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
(1.12)
where we assume that p(t) ≠ q(t) + 1, for In this section we will see that we can easily solve the corresponding second order linear homogeneous equation with constant coefficients
(1.13)
where we assume the constants 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(t) ≠ 1 + q(t), , and Then the IVP
(1.14)
has a unique solution y(t) on
Proof.
Expanding equation (1.12) out we have first solving for y(t + 2) and then solving for y(t) that
(1.15)
and, since p(t) ≠ 1 + q(t), ,
(1.16)
If we let t = t 0 in (1.15), then equation (1.12) holds at t = t 0 iff
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 . On the other hand, if t 0 > a, then using equation (1.16) with t = t 0 − 1, we have that
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 Hence the result follows. □
Remark 1.30.
It follows from Theorem 1.29, that if p(t) ≠ 1 + q(t), , then the general solution of the linear homogeneous equation
is given by
where y 1(t), y 2(t) are any two linearly independent solutions of (1.13) on
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 (possibly complex) are solutions (called the characteristic values of (1.13) ) of the characteristic equation
Then
is a general solution of (1.13) .
Proof.
Assume are characteristic values of (1.13). Then the characteristic equation of (1.13) is given by
It follows that and . Hence , since p ≠ q + 1. Hence, we have that and so , i = 1, 2, are well defined. Since
we have that , i = 1, 2, are solutions of (1.13). Since these two solutions are linearly independent on , we have that
is a general solution of (1.13) on . □
Example 1.32 (Fibonacci Numbers).
The Fibonacci numbers F(t), t = 1, 2, 3, ⋯ are defined recursively by
The Fibonacci sequence is given by
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
To solve this IVP we first get that the characteristic equation is
Hence, the characteristic values are
It follows that
(1.17)
Applying the initial conditions we get the system
Solving this system for c 1 and c 2, using (1.17) and simplifying we get
for Note F(t) is the integer nearest to . It follows that F(20) is the integer nearest
which gives us that F(20) = 6765.
Usually, we want to find all real-valued solutions of (1.13). When a characteristic root is complex, 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 .
Theorem 1.33 (Complex Roots).
If the characteristic values are , β > 0 and α ≠ − 1, then a general solution of (1.13) is given by
where
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
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
we get that
It follows from the (delta) Euler’s formula (1.11) that
is a solution of (1.13). But since p and q are real, we have that the real part, , and the imaginary part, , of y(t) are solutions of (1.13). Since p ≠ q + 1 and y 1(t), y 2(t) are linearly independent on , we get from Remark 1.30 that
is a general solution of (1.13) on . □
Example 1.34.
Solve the difference equation
(1.18)
The characteristic equation is
and so the characteristic values are . Hence using Theorem 1.33, we get
is a general solution of (1.18) on .
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
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
It follows that
are solutions of (1.13). Since these solutions are linearly independent on , we have that
is a general solution of (1.13). □
Example 1.36.
Solve the delta linear difference equation
The characteristic equation is , so the characteristic values are . It follows from Theorem 1.35 that
is a general solution on
Theorem 1.37 (Double Root).
Assume p ≠ 1 + q, and is a double root of the characteristic equation. Then
is a general solution of (1.13) .
Proof.
Since is a characteristic value, we have y 1(t) = e r (t, a) is a solution of (1.13). Since , we have that the characteristic equation for (1.13) is
Hence, in this case, (1.13) has the form
From Exercise 1.32, we have that y 2(t) = (t − a)e r (t, a) is a second solution of (1.13) on Since these two solutions are linearly independent on , we have that
is a general solution of (1.13). □
Example 1.38.
Evaluate the t by t determinant of the following tridiagonal matrix
for For example, we have that
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
Note that D(1) = 4 and D(2) = 12. It follows that if we define D(0) = 1, then we have
It then follows that D(t) is the solution of the IVP
The characteristic equation is , and thus are the characteristic values. Hence by Theorem 1.37,
Using the initial conditions we get the system of equations
Solving this system we get that c 1 = c 2 = 1 and hence
The reader should check this answer for a few values of t.
Example 1.39.
Find the determinant D(t), 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
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
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
Applying the initial conditions we get
for
Sometimes it is convenient to know how to solve the second order linear homogeneous difference equation when it is of the form
(1.19)
where 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
Then
is a general solution of (1.19) .
Example 1.41.
Solve the difference equation
(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
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
is a general solution of (1.19) .
Example 1.43.
Solve the difference equation
(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
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
where and if α ≠ 0 and 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 if α ≠ 0 and let if α = 0. Then if we have that
Since the real and imaginary parts of y(t) are linearly independent solutions of (1.19), we have that
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 we define the operator E by
Then E n : = E ⋅ E n−1 for We say the polynomial in E,
annihilates provided p(E)f(t) = 0 for Similarly we say the polynomial in the operator ,
annihilates provided
Example 1.45.
Here are some simple annihilators for various functions:
(i)
(E − rI)r t = 0;
(ii)
(iii)
(E − rI)2 tr t = 0;
(iv)
(v)
(vi)
(vii)
(viii)
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
(1.22)
First we write this equation in the form
Since E − 5I annihilates the right-hand side, we multiply each side by the operator E − 5I to get
It follows that a solution of (1.22) must be of the form
Substituting this into equation (1.22) we get
Hence we see that we must have . It follows that
is a general solution of (1.22).
Example 1.47.
Solve the second order linear nonhomogeneous difference equation
(1.23)
by the annihilator method. The difference equation (1.23) can be written in the form
Multiplying both sides by the operator E − 4I we get that
Hence y(t) must have the form
Substituting this into the difference equation (1.23) we get after simplification that 8c 3 = 16. Hence c 3 = 2 and we have that
is a general solution of (1.23).
Also we can use the method of annihilators to solve certain nonhomogeneous equations of the form
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
(1.24)
The equation (1.24) can be written in the form
Multiplying both sides by the operator we get that solutions of (1.24) are solutions of
The values of the characteristic equation are . Hence all solutions of (1.24) are of the form
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