2.3. Difference Equations

For some sequences, their algebraic expressions are directly given, such as the sequences we defined in (2.6) and (2.7). While for some other sequences, a member of a sequence \(y_n\) is determined by one or several previous members of the sequence, e.g.

(2.8)\[\begin{align} & y_n = y_{n-1}, \\ & y_n = y_{n-1}+3, \\ & y_n = 2y_{n-1}, \\ & y_n = y_{n-1} + y_{n-2}. \end{align}\]

We can call these algebraic expressions Difference Equations.

Definition 2.6 (Difference Equation)

A difference equation is an equation that defines a sequence recursively: each member of the sequence is defined as a function of previous terms of the sequence

(2.9)\[ y_{n+k} = F(n, y_{n}, y_{n+1}, y_{n+2} ..., y_{n+k-2}, y_{n+k-1}), \quad n \in \mathbb{N}_0 \]

where \(y_{n+k}\) is the current value in the sequence, \(F\) is a function, and \(k\) is the order of the equation. The subscript \(n+k\) is an index representing the current step, and the variables \(y_{n}\), \(y_{n+1}\), \(\ldots\), \(y_{n+k-1}\) represent previous members in the sequence.

Remark 2.2 (Order of Difference Equations)

The order of a difference equation \(k\) is calculated as the difference between the highest and lowest (subscript) indices that appear in the difference equation.

If the highest subscript in a difference equation is \(n+5\) and the lowest subscript in the equation is \(n-3\), then the order of the difference equation is \(k=8\).

Remark 2.3 (Equivalent Difference Equations)

For a given difference equation, if we increase or decrease the subscript of each term by the same amount, the new equation actually represent the same sequence as the old equation does. Also, the properties of the new equation remain the same as the old one. So the new difference equation is equivalent to the old one (the roots of the two equations are the same). For example, we can increase the subscripts in the equation

\[y_n = y_{n-1} + y_{n-2}\]

by \(2\) to get an equivalent equation

\[y_{n+2} = y_{n+1} + y_{n}\]

Theorem 2.1 (Existence and Uniqueness)

Equation (2.9) has one and only one solution corresponding to each arbitrary selection of the \(k\) initial values \(y_0, y_1, y_2, \ldots, y_{k-1}\).

Definition 2.7 (Linear Difference Equation)

A difference equation is linear if it can be expressed in the form

(2.10)\[ a_0(n) y_n + a_1(n) y_{n+1} + a_2(n) y_{n+2} + \ldots + a_k(n) y_{n+k} = R(n), \]

where \(a_p(n), p=1,2,\ldots, k\) and \(R(n)\) are given functions of \(n\).

If a difference equation involves non-linear terms of \(y_n\):

• power \(y_n^2\), \(y_n^3\), …

• exponent \(2^{y_n}\), \(e^{y_n}\), …

• trigonometry \(\sin (y_n)\), \(\cos (y_n)\), \(\tan (y_n)\) …

then it is a non-linear difference equation.

Definition 2.8 (Homogeneous Linear Difference Equation)

When the right hand side of Eq. (2.10) vanishes i.e. \(R(n)=0\), we get

(2.11)\[ a_0(n) y_n + a_1(n) y_{n+1} + a_2(n) y_{n+2} + \ldots + a_k(n) y_{n+k} = 0, \]

which is called a homogeneous linear difference equation.

Example 2.2

Identify the order, linearity and homogeneity of the following difference equations:

• \(y_{n+1} - 2 y_n = 0\)

• \(y_{n+1} - 3 y_n + y_{n-1} = e^{-n}\)

• \(y_{n+3} - y_n^2 = n 2^n\)

Solution

\[\begin{split} \begin{align*} &y_{n+1} - 2 y_n = 0 & \text{(first order, linear, homogeneous)} \\ &y_{n+1} - 3 y_n + y_{n-1} = e^{-n} & \text{(second order, linear, nonhomogeneous)} \\ &y_{n+3} - y_n^2 = n 2^n & \text{(third order, nonlinear, nonhomogeneous)} \end{align*} \end{split}\]

Definition 2.9 (Shift Operator)

We define \(\E\) as an operator shifting a member in a sequence to the next member

\[ \E y_n = y_{n+1}. \]

If we apply the shift operator consecutively for \(2\) times, we write it as

\[ \E ^2 y_n = \E (\E y_n) = \E y_{n+1} = y_{n+2}. \]

If we apply the shift operator consecutively for \(k\) times, we write it as

\[ \E ^k y_n = y_{n+k}. \]

Example 2.3

• Use the definition of the shift Operator \(\E\) to expand the following expression
  \(\left(\E ^4 - \E ^2 +1 \right) y_n \)

• Write the following expression into a shift operator \(\E\) form
  \( y_{n+2} - 2y_{n+1} + y_n \)

Solution

• \(\left(\E ^4 - \E ^2 +1 \right) y_n = y_{n+4} - y_{n+2} + y_n\)

• \( y_{n+2} - 2y_{n+1} + y_n = \left( \E ^2 - 2\E + 1\right) y_n \)