## R1 11 v

Note the restriction «i; ^ 0. If this were not true, then y,+ l would not depend on .v,7 and we could then solve it directly as a single, lirst-order difference equation. Using this to substitute for.v, gives

/»+1 — «11 y, \ Vi+2 = «11 V/H-t 4-0|2«2l>'i -FapiJ22( -

Alter simplifying and rearranging this becomes yl+2 - (fl|| -F «22)Jin-1 + (fl|ifl22 - Oi2«21 ).V, = 0

which is a homogeneous, linear, second-order difference equation with constant coefficients. Equations like this were solved in chapter 20; theorem 20.2 gives the solution (for the case of real and distinct mots):

where C\ and C; are arbitrary constants (the values of which arc determined from initial conditions) and r\ and r2 are the roots of the characteristic equation for this second-order difference equation, which is r2 — («h -F u22)r -F («IIÜ22 - rti2«2i) = 0

To lind x,, we need only substitute the solution for y, back into equation (24.34). This gives

a 12

Simplifying this expression gives x, =-C,r{ + ——— Ciri (24.36)

a 12 fli2

Together, equations (24.35) and (24.36) give the solutions to the homogeneous difference equation system in (24.32) and (24.33) for the case of real and distinct roots.

If the roots turn out to be real and equal or complex valued, we would use the appropriate solution to the second-order differential equation for y,, followed by substitution to obtain the solution for x,.

Example 24.18 Solve the following system of homogeneous difference equations:

Solution

The first difference equation implies that yi+2

Substitute for.r,+| to get yt+2 = 6y,+i + 8(y, + x,)

Use the first equation again to get the following expression for x,

and use it to substitute for x,. This gives

Rearrange this equation to get

_v,+2 - 7.v,-i-1 - 12y, = 0 The characteristic equation is r2 - Ir - 2 = 0 with real and distinct roots rt, r2 = 7/2 ± v/57/2. The solution is y, =Cyr\ + C2r'2

I"he solution for x, is obtained by substituting the solution lor y, back into the expression for*,. This gives

After simplifying, this equations becomes n — 6 „ . r-> — 6 „ , = —C""i + —^

The Direct Method

The direct method offers an approach that can be extended quite easily to the case of more than two difference equations.

In matiix form the system ofhoinogeneous difference equations is yr+l = Ay, where A is an n x n matrix of constant coefficients and y,+ i and y, are vectors of n variables. By analogy with the solution for a single difference equation, we postulate the solution to be

where k is a vector of arbitrary constants and r is a scalar. If correct, this solution must satisfy the system of homogeneous difference equations. Let us substitute the proposed solution into the difference equations and see if it does indeed satisfy them:

After simplifying and ruling out the trivial solution r — 0, this becomes where / is the identity matrix and 0 is the zero-vector. If we choose r to solve this system of equations, then our proposed solution works! As in the case of differential equations, the solution values of r are found by solving which is a polynomial of degree n in the unknown number r. As before, this is the characteristic equation of malrtx A and its solutions are the characteristic roots (eigenvalues) of matrix A. A nonzero vector. k|. which is a solution of equation (24.37) for a particular eigenvalue, /■,, is called the eigenvector of matrix A corresponding to the eigenvalue r,.

In general, there are n equations and n characteristic roots: therefore, there are n solutions to the system of difference equations. The general solution to the homogeneous form is a linear combination of n distinct solutions.

Example 24.19 Solve the following system of homogeneous difference equations

Solution

The characteristic roots are tire solution to

0 0