . Then we define the μ-th order nabla fractional Taylor monomial, H μ (t, a), by

 $$\displaystyle{H_{\mu }(t,a) = \frac{(t - a)^{\overline{\mu }}} {\Gamma (\mu +1)},}$$

whenever the right-hand side of this equation is sensible.

In the next theorem we collect some of the properties of fractional nabla Taylor monomials.

Theorem 3.57.

The following hold:

(i)

H μ (a,a) = 0;

(ii)

 $$\nabla H_{\mu }(t,a) = H_{\mu -1}(t,a);$$

(iii)

 $$\int _{a}^{t}H_{\mu }(s,a)\nabla s = H_{\mu +1}(t,a);$$

(iv)

 $$\int _{a}^{t}H_{\mu }(t,\rho (s))\nabla s = H_{\mu +1}(t,a);$$

(v)

for  $$k \in \mathbb{N}_{1}$$ ,  $$H_{-k}(t,a) = 0$$ ,  $$t \in \mathbb{N}_{a},$$

provided the expressions in this theorem are well defined.

Proof.

Part (i) follows immediately from the definition of H μ (t, a). The proofs of parts (ii)–(iii) of this theorem are the same as the proof of Theorem 3.47, where we used the fractional power rules instead of the integer power rules. Finally, part (v) follows since

 $$\displaystyle{H_{-k}(t,a) = \frac{(t - a)^{\overline{ - k}}} {\Gamma (-k + 1)} = 0}$$

by our earlier convention when the denominator is undefined but the numerator is defined. □ 

Now we can define the fractional nabla sum in terms of the nabla fractional Taylor monomial as follows.

Definition 3.58 (Nabla Fractional Sum).

Let  $$f: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$$ be given and assume μ > 0. Then

 $$\displaystyle{\nabla _{a}^{-\mu }f(t):=\int _{ a}^{t}H_{\mu -1}(t,\rho (s))f(s)\nabla s,}$$

for  $$t \in \mathbb{N}_{a}$$ , where by convention  $$\nabla _{a}^{-\mu }f(a) = 0$$ .

The following example appears in Hein et al. [119].

Example 3.59.

Use the definition (Definition 3.58) of the fractional sum to find  $$\nabla _{a}^{-\mu }1$$ . By definition

 $$\displaystyle\begin{array}{rcl} \nabla _{a}^{-\mu }1& =& \int _{ a}^{t}H_{\mu -1}(t,\rho (s)) \cdot 1\nabla s {}\\ & =& \int _{a}^{t}H_{\mu -1}(t,\rho (s))\nabla s {}\\ & =& H_{\mu }(t,a),\quad t \in \mathbb{N}_{a} {}\\ \end{array}$$

by part (iv) of Theorem 3.57.

For those readers that have read Chap. 2 we gave a relationship between a certain delta fractional sum and a certain nabla fractional sum. This formula is sometimes useful for obtaining results for the nabla fractional calculus from the delta fractional calculus. Since we want this chapter to be self-contained we will not use this formula in this chapter.

Theorem 3.60.

Assume  $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ and ν > 0. Then

 $$\displaystyle{\Delta _{a}^{-\nu }f(t+\nu ) = \nabla _{ a}^{-\nu }f(t) + H_{\nu -1}(t,\rho (a))f(a),}$$

for  $$t \in \mathbb{N}_{a}$$ . In particular, if f(a) = 0, then

 $$\displaystyle{\Delta _{a}^{-\nu }f(t+\nu ) = \nabla _{ a}^{-\nu }f(t),}$$

for  $$t \in \mathbb{N}_{a}$$ .

Proof.

Note that  $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ implies  $$\Delta _{a}^{-\nu }f(t+\nu )$$ is defined for  $$t \in \mathbb{N}_{a}$$ . Using the definition of the ν-th order fractional sum (Definition 3.58) we find that

 $$\displaystyle\begin{array}{rcl} \Delta _{a}^{-\nu }f(t+\nu )& =& \int _{ a}^{t+1}h_{\nu -1}(t+\nu,\sigma (\tau ))f(\tau )\nabla \tau {}\\ & =& \sum _{\tau =a}^{t}h_{\nu -1}(t+\nu,\sigma (\tau ))f(\tau ) {}\\ & =& \sum _{\tau =a}^{t}\frac{(t +\nu -\rho (\tau ))^{\underline{\nu -1}}} {\Gamma (\nu )} f(\tau ) {}\\ & =& \sum _{\tau =a}^{t} \frac{\Gamma (t +\nu -\tau )} {\Gamma (\nu )\Gamma (t +\nu -\tau )} {}\\ & =& \sum _{\tau =a}^{t}\frac{(t -\tau +1)^{\overline{\nu - 1}}} {\Gamma (\nu )} f(\tau ) {}\\ & =& \nabla _{a}^{-\nu }f(t) {}\\ \end{array}$$

for  $$t \in \mathbb{N}_{a}$$ . □ 

We next define the nabla fractional difference (nabla Riemann–Liouville fractional difference) in terms of a nabla fractional sum. The Caputo fractional sum (Definition 3.117) will be considered in Sect. 3.18.

Definition 3.61 (Nabla Fractional Difference).

Let  $$f: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$$ ,  $$\nu \in \mathbb{R}^{+}$$ and choose N such that N − 1 < ν ≤ N. Then we define the ν-th order nabla fractional difference,  $$\nabla _{a}^{\nu }f(t)$$ , by

 $$\displaystyle{\nabla _{a}^{\nu }f(t):= \nabla ^{N}\nabla _{ a}^{-(N-\nu )}f(t)\quad \mbox{ for}\quad t \in \mathbb{N}_{ a+N}.}$$

We now have a definition for both fractional sums and fractional differences; however, they may still be unified to a similar form. We will show here that the traditional definition of a fractional difference can be rewritten in a form similar to the definition for a fractional sum. The following result appears in Ahrendt et al. [3].

Theorem 3.62.

Assume  $$f: \mathbb{N}_{a} \rightarrow \mathbb{R},$$ ν > 0,  $$\nu \not\in \mathbb{N}_{1}$$ , and choose  $$N \in \mathbb{N}_{1}$$ such that N − 1 < ν < N. Then

 $$\displaystyle{ \nabla _{a}^{\nu }f(t) =\int _{ a}^{t}H_{ -\nu -1}(t,\rho (\tau ))f(\tau )\nabla \tau, }$$

(3.32)

for  $$t \in \mathbb{N}_{a+1}$$ .

Proof.

Note that

 $$\displaystyle\begin{array}{rcl} & & \nabla _{a}^{\nu }f(t) = \nabla ^{N}\nabla _{ a}^{-(N-\nu )}f(t) {}\\ & & = \nabla ^{N}\left (\int _{ a}^{t}H_{ N-\nu -1}(t,\rho (\tau ))f(\tau )\nabla \tau \right ) {}\\ & & = \nabla ^{N-1}\nabla \left (\int _{ a}^{t}H_{ N-\nu -1}(t,\rho (\tau ))f(\tau )\nabla \tau \right ) {}\\ & & = \nabla ^{N-1}\left (\int _{ a}^{t}H_{ N-\nu -2}(t,\rho (\tau ))f(\tau )\nabla \tau + H_{N-\nu -1}(\rho (t),\rho (t))f(t)\right ) {}\\ & & = \nabla ^{N-1}\int _{ a}^{t}H_{ N-\nu -2}(t,\rho (\tau ))f(\tau )\nabla \tau. {}\\ \end{array}$$

By applying Leibniz’s Rule N − 1 more times, we deduce that

 $$\displaystyle{ \nabla _{a}^{\nu }f(t) =\int _{ a}^{t}H_{ -\nu -1}(t,\rho (\tau ))f(\tau )\nabla \tau, }$$

which is the desired result. □ 

In the following theorem we show that the nabla fractional difference, for each fixed  $$t \in \mathbb{N}_{a}$$ , is a continuous function of ν for ν > 0. The following theorem appears in Ahrendt et al. [3].

Theorem 3.63 (Continuity of the Nabla Fractional Difference).

Assume  $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ . Then the fractional difference  $$\nabla _{a}^{\nu }f$$ is continuous with respect to ν for ν > 0.

Proof.

It is sufficient for this proof to show that for  $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ , N − 1 < ν ≤ N, and  $$m \in \mathbb{N}_{0}$$ , the following hold:

 $$\displaystyle{ \nabla _{a}^{\nu }f(a + N + m)\mbox{ is continuous with respect to $\nu $ on $(N - 1,N)$,} }$$

(3.33)

 $$\displaystyle{ \nabla _{a}^{\nu }f(a + N + m) \rightarrow \nabla ^{N}f(a + N + m)\text{ as }\nu \rightarrow N^{-}, }$$

(3.34)

and

 $$\displaystyle{ \nabla _{a}^{\nu }f(a + N + m) \rightarrow \nabla ^{N-1}f(a + N + m)\text{ as }\nu \rightarrow (N - 1)^{+}. }$$

(3.35)

Let ν be fixed such that N − 1 < ν < N. We now show that (3.33) holds. To see this note that we have the following:

 $$\displaystyle\begin{array}{rcl} & & \nabla _{a}^{\nu }f(a + N + m) =\int _{ a}^{t}H_{ -\nu -1}(t,\rho (\tau ))\nabla \tau {}\\ & =& \frac{1} {\Gamma (-\nu )}\sum _{\tau =a+1}^{t}(t -\rho (\tau ))^{\overline{ -\nu -1}}\bigg\vert _{ t=a+N+m}f(\tau ) {}\\ & =& \frac{1} {\Gamma (-\nu )}\sum _{\tau =a+1}^{a+N+m}(a + N + m -\rho (\tau ))^{\overline{ -\nu -1}}f(\tau ) {}\\ & =& \sum _{\tau =a+1}^{a+N+m}\frac{\Gamma (a + N + m -\tau +1 -\nu -1)} {\Gamma (a + N + m -\tau +1)\Gamma (-\nu )} f(\tau ) {}\\ & =& \sum _{\tau =a+1}^{a+N+m}\frac{(a + N + m -\tau -\nu - 1)\cdots (-\nu )\Gamma (-\nu )} {(a + N + m-\tau )!\Gamma (-\nu )} f(\tau ) {}\\ & =& \sum _{\tau =a+1}^{a+N+m-1}\frac{(a + N + m -\tau -\nu - 1)\cdots (-\nu )} {(a + N + m-\tau )!} f(\tau ) + f(a + N + m). {}\\ \end{array}$$

Letting  $$i:= a + N + m-\tau$$ , we get

 $$\displaystyle\begin{array}{rcl} & & \nabla _{a}^{\nu }f(a + N + m) {}\\ & =& \sum _{i=1}^{N+m}\frac{(i - 1-\nu )\cdots (1-\nu )(-\nu )} {i!} f(a + N + M - i) + f(a + N + M). {}\\ \end{array}$$

This shows that the ν-th order fractional difference is continuous on N − 1 < ν < N, showing (3.33) holds.

Now we consider the case ν → N in order to show that (3.34) holds:

 $$\displaystyle\begin{array}{rcl} & & \lim _{\nu \rightarrow N^{-}}\nabla _{a}^{\nu }f(a + N + m) {}\\ & & \quad =\lim _{\nu \rightarrow N^{-}}\sum _{i=1}^{N+m}\frac{(i - 1-\nu )\cdots (1-\nu )(-\nu )} {i!} f(a + N + M - i) {}\\ & & \qquad +\, f(a + N + M) {}\\ & & \quad =\sum _{ i=1}^{N+m}\left (\frac{(i - 1 - N)\cdots (-N)} {i!} f(a + N + m - i)\right ) + f(a + N + m) {}\\ & & \quad =\sum _{ i=0}^{N+m}\left ((-1)^{i}\frac{(N + 1 - i)\cdots (N)} {i!} f(a + N + m - i)\right ) {}\\ & & \quad =\sum _{ i=0}^{N+m}(-1)^{i}{N\choose i}f(a + N + m - i) {}\\ & & \quad =\sum _{ i=0}^{N}(-1)^{i}{N\choose i}f(a + m + N - i) {}\\ & & \quad = \nabla ^{N}f(a + N + m). {}\\ \end{array}$$

Finally we want to show (3.35) holds. So we write

 $$\displaystyle\begin{array}{rcl} & & \lim _{\nu \rightarrow (N-1)^{+}}\nabla _{a}^{\nu }f(a + N + m) {}\\ & & \quad =\lim _{\nu \rightarrow (N-1)^{+}}\sum _{i=1}^{N+m}\frac{(i - 1-\nu )\cdots (1-\nu )(-\nu )} {i!} f(a + N + M - i) {}\\ & & \qquad +\, f(a + N + M) {}\\ & & \quad =\sum _{ i=1}^{N+m}\frac{(i - N)(i - N - 1)\cdots (1 - N)} {i!} f(a + N + m - i) {}\\ & & \qquad +\, f(a + N + m) {}\\ & & \quad =\sum _{ i=0}^{N+m}(-1)^{i}\frac{(N - i)(N + 1 - i)\cdots (N - 1)} {i!} f(a + N + m - i) {}\\ & & \quad =\sum _{ i=0}^{N+m}(-1)^{i}{N - 1\choose i}f(a + m + 1 + N - 1 - i) {}\\ & & \quad = \nabla ^{N-1}f(a + N + m). {}\\ \end{array}$$

This completes the proof. □ 

To prove various properties for the nabla fractional sums and differences it is convenient to develop the theory of the nabla Laplace transform, which we do in the next section.