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for llic second-order difference equation? Thus suppose that we try y, = Ar'

as a solution to equation (20.2). If this is a solution, then it has to satisfy equation (20.2). To see if it does, first note that the proposed solution implies that y,_i = Ar'+] and y,+2 = Ar'+2. Substituting these and our proposed expression for y, into equation (20.2) gives

Simplifying gives

Our proposed solution will therefore work provided that we choose values for r that satisfy the quadratic equation (r~ -Fair +«2) - 0 (since we rule out the trivial solutions r = 0 and A = 0). This quadratic equation is known as the characteristic equation of difference equation (20.2).

Definition 20.3

The characteristic equation of the linear, second-order difference equation with constant coefficients is r- -F atr -F «2 = 0

The values of r that solve the characteristic equation are known as the roots (or eigenvalues or characteristic roots) of the characteristic equation. There are two roots that solve this equation; we will call them /-| and r2. Their values are given in equation (20.6).

Suppose the two roots that solve the characteristic equation are real valued and different. Then we have actually found two solutions that satisfy equation (20.2). They are y!'' = A, r\ and y,'I> = A 2r'2 (20.71

Let's confirm Utat yis a solution to equation (20.2) (and we leave it to the reader to do the same for v,12'). Given y''1 in equation (20.7). then

Substituting these values into equation (20.2) gives yiii+ai^, +a2y,m = A^ + mAtf1 +a2Alr[ = A|r[(/-f + t/,ri + a2) = 0

The final equality follows because we know ihat n solves the characteristic equation. Therefore y,n) satisfies equation (20.2) and is a solution.

Although having two solutions may appear to present a problem, since we are. after all, looking for one general solution, in fact the general solution is a linear combination of two linearly independent solutions. As a result we actually require two solutions. Intuitively the reason for this is that two linearly independent solutions are required in order to recover the two constants that are lost in taking the first and then the second difference of the underlying equation for v,. The real problem arises therefore in the case of real-valued but equal roots. That is, when a\ - 4</: = 0, so that n - r2, we appear to have just one solution. However, it is possible even in this case to find a second distinct solution. Rather than derive it. we state it and then verify thai it is correct.

I(Yi = r2 = r. the two distinct solutions are y!" = A,r' and yf = A2tr> (20.8)

These are linearly independent of one another (and are therefore distinct) because one cannot be made equal to the other by multiplying ii by any constant. It is possible to verify that both of these are solutions to equation (20.2) by substitution as was done above. We do this for the second Solution.

I'll\ = Aiit + 1 )r'+l and = A2it + 2 >r'+: Substituting these values into equation (20.2) gives

+ 1 + «2>V2' = A2(t + 2)r'+1 + ,hA2(i -h I >r'+l +a2A:lr' = Air11 (r +2)r2 + fl|(/ + 1 )r + cr>r| = A2r'fr(r + a,r-h(i2) 4- r(2r + a{ )] = A2r'\0 4- /•<-«, 4-fl()l = 0

The second-to-last equality follows from the previous one because r solves the characteristic equation and because the case of equal roots arises only when (if — 4(7i = 0. which means that r = -a\/2.

We now have derived two solutions to equation (20.2) for the case of real-valued, different roots and for the case of real-valued but equal roots. As we mentioned above, the general solution is obtained by taking a linear combination of the two distinct solutions. We slate this formally:

Theorem 20.3

Lety»'" and y'2' be the two solutions to equation (20.2) given by equation (20.7)

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