Using Laplace Transforms to Solve Fractional Equations - Discrete Delta Fractional Calculus and Laplace Transforms - Discrete Fractional Calculus

Discrete Fractional Calculus (2015)

2. Discrete Delta Fractional Calculus and Laplace Transforms

2.7. Using Laplace Transforms to Solve Fractional Equations

When solving certain summation equations one uses the formula

$$\displaystyle{ \mathcal{L}_{a}\left \{\Delta _{a}^{-N}f\right \}(s) = \frac{F_{a}(s)} {s^{N}}, }$$

(2.46)

where N is a positive integer. Since the summation equation (2.5) can be written in the form

$$\displaystyle{ y(t) = 2 \cdot 4^{t} + 2\int _{ 0}^{t}y(s)\ \Delta s,\quad t \in \mathbb{N}_{ 0}, }$$

this is an example of a summation equation for which we want to use the formula (2.46) with N = 1.

We will now set out to generalize formulas (2.4) and (2.46) to the fractional case so that we can solve fractional difference and summation equations using Laplace transforms.

We will show (see Theorem 2.65) that if $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ is of exponential order, then $$\Delta _{a}^{-\nu }f$$ and $$\Delta _{a}^{\nu }f$$ are of a certain exponential order and hence their Laplace transforms will exist. We will use the following lemma, which gives an estimate for t ν in the proof of Theorem 2.65.

Lemma 2.63.

Assume ν > −1 and N − 1 < ν ≤ N. Then

$$\displaystyle{ t^{\underline{\nu }} \leq t^{N},\quad \mbox{ for $t$ sufficiently large}. }$$

(2.47)

Proof.

In this proof we use the fact that $$\Gamma (x) > 0$$ for x > 0 and $$\Gamma (x)$$ is strictly increasing for x ≥ 2. First consider the case − 1 < ν ≤ 0. Then, since $$t + 1-\nu \geq t + 1$$ , we have for large t

$$\displaystyle\begin{array}{rcl} t^{\underline{\nu }}& =& \frac{\Gamma (t + 1)} {\Gamma (t + 1-\nu )} {}\\ & \leq & 1 = t^{0} = t^{N}. {}\\ \end{array}$$

Next, consider the case ν > 0. Then for large t we have

$$\displaystyle{ t^{\underline{\nu }} = \frac{\Gamma (t + 1)} {\Gamma (t + 1-\nu )} \leq \frac{\Gamma (t + 1)} {\Gamma (t + 1 - N)} = t\left (t - 1\right )\cdots \left (t -\left (N - 1\right )\right ) \leq t^{N}. }$$

This completes the proof. □ 

Remark 2.64.

Thus far whenever we have considered a function $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ , we have always taken the domain of $$\Delta _{a}^{-\nu }f$$ to be the set $$\mathbb{N}_{a+\nu }$$ . However, it is sometimes convenient to take the domain of $$\Delta _{a}^{-\nu }f$$ to be the set $$\mathbb{N}_{a+\nu -N}$$ , where ν > 0, and N − 1 < ν ≤ N. By our convention on sums we see that

$$\displaystyle{\Delta _{a}^{-\nu }f(a +\nu -N + k) = 0,\quad \mbox{ for}\quad 0 \leq k \leq N - 1.}$$

Later (see, for example, Theorem 2.67) we will consider both of the

$$\displaystyle{\mathcal{L}_{a+\nu }\{\Delta _{a}^{-\nu }f\}(s)\quad \mbox{ and}\quad \mathcal{L}_{ a+\nu -N}\{\Delta _{a}^{-\nu }f\}(s).}$$

Note that $$\Delta _{a}^{-\nu }f: \mathbb{N}_{a+\nu } \rightarrow \mathbb{R}$$ and $$\Delta _{a}^{-\nu }f: \mathbb{N}_{a+\nu -N} \rightarrow \mathbb{R}$$ are of the same exponential order. Theorem 2.67 will give a relationship between these two Laplace transforms.

Theorem 2.65.

Suppose that $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ is of exponential order r ≥ 1, and let ν > 0, N − 1 < ν ≤ N, be given. Then for each fixed ε > 0, $$\Delta _{a}^{-\nu }f: \mathbb{N}_{a+\nu } \rightarrow \mathbb{R}$$ , $$\Delta _{a}^{-\nu }f: \mathbb{N}_{a+\nu -N} \rightarrow \mathbb{R}$$ , and $$\Delta _{a}^{\nu }f: \mathbb{N}_{a+N-\nu }\rightarrow \mathbb{R}$$ are of exponential order r + ε.

Proof.

First we show if $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ is of exponential order r = 1, then $$\Delta _{a}^{-\nu }f: \mathbb{N}_{a+\nu } \rightarrow \mathbb{R}$$ is of exponential order $$r = 1+\epsilon,$$ for each ε > 0. By Exercise 2.1 it suffices to show that f is bounded on $$\mathbb{N}_{a}$$ implies $$\Delta _{a}^{-\nu }f: \mathbb{N}_{a+\nu } \rightarrow \mathbb{R}$$ is of exponential order $$r = 1+\epsilon,$$ for each ε > 0. To this end assume

$$\displaystyle{\vert f(t)\vert \leq N,\quad t \in \mathbb{N}_{a}.}$$

Then, for $$t \in \mathbb{N}_{a+\nu },$$

$$\displaystyle\begin{array}{rcl} \vert \Delta _{a}^{-\nu }f(t)\vert & =& \left \vert \int _{ a}^{t-\nu +1}h_{\nu -1}(t,\sigma (s))f(s)\Delta s\right \vert {}\\ &\leq & \int _{a}^{t-\nu +1}h_{\nu -1}(t,\sigma (s))\vert f(s)\vert \Delta s {}\\ & \leq & N\int _{a}^{t-\nu +1}h_{\nu -1}(t,\sigma (s))\Delta s {}\\ & =& -Nh_{\nu }(t,s)\vert _{s=a}^{s=t-\nu +1},\quad \quad \quad \quad \quad \mbox{ by Theorem <InternalRef RefID="FPar27">2.27</InternalRef>7, part (v)} {}\\ & =& -Nh_{\nu }(t,t -\nu +1) + Nh_{\nu }(t,a) {}\\ & =& Nh_{\nu }(t,a). {}\\ \end{array}$$

Since, by Theorem 2.56, h ν (t, a) is of exponential order 1 +ε for each ε > 0, it follows that $$\Delta _{a}^{-\nu }f: \mathbb{N}_{a+\nu } \rightarrow \mathbb{R}$$ is of exponential order 1 +ε, for each ε > 0.

Next assume f is of exponential order r > 1, there exist an A > 0 and a $$T \in \mathbb{N}_{a}$$ such that

$$\displaystyle{ \left \vert f(t)\right \vert \leq Ar^{t},\quad \text{ for all }\quad t \in \mathbb{N}_{ T}. }$$

(2.48)

For $$t \in \mathbb{N}_{T+\nu }$$ , sufficiently large, consider

$$\displaystyle\begin{array}{rcl} & & \left \vert \Delta _{a}^{-\nu }f(t)\right \vert = \left \vert \sum _{ s=a}^{t-\nu }h_{\nu -1}(t,\sigma (s))f(s)\right \vert {}\\ & &\leq \sum _{s=a}^{t-\nu }h_{\nu -1}(t,\sigma (s))\vert f(s)\vert {}\\ & & =\sum _{ s=a}^{T-1}h_{\nu -1}(t,\sigma (s))\vert f(s)\vert +\sum _{ s=T}^{t-\nu }h_{\nu -1}(t,\sigma (s))\vert f(s)\vert {}\\ & &\leq \left (\sum _{s=a}^{T-1}\frac{\vert f(s)\vert } {\Gamma (\nu )} \right )(t - a)^{N-1} + \frac{A(t - a)^{N-1}} {\Gamma (\nu )} \int _{T}^{t-\nu +1}r^{s}\Delta s {}\\ & & = \left (\sum _{s=a}^{T-1}\frac{\vert f(s)\vert } {\Gamma (\nu )} \right )(t - a)^{N-1} + \frac{A(t - a)^{N-1}} {\Gamma (\nu )} \left [ \frac{r^{s}} {r - 1}\right ]_{s=T}^{s=t-\nu +1} {}\\ & & = \left (\sum _{s=a}^{T-1}\frac{\vert f(s)\vert } {\Gamma (\nu )} \right )(t - a)^{N-1} + \frac{A(t - a)^{N-1}} {(r - 1)\Gamma (\nu )} [r^{t-\nu +1} - r^{T}] {}\\ & & \leq \left (\sum _{s=a}^{T-1}\frac{\vert f(s)\vert } {\Gamma (\nu )} \right )(t - a)^{N-1} + \frac{A(t - a)^{N-1}r^{1-\nu }} {(r - 1)\Gamma (\nu )} r^{t} {}\\ & & = B(t - a)^{N-1} + C(t - a)^{N-1}r^{t}, {}\\ \end{array}$$

where B and C are constants. But for any fixed ε > 0 we get by applying L’Hôpital’s rule, that

$$\displaystyle{\lim _{t\rightarrow \infty }\frac{B(t - a)^{N-1} + C(t - a)^{N-1}r^{t}} {(r+\epsilon )^{t}} = 0.}$$

Therefore, $$\Delta _{a}^{-\nu }f: \mathbb{N}_{a+\nu } \rightarrow \mathbb{R}$$ is of exponential order r +ε for each fixed ε > 0. By Remark 2.64, we also have $$\Delta _{a}^{-\nu }f: \mathbb{N}_{a+\nu -N} \rightarrow \mathbb{R}$$ is of exponential order r +ε for each fixed ε > 0.

Finally, we show $$\Delta _{a}^{\nu }f: \mathbb{N}_{a+N-\nu }\rightarrow \mathbb{R}$$ , where N − 1 < ν ≤ N, is of exponential order r +ε for each fixed ε > 0. Since

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

and by the first part of the proof, $$\Delta _{a}^{-(N-\nu )}f(t)$$ is of exponential order r +ε, we have by Exercise 2.2 that $$\Delta _{a}^{\nu }f$$ is of exponential order r +ε. □ 

Corollary 2.66.

Suppose that $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ is of exponential order r ≥ 1 and let ν > 0 be given with N − 1 < ν ≤ N. Then

$$\displaystyle{ \mathcal{L}_{a+\nu }\left \{\Delta _{a}^{-\nu }f\right \}(s),\quad \mathcal{L}_{ a+\nu -N}\left \{\Delta _{a}^{-\nu }f\right \}(s),\quad \text{ and }\quad \mathcal{L}_{ a+N-\nu }\left \{\Delta _{a}^{\nu }f\right \}(s) }$$

converge for all |s + 1| > r.

Proof.

Suppose f, r, and ν are as in the statement of this corollary and fix s 0 so that | s 0 + 1 |  > r. Then there is an ε 0 > 0 so that $$\vert s_{0} + 1\vert > r +\epsilon _{0}.$$ Since we know by Theorem 2.65 that $$\Delta _{a}^{-\nu }f: \mathbb{N}_{a+\nu } \rightarrow \mathbb{R}$$ , $$\Delta _{a}^{-\nu }f: \mathbb{N}_{a+\nu -N} \rightarrow \mathbb{R}$$ , and $$\Delta _{a}^{\nu }f: \mathbb{N}_{a+N-\nu }\rightarrow \mathbb{R}$$ are of exponential order r +ε 0, it follows from Theorem 2.4 that $$\mathcal{L}_{a+\nu }\left \{\Delta _{a}^{-\nu }f\right \}(s_{0})$$ , $$\mathcal{L}_{a+\nu -N}\left \{\Delta _{a}^{-\nu }f\right \}\left (s_{0}\right )$$ , and $$\mathcal{L}_{a+N-\nu }\left \{\Delta _{a}^{\nu }f\right \}\left (s_{0}\right )$$ converge. Since | s 0 + 1 |  > r is arbitrary, we have that

$$\displaystyle{ \mathcal{L}_{a+\nu }\left \{\Delta _{a}^{-\nu }f\right \}(s),\quad \mathcal{L}_{ a+\nu -N}\left \{\Delta _{a}^{-\nu }f\right \}(s),\quad \text{ and }\quad \mathcal{L}_{ a+N-\nu }\left \{\Delta _{a}^{\nu }f\right \}(s) }$$

all converge for all | s + 1 |  > r.  □