is an example of a first-order difference equation. The symbols L 1 (), L 2 () are operator
symbols representing linear operators. Using the operator E, the above equations
can be written as
L 2 (yn) = ∆^2 yn= (E−1)^2 yn=yn+2 − 2 yn+1 +yn= 0 and
L 1 (yn) = ∆ yn− 3 yn= (E−1)yn− 3 yn=yn+1 − 4 yn= 0,respectively.
There are many instances where variable quantities are assigned values at uni-
formly spaced time intervals. Let us study these discrete variable quantities by
using differences and difference equations. An equation which relates values of a
function yand one or more of its differences is called a difference equation. In deal-
ing with difference equations one assumes that the function y and its differences
∆yn, ∆^2 yn,.. ., evaluated at xn, are all defined for every number x in some set of
values {x 0 , x 0 +h, x 0 + 2h,... , x 0 +nh,.. .}.A difference equation is called linear and
of order mif it can be written in the form
L(yn) = a 0 (n)yn+m+a 1 (n)yn+m− 1 +··· +am− 1 (n)yn+1 +am(n)yn=g(n), (10 .54)where the coefficients ai(n), i = 0, 1 , 2 ,... , m, and the right-hand side g(n)are known
functions of n. If g(n)= 0, the difference equation is said to be nonhomogeneous and
if g(n) = 0 , the difference equation is called homogeneous.
The difference equation (10.54) can be written in the operator form
L(yn) = [a 0 (n)Em+a 1 (n)Em−^1 +···+am− 1 (n)E+am(n)]yn=g(n),where Eis the stepping operator.
A mth-order linear initial value problem associated with a mth-order linear
difference equation consists of a linear difference equation of the form given in the
equation (10.54) together with a set of minitial values of the type
y 0 =α 0 , y 1 =α 1 , y 2 =α 2 ,... , y m− 1 =αm− 1 ,where α 0 , α 1 ,... , α m− 1 are specified constants.
Example 10-34.
Show ∆ak= (a−1)ak, for aconstant and kan integer.
Solution: Let yk=ak,then by definition
∆yk=yk+1 −yk=ak+1 −ak= (a−1)ak.