Discrete Fractional Calculus (2015)
3. Nabla Fractional Calculus
3.10. Nabla Laplace Transforms
Having established the necessary preliminaries, we are now ready to discuss an important application of this material: the Laplace transform. The Laplace transform, as in the standard calculus, will provide us with an elegant way to solve initial value problems for a fractional nabla difference equation. In this section, we will lay the groundwork for this method, prove the basic properties, and establish a means in which to solve various initial value (nabla) fractional difference equations. We begin this section by defining the nabla Laplace transform operator (based at a) as follows:
Definition 3.64.
Assume . Then the nabla Laplace transform of f is defined by
for those values of s ≠ 1 such that this improper integral converges.
In the following theorem we give another formula for the Laplace transform, which is often more convenient to use.
Theorem 3.65.
Assume . Then
(3.36)
for those values of s such that this infinite series converges.
Proof.
Assume . Then
for those values of s such that this infinite series converges. □
In the definition of the nabla Laplace transform we assumed s ≠ 1 because we do not define . But the formula of the nabla Laplace transform (3.36) is well defined when s = 1. From now on we will always include s = 1 in the domain of convergence for the nabla Laplace transform although in the proofs we will often assume s ≠ 1. In fact the formula (3.36) for any gives us that
Example 3.66.
We use the last theorem to find . By Theorem 3.65 we obtain
That is
Theorem 3.67.
For all nonnegative integers n, we have that
Proof.
The proof is by induction on n. The result is true for n = 0 by the previous example. Suppose now that for some fixed n ≥ 0 and | s − 1 | < 1. Then consider
We will apply the first integration by parts formula (3.19) with
It follows that
Hence by the integration by parts formula (3.19)
Using the nabla form of L’Hôpital’s rule (Exercise 3.19) we calculate
since | s − 1 | < 1. Thus we have that
completing the proof. □
Definition 3.68.
A function is said to be of exponential order r > 0 if there exist a constant M > 0 and a number such that
Theorem 3.69.
For , the Taylor monomials H n (t,a) are of exponential order 1 + ε for all ε > 0. Also, H 0 (t,a) is of exponential order 1.
Proof.
Since
H 0(t, a) is of exponential order 1. Next, assume and ε > 0 is fixed. Using repeated applications of the nabla L’Hôpital’s rule, we get
It follows from this that each H n (t, a), , is of exponential order 1 +ε for all ε > 0. □
Theorem 3.70 (Existence of Nabla Laplace Transform).
If is a function of exponential order r > 0, then its Laplace transform exists for .
Proof.
Let f be a function of exponential order r. Then there is a constant M > 0 and a number such that | f(t) | ≤ Mr t for all . Pick K so that , then we have that
We now show that
converges for . To see this, consider
which converges since r | s − 1 | < 1. It follows that converges absolutely for . □
Theorem 3.71.
The Laplace transform of the Taylor monomial, H n (t,a), , exists for |s − 1| < 1.
Proof.
The proof of this theorem follows from Theorems 3.69 and 3.70. □
Similarly, by Exercise 3.30 each of the functions E p (t, a), Cosh p (t, a), Sinh p (t, a), Cos p (t, a), and Sin p (t, a) is of exponential order | 1 + p | , and hence by Theorem 3.70 their Laplace transforms exist for .
Theorem 3.72 (Uniqueness Theorem).
Assume . Then f(t) = g(t), if and only if
for some r > 0.
Proof.
Since is a linear operator it suffices to show that f(t) = 0 for if and only if for | s − 1 | < r for some r > 0. If f(t) = 0 for , then trivially for all . Conversely, assume that for | s − 1 | < r for some r > 0. In this case we have that
This implies that
This completes the proof. □