Difference Equations Assignment Help

Difference Equation Introduction

The difference equation is a formula for computing an output sample at time n based on past and present input samples and past output samples in the time domain.

Classication of difference Equations:

• First-order equations
• Second-order equations

First-order difference equations:

A general first-order difference equation takes the form

x_t = f (t, x_t1) for all t.

We can solve such an equation by successive calculation: given x₀ we have

x1 = f (1, x0)
x2 = f (2, x1) = f (2, f (1, x0))
and so on.

First-order linear difference equations with constant coefficient

A first-order difference equation with constant coefficient takes the form x_t = ax_t1 + b_t where b_t for t = 1, are constants.

Equilibrium

In general, the starting point x₀, the value of x_t changes with t.

x* = b/(1 a)

We call x* the equilibrium value of x. We can write the solution as x_t = a^t(x0 x*) + x*.

Qualitative behaviour

The qualitative behaviour of the solution path depends on the value of a.

|a| < 1

x_t converges to x*: the solution is stable. There are two sub cases:

0 < a < 1

Monotonic convergence.

1 < a < 0

Damped oscillations.

|a| > 1

Divergence:

a > 1

Explosion.

a < 1

Explosive oscillations.

First-order linear difference equations with variable coefficient

The solution of the equation x_t = a_tx_t1 + bt. can be found, as before, by successive calculation. We obtain

x_t = (_s=1_ta_s)x0 + k=1^t(_s=k+1^ta_s)b_k

Where a product with no terms (e.g. from t+1 to t) is 1.

Second-order difference equations

A general second-order difference equation takes the form x_t+2 = f (t, x_t, x_t+1).

As for a first-order equation, a second-order equation has a unique solution: by successive calculation we can see that given x0 and x1 there exists a uniquely determined value of x_t for all t 2.

Example

Consider the equation

xt+2 5xt+1 + 6xt = 2t 3.

The associated homogeneous equation is

xt+2 5xt+1 + 6xt = 0.

Two solutions of this equation are ut = 2t and vt = 3t. These are linearly independent:

To find a solution of the original equation we can takes the form u*t = at + b.

In order for u*t to be a solution we need

a(t+2) + b 5[a(t+1) + b] + 6(at + b) = 2t3.

Equating coefficients, we have a = 1 and b = 0.

Thus u*t = t is a solution.

We conclude that the general solution of the equation is xt = A2t + B3t + t.

Second-order linear equations with constant coefficients:

A second-order linear equation with constant coefficients takes the form

x_t+2 + ax_t+1 + bx_t = c_t.

The strategy for solving such an equation is very similar to the strategy for solving a second-order linear differential equations with constant coefficients. We first consider the associated homogeneous equation

xt+2 + axt+1 + bxt = 0.

The homogeneous equation:

We need to find two solutions of the homogeneous equation

xt+2 + axt+1 + bxt = 0.

We can guess that a solution takes the form u_t = m^t. In order for ut to be a solution, we need

m^t(m² + am + b) = 0

or, if m 0,

m² + am + b = 0.

This is the characteristic equation of the difference equation. Its solutions are

(1/2)a ((1/4)a² b).