First Order Linear Difference Equations - Basic Difference Calculus - Discrete Fractional Calculus

Discrete Fractional Calculus (2015)

1. Basic Difference Calculus

1.7. First Order Linear Difference Equations

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

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

(1.28)

where we assume $$p,q: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ and p(t) ≠ − 1 for $$t \in \mathbb{N}_{a}.$$

We will use the following Leibniz formula to find a variation of constants formula for (1.28).

Theorem 1.67 (Leibniz Formula).

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

$$\displaystyle{ \Delta \left (\int _{a}^{t}f(t,s)\Delta s\right ) =\int _{ a}^{t}\Delta _{ t}f(t,s)\Delta s + f(\sigma (t),t),\quad t \in \mathbb{N}_{a}. }$$

(1.29)

Proof.

We have that

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

which completes the proof. □ 

Theorem 1.68 (Variation of Constants Formula).

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

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

is given by

$$\displaystyle{y(t) = Ae_{p}(t,a) +\int _{ a}^{t}e_{ p}(t,\sigma (s))q(s)\Delta s,}$$

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

Proof.

The proof (see Exercise 1.56) of the uniqueness of solutions of IVPs for this case is similar to the proof of Theorem 1.29. Let

$$\displaystyle{y(t) = Ae_{p}(t,a) +\int _{ a}^{t}e_{ p}(t,\sigma (s))q(s)\Delta s,\quad t \in \mathbb{N}_{a}.}$$

Using the Leibniz formula (1.29), we get

$$\displaystyle\begin{array}{rcl} \Delta y(t)& =& Ap(t)e_{p}(t,a) +\int _{ a}^{t}p(t)e_{ p}(t,\sigma (s))q(s)\Delta s + e_{p}(\sigma (t),\sigma (t))q(t) {}\\ & =& p(t)\left [Ae_{p}(t,a) +\int _{ a}^{t}e_{ p}(t,\sigma (s))q(s)\Delta s\right ] + q(t) {}\\ & =& p(t)y(t) + q(t). {}\\ \end{array}$$

Also y(a) = A.  □ 

Of course, it is always possible to compute solutions of difference equations by direct step by step computation from the difference equation. We next give an interesting example due to Gautschi [87] (and appearing in Kelley and Peterson [134, 135]) that illustrates that round off error can be a serious problem.

Example 1.69 (Gautschi [87]).

First we solve the IVP

$$\displaystyle\begin{array}{rcl} & & \Delta y(t) = (t - 1)y(t) + 1,\quad t \in \mathbb{N}_{1} {}\\ & & \phantom{\Delta y(t)}y(1) = 1 - e. {}\\ \end{array}$$

Note that p(t): = t − 1 is a regressive function on $$\mathbb{N}_{1}$$ . Using the variation of constants formula in Theorem 1.68, we get that the solution of our given IVP is given by

$$\displaystyle\begin{array}{rcl} y(t)& =& (1 - e)e_{t-1}(t,1) +\int _{ 1}^{t}e_{ t-1}(t,\sigma (s)) \cdot 1\Delta s {}\\ & =& e_{t-1}(t,1)\left [1 - e +\int _{ 1}^{t}e_{ t-1}(1,\sigma (s))\Delta s\right ] {}\\ & =& e_{t-1}(t,1)\left [1 - e +\int _{ 1}^{t} \frac{1} {e_{t-1}(\sigma (s),1)}\Delta s\right ]. {}\\ \end{array}$$

From Example 1.13, we have that e t−1(t, 1) = (t − 1)! . Hence

$$\displaystyle\begin{array}{rcl} y(t)& =& (t - 1)!\left [1 - e +\int _{ 1}^{t} \frac{1} {(\sigma (s) - 1)!}\Delta s\right ] {}\\ & =& (t - 1)!\left [1 - e +\sum _{ s=1}^{t-1} \frac{1} {s!}\right ] {}\\ & =& -(t - 1)!\sum _{k=t}^{\infty }\frac{1} {k!}. {}\\ \end{array}$$

Note that this solution is negative on $$\mathbb{N}_{1}$$ . Now if one was to approximate the initial value 1 − e in this IVP by a finite decimal expansion, it can be shown that the solution z(t) of this new IVP satisfies $$\lim _{t\rightarrow \infty }z(t) = \infty $$ and hence z(t) is not a good approximation for the actual solution. For example, if z(t) solves the IVP

$$\displaystyle\begin{array}{rcl} \Delta z(t)& =& (1 - t)z(t) + 1,\quad t \in \mathbb{N}_{1} {}\\ z(1)& =& -1.718, {}\\ \end{array}$$

then z(2) = −. 718, z(3) = −. 436, z(4) = −. 308, z(5) = −. 232, z(6) = −. 16, z(7) = . 04 and after that z(t) increases rapidly with $$\lim _{t\rightarrow \infty }z(t) = \infty.$$ Hence z(t) is not a good approximation to the actual solution y(t) of our original IVP.

A general solution of the linear equation (1.28) is given by adding a general solution of the corresponding homogeneous equation $$\Delta y(t) = p(t)y(t)$$ to a particular solution to the nonhomogeneous difference equation (1.28). Hence by Theorem 1.14 and Theorem 1.68

$$\displaystyle{y(t) = ce_{p}(t,a) +\int _{ a}^{t}e_{ p}(t,\sigma (s))q(s)\Delta s}$$

is a general solution of (1.28). We use this fact in the following example.

Example 1.70.

Find a general solution of the linear difference equation

$$\displaystyle{ \Delta y(t) = (\ominus 2)y(t) + t,\quad t \in \mathbb{N}_{0}. }$$

(1.30)

Note that the constant function ⊖ 2 is a regressive function on $$\mathbb{N}_{0}$$ . The general solution of (1.30) is given by

$$\displaystyle\begin{array}{rcl} y(t)& =& ce_{p}(t,a) +\int _{ a}^{t}e_{ p}(t,\sigma (s))q(s)\Delta s {}\\ & =& ce_{\ominus 2}(t,0) +\int _{ 0}^{t}s\;e_{ \ominus 2}(t,\sigma (s))\Delta s {}\\ & =& ce_{\ominus 2}(t,0) +\int _{ 0}^{t}s\;e_{ 2}(\sigma (s),t)\Delta s {}\\ & =& ce_{\ominus 2}(t,0) + 3\int _{0}^{t}s\;e_{ 2}(s,t)\Delta s. {}\\ \end{array}$$

Integrating by parts we get

$$\displaystyle\begin{array}{rcl} y(t)& =& ce_{\ominus 2}(t,0) + \frac{3} {2}se_{2}(s,t)\big\vert _{s=0}^{t} -\frac{3} {2}\int _{0}^{t}e_{ 2}(\sigma (s),t)\Delta s {}\\ & =& ce_{\ominus 2}(t,0) + \frac{3} {2}t -\frac{9} {2}\int _{0}^{t}e_{ 2}(s,t)\Delta s {}\\ & =& ce_{\ominus 2}(t,0) + \frac{3} {2}t -\frac{9} {4}e_{2}(s,t)\big\vert _{0}^{t} {}\\ & =& ce_{\ominus 2}(t,0) + \frac{3} {2}t -\frac{9} {4} + \frac{9} {4}e_{2}(0,t) {}\\ & =& \alpha e_{\ominus 2}(t,0) + \frac{3} {2}t -\frac{9} {4} {}\\ & =& \alpha \left (\frac{1} {3}\right )^{t} + \frac{3} {2}t -\frac{9} {4}. {}\\ & & {}\\ \end{array}$$