Composition Rules - Discrete Delta Fractional Calculus and Laplace Transforms - Discrete Fractional Calculus

Discrete Fractional Calculus (2015)

2. Discrete Delta Fractional Calculus and Laplace Transforms

2.5. Composition Rules

Theorem 2.46 (Composition of Fractional Sums).

Assume f is defined on  $$\mathbb{N}_{a}$$ and μ, ν are positive numbers. Then

 $$\displaystyle{\left [\Delta _{a+\nu }^{-\mu }\left (\Delta _{ a}^{-\nu }f\right )\right ](t) = \left (\Delta _{ a}^{-(\mu +\nu )}f\right )(t) = \left [\Delta _{ a+\mu }^{-\nu }\left (\Delta _{ a}^{-\mu }f\right )\right ](t)}$$

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

Proof.

For  $$t \in \mathbb{N}_{a+\mu +\nu }$$ , consider

 $$\displaystyle\begin{array}{rcl} & & \left [\Delta _{a+\nu }^{-\mu }\left (\Delta _{ a}^{-\nu }f\right )\right ](t) =\sum _{ s=a+\nu }^{t-\mu }h_{\mu -1}(t,\sigma (s))\left (\Delta _{a}^{-\nu }f\right )(s) {}\\ & =& \sum _{s=a+\nu }^{t-\mu }h_{\mu -1}(t,\sigma (s))\sum _{r=a}^{s-\nu }h_{\nu -1}(s,\sigma (r))f(r) {}\\ & =& \frac{1} {\Gamma (\mu )\Gamma (\nu )}\sum _{s=a+\nu }^{t-\mu }\sum _{ r=a}^{s-\nu }(t -\sigma (s))^{\underline{\mu -1}}(s -\sigma (r))^{\underline{\nu -1}}f(r) {}\\ & =& \frac{1} {\Gamma (\mu )\Gamma (\nu )}\sum _{r=a}^{t-(\mu +\nu )}\sum _{ s=r+\nu }^{t-\mu }(t -\sigma (s))^{\underline{\mu -1}}(s -\sigma (r))^{\underline{\nu -1}}f(r), {}\\ \end{array}$$

where in the last step we interchanged the order of summation. Letting  $$x = s -\sigma (r)$$ we obtain

 $$\displaystyle\begin{array}{rcl} & & \left [\Delta _{a+\nu }^{-\mu }\left (\Delta _{ a}^{-\nu }f\right )\right ](t) {}\\ & =& \frac{1} {\Gamma (\mu )\Gamma (\nu )}\sum _{r=a}^{t-(\mu +\nu )}\left [\sum _{ x=\nu -1}^{t-\mu -r-1}(t - x - r - 2)^{\underline{\mu -1}}x^{\underline{\nu -1}}\right ]f(r) {}\\ & =& \frac{1} {\Gamma (\nu )}\sum _{r=a}^{t-(\mu +\nu )}\left [ \frac{1} {\Gamma (\mu )}\sum _{x=\nu -1}^{(t-r-1)-\mu }(t - r - 1 -\sigma (x))^{\underline{\mu -1}}x^{\underline{\nu -1}}\right ]f(r) {}\\ & =& \frac{1} {\Gamma (\nu )}\sum _{r=a}^{t-(\mu +\nu )}\left [\Delta _{\nu -1}^{-\mu }t^{\underline{\nu -1}}\right ]_{ t\rightarrow t-r-1}f(r). {}\\ \end{array}$$

But by Theorem 2.38

 $$\displaystyle{\Delta _{\nu -1}^{-\mu }t^{\underline{\nu -1}} = \frac{\Gamma (\nu )} {\Gamma (\nu +\mu )}t^{\underline{\mu +\nu -1}}}$$

and therefore

 $$\displaystyle\begin{array}{rcl} \left [\Delta _{a+\nu }^{-\mu }\left (\Delta _{ a}^{-\nu }f\right )\right ](t)& =& \frac{1} {\Gamma (\nu )}\sum _{r=a}^{t-(\mu +\nu )} \frac{\Gamma (\nu )} {\Gamma (\mu +\nu )}(t - r - 1)^{\underline{\mu +\nu -1}}f(r) {}\\ & =& \frac{1} {\Gamma (\mu +\nu )}\sum _{r=a}^{t-(\mu +\nu )}(t -\sigma (r))^{\underline{\mu +\nu -1}}f(r) {}\\ & =& \left (\Delta _{a}^{-(\mu +\nu )}f\right )(t), {}\\ \end{array}$$

 $$t \in \mathbb{N}_{a+\nu +\mu },$$ which is one of the desired conclusions. Interchanging μ and ν in the above formula we also get the result

 $$\displaystyle{\left [\Delta _{a+\mu }^{-\nu }\left (\Delta _{ a}^{-\mu }f\right )\right ](t) = \left (\Delta _{ a}^{-(\mu +\nu )}f\right )(t)}$$

for  $$t \in \mathbb{N}_{a+\mu +\nu }.$$  □ 

In the next lemma we give composition rules for an integer difference with a fractional sum and with a fractional difference. Atici and Eloe proved (2.26) with the additional assumption that 0 < k < ν and Holm [123, 125] proved (2.26) in this more general setting.

Lemma 2.47.

Assume  $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ , ν > 0, N − 1 < ν ≤ N. Then

 $$\displaystyle{ \left [\Delta ^{k}\left (\Delta _{ a}^{-\nu }f\right )\right ](t) = \left (\Delta _{ a}^{k-\nu }f\right )(t),\qquad t \in \mathbb{N}_{ a+\nu }. }$$

(2.26)

and

 $$\displaystyle{ \left [\Delta ^{k}\left (\Delta _{ a}^{\nu }f\right )\right ](t) = \left (\Delta _{ a}^{k+\nu }f\right )(t),\qquad t \in \mathbb{N}_{ a+N-\nu }. }$$

(2.27)

Proof.

First we prove that

 $$\displaystyle{ \left [\Delta ^{k}\left (\Delta _{ a}^{-k}f\right )\right ](t) = f(t),\qquad t \in \mathbb{N}_{ a+k} }$$

(2.28)

by induction for  $$k \in \mathbb{N}_{1}$$ . For the base case we have

 $$\displaystyle{\Delta \Delta _{a}^{-1}f(t) = \Delta \bigg[\int _{ a}^{t}f(\tau )\Delta \tau \bigg] = f(t)}$$

for  $$t \in \mathbb{N}_{a+1}.$$ Now assume k ≥ 1 and (2.28) holds. Then

 $$\displaystyle\begin{array}{rcl} \Delta ^{k+1}\Delta _{ a}^{k+1}f(t)& & = \Delta ^{k+1}\Delta _{ a+k}^{-1}\Delta _{ a}^{-k}f(t)\qquad \mbox{ using Theorem <InternalRef RefID="FPar4">2.4</InternalRef>6} {}\\ & & = \Delta ^{k}[\Delta \Delta _{ a+k}^{-1}]\Delta _{ a}^{-k}f(t) {}\\ & & = \Delta ^{k}\Delta _{ a}^{-k}f(t)\qquad \mbox{ by the base case with base $a + k$} {}\\ & & = f(t)\qquad \mbox{ by the induction assumption (<InternalRef RefID="Equ88">2.28</InternalRef>)} {}\\ \end{array}$$

for  $$t \in \mathbb{N}_{a+k+1}.$$ Therefore, for k ≥ N

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

and for k < N

 $$\displaystyle{\Delta ^{k}\Delta _{ a}^{-N}f(t) = \Delta ^{k}\Delta _{ a+N-k}^{-k}[\Delta _{ a}^{-(N-k)}]f(t) = \Delta _{ a}^{-(N-k)}f(t) = \Delta _{ a}^{k-N}f(t)}$$

for  $$t \in \mathbb{N}_{a+N}.$$ Hence for all  $$k \in \mathbb{N}_{1}$$ we have that (2.26) holds for the case ν = N. It is also true that (2.27) holds when ν = N. Assume for the rest of this proof that N − 1 < ν < N. We will now show by induction that (2.27) holds for  $$k \in \mathbb{N}_{1}.$$ For the base case k = 1 we have using the Leibniz rule (2.11)

 $$\displaystyle\begin{array}{rcl} & & \Delta \Delta _{a}^{\nu }f(t) {}\\ & & = \Delta \bigg[\int _{a}^{t+\nu +1}h_{ -\nu -1}(t,\sigma (\tau ))f(\tau )\Delta \tau \bigg] {}\\ & & =\int _{ a}^{t+\nu +1}h_{ -\nu -2}(t,\sigma (\tau ))f(\tau )\Delta \tau + h_{\nu -1}(\sigma (t),t +\nu +1)f(t +\nu +1) {}\\ & & =\int _{ a}^{t+\nu +1}h_{ -\nu -2}(t,\sigma (\tau ))f(\tau )\Delta \tau + f(t +\nu +1) {}\\ & & =\int _{ a}^{t+\nu +2}h_{ -\nu -2}(t,\sigma (\tau ))f(\tau )\Delta \tau {}\\ & & = \Delta _{a}^{-(-\nu -1)}f(t) {}\\ & & = \Delta _{a}^{1+\nu }f(t). {}\\ \end{array}$$

Hence the base case

 $$\displaystyle{\Delta \Delta _{a}^{\nu }f(t) = \Delta _{ a}^{1+\nu }f(t)}$$

holds. Now assume k ≥ 1 and

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

(2.29)

holds. It follows from the induction hypothesis (2.29) and the base case that

 $$\displaystyle\begin{array}{rcl} \Delta ^{k+1}\Delta _{ a}^{\nu }f(t)& & = \Delta \Delta ^{k}\Delta _{ a}^{1+\nu }f(t) {}\\ & & = \Delta \Delta _{a}^{k+\nu }f(t) {}\\ & & = \Delta _{a}^{k+1+\nu }f(t). {}\\ \end{array}$$

Hence (2.27) holds for all  $$k \in \mathbb{N}_{1}.$$ The proof of (2.26) is very similar and is left as an exercise (Exercise 2.23). □ 

We now prove a composition rule that appears in Holm [125] for a fractional difference with a fractional sum.

Theorem 2.48.

Assume  $$f: \mathbb{N}_{a} \rightarrow \mathbb{R},$$ ν,μ > 0 and N − 1 < ν ≤ N,  $$N \in \mathbb{N}_{1}.$$ Then

 $$\displaystyle{ \Delta _{a+\mu }^{\nu }\Delta _{ a}^{-\mu }f(t) = \Delta _{ a}^{\nu -\mu }f(t),\qquad t \in \mathbb{N}_{ a+\mu +N-\nu }. }$$

(2.30)

Proof.

Note that for  $$t \in \mathbb{N}_{a+\mu +N-\nu }$$ ,

 $$\displaystyle\begin{array}{rcl} \Delta _{a+\mu }^{\nu }\Delta _{ a}^{-\mu }f(t)& =& \Delta ^{N}\Delta _{ a+\mu }^{-(N-\nu )}\Delta _{ a}^{-\mu }f(t) {}\\ & =& \Delta ^{N}\Delta _{ a}^{-(N-\nu +\mu )}f(t)\qquad \mbox{ by Theorem <InternalRef RefID="FPar4">2.4</InternalRef>6} {}\\ & =& \Delta _{a}^{N-(N-\nu +\mu )}f(t)\qquad \mbox{ by (<InternalRef RefID="Equ86">2.26</InternalRef>)} {}\\ & =& \Delta _{a}^{\nu -\mu }f(t). {}\\ \end{array}$$

Hence (2.30) holds. □ 

Remark 2.49.

From Theorem 2.46 we saw that we can take fractional sums of fractional sums by adding exponents and by Theorem 2.48 we can take fractional differences of fractional sums by adding exponents. The fundamental theorem of calculus gives us that

 $$\displaystyle{\Delta _{a}^{-1}\Delta f(\tau ) =\int _{ a}^{t}\Delta f(\tau ) = f(t) - f(a) = \Delta _{ a}^{0}f(t) - f(a).}$$

Hence we should not expect the fractional sum of a fractional difference can be obtained by adding exponents.

In the next theorem we give a formula for a fractional sum of an integer difference. The first formula in the following Theorem 2.50 is given in Atici et al. [34] and the second formula appears in Holm [125].

Theorem 2.50.

Assume  $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ ,  $$k \in \mathbb{N}_{0}$$ and ν,μ > 0 with N − 1 < μ ≤ N. Then

 $$\displaystyle{ \Delta _{a}^{-\nu }\Delta ^{k}f(t) = \Delta _{ a}^{k-\nu }f(t) -\sum \limits _{ j=0}^{k-1}h_{\nu -k+j}(t,a)\Delta ^{j}f(a), }$$

(2.31)

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

 $$\displaystyle\begin{array}{rcl} \Delta _{a+N-\mu }^{-\nu }\Delta _{ a}^{\mu }f(t)& & = \Delta _{ a}^{\mu -\nu }f(t) \\ & & \quad \quad -\sum _{j=0}^{N-1}h_{\nu -N+j}(t - N+\nu,a)\Delta _{a}^{j-(N-\mu )}f(a + N-\mu ),{}\end{array}$$

(2.32)

for  $$t \in \mathbb{N}_{a+N-\mu +\nu }.$$

Proof.

We first prove that (2.31) holds by induction for  $$k \in \mathbb{N}_{1}.$$ For the base case k = 1 we have using integration by parts and

 $$\displaystyle{h_{\nu -1}(t,t -\nu +1) = 1 = h_{\nu -2}(t,t -\nu +2)}$$

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

 $$\displaystyle\begin{array}{rcl} \Delta _{a}^{-\nu }\Delta f(t)& & =\int _{ a}^{t-\nu +1}h_{\nu -1}(t,\sigma (\tau ))\Delta f(\tau )\Delta \tau {}\\ & & = h_{\nu -1}(t,\tau )f(t)\bigg\vert _{\tau =a}^{t-\nu +1} +\int _{ a}^{t-\nu +1}h_{\nu -2}(t,\sigma (\tau ))f(\tau )\Delta \tau {}\\ & & = h_{\nu -1}(t,t -\nu +1)f(t -\nu +1) - h_{\nu -1}(t,a)f(a) {}\\ & & \qquad \qquad \qquad \qquad +\int _{ a}^{t-\nu +1}h_{\nu -2}(t,\sigma (\tau ))\Delta \tau {}\\ & & = f(t -\nu +1) - h_{\nu -1}(t,a)f(a) +\int _{ a}^{t-\nu +1}h_{\nu -2}(t,\sigma (\tau ))f(\tau )\Delta \tau {}\\ & & =\int _{ a}^{t-\nu +2}h_{\nu -2}(t,\sigma (\tau ))f(\tau )\Delta \tau - h_{\nu -1}(t,a)f(a) {}\\ & & = \Delta _{a}^{1-\nu }f(t) - h_{\nu -1}(t,a)f(a) {}\\ \end{array}$$

which proves (2.31) for the base case k = 1. Now assume k ≥ 1 and (2.31) holds for that k. Then we have that

 $$\displaystyle\begin{array}{rcl} \Delta _{a}^{-\nu }\Delta ^{k+1}f(t)& & = \Delta _{ a}^{-\nu }\Delta ^{k}\Delta f(t) {}\\ & & = \Delta _{a}^{k-\nu }\Delta f(t) -\sum _{ j=0}^{k-1}h_{\nu -k-j}(t,a)\Delta ^{j+1}f(a)\quad (\mbox{ by (<InternalRef RefID="Equ95">2.31</InternalRef>)}) {}\\ & & = \Delta _{a}^{k-\nu }f(t) -\sum _{ j=0}^{k-1}h_{\nu -k+j}(t,a)\Delta ^{j+1}f(a) - h_{\nu -k-1}(t,a)f(a) {}\\ & & = \Delta _{a}^{-\nu }\Delta _{ a}^{k+1-\nu }f(t) -\sum \limits _{ j=0}^{k}h_{\nu -k-1+j}(t,a)\Delta ^{j}f(a), {}\\ \end{array}$$

for  $$t \in \mathbb{N}_{a+N-\nu }.$$ Hence (2.31) holds. Next we show that (2.32) holds. To see this suppose now that ν > 0 and μ > 0 with N − 1 < μ ≤ N. Letting  $$g(t) = \Delta _{a}^{-(N-\mu )}f(t)$$ and  $$b = a + N-\mu$$ (the first point in the domain of g), we have for  $$t \in \mathbb{N}_{a+N-\mu +\nu },$$

 $$\displaystyle\begin{array}{rcl} & & \Delta _{a+N-\mu }^{-\nu }\Delta _{ a}^{\mu }f(t) {}\\ & & = \Delta _{a+N-\mu }^{-\nu }\Delta ^{N}\left (\Delta _{ a}^{-(N-\mu )}f(t)\right ) {}\\ & & = \Delta _{a+N-\mu }^{-\nu }\Delta ^{N}g(t) {}\\ & & = \Delta _{a+N-\mu }^{N-\nu }g(t) -\sum \limits _{ j=0}^{N-1}h_{\nu -N+j}(t,b)\Delta ^{j}g(b)\quad \quad \text{by (<InternalRef RefID="Equ96">2.32</InternalRef>)} {}\\ & & = \Delta _{a+N-\mu }^{N-\nu }\Delta _{ a}^{-(N-\mu )}f(t) -\sum \limits _{ j=0}^{N-1}h_{\nu -N+j}(t,b)\Delta ^{j}\Delta _{ a}^{-(N-\mu )}f(b) {}\\ & & = \Delta _{a}^{\mu -\nu }f(t) -\sum \limits _{ j=0}^{N-1}h_{\nu -N+j}(t - N+\nu,a)\Delta _{a}^{j-N+\mu }f(a + N-\mu ), {}\\ \end{array}$$

where in this last step, we applied 2.31. □ 

Finally, we give a composition formula for composing two fractional differences. Note that the rule for this composition is nearly identical to the rule (2.32) for the composition  $$\Delta _{a+M-\mu }^{-\nu }\Delta _{a}^{\mu }.$$ Theorem 2.51 is given for the specific case  $$\mu \in \mathbb{N}_{0}$$ by Atici and Eloe in [34].

Theorem 2.51.

Let  $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ be given and suppose ν,μ > 0, with N − 1 < ν ≤ N and M − 1 < μ ≤ M. Then for  $$t \in \mathbb{N}_{a+M-\mu +N-\nu },$$

 $$\displaystyle\begin{array}{rcl} & & \Delta _{a+M-\mu }^{\nu }\Delta _{ a}^{\mu }f(t) = \Delta _{ a}^{\nu +\mu }f(t) - \\ & &\qquad \qquad \qquad \sum _{j=0}^{M-1}h_{ -\nu -M+j}(t - M+\mu,a)\Delta _{a}^{j-M+\mu }f(a + M-\mu ){}\end{array}$$

(2.33)

for N − 1 < ν < N. If ν = N, then ( 2.33) simplifies to

 $$\displaystyle{\Delta _{a+M-\mu }^{\nu }\Delta _{ a}^{\mu }f(t) = \Delta _{ a}^{\nu +\mu }f(t),\quad t \in \mathbb{N}_{ a+M-\mu }.}$$

Proof.

Let f, ν, and μ  be given as in the statement of the theorem. Lemma 2.47 has already proven the case when ν = N. 

If N − 1 < ν < N, then for  $$t \in \mathbb{N}_{a+M-\mu +N-\nu }$$ , we have

 $$\displaystyle\begin{array}{rcl} & & \Delta _{a+M-\mu }^{\nu }\Delta _{ a}^{\mu }f(t) {}\\ & & = \Delta ^{N}\left [\Delta _{ a+M-\mu }^{-(N-\nu )}\Delta _{ a}^{\mu }f(t)\right ]\text{, and now using (<InternalRef RefID="Equ137">2.50</InternalRef>),} {}\\ & & = \Delta ^{N}\Bigg[\Delta _{ a}^{-N+\nu +\mu }f(t) {}\\ & & \quad -\sum \limits _{j=0}^{M-1}\Delta _{ a}^{j-M+\mu }f(a + M-\mu )h_{ N-\nu -M+j}(t - M+\mu,a)\Bigg] {}\\ & & = \Delta _{a}^{\nu +\mu }\Delta ^{N}h_{ N-\nu -M+j}(t - M+\mu )f(t) - {}\\ & &\quad \sum \limits _{j=0}^{M-1}\Delta _{ a}^{j-M+\mu }f(a + M-\mu )\Delta ^{N}h_{ N-\nu -M+j}(t - M+\mu,a)\text{ (Lemma <InternalRef RefID="FPar47">2.47</InternalRef>) } {}\\ & & = \Delta _{a}^{\nu +\mu }f(t) - {}\\ & &\quad \sum \limits _{j=0}^{M-1}\Delta _{ a}^{j-M+\mu }f(a + M-\mu )h_{ -\nu -M+j}(t - M+\mu,a) {}\\ & & = \Delta _{a}^{\nu +\mu }f(t) {}\\ & & \quad -\sum _{j=0}^{M-1}\Delta _{ a}^{j-M+\mu }h_{ -\nu -M+\mu }(t - M+\mu )f(a + M-\mu ). {}\\ \end{array}$$

 □ 

Theorem 2.52 (Variation of Constants Formula).

Assume N ≥ 1 is an integer and N − 1 < ν ≤ N. If  $$f: \mathbb{N}_{0} \rightarrow \mathbb{R}$$ , then the solution of the IVP

 $$\displaystyle\begin{array}{rcl} \Delta _{\nu -N}^{\nu }y(t)& =& f(t),\quad t \in \mathbb{N}_{ 0}{}\end{array}$$

(2.34)

 $$\displaystyle\begin{array}{rcl} y(\nu -N + i)& =& 0,\quad 0 \leq i \leq N - 1{}\end{array}$$

(2.35)

is given by

 $$\displaystyle{y(t) = \Delta _{0}^{-\nu }f(t) =\sum _{ s=0}^{t-\nu }h_{\nu -1}(t,\sigma (s))f(s),\quad t \in \mathbb{N}_{\nu -N}.}$$

Proof.

Let

 $$\displaystyle{y(t) = \Delta _{0}^{-\nu }f(t) =\sum _{ s=0}^{t-\nu }h_{\nu -1}(t,\sigma (s))\;f(s).}$$

Then by our convention on sums

 $$\displaystyle{y(\nu -N + i) =\sum _{ s=0}^{-N+i}h_{\nu -1}(\nu -N + i,\sigma (s))\;f(s) = 0}$$

for 0 ≤ i ≤ N − 1, and hence the initial conditions (2.35) are satisfied.

Also, for  $$t \in \mathbb{N}_{0}$$ ,

 $$\displaystyle\begin{array}{rcl} \Delta _{\nu -N}^{\nu }y(t)& =& \Delta ^{N}\Delta _{\nu -N}^{-(N-\nu )}y(t) {}\\ & =& \Delta ^{N}\sum _{ s=\nu -N}^{t-(N-\nu )}h_{ N-\nu -1}(t,\sigma (s))\;y(s) {}\\ & =& \Delta ^{N}\sum _{ s=\nu }^{t-(N-\nu )}h_{ N-\nu -1}(t,\sigma (s))\;y(s), {}\\ & & {}\\ \end{array}$$

where in the last step we used the initial conditions (2.35). Hence,

 $$\displaystyle\begin{array}{rcl} \Delta _{\nu -N}^{\nu }y(t)& =& \Delta ^{N}\Delta _{\nu }^{-(N-\nu )}y(t) {}\\ & =& \Delta ^{N}\Delta _{ 0+\nu }^{-(N-\nu )}\Delta _{ 0}^{-\nu }f(t) {}\\ & =& \Delta ^{N}\Delta _{ 0}^{-N}f(t) {}\\ & =& f(t). {}\\ \end{array}$$

Therefore y is a solution of the fractional difference equation (2.34) on  $$\mathbb{N}_{0}.$$  □ 

Next we use the fractional variation of constants formula to solve a simple fractional IVP.

Example 2.53.

Use the variation of constants formula in Theorem 2.52 to solve the fractional IVP

 $$\displaystyle\begin{array}{rcl} \Delta _{-\frac{1} {2} }^{ \frac{1} {2} }y(t)& =& 5,\quad t \in \mathbb{N}_{0} {}\\ y\left (-\frac{1} {2}\right )& =& 3\sqrt{\pi }. {}\\ \end{array}$$

The solution of this IVP is defined on  $$\mathbb{N}_{-\frac{1} {2} }.$$ Note that the corresponding homogeneous fractional difference equation

 $$\displaystyle{\Delta _{-\frac{1} {2} }^{ \frac{1} {2} }y(t) = 0,\quad t \in \mathbb{N}_{0}}$$

has the general fractional equation form

 $$\displaystyle{\Delta _{a+\nu -N}^{\nu }y(t) = 0,\quad t \in \mathbb{N}_{ a}}$$

in Theorem 2.43, where

 $$\displaystyle{a = 0,\quad \nu = \frac{1} {2}\quad N = 1,\quad a +\nu -N = -\frac{1} {2}.}$$

Hence  $$t^{\underline{-\frac{1} {2} }}$$ is a solution of the homogeneous equation  $$\Delta _{-\frac{1} {2} }^{ \frac{1} {2} }y(t) = 0$$ and hence (using Theorem 2.52) a general solution of  $$\Delta _{-\frac{1} {2} }^{ \frac{1} {2} }y(t) = 5$$ is given by

 $$\displaystyle\begin{array}{rcl} y(t)& =& ct^{\underline{-\frac{1} {2} }} + \Delta _{0}^{-\frac{1} {2} }5 \\ & =& ct^{\underline{-\frac{1} {2} }} + 5\Delta _{0}^{-\frac{1} {2} }1{}\end{array}$$

(2.36)

By formula (2.16) we have that

 $$\displaystyle\begin{array}{rcl} \Delta _{0}^{-\frac{1} {2} }1& =& \Delta _{0}^{-\frac{1} {2} }t^{\underline{0}} = \frac{\Gamma (1)} {\Gamma (\frac{3} {2})}\;t^{\underline{\frac{1} {2} }} = \frac{2} {\sqrt{\pi }}\;t^{\underline{\frac{1} {2} }}, {}\\ \end{array}$$

which is the expression that we got for  $$\Delta _{0}^{-\frac{1} {2} }1$$ in Example 2.28. It follows from (2.36) that

 $$\displaystyle{y(t) = ct^{\underline{-\frac{1} {2} }} + \frac{10} {\sqrt{\pi }} t^{\underline{\frac{1} {2} }}.}$$

Using the initial condition  $$y\left (-\frac{1} {2}\right ) = 3\sqrt{\pi }$$ we get that c = 3. Therefore, the solution of the given IVP is

 $$\displaystyle{y(t) = 3t^{\underline{-\frac{1} {2} }} + \frac{10} {\sqrt{\pi }}\;t^{\underline{\frac{1} {2} }},}$$

for  $$t \in \mathbb{N}_{-\frac{1} {2} }.$$

Also, it is often necessary to know how a shifted Laplace transform with respect to its base relates to the original Laplace transform with base a, as is described in the following theorem.

Theorem 2.54.

Let  $$m \in \mathbb{N}_{0}$$ be given and suppose  $$f: \mathbb{N}_{a-m} \rightarrow \mathbb{R}$$ and  $$g: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ are of exponential order r > 0. Then for |s + 1| > r,

 $$\displaystyle{ \mathcal{L}_{a-m}\left \{f\right \}(s) = \frac{1} {\left (s + 1\right )^{m}}\mathcal{L}_{a}\left \{f\right \}(s) +\sum \limits _{ k=0}^{m-1}\frac{f\left (a + k - m\right )} {\left (s + 1\right )^{k+1}} }$$

(2.37)

and

 $$\displaystyle{ \mathcal{L}_{a+m}\left \{g\right \}(s) = \left (s + 1\right )^{m}\mathcal{L}_{ a}\left \{g\right \}(s) -\sum \limits _{k=0}^{m-1}\left (s + 1\right )^{m-1-k}g\left (a + k\right ). }$$

(2.38)

Proof.

Let f, g, r, and m be given as in the statement of this theorem. Then for | s + 1 |  > r,

 $$\displaystyle\begin{array}{rcl} \mathcal{L}_{a-m}\left \{f\right \}\left (s\right )& =& \sum \limits _{k=0}^{\infty }\frac{f\left (a - m + k\right )} {\left (s + 1\right )^{k+1}} {}\\ & =& \sum \limits _{k=m}^{\infty }\frac{f\left (a - m + k\right )} {\left (s + 1\right )^{k+1}} +\sum \limits _{ k=0}^{m-1}\frac{f\left (a - m + k\right )} {\left (s + 1\right )^{k+1}} {}\\ & =& \sum \limits _{k=0}^{\infty } \frac{f\left (a + k\right )} {\left (s + 1\right )^{k+m+1}} +\sum \limits _{ k=0}^{m-1}\frac{f\left (a + k - m\right )} {\left (s + 1\right )^{k+1}} {}\\ & =& \frac{1} {\left (s + 1\right )^{m}}\mathcal{L}_{a}\left \{f\right \}(s) +\sum \limits _{ k=0}^{m-1}\frac{f\left (a + k - m\right )} {\left (s + 1\right )^{k+1}}, {}\\ \end{array}$$

and hence (2.37) holds.

Next, consider

 $$\displaystyle\begin{array}{rcl} \mathcal{L}_{a+m}\left \{g\right \}\left (s\right )& =& \sum \limits _{k=0}^{\infty }\frac{g\left (a + m + k\right )} {\left (s + 1\right )^{k+1}} {}\\ & =& \sum \limits _{k=m}^{\infty } \frac{g\left (a + k\right )} {\left (s + 1\right )^{k-m+1}} {}\\ & =& \sum \limits _{k=0}^{\infty } \frac{g\left (a + k\right )} {\left (s + 1\right )^{k-m+1}} -\sum \limits _{k=0}^{m-1} \frac{g\left (a + k\right )} {\left (s + 1\right )^{k-m+1}} {}\\ & =& \left (s + 1\right )^{m}\mathcal{L}_{ a}\left \{g\right \}(s) -\sum \limits _{k=0}^{m-1}\left (s + 1\right )^{m-1-k}g\left (a + k\right ), {}\\ \end{array}$$

and thus (2.38) holds. □ 

We leave it as an exercise to verify that applying formulas (2.37) and (2.38) yields

 $$\displaystyle{ \mathcal{L}_{\left (a+m\right )-m}\left \{f\right \}(s) = \mathcal{L}_{\left (a-m\right )+m}\left \{f\right \}(s) = \mathcal{L}_{a}\left \{f\right \}(s), }$$

for | s + 1 |  > r.

Recall the definition of the fractional Taylor monomials (Definition 2.24).

Definition 2.55.

For each  $$\mu \in \mathbb{R}\setminus \left (-\mathbb{N}_{1}\right )$$ , define the μ-th order Taylor monomial, h μ (t, a), by

 $$\displaystyle{ h_{\mu }(t,a):= \frac{\left (t - a\right )^{\underline{\mu }}} {\Gamma (\mu +1)},\quad \text{ for }\quad t \in \mathbb{N}_{a}. }$$

Theorem 2.56.

If μ ≤ 0 and  $$\mu \notin (-\mathbb{N}_{1})$$ , then h μ (t,a) is bounded (and hence is of exponential order r = 1). If μ > 0, then for every r > 1, h μ (t,a) is of exponential order r.

Proof.

First consider the case that μ ≤ 0 with  $$\mu \not\in \left (-\mathbb{N}_{0}\right )$$ . Then for all large  $$t \in \mathbb{N}_{a},$$

 $$\displaystyle{ h_{\mu }\left (t,a\right ) = \frac{\Gamma (t - a + 1)} {\Gamma (\mu +1)\Gamma (t - a + 1-\mu )} \leq \frac{1} {\Gamma (\mu +1)}, }$$

implying that h μ is of exponential order one (i.e., bounded).

Next assume that μ > 0, with  $$N \in \mathbb{N}_{0}$$ chosen so that N − 1 < μ ≤ N.  Then for any fixed r > 1,

 $$\displaystyle\begin{array}{rcl} h_{\mu }\left (t,a\right ) = \frac{\left (t - a\right )^{\underline{\mu }}} {\Gamma (\mu +1)} & =& \frac{\Gamma (t - a + 1)} {\Gamma (\mu +1)\Gamma (t - a + 1-\mu )} {}\\ & \leq & \frac{\Gamma (t - a + 1)} {\Gamma (\mu +1)\Gamma (t - a + 1 - N)} {}\\ & =& \frac{\left (t - a\right ) \cdot \cdot \cdot \left (t - a - N + 1\right )} {\Gamma (\mu +1)} {}\\ & \leq & \frac{\left (t - a\right )^{N}} {\Gamma (\mu +1)} {}\\ & \leq & \frac{r^{t}} {\Gamma (\mu +1)}\text{,} {}\\ \end{array}$$

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

Therefore, h μ (t, a) is of exponential order r for each  $$\mu \in \mathbb{R}\setminus \left (-\mathbb{N}_{1}\right )$$ and r > 1. It follows from Theorem 2.4 that  $$\mathcal{L}_{a}\left \{h_{\mu }\left (t,a\right )\right \}(s)$$ exists for | s + 1 |  > 1.  □ 

Remark 2.57.

Note that the fractional Taylor monomials, h μ (t, a) for μ > 0 are examples of functions that are of order r for all r > 1, but are not of order 1 (see Exercise 2.4).

Theorem 2.58.

Let  $$\mu \in \mathbb{R}\setminus \left (-\mathbb{N}_{1}\right ).$$ Then

 $$\displaystyle{ \mathcal{L}_{a+\mu }\left \{h_{\mu }\left (t,a\right )\right \}\left (s\right ) = \frac{\left (s + 1\right )^{\mu }} {s^{\mu +1}} }$$

(2.39)

for |s + 1| > 1.

Proof.

For | s + 1 |  > 1, consider

 $$\displaystyle\begin{array}{rcl} \frac{(s + 1)^{\mu }} {s^{\mu +1}} & & = \frac{1} {s + 1}\left (\frac{s + 1} {s} \right )^{\mu +1} = \frac{1} {s + 1}\left (1 - \frac{1} {s + 1}\right )^{-\mu -1}. {}\\ \end{array}$$

Since  $$\vert \frac{1} {s+1}\vert < 1$$ , we have by the binomial theorem that

 $$\displaystyle\begin{array}{rcl} \frac{(s + 1)^{\mu }} {s^{\mu +1}} & =& \frac{1} {s + 1}\sum \limits _{k=0}^{\infty }(-1)^{k}\binom{-\mu - 1}{k}\left ( \frac{1} {s + 1}\right )^{k} \\ & =& \sum \limits _{k=0}^{\infty }(-1)^{k}\binom{-\mu - 1}{k} \frac{1} {(s + 1)^{k+1}}.{}\end{array}$$

(2.40)

But

 $$\displaystyle\begin{array}{rcl} (-1)^{k}\binom{-\mu - 1}{k}& =& (-1)^{k}\frac{(-\mu - 1)^{\underline{k}}} {k!} \\ & =& (-1)^{k}\frac{(-\mu - 1)(-\mu - 2)\cdots (-\mu - k)} {k!} \\ & =& \frac{(\mu +k)(\mu +k - 1)\cdots (\mu +1)} {k!} \\ & =& \frac{(\mu +k)^{\underline{k}}} {k!} \\ & =& \binom{\mu +k}{k} = \binom{\mu +k}{\mu }\quad \mbox{ by Exercise 1.12, (v)} \\ & =& \frac{(\mu +k)^{\underline{\mu }}} {\Gamma (\mu +1)} \\ & =& \frac{[(a +\mu +k) - a]^{\underline{\mu }}} {\Gamma (\mu +1)} \\ & =& h_{\mu }(a +\mu +k,a). {}\end{array}$$

(2.41)

Using (2.40) and (2.41), we have that

 $$\displaystyle\begin{array}{rcl} \frac{(s + 1)^{\mu }} {s^{\mu +1}} & =& \sum \limits _{k=0}^{\infty }\frac{h_{\mu }(a +\mu +k,a)} {(s + 1)^{k+1}} {}\\ & =& \mathcal{L}_{a+\mu }\left \{h_{\mu }\left (t,a\right )\right \}(s), {}\\ \end{array}$$

for | s + 1 |  > 1.  □