First Order Linear Difference Equations - Nabla Fractional Calculus

Discrete Fractional Calculus (2015)

3. Nabla Fractional Calculus

3.7. First Order Linear Difference Equations

In this section we show how to solve the first order nabla linear equation

$\displaystyle{ \nabla y(t) = p(t)y(t) + q(t),\quad t \in \mathbb{N}_{a+1}, }$

(3.21)

where we assume $p,q: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$ and $p \in \mathcal{R}$ . At the end of this section we will then show how to use the fact that we can solve the first order nabla linear equation (3.21) to solve certain nabla second order linear equations with variable coefficients (3.13) by the method of factoring.

We begin by using one of the following nabla Leibniz’s formulas to find a variation of constants formula for (3.21).

Theorem 3.41 (Nabla Leibniz Formulas).

Assume $f: \mathbb{N}_{a} \times \mathbb{N}_{a+1} \rightarrow \mathbb{R}$ . Then

$\displaystyle{ \nabla \left (\int _{a}^{t}f(t,\tau )\nabla \tau \right ) =\int _{ a}^{t}\nabla _{ t}f(t,\tau )\nabla \tau + f(\rho (t),t), }$

(3.22)

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

$\displaystyle{ \nabla \left (\int _{a}^{t}f(t,\tau )\nabla \tau \right ) =\int _{ a}^{t-1}\nabla _{ t}f(t,\tau )\nabla \tau + f(t,t), }$

(3.23)

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

Proof.

The proof of (3.22) follows from the following:

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

for $t \in \mathbb{N}_{a+1}$ . The proof of (3.23) is Exercise 3.22. □

Theorem 3.42 (Variation of Constants Formula).

Assume $p,q: \mathbb{N}_{a+1} \rightarrow \mathbb{R}$ and $p \in \mathcal{R}$ . Then the unique solution of the IVP

$\displaystyle\begin{array}{rcl} \nabla y(t)& =& p(t)y(t) + q(t),\quad t \in \mathbb{N}_{a+1} {}\\ y(a)& =& A {}\\ \end{array}$

is given by

$\displaystyle{y(t) = AE_{p}(t,a) +\int _{ a}^{t}E_{ p}(t,\rho (s))q(s)\nabla s,\quad t \in \mathbb{N}_{a}.}$

Proof.

The proof of uniqueness is left to the reader. Let

$\displaystyle{ y(t):= AE_{p}(t,a) +\int _{ a}^{t}E_{ p}(t,\rho (s))q(s)\nabla s,\quad t \in \mathbb{N}_{a}. }$

Using the nabla Leibniz formula (3.22), we obtain

$\displaystyle\begin{array}{rcl} \nabla y(t)& =& Ap(t)E_{p}(t,a) +\int _{ a}^{t}p(t)E_{ p}(t,\rho (s))q(s)\nabla s + E_{p}(\rho (t),\rho (t))q(t) {}\\ & =& p(t)\left [AE_{p}(t,a) +\int _{ a}^{t}E_{ p}(t,\rho (s))q(s)\nabla s\right ] + q(t) {}\\ & =& p(t)y(t) + q(t) {}\\ \end{array}$

for $t \in \mathbb{N}_{a+1}$ . We also see that y(a) = A. And this completes the proof. □

Example 3.43.

Assuming $r \in \mathcal{R}$ , solve the IVP

$\displaystyle{ \nabla y(t) = r(t)y(t) + E_{r}(t,a),\quad t \in \mathbb{N}_{a+1} }$

(3.24)

$\displaystyle{ y(a) = 0. }$

(3.25)

Using the variation of constants formula in Theorem 3.42, we have

$\displaystyle\begin{array}{rcl} y(t)& =& \int _{a}^{t}E_{ r}(t,\rho (s))E_{r}(s,a)\nabla s {}\\ & =& E_{r}(t,a)\int _{a}^{t}E_{ r}(a,\rho (s))E_{r}(s,a)\nabla s {}\\ & =& E_{r}(t,a)\int _{a}^{t} \frac{E_{r}(s,a)} {E_{r}(\rho (s),a)}\nabla s {}\\ & =& E_{r}(t,a)\int _{a}^{t} \frac{E_{r}(s,a)} {[1 - r(s)]E_{r}(s,a)}\nabla s {}\\ & =& E_{r}(t,a)\int _{a}^{t} \frac{1} {1 - r(s)}\nabla s. {}\\ \end{array}$

If we further assume r(t) = r ≠ 1 is a constant, then we obtain that the function $\frac{1} {1-r}(t - a)E_{r}(t,a)$ is the solution of the IVP

$\displaystyle{\nabla y(t) = ry(t) + E_{r}(t,a),\quad y(a) = 0.}$

A general solution of the linear equation (3.21) is given by adding a general solution of the corresponding homogeneous equation ∇y(t) = p(t)y(t) to a particular solution to the nonhomogeneous difference equation (3.21). Hence,

$\displaystyle{y(t) = cE_{p}(t,a) +\int _{ 0}^{t}E_{ p}(t,\rho (s))q(s)\nabla s}$

is a general solution of (3.21). We use this fact in the following example.

Example 3.44.

Find a general solution of the linear difference equation

$\displaystyle{ \nabla y(t) = (\boxminus 3)y(t) + 3t,\quad t \in \mathbb{N}_{1}. }$

(3.26)

Note that the constant function $p(t):= \boxminus 3$ is a regressive function on $\mathbb{N}_{1}$ . Hence, the general solution of (3.26) is given by

$\displaystyle\begin{array}{rcl} y(t)& =& cE_{p}(t,a) +\int _{ a}^{t}E_{ p}(t,\rho (s))q(s)\nabla s {}\\ & =& cE_{\boxminus 3}(t,0) + 3\int _{0}^{t}sE_{ \boxminus 3}(t,\rho (s))\nabla s {}\\ & =& cE_{\boxminus 3}(t,0) + 3\int _{0}^{t}sE_{ 3}(\rho (s),t)\nabla s {}\\ & =& cE_{\boxminus 3}(t,0) - 6\int _{0}^{t}sE_{ 3}(s,t)\nabla s, {}\\ \end{array}$

for $t \in \mathbb{N}_{0}$ . Integrating by parts we get

$\displaystyle\begin{array}{rcl} y(t)& =& cE_{\ominus 3}(t,0) - 2sE_{3}(s,t)\big\vert _{s=0}^{t} + 2\int _{ 0}^{t}E_{ 3}(\rho (s),t)\nabla s {}\\ & =& cE_{\boxminus 3}(t,0) - 2t - 4\int _{0}^{t}E_{ 3}(s,t)\nabla s {}\\ & =& cE_{\boxminus 3}(t,0) - 2t -\frac{4} {3}E_{3}(s,t)\big\vert _{0}^{t} {}\\ & =& cE_{\boxminus 3}(t,0) - 2t -\frac{4} {3} + \frac{4} {3}E_{3}(0,t) {}\\ & =& \alpha E_{\boxminus 3}(t,0) - 2t -\frac{4} {3} {}\\ & =& \alpha (-2)^{t} - 2t -\frac{4} {3}{}\\ && {}\\ \end{array}$

for $t \in \mathbb{N}_{0}$ .

Example 3.45.

Assuming r ≠ 1, use the method of factoring to solve the nabla difference equation

$\displaystyle{ \nabla ^{2}y(t) - 2r\nabla y(t) + r^{2}y(t) = 0,\quad t \in \mathbb{N}_{ a}. }$

(3.27)

A factored form of (3.27) is

$\displaystyle{ (\nabla - rI)(\nabla - rI)y(t) = 0,\quad t \in \mathbb{N}_{a}. }$

(3.28)

It follows from (3.28) that any solution of $(\nabla - rI)y(t) = 0$ is a solution of (3.27). Hence y ₁(t) = E _r(t, a) is a solution of (3.27). It also follows from the factored equation (3.28) that the solution y(t) of the IVP

$\displaystyle{(\nabla - rI)y(t) = E_{r}(t,a),\quad y(a) = 0}$

is a solution of (3.27). Hence, by the variation of constants formula in Theorem 3.42,

$\displaystyle\begin{array}{rcl} y(t)& =& \int _{a}^{t}E_{ r}(t,\rho (s))E_{r}(s,a)\nabla s = E_{r}(t,a)\int _{a}^{t}E_{ r}(a,\rho (s))E_{r}(s,a)\nabla s {}\\ & =& E_{r}(t,a)\int _{a}^{t}E_{ \ominus r}(\rho (s),a)E_{r}(s,a)\nabla s {}\\ & =& E_{r}(t,a)\int _{a}^{t}[1 -\ominus r]E_{ \ominus r}(s,a)E_{r}(s,a)\nabla s {}\\ & =& E_{r}(t,a)\int _{a}^{t}[1 -\ominus r]\nabla s = E_{ r}(t,a)\int _{a}^{t} \frac{1} {1 - r}\nabla s {}\\ & =& \frac{1} {1 - r}(t - a)E_{r}(t,a) {}\\ \end{array}$

is a solution of (3.27). But this implies that $y_{2}(t) = (t - a)E_{r}(t,a)$ is a solution of (3.27). Since y ₁(t) and y ₂(t) are linearly independent on $\mathbb{N}_{a},$

$\displaystyle{y(t) = c_{1}E_{r}(t,a) + c_{2}(t - a)E_{r}(t,a)}$

is a general solution of (3.27) on $\mathbb{N}_{a}$ .