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HW8_solutions

# HW8_solutions - Math 167 homework 8 solutions 5.3.2...

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Unformatted text preview: Math 167 homework 8 solutions November 27, 2010 5.3.2: Bernadelli studied a beetle "which lives three years only, and propagates in its third year." They survive the first year with probability 1 , and 2 the second with probability 1 , and then produce six females on the way out: 3 0 0 6 Beetle matrix A = 1 0 0 . 2 1 0 3 0 Show that A3 = I, and follow the distribution of 3000 beetles for six years. Since 0 2 0 A2 = 0 0 3 , 1 0 0 6 we have A3 = A2 A = I. After 6 years the population distribution for the 3000 beetles remains unchanged. This follows from A6 = I. 5.3.4: Suppose each "Gibonacci" number Gk+2 is the average of the two previous numbers Gk+1 and Gk . Then Gk+2 = 1 (Gk+1 + Gk ): 2 Gk+2 = 1 (Gk+1 + Gk ) 2 Gk+1 = Gk+1 is Gk+2 Gk+1 =A Gk+1 Gk . (a) Find the eigenvalues and eigenvectors of A. The matrix A= 1 1 2 1 2 1 0 has eigenvalues 1 = 1 and 2 = - 1 with eigenvectors 2 1 1 respectively. (b) Find the limit as n of the matrices An = Sn S -1 . Since 2 1 1 0 1 1 3 3 An = , (-1)n 1 -1 1 -2 0 2n 3 3 it follows that lim A = n and 1 -2 , 1 1 1 0 0 0 n 1 -2 2 3 1 3 1 3 -1 3 = 2 3 2 3 1 3 1 3 . 2 (c) If G0 = 0 and G1 = 1, show that the Gibonacci numbers approach 3 . Since Gn+1 Gn =A Gn Gn-1 = A2 Gn-1 Gn-2 = = An G1 G0 , it follows from part (b) that lim Gn+1 Gn = lim A n n G1 G0 n = 2 3 2 3 1 3 1 3 1 0 = 2 3 2 3 , and thus limn Gn = 2 . 3 5.3.8: Suppose there is an epidemic in which every month half of those who are well become sick, and a quarter of those who are sick become dead. Find the steady state for the corresponding Markov process 1 1 0 dk+1 dk 4 1 sk+1 = 0 3 2 sk . 4 wk+1 0 0 1 2 wk 2 The eigenvalues for the Markov matrix are are 1 = 1, 2 = 3/4, and 3 = 1/2 with eigenvectors 1 -1 1 x1 = 0 , x2 = 1 , and x3 = -2 , 0 0 1 respectively. Thus the steady state solution is d s = c1 x1 , w where c1 is the first component of the solution c of the linear system 1 -1 1 c1 d0 0 c2 = s0 . x1 x2 x3 c = 1 -2 0 0 1 c3 w0 Solving this system gives c1 = d0 + s0 + w0 , and thus d = d0 + s0 + w0 , s = 0, and w = 0. 5.3.10: Find the limiting values of yk and zk (k ) if yk+1 = .8yk + .3zk zk+1 = .2yk + .7zk y0 = 0 z0 = 5. Also find formulas for yk and zk from Ak = Sk S -1 . .8 .3 . The matrix A has eigenvalues 1 = 1 and 2 = .5 with Let A = .2 .7 eigenvectors -1 3 , and 1 2 respectively. It follows that Ak = and thus 1 5 3+ 2- 1 2k-1 1 2k-1 3- 2+ 3 2k 3 2k , 1 3 3 y y = lim Ak 0 = z0 z k 5 2 2 3 3 0 . = 2 5 0 -1 u (eigenvalues i and -i) 1 0 goes around in a circle: u = (cos t, sin t). Suppose we approximate du/dt by forward, backward, and centered differences F, B, C: 5.3.16: The solution to du/dt = Au = (F) un+1 - un = Aun or un+1 = (I + A)un (this is Euler's method). (B) un+1 - un = Aun+1 or un+1 = (I - A)-1 un (backwards Euler). 1 (C) un+1 - un = 1 A(un+1 + un ) or un+1 = (I - 1 A)-1 (I + 2 A)un . 2 2 Find the eigenvalues of I + A, (I - A)-1 , and (I - 1 A)-1 (I + 1 A). For which 2 2 difference equation does the solution un stay on a circle? (F): The eigenvalues of (I + A) are 1 = 1 + i and 2 = 1 - i. Note that the eigenvalues imply that (I + A) cannot be an orthogonal matrix. The magnitude of un must grow with n, thus the un 's are not on a circle. 1 (B): The eigenvalues of (I - A)-1 are 1 = 2 (1 + i) and 2 = 1 (1 - i). The 2 eigenvalues imply that the magnitude of un is decreasing with n. Thus the un 's are not on a circle. 1 (C): The eigenvalues of (I - 1 A)-1 (I + 2 A) are 1 = .6+.8i and 2 = .6-.8i. 2 Note that the matrix is an orthogonal matrix, thus the magnitude of the un 's is constant. This means that the un 's stay on a circle. 5.3.22: What are the limits as k (the steady states) of the following? .4 .2 .6 .8 Note that the matrix A= k 1 0 , .4 .2 .6 .8 k 0 1 , .4 .2 .6 .8 k . .4 .2 .6 .8 has the eigenvalues 1 = .2 and 2 = 1 with eigenvectors -1 1 respectively. It follows that Ak = 1 4 -1 1 1 3 .2k 0 0 1 4 -3 1 1 1 , and 1 3 , and thus k lim Ak = 1 4 -1 1 1 3 0 0 0 1 1 0 0 1 -3 1 1 1 1 4 1 4 1 3 1 3 , . = 1 4 1 1 3 3 . In particular, k lim Ak = and k lim Ak = 5.4.2: For the matrix A= write the general solution to du/dt = Au, and the specific solution that matches u(0) = (3, 1). What is the steady state as t ? (This is a continuous Markov process; = 0 in a differential equation corresponds to = 1 in a difference equation, since e0t = 1.) The eigenvalues of A are 1 = 0 and 2 = -2 with eigenvectors x1 = 1 1 and x2 = 1 -1 , -1 1 1 -1 respectively. Thus the general solution u(t) is given by u(t) = c1 e1 t x1 + c2 e2 t x2 = c1 + c2 e-2t c1 - c2 e-2t , where c1 , c2 R. If u(0) = (3, 1) is the initial condition, then c1 = 2 and c2 = 1, and so the particular solution is u(t) = The steady state solution is t 2 + e-2t 2 - e-2t 2 2 . lim u(t) = . 5.4.4: If P is a projection matrix, show from the infinite series that eP I + 1.718 P. 5 As a projection matrix, P satisfies the relation P 2 = P , which implies P n = P for all n 2. It follows that eP = I + =I =I =I =I P P2 P3 Pn + + ++ + 1! 2! 3! n! P P P P + + + ++ + 1! 2! 3! n! 1 1 1 1 + + ++ + P + 1! 2! 3! n! 1 2 1 1 13 1n + -1 + 1 + + + ++ + 1! 2! 3! n! + -1 + e1 P I + 1.718 P. P 5.4.6: The higher order equation y + y = 0 can be written as a first-order system by introducing the velocity as y another unknown : d dt y y = y y = y -y . If this is du/dt = Au, what is the 2 by 2 matrix A? Find its eigenvalues and eigenvectors, and compute the solution that starts from y(0) = 2, and y (0) = 0. The matrix is 0 1 . A= -1 0 It has the eigenvalues 1 = i and 2 = -i with eigenvectors x1 = 1 i and x2 = i , 1 respectively. For the initial conditions y(0) = 2 and y (0) = 0, the solution is given by y(t) y (t) = 1 2 1 i i 1 eit 0 0 e-it 1 -i -i 1 2 0 = 2 cos t -2 sin t . 5.9: What happens to the Fibonacci sequence if we go backward in time, and how is F-k related to Fk ? The law Fk+2 = Fk+1 + Fk is still in force, so F-1 = 1. 6 Since the law Fk+2 = Fk+1 + Fk is assumed to hold true for all integers k, the relation Fk+1 Fk 1 1 =A , where A := , Fk-1 1 0 Fk holds true for all integers k. In particular, replacing k by -k and multiplying from the left by A-1 it follows from the above relation that F-k F = A-1 -(k-1) , F-(k+1) F-k This in turn implies that F-(k+1) F-k = A-1 k k = 1, 2, 3, . . . . F0 , F-1 k = 1, 2, 3, . . . , 1+ 5 2 where F0 = 0 and F-1 = 1. Since the matrix A has the eigenvalues 1 = and 2 = 1-2 5 with eigenvectors x1 = 1 1 and x2 = 2 , 1 1 1 respectively, the matrix A-1 has the eigenvalues 1 = 1 and 2 = 2 with the same eigenvectors x1 and x2 . Just as we had done in class for the "forward" Fibonacci recursion, it follows that 1 F-k = (1 )k - (2 )k , 5 Since 1 = - it follows that 1 1- 5 F-k = (-1)k 2 5 k k = 1, 2, 3, . . . . 1+ 5 2 1- 5 2 and 2 = - , k - k 1+ 5 2 1- 5 2 k 1+ 5 1 = (-1)k+1 2 5 - 7 , k = 1, 2, 3, . . . . Comparing this with the formula k 1 1+ 5 Fk = - 2 5 F-k = (-1)k+1 Fk 1- 5 2 k for the Fibonacci numbers, which we had derived in class, it follows that for all k = 0, 1, 2, . . . . , k = 0, 1, 2, . . . , Can you find a time T at which the solution u(T ) is guaranteed to return to the initial value u(0)? The matrix A has eigenvalues 1 = 0, 2 = i 2, and 3 = -i 2. It follows that 1 0 0 0 S -1 , eAt = S 0 ei 2t 0 0 e-i 2t where S is a matrix the columns of which are eigenvectors corresponding to 1 , 2 and 3 . We wish to find a T sothat eAT = eA0 = I. The last relation is satisfied if, and only if, ei 2T = e-i 2T = 1. All T > 0 with this property are given by 2j T = , j = 1, 2, 3, . . . . 2 5.17: (a) Find the eigenvalues and eigenvectors of A = 0 4 . 1 0 4 The eigenvalues of A are 1 = 1 and 2 = -1 with eigenvectors x1 = respectively. 8 4 1 and x2 = 4 -1 , 5.10: Find the general solution to du/dt = Au if 0 -1 0 A= 1 0 -1 . 0 1 0 (b) Solve du/dt = Au starting from u(0) = (100, 100). The solution u(t) is given by u(t) = c1 et x1 + c2 e-t x2 where c1 and c2 are determined by 4 4 1 -1 c1 100 = . c2 100 It follows that c1 = 125/2 and c2 = -75/2, and thus u(t) = 25 2 20et - 12e-t 5et + 3e-t . (c) If v(t) = income to stockbrokers and w(t) = income to client, and they 1 help each other by dv/dt = 4w and dw/dt = 4 v, what does the ratio v/w approach as t ? Since 4c1 et + 4c2 e-t v(t) , = u(t) = c1 et x1 + c2 e-t x2 = c1 et - c2 e-t w(t) it follows that v(t) 4c1 et + 4c2 e-t = lim = 4, t w(t) t c1 et - c2 e-t lim provided that c1 = 0. 9 ...
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