Second Order Linear Equations with Constant Coefficients - Basic Difference Calculus - Discrete Fractional Calculus

Discrete Fractional Calculus (2015)

1. Basic Difference Calculus

1.4. Second Order Linear Equations with Constant Coefficients

The nonhomogeneous second order linear difference equation is given by

$$\displaystyle{ \Delta ^{2}y(t) + p(t)\Delta y(t) + q(t)y(t) = f(t),\quad t \in \mathbb{N}_{ a}, }$$

(1.12)

where we assume that p(t) ≠ q(t) + 1, for $$t \in \mathbb{N}_{a}.$$ In this section we will see that we can easily solve the corresponding second order linear homogeneous equation with constant coefficients

$$\displaystyle{ \Delta ^{2}y(t) + p\Delta y(t) + qy(t) = 0,\quad t \in \mathbb{N}_{ a}, }$$

(1.13)

where we assume the constants $$p,q \in \mathbb{R}$$ satisfy p ≠ 1 + q.

First we prove an existence-uniqueness theorem for solutions of initial value problems (IVPs) for (1.12).

Theorem 1.29.

Assume that $$p,q,f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ , p(t) ≠ 1 + q(t), $$t \in \mathbb{N}_{a}$$ , $$A,B \in \mathbb{R},$$ and $$t_{0} \in \mathbb{N}_{a}.$$ Then the IVP

$$\displaystyle{ \Delta ^{2}y(t) + p(t)\Delta y(t) + q(t)y(t) = f(t),\;\;t \in \mathbb{N}_{ a},\;\;y(t_{0}) = A,\;\;y(t_{0} + 1) = B, }$$

(1.14)

has a unique solution y(t) on $$\mathbb{N}_{a}.$$

Proof.

Expanding equation (1.12) out we have first solving for y(t + 2) and then solving for y(t) that

$$\displaystyle{ y(t + 2) = [2 - p(t)]y(t + 1) - [1 - p(t) + q(t)]y(t) + f(t) }$$

(1.15)

and, since p(t) ≠ 1 + q(t), $$t \in \mathbb{N}_{a}$$ ,

$$\displaystyle{ y(t) = \frac{2 - p(t)} {1 - p(t) + q(t)}y(t + 1) - \frac{1} {1 - p(t) + q(t)}y(t + 2) - \frac{f(t)} {1 - p(t) + q(t)}. }$$

(1.16)

If we let t = t 0 in (1.15), then equation (1.12) holds at t = t 0 iff

$$\displaystyle{y(t_{0} + 2) = [2 - p(t_{0})]B - [1 - p(t_{0}) + q(t_{0})]A + f(t_{0}).}$$

Hence, the solution of the IVP (1.14) is uniquely determined at t 0 + 2. But using the equation (1.15) evaluated at t = t 0 + 1, we have that the unique values of the solution at y(t 0 + 1) and y(t 0 + 2) uniquely determines the value of the solution at t 0 + 3. By induction we get that the solution of the IVP (1.14) is uniquely determined on $$\mathbb{N}_{t_{0}}$$ . On the other hand, if t 0 > a, then using equation (1.16) with t = t 0 − 1, we have that

$$\displaystyle{y(t_{0} - 1) = \frac{1} {1 - p(t_{0} - 1) + q(t_{0} - 1)}\bigg[[2 - p(t_{0} - 1)]A - B - f(t_{0} - 1)\bigg].}$$

Hence the solution of the IVP (1.14) is uniquely determined at t 0 − 1. Proceeding in this manner we have by mathematical induction that the solution of the IVP (1.14) is uniquely determined on $$\mathbb{N}_{a}^{t_{0}}.$$ Hence the result follows. □ 

Remark 1.30.

It follows from Theorem 1.29, that if p(t) ≠ 1 + q(t), $$t \in \mathbb{N}_{a}$$ , then the general solution of the linear homogeneous equation

$$\displaystyle{\Delta ^{2}y(t) + p(t)\Delta y(t) + q(t)y(t) = 0}$$

is given by

$$\displaystyle{y(t) = c_{1}y_{1}(t) + c_{2}y_{2}(t),\quad t \in \mathbb{N}_{a},}$$

where y 1(t), y 2(t) are any two linearly independent solutions of (1.13) on $$\mathbb{N}_{a}.$$

We now show we can solve the second order linear homogeneous equation (1.13) with constant coefficients.

Theorem 1.31 (Distinct Roots).

Assume p ≠ 1 + q and $$\lambda _{1}\neq \lambda _{2}$$ (possibly complex) are solutions (called the characteristic values of (1.13) ) of the characteristic equation

$$\displaystyle{\lambda ^{2} + p\lambda + q = 0.}$$

Then

$$\displaystyle{y(t) = c_{1}e_{\lambda _{1}}(t,a) + c_{2}e_{\lambda _{2}}(t,a)}$$

is a general solution of (1.13) .

Proof.

Assume $$\lambda _{1},$$ $$\lambda _{2}$$ are characteristic values of (1.13). Then the characteristic equation of (1.13) is given by

$$\displaystyle{\lambda ^{2} - (\lambda _{ 1} +\lambda _{2})\lambda +\lambda _{1}\lambda _{2} = 0.}$$

It follows that $$p = -(\lambda _{1} +\lambda _{2})$$ and $$q =\lambda _{1}\lambda _{2}$$ . Hence $$q + 1 - p = (\lambda _{1} + 1)$$ $$(\lambda _{2} + 1)\neq 0$$ , since p ≠ q + 1. Hence, we have that $$\lambda _{1},\lambda _{2}\neq - 1$$ and so $$e_{\lambda _{i}}(t,a)$$ , i = 1, 2, are well defined. Since

$$\displaystyle{\Delta ^{2}e_{\lambda _{ i}}(t,a) + p\;\Delta e_{\lambda _{i}}(t,a) + q\;e_{\lambda _{i}}(t,a) = [\lambda _{i}^{2} + p\lambda _{ i} + q]e_{\lambda _{i}}(t,a) = 0,}$$

we have that $$e_{\lambda _{i}}(t,a)$$ , i = 1, 2, are solutions of (1.13). Since these two solutions are linearly independent on $$\mathbb{N}_{a}$$ , we have that

$$\displaystyle{y(t) = c_{1}e_{\lambda _{1}}(t,a) + c_{2}e_{\lambda _{2}}(t,a)}$$

is a general solution of (1.13) on $$\mathbb{N}_{a}$$ . □ 

Example 1.32 (Fibonacci Numbers).

The Fibonacci numbers F(t), t = 1, 2, 3, ⋯ are defined recursively by

$$\displaystyle\begin{array}{rcl} F(t + 2)& =& F(t) + F(t + 1),\quad t \in \mathbb{N}_{1}, {}\\ F(1)& =& 1 = F(2). {}\\ \end{array}$$

The Fibonacci sequence is given by

$$\displaystyle{1,\quad 1,\quad 2,\quad 3,\quad 5,\quad 8,\quad 13,\quad 21,\quad 34,\quad \cdots \,.}$$

Fibonacci used this to model the population of pairs of rabbits under certain assumptions. To find F(t), note that F(t) is the solution of the IVP

$$\displaystyle\begin{array}{rcl} \Delta ^{2}F(t)& +& \Delta F(t) - F(t) = 0,\quad t \in \mathbb{N}_{ 1}, {}\\ F(1)& =& 1 = F(2). {}\\ \end{array}$$

To solve this IVP we first get that the characteristic equation is

$$\displaystyle{\lambda ^{2} +\lambda -1 = 0.}$$

Hence, the characteristic values are

$$\displaystyle{\lambda _{1} = \frac{-1 + \sqrt{5}} {2},\quad \lambda _{2} = \frac{-1 -\sqrt{5}} {2}.}$$

It follows that

$$\displaystyle\begin{array}{rcl} F(t)& =& c_{1}e_{\lambda _{1}}(t,1) + c_{2}e_{\lambda _{2}}(t,1) \\ & =& c_{1}\left (\frac{1 + \sqrt{5}} {2} \right )^{t-1} + c_{ 2}\left (\frac{1 -\sqrt{5}} {2} \right )^{t-1}.{}\end{array}$$

(1.17)

Applying the initial conditions we get the system

$$\displaystyle\begin{array}{rcl} F(1)& =& 1 = c_{1} + c_{2} {}\\ F(2)& =& 1 = c_{1}\left (\frac{1 + \sqrt{5}} {2} \right ) + c_{2}\left (\frac{1 -\sqrt{5}} {2} \right ). {}\\ \end{array}$$

Solving this system for c 1 and c 2, using (1.17) and simplifying we get

$$\displaystyle{F(t) = \frac{1} {\sqrt{5}}\left (\frac{1 + \sqrt{5}} {2} \right )^{t} - \frac{1} {\sqrt{5}}\left (\frac{1 -\sqrt{5}} {2} \right )^{t},}$$

for $$t \in \mathbb{N}_{1}.$$ Note F(t) is the integer nearest to $$\frac{1} {\sqrt{5}}\left (\frac{1+\sqrt{5}} {2} \right )^{t}$$ . It follows that F(20) is the integer nearest

$$\displaystyle{ \frac{1} {\sqrt{5}}\left (\frac{1 + \sqrt{5}} {2} \right )^{20} \approx 6765.00003,}$$

which gives us that F(20) = 6765. 

Usually, we want to find all real-valued solutions of (1.13). When a characteristic root $$\lambda _{1}$$ is complex, $$e_{\lambda _{1}}(t,a)$$ is a complex-valued solution. In the next theorem we show how to use this complex-valued solution to find two linearly independent real-valued solutions on $$\mathbb{N}_{a}$$ .

Theorem 1.33 (Complex Roots).

If the characteristic values are $$\lambda =\alpha \pm i\beta$$ , β > 0 and α ≠ − 1, then a general solution of (1.13) is given by

$$\displaystyle{y(t) = c_{1}e_{\alpha }(t,a)\cos _{\gamma }(t,a) + c_{2}e_{\alpha }(t,a)\sin _{\gamma }(t,a),}$$

where $$\gamma:= \frac{\beta } {1+\alpha }.$$

Proof.

First we show that if α ± i β, β > 0 are complex characteristic values of (1.13), then the condition p ≠ q + 1 is satisfied. In this case the characteristic equation for (1.13) is given by

$$\displaystyle{\lambda ^{2} - 2\alpha \lambda +\alpha ^{2} +\beta ^{2} = 0.}$$

It follows that p = −2α and q = α 2 +β 2. Therefore, 1 + q − p = (α + 1)2 +β 2 ≠ 0. This implies that p ≠ q + 1. By Theorem 1.31, we have that y(t) = e α+i β (t, a) is a complex-valued solution of (1.13). Using

$$\displaystyle{\alpha +i\beta =\alpha \oplus i \frac{\beta } {1+\alpha } =\alpha \oplus i\gamma,}$$

we get that

$$\displaystyle{y(t) = e_{\alpha +i\beta }(t,a) = e_{\alpha \oplus i\gamma }(t,a) = e_{\alpha }(t,a)e_{i\gamma }(t,a).}$$

It follows from the (delta) Euler’s formula (1.11) that

$$\displaystyle\begin{array}{rcl} y(t)& =& e_{\alpha }(t,a)e_{i\gamma }(t,a) {}\\ & =& e_{\alpha }(t,a)[\cos _{\gamma }(t,a) + i\sin _{\gamma }(t,a)] {}\\ & =& y_{1}(t) + iy_{2}(t) {}\\ \end{array}$$

is a solution of (1.13). But since p and q are real, we have that the real part, $$y_{1}(t) = e_{\alpha }(t,a)\cos _{\gamma }(t,a)$$ , and the imaginary part, $$y_{2}(t) = e_{\alpha }(t,a)\sin _{\gamma }(t,a)$$ , of y(t) are solutions of (1.13). Since p ≠ q + 1 and y 1(t), y 2(t) are linearly independent on $$\mathbb{N}_{a}$$ , we get from Remark 1.30 that

$$\displaystyle{y(t) = c_{1}e_{\alpha }(t,a)\cos _{\gamma }(t,a) + c_{2}e_{\alpha }(t,a)\sin _{\gamma }(t,a)}$$

is a general solution of (1.13) on $$\mathbb{N}_{a}$$ . □ 

Example 1.34.

Solve the difference equation

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

(1.18)

The characteristic equation is

$$\displaystyle{\lambda ^{2} - 2\lambda + 2 = 0}$$

and so the characteristic values are $$\lambda = 1 \pm i$$ . Hence using Theorem 1.33, we get

$$\displaystyle{y(t) = c_{1}e_{1}(t,a)\cos _{\frac{1} {2} }(t,a) + c_{2}e_{1}(t,a)\sin _{\frac{1} {2} }(t,a)}$$

is a general solution of (1.18) on $$\mathbb{N}_{a}$$ .

The previous theorem (Theorem 1.33) excluded the case when the characteristic values of (1.13) are − 1 ± i β, where β > 0. The next theorem considers this case.

Theorem 1.35.

If the characteristic values of (1.13) are − 1 ± iβ, where β > 0, then a general solution of (1.13) is given by

$$\displaystyle{y(t) = c_{1}\beta ^{t-a}\cos \left [ \frac{\pi } {2}(t - a)\right ] + c_{2}\beta ^{t-a}\sin \left [ \frac{\pi } {2}(t - a)\right ],}$$

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

Proof.

First note that by the first part of the proof of Theorem 1.33 we have that p ≠ q + 1. Since − 1 + i β is a characteristic root of (1.13), we have that y(t) = e −1+i β (t, a) is a complex-valued solution of (1.13). Now

$$\displaystyle\begin{array}{rcl} y(t)& =& e_{-1+i\beta }(t,a) {}\\ & =& (i\beta )^{t-a} {}\\ & =& \left (\beta e^{i \frac{\pi }{2} }\right )^{t-a} {}\\ & =& \beta ^{t-a}\cos \left [ \frac{\pi } {2}(t - a)\right ] + i\beta ^{t-a}\sin \left [ \frac{\pi } {2}(t - a)\right ]. {}\\ \end{array}$$

It follows that

$$\displaystyle{y_{1}(t) =\beta ^{t-a}\cos \left [ \frac{\pi } {2}(t - a)\right ],\quad y_{2}(t) =\beta ^{t-a}\sin \left [ \frac{\pi } {2}(t - a)\right ]}$$

are solutions of (1.13). Since these solutions are linearly independent on $$\mathbb{N}_{a}$$ , we have that

$$\displaystyle{y(t) = c_{1}\beta ^{t-a}\cos \left [ \frac{\pi } {2}(t - a)\right ] + c_{2}\beta ^{t-a}\sin \left [ \frac{\pi } {2}(t - a)\right ],}$$

is a general solution of (1.13). □ 

Example 1.36.

Solve the delta linear difference equation

$$\displaystyle{\Delta ^{2}y(t) + 2\Delta y(t) + 5y(t) = 0,\quad t \in \mathbb{N}_{ 0}.}$$

The characteristic equation is $$\lambda ^{2} + 2\lambda + 5 = 0$$ , so the characteristic values are $$\lambda = -1 \pm 2i$$ . It follows from Theorem 1.35 that

$$\displaystyle{y(t) = c_{1}2^{t}\cos \left ( \frac{\pi } {2}t\right ) + c_{2}2^{t}\sin \left ( \frac{\pi } {2}t\right ),}$$

is a general solution on $$\mathbb{N}_{0}.$$

Theorem 1.37 (Double Root).

Assume p ≠ 1 + q, and $$\lambda _{1} =\lambda _{2} = r$$ is a double root of the characteristic equation. Then

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

is a general solution of (1.13) .

Proof.

Since $$\lambda _{1} = r$$ is a characteristic value, we have y 1(t) = e r (t, a) is a solution of (1.13). Since $$\lambda _{1} =\lambda _{2} = r$$ , we have that the characteristic equation for (1.13) is

$$\displaystyle{(\lambda -r)^{2} =\lambda ^{2} - 2r\lambda + r^{2} = 0.}$$

Hence, in this case, (1.13) has the form

$$\displaystyle{\Delta ^{2}y(t) - 2r\Delta y(t) + r^{2}y(t) = 0.}$$

From Exercise 1.32, we have that y 2(t) = (t − a)e r (t, a) is a second solution of (1.13) on $$\mathbb{N}_{a}.$$ Since these two solutions are linearly independent on $$\mathbb{N}_{a}$$ , we have that

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

is a general solution of (1.13). □ 

Example 1.38.

Evaluate the t by t determinant of the following tridiagonal matrix

$$\displaystyle{M(t):= \left [\begin{array}{cccccccc} 4&4&0&0&\cdots &0&0&0\\ 1 &4 &4 &0 &\cdots &0 &0 &0 \\ 0&1&4&4&\cdots &0&0&0\\ 0 &0 &1 &4 &\cdots &0 &0 &0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\ 0&0&0&0&\cdots &4&4&0\\ 0 &0 &0 &0 &\cdots &1 &4 &4 \\ 0&0&0&0&\cdots &0&1&4 \end{array} \right ]}$$

for $$t \in \mathbb{N}_{1}.$$ For example, we have that

$$\displaystyle{M(1):= \left [\begin{array}{*{10}c} 4 \end{array} \right ],\ M(2):= \left [\begin{array}{*{10}c} 4&4\\ 1 &4\\ \end{array} \right ],\ \text{and}\ \ M(3):= \left [\begin{array}{*{10}c} 4&4&0\\ 1 &4 &4 \\ 0&1&4\\ \end{array} \right ].}$$

Let D(t) be the value of the determinant of M(t). Expanding the t + 2 by t + 2 determinant D(t + 2) along its first row we get

$$\displaystyle{D(t + 2) = 4D(t + 1) - 4D(t),\quad t \in \mathbb{N}_{1}.}$$

Note that D(1) = 4 and D(2) = 12. It follows that if we define D(0) = 1, then we have

$$\displaystyle{D(t + 2) - 4D(t + 1) + 4D(t) = 0,\quad t \in \mathbb{N}_{0}.}$$

It then follows that D(t) is the solution of the IVP

$$\displaystyle\begin{array}{rcl} & & \Delta ^{2}D(t) - 2\Delta D(t) + D(t) = 0,\quad t \in \mathbb{N}_{ 0} {}\\ & & D(0) = 1,\quad D(1) = 4. {}\\ \end{array}$$

The characteristic equation is $$\lambda ^{2} - 2\lambda + 1 = 0$$ , and thus $$\lambda _{1} =\lambda _{2} = 1$$ are the characteristic values. Hence by Theorem 1.37,

$$\displaystyle\begin{array}{rcl} D(t)& =& c_{1}e_{1}(t,0) + c_{2}(t - 0)e_{1}(t,0) {}\\ & =& c_{1}2^{t} + c_{ 2}t2^{t}. {}\\ \end{array}$$

Using the initial conditions we get the system of equations

$$\displaystyle\begin{array}{rcl} D(0)& =& 1 = c_{1} {}\\ D(1)& =& 4 = 2c_{1} + 2c_{2}. {}\\ \end{array}$$

Solving this system we get that c 1 = c 2 = 1 and hence

$$\displaystyle{D(t) = 2^{t} + t2^{t}.}$$

The reader should check this answer for a few values of t.

Example 1.39.

Find the determinant D(t),  $$t \in \mathbb{N}_{1}$$ of the t × t matrix that has zeros down the diagonal, 2’s down the superdiagonal, 8’s down the subdiagonal. This leads to solving the IVP

$$\displaystyle\begin{array}{rcl} & & D(t + 2) + 16D(t) = 0,\quad t \in \mathbb{N}_{0} {}\\ & & \quad D(0) = 1,\quad D(1) = 0. {}\\ \end{array}$$

If we tried to use Theorem 1.33 to solve the difference equation D(t + 2) + 16D(t) = 0 we would write the equation D(t + 2) + 16D(t) = 0 in the form

$$\displaystyle{\Delta ^{2}D(t) + 2\Delta D(t) + 17D(t) = 0.}$$

The characteristic values for this equation are − 1 ± 4i. But Theorem 1.33 does not apply since the real part of − 1 ± 4i is − 1. Applying Theorem 1.44 we get that a general solution of D(t + 2) + 16D(t) = 0 is given by

$$\displaystyle{D(t) = c_{1}4^{t}\cos \left ( \frac{\pi } {2}t\right ) + c_{2}4^{t}\cos \left ( \frac{\pi } {2}t\right ),\quad t \in \mathbb{N}_{0}.}$$

Applying the initial conditions we get

$$\displaystyle{D(t) = 4^{t}\cos \left ( \frac{\pi } {2}t\right ),}$$

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

Sometimes it is convenient to know how to solve the second order linear homogeneous difference equation when it is of the form

$$\displaystyle{ y(t + 2) + cy(t + 1) + dy(t) = 0, }$$

(1.19)

where $$c,d \in \mathbb{R}$$ and d ≠ 0 without first writing (1.19) (as we did in Examples 1.32 and 1.38) in the form (1.13). Exercise 1.36 shows that the difference equation 1.13 with p ≠ 1 + q is equivalent to the difference equation 1.19 with d≠ 0. Similar to the proof of Theorem 1.31 we can prove (Exercise 1.29) the theorem.

Theorem 1.40 (Distinct Roots).

Assume d ≠ 0 and r 1 , r 2 are distinct roots of the equation

$$\displaystyle{r^{2} + cr + d = 0.}$$

Then

$$\displaystyle{y(t) = c_{1}r_{1}^{t} + c_{ 2}r_{2}^{t}}$$

is a general solution of (1.19) .

Example 1.41.

Solve the difference equation

$$\displaystyle{ u(t + 2) - 5u(t + 1) + 6u(t) = 0. }$$

(1.20)

Solving r 2 − 5r + 6 = (r − 2)(r − 3) = 0, we get r 1 = 2, r 2 = 3. It follows from Theorem 1.40 that

$$\displaystyle{u(t) = c_{1}2^{t} + c_{ 2}3^{t}}$$

is a general solution of (1.20).

Similar to the proof of Theorem 1.37 one could prove (see Exercise 1.30) the following theorem.

Theorem 1.42 (Double Root).

Assume d ≠ 0 and r is a double root of r 2 + cr + d = 0. Then

$$\displaystyle{y(t) = c_{1}r^{t} + c_{ 2}tr^{t}}$$

is a general solution of (1.19) .

Example 1.43.

Solve the difference equation

$$\displaystyle{ u(t + 2) + 4u(t + 1) + 4u(t) = 0. }$$

(1.21)

Solving the equation r 2 + 4r + 4 = (r + 2)2 = 0, we get r 1 = r 2 = −2. It follows from Theorem 1.42 that

$$\displaystyle{u(t) = c_{1}(-2)^{t} + c_{ 2}t(-2)^{t}}$$

is a general solution of (1.21).

Theorem 1.44 (Complex Roots).

Assume d ≠ 0 and α ± iβ, β > 0 are complex roots of r 2 + cr + d = 0. Then

$$\displaystyle{y(t) = c_{1}r^{t}\cos (\theta t) + c_{ 2}r^{t}\sin (\theta t),}$$

where $$r = \sqrt{\alpha ^{2 } +\beta ^{2}}$$ and $$\theta = Tan^{-1}\frac{\beta } {\alpha }$$ if α ≠ 0 and $$\theta = \frac{\pi } {2}$$ if α = 0 is a general solution of (1.19) .

Proof.

Since r = α + i β, β > 0 is a solution of r 2 + cr + d = 0, we have by Theorem 1.40 that y(t) = (α + i β) t is a (complex-valued) solution of (1.19). Let $$\theta:= Tan^{-1}\frac{\beta } {\alpha }$$ if α ≠ 0 and let $$\theta = \frac{\pi } {2}$$ if α = 0. Then if $$r:= \sqrt{\alpha ^{2 } +\beta ^{2}}$$ we have that

$$\displaystyle\begin{array}{rcl} y(t)& =& (\alpha +i\beta )^{t} {}\\ & =& \left (re^{i\theta }\right )^{t} {}\\ & =& r^{t}e^{i\theta t} {}\\ & =& r^{t}\cos (\theta t) + ir^{t}\sin (\theta t). {}\\ \end{array}$$

Since the real and imaginary parts of y(t) are linearly independent solutions of (1.19), we have that

$$\displaystyle{y(t) = c_{1}r^{t}\cos (\theta t) + c_{ 2}r^{t}\sin (\theta t)}$$

is a general solution of (1.19). □ 

Next we briefly discuss the method of annihilators for solving certain nonhomogeneous difference equations. For an arbitrary function $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ we define the operator E by

$$\displaystyle{Ef(t) = f(t + 1),\quad t \in \mathbb{N}_{a}.}$$

Then E n : = E ⋅ E n−1 for $$n \in \mathbb{N}_{1}.$$ We say the polynomial in E, 

$$\displaystyle{p(E):= E^{n} + a_{ 1}E^{n-1} +\ldots +a_{ n}I,}$$

annihilates $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ provided p(E)f(t) = 0 for $$t \in \mathbb{N}_{a}.$$ Similarly we say the polynomial in the operator $$\Delta $$ ,

$$\displaystyle{p(\Delta ):= \Delta ^{n} + a_{ 1}\Delta ^{n-1} +\ldots +a_{ n}I,}$$

annihilates $$f: \mathbb{N}_{a} \rightarrow \mathbb{R}$$ provided $$p(\Delta )f(t) = 0$$

Example 1.45.

Here are some simple annihilators for various functions:

(i)

(E − rI)r t  = 0; 

(ii)

$$(\Delta - rI)e_{r}(t,a) = 0;$$

(iii)

(E − rI)2 tr t  = 0; 

(iv)

$$(\Delta ^{2} - p^{2})\cosh _{p}(t,a) = 0;$$

(v)

$$(\Delta ^{2} - p^{2})\sinh _{p}(t,a) = 0;$$

(vi)

$$(\Delta ^{2} + p^{2})\cosh _{p}(t,a) = 0;$$

(vii)

$$(\Delta ^{2} + p^{2})\sinh _{p}(t,a) = 0;$$

(viii)

$$\Delta ^{n}(t - a)^{k} = 0,\quad \mbox{ for integers}\quad n \geq k \geq 0.$$

We now give some simple examples where we use the method of annihilators to solve various difference equations.

Example 1.46.

Solve the first order linear equation

$$\displaystyle{ y(t + 1) - 3y(t) = 5^{t},\quad t \in \mathbb{Z}. }$$

(1.22)

First we write this equation in the form

$$\displaystyle{(E - 3I)y(t) = 5^{t}.}$$

Since E − 5I annihilates the right-hand side, we multiply each side by the operator E − 5I to get

$$\displaystyle{(E - 5I)(E - 3I)y(t) = 0,\quad t \in \mathbb{Z}.}$$

It follows that a solution of (1.22) must be of the form

$$\displaystyle{y(t) = c_{1}3^{t} + c_{ 2}5^{t},\quad t \in \mathbb{Z}.}$$

Substituting this into equation (1.22) we get

$$\displaystyle{c_{1}3^{t+1} + c_{ 2}5^{t+1} - c_{ 1}3^{t+1} - 3c_{ 2}5^{t} = 5^{t},\quad t \in \mathbb{Z}.}$$

Hence we see that we must have $$c_{2} = \frac{1} {2}$$ . It follows that

$$\displaystyle{y(t) = c_{1}3^{t} + \frac{1} {2} \cdot 5^{t},\quad t \in \mathbb{Z}}$$

is a general solution of (1.22).

Example 1.47.

Solve the second order linear nonhomogeneous difference equation

$$\displaystyle{ y(t + 2) - 6y(t + 1) + 8y(t) = 16 \cdot 4^{t},\quad t \in \mathbb{N}_{ 0}. }$$

(1.23)

by the annihilator method. The difference equation (1.23) can be written in the form

$$\displaystyle{(E - 2I)(E - 4I)y(t) = 16 \cdot 4^{t},\quad t \in \mathbb{N}_{ 0}.}$$

Multiplying both sides by the operator E − 4I we get that

$$\displaystyle{(E - 2I)(E - 4I)^{2}y(t) = 0.}$$

Hence y(t) must have the form

$$\displaystyle{y(t) = c_{1}2^{t} + c_{ 2}4^{t} + c_{ 3}t4^{t},\quad t \in \mathbb{N}_{ 0}.}$$

Substituting this into the difference equation (1.23) we get after simplification that 8c 3 = 16. Hence c 3 = 2 and we have that

$$\displaystyle{y(t) = c_{1}2^{t} + c_{ 2}4^{t} + 2t4^{t},\quad t \in \mathbb{N}_{ 0}}$$

is a general solution of (1.23).

Also we can use the method of annihilators to solve certain nonhomogeneous equations of the form

$$\displaystyle{\Delta ^{2}y(t) + p\Delta y(t) + qy(t) = f(t),}$$

where p, q are real constants with p ≠ 1 + q, as is shown in the following example.

Example 1.48.

Use the method of annihilators to solve the nonhomogeneous equation

$$\displaystyle{ \Delta ^{2}y(t) - 3\Delta y(t) + 2y(t) = 4^{t-a},\quad t \in \mathbb{N}_{ a}. }$$

(1.24)

The equation (1.24) can be written in the form

$$\displaystyle{(\Delta - I)(\Delta - 2I)y(t) = e_{3}(t,a).}$$

Multiplying both sides by the operator $$(\Delta - 3I)$$ we get that solutions of (1.24) are solutions of

$$\displaystyle{(\Delta - I)(\Delta - 2I)(\Delta - 3I)y(t) = 0.}$$

The values of the characteristic equation $$(\lambda -1)(\lambda -2)(\lambda -3) = 0$$ are $$\lambda _{1} = 1,$$ $$\lambda _{2} = 2,$$ $$\lambda _{3} = 3$$ . Hence all solutions of (1.24) are of the form

$$\displaystyle{y(t) = c_{1}e_{1}(t,a) + c_{2}e_{2}(t,a) + c_{3}e_{3}(t,a).}$$

Substituting this into the equation we get that we must have 2c 3 e 3(t, a) = e 3(t, a) which gives us that c 3 = 2. Hence the general solution of (1.24) is given by

$$\displaystyle{y(t) = c_{1}e_{1}(t,a) + c_{2}e_{2}(t,a) + 2e_{3}(t,a),\quad t \in \mathbb{N}_{a}.}$$