The Computable Differential Equation Lecture ... - Bruce E. Shapiro

120 CHAPTER 6. LINEAR MULTISTEP METHODS
Difference Equations
Any linear equation relating variables y 0 , y 1 , ... of the form
k∑
a j y j = g (6.27)
j=0
is called an (inhomogeneous) difference equation. The corresponding homogeneous
difference equation is
k∑
a j y j = 0 (6.28)
j=0
There are k linearly independent solutions to 6.28; the general solution of the homogeneous
equation is the sum of these linearly independent terms. To find the terms
of the homogeneous equation we substitute y j = r j to obtain the characteristic
equation of the difference equation
k∑
a j r j = 0 (6.29)
j=0
Corresponding to each root r of multiplicity s is a term
a k = r k (c 1 + c 2 k + c 3 k 2 + · · · + c s k s−1 ) (6.30)
in the homogeneous solution. The solution to the inhomogeneous equation is the
sum of the homogeneous solution and a particular solution to the full equation.
Example 6.1. Solve a k+1 − 2a k = 3 k + 5k.
Solution. The homogeneous (linear) part of the equation is a k+1 − 2a k = 0; the
characteristic polynomial is r k+1 − 2r k = 0, which has a single nonzero root of
r = 2. (We ignore the zero roots because we always renormalize to k = 0 in the
characteristic equation; if zero-valued roots still remained after this division by r k
we would have to include them in the solution). For a particular solution we guess
a k = A(3) k +5Bk+C where A, B, C are unknown constants. We guess a polynomial
to satisfy the 5k term and the exponential 3 k to satisfy the corresponding term in
the difference equation. To determine the coefficients we substitute
3 k + 5k = a k+1 − a k = (3 k+1 A + 5B(k + 1) + C) − 2(3 k A + 5Bk + C) (6.31)
= 3 k A − 5Bk + 5B − C (6.32)
Equating coefficients of linearly independent terms gives A = 1, B = −1, and
C = −5, so the solution is a k = 2 k + 3 k − 5k − 5.
Math 582B, Spring 2007
California State University Northridge
c○2007, B.E.Shapiro
Last revised: May 23, 2007