Linear Difference Equations

[alext87]

I am having a bit of problem with finding the particular function for linear difference equations. The method of finding the particular function is similar to the method for finding the particular integral in linear differiational equations. Can anybody help?

[the tree]

I'm assuming from context that you mean differencial equations. What problem are you having with what functions?

[woelen]

Could you provide us with an example of a difference equation which needs to be solved?

 

Indeed, the methodology for solving linear difference equations is somewhat similar to the methodology for solving linear differential equations. Main difference is the stability criterion, which for linear differential equations is that the real part of the eigenvalues must be less than 0, for the difference equations the absolute value of the eigenvalues must be less than 1.

 

Also, the solutions for homogeneous differential equations are exponentials of the eigenvalues, while for difference equations you get powers of the eigenvalues.

[alext87]

Yx - aYx-1 = cx + d

 

This is one of the different equations and the particular function is rather complicated! Have a go at solving it please. Thanks!

[woelen]

This particular function indeed is quite complex, but the mechanism of solving it is not that difficult.

 

First take the Z-transform of the equation, regarding x as discrete time:

 

Y - aY/z = cz/(z-1)² + dz/(z-1)

 

Y can be expressed as function of z:

 

Y = c/((z-1)²(z-a)) + d/((z-1)(z-a))

 

This can be written as (c- d + dz)/((z-1)²(z-a)), which in turn can be written as P(z)/(z-1)² + Q(z)/(z-a). Here P and Q are polynomials in z. P is a quadratic polynomial in z, Q is first degree in z.

 

So, in the Z-domain your solution Y(z) has only terms of the form C1/(z-1)², C2z/(z-1)² and C3z²/(z-1)² and terms of the form C4/(z-a) and C5z/(z-a). Here, the C's are all constant (of course, different for each term).

 

Now you need to transform back to the time domain. The value z/(z-1)² has backtransform x. Then of course, the values multiplied with powers of z are simple. These simply are shifted versions of x in time. E.g. 1/(z-1)² has backtransform x, but at time x - 1. This simply is x - 1, and similarly z²/(z-1)² has backtransform x + 1.

 

The term z/(z-a) has backtransform a^x. Again, 1/(z-a) then has the same backtransform, one timestep shifted, that is a^(x-1).

 

Now it is just plain algebra, collecting all the terms. That exercise I leave to you .