Second Order Linear Equations with Constant Coefficients - Nabla Fractional Calculus - Discrete Fractional Calculus

Discrete Fractional Calculus (2015)

3. Nabla Fractional Calculus

3.5. Second Order Linear Equations with Constant Coefficients

The nonhomogeneous second order linear nabla difference equation is given by

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

(3.12)

where we assume $$p,g,f: \mathbb{N}_{a+2} \rightarrow \mathbb{R}$$ and $$1 + p(t) + q(t)\neq 0$$ for $$t \in \mathbb{N}_{a+2}$$ . In this section we will see that we can easily solve the corresponding second order linear homogeneous nabla difference equation with constant coefficients

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

(3.13)

where we assume the constants $$p,q \in \mathbb{R}$$ satisfy $$1 + p + q\neq 0$$ .

First we prove an existence-uniqueness theorem for solutions of initial value problems (IVPs) for (3.12).

Theorem 3.20.

Assume that $$p,q,f: \mathbb{N}_{a+2} \rightarrow \mathbb{R}$$ , $$1 + p(t) + q(t)\neq 0$$ , $$t \in \mathbb{N}_{a+2}$$ , $$A,B \in \mathbb{R},$$ and $$t_{0} \in \mathbb{N}_{a+1}$$ . Then the IVP

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

(3.14)

$$\displaystyle{ y(t_{0} - 1) = A,\quad y(t_{0}) = B, }$$

(3.15)

where $$t_{0} \in \mathbb{N}_{a+1}$$ and $$A,B \in \mathbb{R}$$ has a unique solution y(t) on $$\mathbb{N}_{a}$$ .

Proof.

Expanding equation (3.14) we have by first solving for y(t) and then solving for y(t − 2) that, since $$1 + p(t) + q(t)\neq 0$$ ,

$$\displaystyle\begin{array}{rcl} y(t)& =& \frac{2 + p(t)} {1 + p(t) + q(t)}y(t - 1) \\ & & - \frac{1} {1 + p(t) + q(t)}y(t - 2) + \frac{f(t)} {1 + p(t) + q(t)}{}\end{array}$$

(3.16)

and

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

(3.17)

If we let $$t = t_{0} + 1$$ in (3.16), then equation (3.14) holds at $$t = t_{0} + 1$$ iff

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

Hence, the solution of the IVP (3.14), (3.15) is uniquely determined at t 0 + 1. But using the equation (3.16) evaluated at $$t = t_{0} + 2$$ , we have that the unique values of the solution at t 0 and t 0 + 1 uniquely determine the value of the solution at t 0 + 2. By induction we get that the solution of the IVP (3.14), (3.15) is uniquely determined on $$\mathbb{N}_{t_{0}-1}$$ . On the other hand if t 0 ≥ a + 2, then using equation (3.17) with t = t 0, we have that

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

Hence the solution of the IVP (3.14), (3.15) is uniquely determined at t 0 − 2. Similarly, if t 0 − 3 ≥ a, then the value of the solution at t 0 − 2 and at t 0 − 1 uniquely determines the value of the solution at t 0 − 3. Proceeding in this manner we have by mathematical induction that the solution of the IVP (3.14), (3.15) is uniquely determined on $$\mathbb{N}_{a}^{t_{0}-1}$$ . Hence the result follows. □ 

Remark 3.21.

Note that the so-called initial conditions in (3.15)

$$\displaystyle{y(t_{0} - 1) = A,\quad y(t_{0}) = B}$$

hold iff the equations y(t 0) = C, $$\nabla y(t_{0}) = D:= B - A$$ are satisfied. Because of this we also say that

$$\displaystyle{y(t_{0}) = C,\quad \nabla y(t_{0}) = D}$$

are initial conditions for solutions of equation (3.12). In particular, Theorem 3.20 holds if we replace the conditions (3.15) by the conditions

$$\displaystyle{y(t_{0}) = C,\quad \nabla y(t_{0}) = D.}$$

Remark 3.22.

From Exercise 3.21 we see that if $$1 + p(t) + q(t)\neq 0$$ , $$t \in \mathbb{N}_{a+2}$$ , then the general solution of the linear homogeneous equation

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

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 (3.13) on $$\mathbb{N}_{a}$$ .

Next we show we can solve the second order linear nabla difference equation with constant coefficients (3.13). We say the equation

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

is the characteristic equation of the nabla linear difference equation (3.13) and the solutions of this characteristic equation are called the characteristic values of (3.13).

Theorem 3.23 (Distinct Roots).

Assume $$1 + p + q\neq 0$$ and $$\lambda _{1}\neq \lambda _{2}$$ (possibly complex) are the characteristic values of (3.13) . Then

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

is a general solution of (3.13) on $$\mathbb{N}_{a}$$ .

Proof.

Since $$\lambda _{1}$$ , $$\lambda _{2}$$ satisfy the characteristic equation for (3.13), we have that the characteristic polynomial for (3.13) is given by

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

and hence

$$\displaystyle{p = -\lambda _{1} -\lambda _{2},\quad q =\lambda _{1}\lambda _{2}.}$$

Since

$$\displaystyle{1 + p + q = 1 + (-\lambda _{1} -\lambda _{2}) +\lambda _{1}\lambda _{2} = (1 -\lambda _{1})(1 -\lambda _{2})\neq 0,}$$

we have that $$\lambda _{1},\lambda _{2}\neq 1$$ and hence $$E_{\lambda _{1}}(t,a)$$ and $$E_{\lambda _{2}}(t,a)$$ are well defined. Next note that

$$\displaystyle\begin{array}{rcl} & & \nabla ^{2}E_{\lambda _{ i}}(t,a) + p\;\nabla E_{\lambda _{i}}(t,a) + q\;E_{\lambda _{i}}(t,a) {}\\ & & \qquad \qquad \quad = [\lambda _{i}^{2} + p\lambda _{ i} + q]E_{\lambda _{i}}(t,a) {}\\ & & \qquad \qquad \quad = 0, {}\\ \end{array}$$

for i = 1, 2. Hence $$E_{\lambda _{i}}(t,a)$$ , i = 1, 2 are solutions of (3.13). Since $$\lambda _{1}\neq \lambda _{2}$$ , these two solutions are linearly independent on $$\mathbb{N}_{a}$$ , and by Remark 3.22,

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

is a general solution of (3.13) on $$\mathbb{N}_{a}$$ . □ 

Example 3.24.

Solve the nabla linear difference equation

$$\displaystyle{\nabla ^{2}y(t) + 2\nabla y(t) - 8y(t) = 0.\quad t \in \mathbb{N}_{ a+2}.}$$

The characteristic equation is

$$\displaystyle{\lambda ^{2} + 2\lambda - 8 = (\lambda -2)(\lambda +4) = 0}$$

and the characteristic roots are

$$\displaystyle{\lambda _{1} = 2,\quad \lambda _{2} = -4.}$$

Note that $$1 + p + q = -5\neq 0$$ , so we can apply Theorem 3.23. Then we have that

$$\displaystyle\begin{array}{rcl} y(t)& =& c_{1}E_{\lambda _{1}}(t,a) + c_{2}E_{\lambda _{2}}(t,a) {}\\ & =& c_{1}E_{2}(t,a) + c_{2}E_{-4}(t,a) {}\\ & =& c_{1}(-1)^{a-t} + c_{ 2}5^{a-t} {}\\ \end{array}$$

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

Usually, we want to find all real-valued solutions of (3.13). When a characteristic value $$\lambda _{1}$$ of (3.13) 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 3.25 (Complex Roots).

Assume the characteristic values of (3.13) are $$\lambda =\alpha \pm i\beta$$ , β > 0 and α ≠ 1. Then a general solution of (3.13) is given by

$$\displaystyle{y(t) = c_{1}E_{\alpha }(t,a)\mbox{ Cos}_{\gamma }(t,a) + c_{2}E_{\alpha }(t,a)\mbox{ Sin}_{\gamma }(t,a),}$$

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

Proof.

Since the characteristic roots are $$\lambda =\alpha \pm i\beta$$ , β > 0, we have that the characteristic equation is given by

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

It follows that $$p = -2\alpha$$ and $$q =\alpha ^{2} +\beta ^{2}$$ , and hence

$$\displaystyle{1 + p + q = (1-\alpha )^{2} +\beta ^{2}\neq 0.}$$

Hence, Remark 3.22 applies. By the proof of Theorem 3.23, we have that $$y(t) = E_{\alpha +i\beta }(t,a)$$ is a complex-valued solution of (3.13). Using

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

where $$\gamma = \frac{\beta } {1-\alpha }$$ , α ≠ 1, we get that

$$\displaystyle{y(t) = E_{\alpha +i\beta }(t,a) = E_{\alpha \boxplus i\gamma }(t,a) = E_{\alpha }(t,a)E_{i\gamma }(t,a)}$$

is a nontrivial solution. It follows from Euler’s formula (3.11) that

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

is a solution of (3.13). But since p and q are real, we have that the real part, y 1(t) = E α (t, a)Cos γ (t, a), and the imaginary part, y 2(t) = E α (t, a)Sin γ (t, a), of y(t) are solutions of (3.13). But y 1(t), y 2(t) are linearly independent on $$\mathbb{N}_{a}$$ , so we get that

$$\displaystyle{y(t) = c_{1}E_{\alpha }(t,a)\mbox{ Cos}_{\gamma }(t,a) + c_{2}E_{\alpha }(t,a)\mbox{ Sin}_{\gamma }(t,a)}$$

is a general solution of (3.13) on $$\mathbb{N}_{a}$$ . □ 

Example 3.26.

Solve the nabla difference equation

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

(3.18)

The characteristic equation is

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

and so, the characteristic roots are $$\lambda = -1 \pm i$$ . Note that $$1 + p + q = 5\neq 0$$ . So, applying Theorem 3.25, we find that

$$\displaystyle{y(t) = c_{1}E_{-1}(t,a)\mbox{ Cos}_{\frac{1} {2} }(t,a) + c_{2}E_{-1}(t,a)\mbox{ Sin}_{\frac{1} {2} }(t,a)}$$

is a general solution of (3.18) on $$\mathbb{N}_{a}$$ .

The previous theorem (Theorem 3.25) excluded the case when the characteristic roots of (3.13) are 1 ± i β, where β > 0. The next theorem considers this case.

Theorem 3.27.

If the characteristic values of (3.13) are 1 ± iβ, where β > 0, then a general solution of (3.13) is given by

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

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

Proof.

Since 1 − i β is a characteristic value of (3.13), we have that $$y(t) = E_{1-i\beta }(t,a)$$ is a complex-valued solution of (3.13). Now

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

It follows that

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

are solutions of (3.13). Since these solutions are linearly independent on $$\mathbb{N}_{a}$$ , we have that

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

is a general solution of (3.13). □ 

Example 3.28.

Solve the nabla linear difference equation

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

The characteristic equation is $$\lambda ^{2} - 2\lambda + 5 = 0$$ , so the characteristic roots are $$\lambda = 1 \pm 2i$$ . It follows from Theorem 3.27 that

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

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

Theorem 3.29 (Double Root).

Assume $$\lambda _{1} =\lambda _{2} = r\neq 1$$ 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 (3.13) .

Proof.

Since $$\lambda _{1} = r$$ is a double root of the characteristic equation, we have that $$\lambda ^{2} - 2r\lambda + r^{2} = 0$$ is the characteristic equation. It follows that $$p = -2r$$ and q = r 2. Therefore

$$\displaystyle{1 + p + q = 1 - 2r + r^{2} = (1 - r)^{2}\neq 0}$$

since r ≠ 1. Hence, Remark 3.22 applies. Since r ≠ 1 is a characteristic root, we have that y 1(t) = E r (t, a) is a nontrivial solution of (3.13). From Exercise 3.14, we have that $$y_{2}(t) = (t - a)E_{r}(t,a)$$ is a second solution of (3.13) on $$\mathbb{N}_{a}$$ . Since these two solutions are linearly independent on $$\mathbb{N}_{a}$$ , we have from Remark 3.22 that

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

is a general solution of (3.13). □ 

Example 3.30.

Solve the nabla difference equation

$$\displaystyle{\nabla ^{2}y(t) + 12\nabla y(t) + 36y(t) = 0,\quad t \in \mathbb{N}_{ a+2}.}$$

The corresponding characteristic equation is

$$\displaystyle{\lambda ^{2} + 12\lambda + 36 = (\lambda +6)^{2} = 0.}$$

Hence $$r = -6\neq 1$$ is a double root, so by Theorem 3.29 a general solution is given by

$$\displaystyle\begin{array}{rcl} y(t)& =& c_{1}E_{-6}(t,a) + c_{2}(t - a)E_{-6}(t,a) {}\\ & =& c_{1}7^{a-t} + c_{ 2}(t - a)7^{a-t} {}\\ \end{array}$$

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