Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 1 docx

22.5 Homogeneous Constant Coefficient Equations Homogeneous constant coefficient equations have the form a n+N + p N−1 a n+N−1 + ··· + p 1 a n+1 + p 0 a n = 0. The substitution a n = r n yields r N + p N−1 r N−1 + ··· + p 1 r + p 0 = 0 (r −r 1 ) m 1 ···(r −r k ) m k = 0. If r 1 is a distinct root then the associated linearly independent solution is r n 1 . If r 1 is a root of multiplicity m > 1 then the associated solutions are r n 1 , nr n 1 , n 2 r n 1 , . . . , n m−1 r n 1 . Result 22.5.1 Consider the homogeneous constant coefficient difference equation a n+N + p N−1 a n+N−1 + ··· + p 1 a n+1 + p 0 a n = 0. The substitution a n = r n yields the equation (r −r 1 ) m 1 ···(r − r k ) m k = 0. A set of linearly independent solutions is {r n 1 , nr n 1 , . . . , n m 1 −1 r n 1 , . . . , r n k , nr n k , . . . , n m k −1 r n k }. Example 22.5.1 Consider the equation a n+2 − 3a n+1 + 2a n = 0 with the initial conditions a 1 = 1 and a 2 = 3. The substitution a n = r n yields r 2 − 3r + 2 = (r −1)(r − 2) = 0. Thus the general solution is a n = c 1 1 n + c 2 2 n . 1174 The initial conditions give the two equations, a 1 = 1 = c 1 + 2c 2 a 2 = 3 = c 1 + 4c 2 Since c 1 = −1 and c 2 = 1, the solution to the difference equation subject to the initial conditions is a n = 2 n − 1. Example 22.5.2 Consider the gambler’s ruin problem that was introduced in Example 22.1.1. The equation for the probabili ty of the gambler’s ruin at n dollars is a n = pa n+1 + qa n−1 subject to a 0 = 1, a N = 0. We assume that 0 < p < 1. With the substitution a n = r n we obtain r = pr 2 + q. The roots of this equation are r = 1 ± √ 1 − 4pq 2p = 1 ±  1 − 4p(1 − p) 2p = 1 ±  (1 − 2p) 2 2p = 1 ± |1 − 2p| 2p . We will consider the two cases p = 1/2 and p = 1/2. 1175 p = 1/2. If p < 1/2, the ro ots are r = 1 ± (1 − 2p) 2p r 1 = 1 − p p = q p , r 2 = 1. If p > 1/2 the roots are r = 1 ± (2p − 1) 2p r 1 = 1, r 2 = −p + 1 p = q p . Thus the general solution for p = 1/2 is a n = c 1 + c 2  q p  n . The boundary condition a 0 = 1 requires that c 1 + c 2 = 1. From the boundary condition a N = 0 we have (1 − c 2 ) + c 2  q p  N = 0 c 2 = −1 −1 + (q/p) N c 2 = p N p N − q N . Solving for c 1 , c 1 = 1 − p N p N − q N c 1 = −q N p N − q N . 1176 Thus we have a n = −q N p N − q N + p N p N − q N  q p  n . p = 1/2. In this case, the two roots of the polynomial are both 1. The general solution is a n = c 1 + c 2 n. The left boundary condition demands that c 1 = 1. From the right boundary condition we obtain 1 + c 2 N = 0 c 2 = − 1 N . Thus the solution for this case is a n = 1 − n N . As a check that this formula makes sense, we see that for n = N/2 the probability of ruin is 1 − N/2 N = 1 2 . 22.6 Reduction of Order Consider the difference equation (n + 1)(n + 2)a n+2 − 3(n + 1)a n+1 + 2a n = 0 for n ≥ 0 (22.1) We see that one solution to this equation is a n = 1/n!. Analogous to the reduction of order for differential equations, the substitution a n = b n /n! will reduce the order of the difference equation. (n + 1)(n + 2)b n+2 (n + 2)! − 3(n + 1)b n+1 (n + 1)! + 2b n n! = 0 b n+2 − 3b n+1 + 2b n = 0 (22.2) 1177 At first glance it appears th at we have not reduced the order of the equation, but writing it in terms of discrete derivatives D 2 b n − Db n = 0 shows that we now have a first order difference e quation for Db n . The substitution b n = r n in equation 22.2 yields the algebraic equation r 2 − 3r + 2 = (r −1)(r − 2) = 0. Thus the solutions are b n = 1 and b n = 2 n . Only the b n = 2 n solution will give us another linearly independent solution for a n . Thus the second solution for a n is a n = b n /n! = 2 n /n!. The general solution to equation 22.1 is then a n = c 1 1 n! + c 2 2 n n! . Result 22.6.1 Let a n = s n be a homogeneous solution of a linear difference equation. The substitution a n = s n b n will yield a difference equation for b n that is of order one less than the equation for a n . 1178 22.7 Exercises Exercise 22.1 Find a formula for the n th term in the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, . . Hint, Solution Exercise 22.2 Solve the difference equation a n+2 = 2 n a n , a 1 = a 2 = 1. Hint, Solution 1179 22.8 Hints Hint 22.1 The difference equation corresponding to the Fibonacci sequence is a n+2 − a n+1 − a n = 0, a 1 = a 2 = 1. Hint 22.2 Consider this exercise as two first order difference equations; one for the even terms, one for the odd terms. 1180 22.9 Solutions Solution 22.1 We can describe the Fibonacci sequence with the difference equation a n+2 − a n+1 − a n = 0, a 1 = a 2 = 1. With the substitution a n = r n we obtain the equation r 2 − r −1 = 0. This equation has the two distinct roots r 1 = 1 + √ 5 2 , r 2 = 1 − √ 5 2 . Thus the general solution is a n = c 1  1 + √ 5 2  n + c 2  1 − √ 5 2  n . From the initial conditions we have c 1 r 1 +c 2 r 2 = 1 c 1 r 2 1 +c 2 r 2 2 = 1. Solving for c 2 in the first equation, c 2 = 1 r 2 (1 − c 1 r 1 ). We subs titute this into the second equation. c 1 r 2 1 + 1 r 2 (1 − c 1 r 1 )r 2 2 = 1 c 1 (r 2 1 − r 1 r 2 ) = 1 − r 2 1181 c 1 = 1 − r 2 r 2 1 − r 1 r 2 = 1 − 1− √ 5 2 1+ √ 5 2 √ 5 = 1+ √ 5 2 1+ √ 5 2 √ 5 = 1 √ 5 Substitute this result into the equation for c 2 . c 2 = 1 r 2  1 − 1 √ 5 r 1  = 2 1 − √ 5  1 − 1 √ 5 1 + √ 5 2  = − 2 1 − √ 5  1 − √ 5 2 √ 5  = − 1 √ 5 Thus the n th term in the Fibonacci sequence has the formula a n = 1 √ 5  1 + √ 5 2  n − 1 √ 5  1 − √ 5 2  n . It is interesting to note that although the Fibonacci sequence is defined in terms of integers, one cannot express the formula form the n th element in terms of rational numbers. 1182 [...]... yields the difference equation 1 1 an (α) + an 1 (α) = 0 4 4 n(n + 1) α(α − 1) 1 + + an 1 (α) 4 4 16 α(α − 1) + 2n + n(n − 1) + an (α) = − The first few an ’s are a0 , − α(α − 1) + 9 16 a0 , α(α − 1) + 25 16 α(α − 1) + 9 16 a0 , Setting α = 1/ 2, the coefficients for the first solution are a0 , − 5 a0 , 16 10 5 a0 , 16 The second solution has the form ∞ w2 = w1 log z + z 1/ 2 an (1/ 2)z n n=0 Differentiating... yields ∞ n=0 1 z 2 w + (1 + z)w = 0 4 ∞ 1 1 α(α − 1) + 2αn + n(n − 1) an (α)z n+α + an (α)z n+α + 4 n=0 4 12 04 ∞ an (α)z n+α +1 = 0 n=0 Divide by z α and adjust the summation indices ∞ 1 [α(α − 1) + 2αn + n(n − 1) ] an (α)z + 4 n=0 n ∞ 1 an (α)z + 4 n=0 ∞ α(α − 1) a0 + 1 a0 + 4 n =1 ∞ n α(α − 1) + 2n + n(n − 1) + an 1 (α)z n = 0 n =1 1 1 an (α) + an 1 (α) z n = 0 4 4 Equating the coefficient of z 0 to zero... the polynomial are 1 − 3 /4 3 = , 2 4 Thus our two series solutions will be of the form 1 = 1+ α2 = 1 ∞ w1 = z 3 /4 1 − 3 /4 1 = 2 4 ∞ an z n , w2 = z 1 /4 n=0 bn z n n=0 Substituting the first series into the differential equation, ∞ n=0 3 3 3 an z n + − + 2n + n(n − 1) + 16 16 16 ∞ an 1 z n = 0 n =1 Equating powers of z, we see that a0 is arbitrary and an = − 3 an 1 16n(n + 1) for n ≥ 1 This difference... j =1 = a0 3 − 16 = a0 − 3 16 3 16 j(j + 1) n n 1 j(j + 1) j =1 n 1 n!(n + 1) ! for n ≥ 1 Substituting the second series into the differential equation, ∞ − n=0 3 3 3 + 2n + n(n − 1) + bn z n + 16 16 16 12 00 ∞ bn 1 z n = 0 n =1 We see that the difference equation for bn is the same as the equation for an Thus we can write the general solution to the differential equation as ∞ w = c1 z 3 /4 − 1+ n =1 n 3 16 ∞ 1. .. the recurrence relation 0 1 a2 = − ((m + 1) am +1 p0−m + am q0−m ) 1 · 2 m=0 1 = − (1 · 1 · 1 + 1 · 1) 2 = 1 119 0 1. 2 1. 1 0.2 0 .4 0.6 0.8 1 1.2 1 .4 0.9 0.8 0.7 Figure 23 .1: Plot of the Numerical Solution and the First Three Terms in the Taylor Series Thus the solution to the problem is y(x) = 1 + x − x2 + O(x3 ) In Figure 23 .1 the numerical solution is plotted in a solid line and the sum of the first three... is graphed in a fine dashed line, and the numerical solution of w1 is graphed in a solid line The same is done for w2 11 96 1. 5 1. 5 1 1 0.5 0.5 1 2 3 4 5 1 6 -0.5 3 4 5 6 -0.5 -1 2 -1 Figure 23.3: The graph of approximations and numerical solution of w1 and w2 Result 23 .1. 1 Consider the nth order linear homogeneous equation dn w dn 1 w dw + pn 1 (z) n 1 + · · · + p1 (z) + p0 (z)w = 0 dz n dz dz If...Solution 22.2 We can consider 2 an , a1 = a2 = 1 n to be a first order difference equation First consider the odd terms an+2 = a1 = 1 2 a3 = 1 22 a5 = 31 an = 2(n 1) /2 (n − 2)(n − 4) · · · (1) For the even terms, a2 = 1 2 a4 = 2 22 a6 = 42 an = 2(n−2)/2 (n − 2)(n − 4) · · · (2) Thus an = 2(n 1) /2 (n−2)(n 4) ··· (1) 2(n−2)/2 (n−2)(n 4) ···(2) 11 83 for odd n for even n Chapter 23 Series Solutions of... cn cn +4 = − (n + 4) (n + 3) for n ≥ 2 For our first solution we have the difference equation a0 = 1, a1 = 0, a2 = 0, a3 = 0, an +4 = − b0 = 0, b1 = 1, b2 = 0, b3 = 0, bn +4 = − an (n + 4) (n + 3) For our second solution, bn (n + 4) (n + 3) The first few terms in the fundamental set of solutions are w1 = 1 − 1 4 1 8 z + z − ··· , 12 672 w2 = z − 1 5 1 9 z + z − ··· 20 14 4 0 In Figure 23.3 the five term approximation... differential equation as ∞ w = c1 z 3 /4 − 1+ n =1 n 3 16 ∞ 1 zn n!(n + 1) ! + c2 z 1 /4 n =1 ∞ c1 z 3 /4 + c2 z 1 /4 1+ n =1 23.2 .1 − 1+ 3 − 16 n 3 16 n 1 zn n!(n + 1) ! 1 zn n!(n + 1) ! Indicial Equation Now let’s consider the general equation for a regular singular point at z = 0 p(z) q(z) w + 2 w = 0 z z Since p(z) and q(z) are analytic at z = 0 we can expand them in Taylor series w + ∞ ∞ n p(z) = pn z , n=0 Substituting... by 1 − z 2 to obtain (1 − z 2 )w + z (1 − z)w + w = 0 Example 23 .1. 6 Find the series expansions about z = 0 of the fundamental set of solutions for w + z 2 w = 0 11 94 0.2 0 .4 0.6 1 0.8 1. 2 0.9 0.8 0.7 0.6 0.5 0 .4 0.3 Figure 23.2: Plot of the solution and approximations Recall that the fundamental set of solutions {w1 , w2 } satisfy w1 (0) = 1 w1 (0) = 0 Thus if w2 (0) = 0 w2 (0) = 1 ∞ ∞ an z n w1 = and . c 1 r 1 )r 2 2 = 1 c 1 (r 2 1 − r 1 r 2 ) = 1 − r 2 11 81 c 1 = 1 − r 2 r 2 1 − r 1 r 2 = 1 − 1 √ 5 2 1+ √ 5 2 √ 5 = 1+ √ 5 2 1+ √ 5 2 √ 5 = 1 √ 5 Substitute this result into the equation for c 2 . c 2 = 1 r 2  1. + 1) a m +1 p 0−m + a m q 0−m ) = − 1 2 (1 · 1 · 1 + 1 · 1) = 1 119 0 0.2 0 .4 0.6 0.8 1 1.2 1 .4 0.7 0.8 0.9 1. 1 1. 2 Figure 23 .1: Plot of the Numerical Solution and the First Three Terms in the Taylor. have c 1 r 1 +c 2 r 2 = 1 c 1 r 2 1 +c 2 r 2 2 = 1. Solving for c 2 in the first equation, c 2 = 1 r 2 (1 − c 1 r 1 ). We subs titute this into the second equation. c 1 r 2 1 + 1 r 2 (1 − c 1 r 1 )r 2 2 =