Convolution - Nabla Fractional Calculus - Discrete Fractional Calculus

Discrete Fractional Calculus (2015)

3. Nabla Fractional Calculus

3.12. Convolution

We are now ready to investigate one of the most important properties in solving initial-value fractional nabla difference equations: convolution. This definition is motivated by the desire to express the fractional nabla sums and fractional nabla differences as convolutions of arbitrary functions and Taylor monomials. As a consequence, the resulting properties that stem from this definition are, in fact, consistent with the standard convolution. Many of the results in this section appear in Hein et al. [119] and Ahrendt et al. [3].

Definition 3.77.

For  $$f,g: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$$ , we define the nabla convolution product of f and g by

 $$\displaystyle{(f {\ast} g)(t):=\int _{ a}^{t}f(t -\rho (\tau ) + a)g(\tau )\nabla \tau,\quad t \in \mathbb{N}_{ a+1}.}$$

Example 3.78.

Use the definition of the nabla convolution product to find 1 ∗Sin p (⋅ , a), p ≠ 0, ±i. By Definition 3.77,

 $$\displaystyle\begin{array}{rcl} (1 {\ast}\mbox{ Sin}_{p}(\cdot,a))(t)& =& \int _{a}^{t}1 \cdot \mbox{ Sin}_{ p}(\tau,a)\nabla \tau {}\\ & =& \int _{a}^{t}\mbox{ Sin}_{ p}(\tau,a)\nabla \tau {}\\ & =& -\frac{1} {p}\mbox{ Cos}_{p}(\tau,a)\big\vert _{a}^{t} {}\\ & =& \frac{1} {p} -\frac{1} {p}\mbox{ Cos}_{p}(t,a), {}\\ \end{array}$$

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

Example 3.79.

Use the definition of the (nabla) convolution product to find

 $$\displaystyle{\left (E_{p}(\cdot,a) {\ast} E_{q}(\cdot,a)\right )(t),\quad p,q\neq 1,\quad p\neq q.}$$

Assume q ≠ 0. By Definition 3.77 and using the second integration by parts formula, we have that

 $$\displaystyle\begin{array}{rcl} & & \left (E_{p}(\cdot,a) {\ast} E_{q}(\cdot,a)\right )(t) {}\\ & & =\int _{ a}^{t}E_{ p}(t -\rho (\tau ) + a,a)E_{q}(\tau,a)\nabla \tau {}\\ & & = \frac{1} {q}E_{p}(t -\tau +a,a)E_{q}(\tau,a)\big\vert _{\tau =a}^{\tau =t} + \frac{p} {q}\int _{a}^{t}E_{ p}(t -\rho (\tau ) + a,a)E_{q}(\tau,a)\nabla \tau {}\\ & & = \frac{1} {q}E_{q}(t,a) -\frac{1} {q}E_{p}(t,a) + \frac{p} {q}\left (E_{p}(\cdot,a) {\ast} E_{q}(\cdot,a)\right )(t). {}\\ & & {}\\ \end{array}$$

Solving for  $$\left (E_{p}(\cdot,a) {\ast} E_{q}(\cdot,a)\right )(t)$$ , we obtain

 $$\displaystyle{\left (E_{p}(\cdot,a) {\ast} E_{q}(\cdot,a)\right )(t) = \frac{1} {p - q}E_{p}(t,a) + \frac{1} {q - p}E_{q}(t,a)}$$

for  $$t \in \mathbb{N}_{a+1}$$ . We leave it to the reader to show that this last formula is also valid if q = 0.

Theorem 3.80.

Assume  $$\nu \in \mathbb{R}\setminus \{0,-1,-2,\ldots \}$$ and  $$f: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$$ . Then

 $$\displaystyle{\nabla _{a}^{-\nu }f(t) = (H_{\nu -1}(\cdot,a) {\ast} f)(t),}$$

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

Proof.

The result follows from the following:

 $$\displaystyle\begin{array}{rcl} (H_{\nu -1}(\cdot,a) {\ast} f)(t)& =& \int _{a}^{t}H_{\nu -1}(t -\rho (\tau ) + a,a)f(\tau )\nabla \tau {}\\ & =& \int _{a}^{t}\frac{(t -\rho (\tau ) + a - a)^{\overline{\nu - 1}}} {\Gamma (\nu )} f(\tau )\nabla \tau {}\\ & =& \int _{a}^{t}\frac{(t -\rho (\tau ))^{\overline{\nu - 1}}} {\Gamma (\nu )} f(\tau )\nabla \tau {}\\ & =& \int _{a}^{t}H_{\nu -1}(t,\rho (\tau ))f(\tau )\nabla \tau {}\\ & =& \nabla _{a}^{-\nu }f(t), {}\\ \end{array}$$

for  $$t \in \mathbb{N}_{a+1}$$ . □ 

Theorem 3.81 (Nabla Convolution Theorem).

Assume  $$f,g: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$$ and their nabla Laplace transforms converge for |s − 1| < r for some r > 0. Then

 $$\displaystyle{\mathcal{L}_{a}\{f {\ast} g\}(s) = \mathcal{L}_{a}\{f\}(s)\mathcal{L}_{a}\{g\}(s),}$$

for |s − 1| < r.

Proof.

The following proves our result:

 $$\displaystyle\begin{array}{rcl} \mathcal{L}_{a}\{f {\ast} g\}(s)& =& \sum _{k=1}^{\infty }(1 - s)^{k-1}(f {\ast} g)(a + k) {}\\ & =& \sum _{k=1}^{\infty }(1 - s)^{k-1}\int _{ a}^{a+k}f(a + k -\rho (\tau ) + a)g(\tau )\nabla \tau {}\\ & =& \sum _{k=1}^{\infty }(1 - s)^{k-1}\sum _{ \tau =a+1}^{a+k}f(a + k -\rho (\tau ) + a)g(\tau ) {}\\ & =& \sum _{k=1}^{\infty }\sum _{ \tau =1}^{k}(1 - s)^{k-1}f(k -\rho (\tau ) + a)g(\tau +a) {}\\ & =& \sum _{\tau =1}^{\infty }\sum _{ k=\tau }^{\infty }(1 - s)^{k-1}f(k -\rho (\tau ) + a)g(a+\tau ) {}\\ & =& \left (\sum _{\tau =1}^{\infty }(1 - s)^{\tau -1}g(a+\tau )\right )\left (\sum _{ k=1}^{\infty }(1 - s)^{k-1}f(a + k)\right ) {}\\ & =& \mathcal{L}_{a}\{g\}(s)\mathcal{L}_{a}\{f\}(s) {}\\ \end{array}$$

for | s − 1 |  < r. □ 

With the above result and the uniqueness of the Laplace transform, it follows that the convolution product is commutative and associative (see Exercise 3.32).

We next establish properties of the Laplace transform that will be useful in solving initial value problems for integer nabla difference equations.

Theorem 3.82 (Transformation of Fractional Sums).

Assume ν > 0 and the nabla Laplace transform of  $$f: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$$ converges for |s − 1| < r for some r > 0. Then

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

for  $$\vert s - 1\vert <\min \{ 1,r\}$$ .

Proof.

The result follows since

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

for  $$\vert s - 1\vert <\min \{ 1,r\}$$ . □ 

Assuming that ν is a positive integer, this result is consistent with the formula in the continuous case for the Laplace transform of the n-th iterated integral of a function. We want to establish similar properties for fractional differences; however, we will first establish integer-order difference properties.

Theorem 3.83 (Transform of Nabla Difference).

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

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

for |s − 1| < r.

Proof.

Note that since we are assuming that  $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ , we have that  $$\nabla f: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$$ and so we can consider  $$\mathcal{L}_{a}\{\nabla f\}(s)$$ . Since  $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ is of exponential order r > 0, it follows that  $$\nabla f: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$$ is of exponential order r > 0. It follows that for | s − 1 |  < r, 

 $$\displaystyle\begin{array}{rcl} \mathcal{L}_{a}\{\nabla f\}(s)& =& \sum _{k=1}^{\infty }(1 - s)^{k-1}\nabla f(a + k) {}\\ & =& \sum _{k=1}^{\infty }(1 - s)^{k-1}\left [f(a + k) - f(a + k - 1)\right ] {}\\ & =& \mathcal{L}_{a}\{f\}(s) -\sum _{k=1}^{\infty }(1 - s)^{k-1}f(a + k - 1) {}\\ & =& \mathcal{L}_{a}\{f\}(s) -\sum _{k=0}^{\infty }(1 - s)^{k}f(a + k) {}\\ & =& \mathcal{L}_{a+1}\{f\}(s) - f(a) - (1 - s)\sum _{k=1}^{\infty }(1 - s)^{k-1}f(a + k) {}\\ & =& \mathcal{L}_{a}\{f\}(s) - f(a) - (1 - s)\mathcal{L}_{a}\{f\}(s) {}\\ & =& s\mathcal{L}_{a}\{f\}(s) - f(a). {}\\ \end{array}$$

This completes the proof. □ 

The following is a simple example where we will use the Nabla Convolution Theorem and Theorem 3.83 to solve an initial value problem.

Example 3.84.

Use the Nabla Convolution Theorem to help you solve the IVP

 $$\displaystyle\begin{array}{rcl} \nabla y(t) - 3y(t)& =& E_{4}(t,a),\quad t \in \mathbb{N}_{a+1} {}\\ y(a)& =& 0. {}\\ \end{array}$$

If y(t) is the solution of this IVP and its Laplace transform, Y a (s), exists, then we have that

 $$\displaystyle{ sY _{a}(s) - y(a) - 3Y _{a}(s) = \frac{1} {s - 4}. }$$

Using the initial condition and solving for Y a (s) we obtain

 $$\displaystyle{ Y _{a}(s) = \frac{1} {s - 4} \frac{1} {s - 3}. }$$

Using the Nabla Convolution Theorem we see that

 $$\displaystyle{y(t) = \left (E_{4}(\cdot,a) {\ast} E_{3}(\cdot,a)\right )(t).}$$

Using Example 3.79 we find that

 $$\displaystyle\begin{array}{rcl} y(t)& =& E_{4}(t,a) - E_{3}(t,a) {}\\ & =& (-3)^{a-t} - (-2)^{a-t} {}\\ \end{array}$$

is the solution of our given IVP on  $$\mathbb{N}_{a}$$ . Of course, in this simple example one could also use partial fractions to find y(t).

We can then generalize this result for an arbitrary number of nabla differences.

Theorem 3.85 (Transform of n-th-Order Nabla Difference).

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

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

(3.38)

for |s − 1| < r, for each  $$n \in \mathbb{N}_{1}$$ .

Proof.

Note that since f is of exponential order r > 0, ∇ n f is of exponential order r > 0 for each  $$n \in \mathbb{N}_{1}$$ . Hence  $$\mathcal{L}_{a}\{\nabla ^{n}f\}(s)$$ converges for | s − 1 |  < r for each  $$n \in \mathbb{N}_{1}$$ . The proof of (3.38) is by induction for  $$n \in \mathbb{N}_{1}$$ . The base case n = 1 follows from Theorem 3.83. Now assume n ≥ 1 and (3.38) holds for | s − 1 |  < r. Then, using Theorem 3.83, we have that

 $$\displaystyle\begin{array}{rcl} \mathcal{L}_{a}\{\nabla ^{n+1}f\}(s)& =& \mathcal{L}_{ a}\{\nabla \left [\nabla ^{n}f\right ]\}(s) {}\\ & =& s\mathcal{L}_{a}\{\nabla ^{n}f\}(s) -\nabla ^{n}f(a) {}\\ & =& s\left [s^{n}\mathcal{L}_{ a}\{f\}(s) -\sum _{k=1}^{n}s^{n-k}\nabla ^{k-1}f(a)\right ] -\nabla ^{n}f(a) {}\\ & =& s^{n+1}\mathcal{L}_{ a}\{f\}(s) -\sum _{k=1}^{n+1}s^{(n+1)-k}\nabla ^{k-1}f(a). {}\\ \end{array}$$

Hence, (3.38) holds when n is replaced by n + 1 and the proof is complete. □ 

Example 3.86.

Solve the IVP

 $$\displaystyle\begin{array}{rcl} \nabla ^{2}y(t) - 6\nabla y(t) + 8y(t)& =& 0,\quad t \in \mathbb{N}_{ a+1} {}\\ y(a) = 1,\quad \nabla y(a)& =& -1. {}\\ \end{array}$$

If y(t) is the solution of this IVP and we let  $$Y _{a}(s):= \mathcal{L}_{a}\{y\}(s),$$ we have that

 $$\displaystyle{ \left [s^{2}Y _{ a}(s) - sy(a) -\nabla y(a)\right ] - 6\left [sY _{a}(s) - y(a)\right ] + 8Y _{a}(s) = 0. }$$

Using the initial conditions we have

 $$\displaystyle{\left [s^{2}Y _{ a}(s) - s + 1\right ] - 6\left [sY _{a}(s) - 1\right ] + 8Y _{a}(s) = 0.}$$

Solving for Y a (s) we have that

 $$\displaystyle\begin{array}{rcl} Y _{a}(s)& =& \frac{s - 7} {s^{2} - 6s + 8} {}\\ & =& \frac{s - 7} {(s - 2)(s - 4)} {}\\ & =& \frac{5} {2} \frac{1} {s - 2} -\frac{3} {2} \frac{1} {s - 4}. {}\\ \end{array}$$

It follows that

 $$\displaystyle\begin{array}{rcl} y(t)& =& \frac{5} {2}E_{2}(t,a) -\frac{3} {2}E_{4}(t,a) {}\\ & =& \frac{5} {2}(-1)^{a-t} -\frac{3} {2}(-3)^{a-t} {}\\ \end{array}$$

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