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Unformatted text preview: 18.03 Problem Set 9: Solutions
34. [Part I problems: 4] (a) [18] A =
1±3i 2. −3i/2 1 v1 = 0. −9/4 −3i/2 1 1 One choice is v1 = . The normal mode is then e(1+3i)t/2 , which has real and 3i/2 3i/2 cos(3t/2) sin(3t/2) imaginary parts u1 = et/2 and u2 = et/2 . −(3/2) sin(3t/2) (3/2) cos(3t/2) An eigenvector for λ1 =
1+3i 2 0.5 1 −2.25 0.5 has characteristic polynomial pA (λ) = λ2 − λ +2.5, and eigenvalues satisﬁes (A − λ1 I )v1 = 0, that is, The initial condition is u(0) = ﬁnd u(0) = A ˙ 1 , which, conveniently, is satisﬁed by u1 . Since u = Au, we ˙ 0 1 0.5 = . So x(0) = 0.5, and the fox population is increasing at t = 0 ˙ 0 −2.25 while the number of rabbits is decreasing. y (t) = 0 occurs next when 3t/2 = π , or t = 2π/3. The graphs of x(t) = et/2 cos(3t/2) and y (t) = −(3/2) sin(3t/2) are “antidamped” sinusoids, with increasing amplitude. The relevant trajectory is the one crossing the positive x axis half π = way out. The values of u(t) are u − 23 −1 0 e−π/3 , u − π = e−π/6 , 3 0 3/2 0 1 u(0) = , , u π = eπ/6 3 −3/2 0 −1 π u 23 = eπ/3 . 0 1b , pA (λ) = λ2 − 2λ + 1 = (λ − 1)2 , so we have a repeated eigenvalue 01 0b λ1 = 1. To ﬁnd an eigenvector form A − λ1 I = . A nonzero eigenvector is given (for 00 1 any b) by v = . If b = 0, the eigenvectors for value λ1 are exactly the multiples of v 0 (the matrix is defective), but for b = 0, A = I and any vector is an eigenvector (the matrix is c complete). When b = 0, the normal modes are et , for c a real constant. When b = 0, the 0 normal modes are et v for any vector v. When b = 0, we must solve (A − λ1 I )w = v1 , that is, 0b 1 0 w= . The solution is w = , so the extra solution is u2 = eλ1 t (tv1 + w) = 00 0 1/b t et . 1/b (b) [8] With A = 35. [Part I problems: 10 pts] (b) det A = 0 when a = 3. trA = 0 when a = 1. det A = (trA)2 /4 when a2 + 2a − 11 = 0 or √ a = −1 ± 2 3, i.e. a ≃ −4.4641 and a = 2.4641. √ (c) [4] Diagram showing: a < −1 − 2 3—stable node = nodal sink √ a = −1√ 2 3—defective stable node = defective nodal sink − −1 − 2 3 < a < 1—counterclockwise stable spiral = spiral sink a = 1—counterclockwise center √ 1 < a < −1 + 2 3–counterclockwise unstable spiral = spiral source √ a = 1√ 2 3—unstable defective node = defective nodal source + 1 + 2 3 < a < 3—unstable node = nodal source a = 3—unstable degenerate comb 3 < a—saddle (a) [4] trA = a − 1, det A = 3 − a, so trA = 2 − det A. √ 1, (b)(c) [18] Here are pictures for a = −2 3, 0, √ 2, −1 + √ 2 3, 2.75, 3, 4. The picture for some a < √ 1 − 2 3 would − show a nodal sink, and that for a = −1 − 2 3 would show a defective nodal sink. 36. [Part I: 6] a −b , pA (λ) = λ2 − 2aλ + (a2 + b2 ) = (λ − a)2 + b2 , so the eigenvalues ba −bi −b are a ± bi. An eigenvector for λ1 = a + bi is given by v1 such that v1 = 0, and b −bi 1 1 we can take v1 = . The corresponding normal mode is e(a+bi)t . Its real and −i −i cos(bt) sin(bt) imaginary parts give linearly independent real solutions, eat and eat . sin(bt) cos(bt) 10 cos(bt) sin(bt) , . Φ(0) = So a fundamental matrix is given by Φ(t) = eat 0 −1 sin(bt) − cos(bt) 10 cos(bt) − sin(bt) Φ(0)−1 = , so eAt = Φ(t)Φ(0)−1 = eat . 0 −1 sin(bt) cos(bt) (a) [9] With A = A(e(a+bi)t ) = A(eat (cos(bt) + i sin(bt))) = eat cos(bt) − sin(bt) sin(bt) cos(bt) = eA(a+bi)t . (b) [9] s2 + 2s + 2 = (s + 1)2 + 1 so the roots of the characteristic polynomial are −1 ± i. Basic solutions are given by y1 = e−t cos(t) and y2 = e−t sin(t). (I write y instead of x because the problem wrote x for the normalized solutions.) y1 (0) = 1, y1 (0) = −1, y2 (0) = 0, y2 (0) = 1. ˙ ˙ So x1 = y1 + y2 and x2 = y2 form a normalized pair of solutions: x1 (t) = e−t (cos t + sin t), x2 (t) = e−t sin t. The companion matrix is A = 0 1 . Its characteristic polynomial is the same, λ2 + 2λ2 , −2 −2 so its eigenvalues are the same, −1 ± i. An eigenvector for value −1 + i is given by v1 such 1−i 1 1 that v1 = 0. We can take v1 = . The corresponding normal −2 −1 − i −1 + i 1 cos t mode is e(−1+i)t , which has real and imaginary parts u1 = e−t −1 + i − cos t − sin t sin t and u2 = e−t . The top entries of these solutions are x1 (t) and x2 (t). − sin t + cos t 1 1 + c2 e2t 1 2 so c1 + c2 . c1 + 2c2 1 u2 = c1 e3t + 1 = c1 = −1 and c2 = 1: 1 2 (ii) We have just computed the columns of the exponential matrix: 2e3t − e2t −e3t + e2t eAt = . 2e3t − 2e2t −e3t + 2e2t (iii) The matrix A has eigenvalues 3 and 2, with eigenvectors 3 1 1 1 1 = u1 (0) = c1 + c2 0 1 2e3t − e2t . Start again for u2 : Thus c1 = 2 and c2 = −1: u1 = 2e3t − 2e2t 1 0 1 1 c1 + c2 c2 e2t so = u2 (0) = c1 + c2 . Thus = 2 1 1 2 c1 + 2c2 −e3t + e2t u2 = . −e3t + 2e2t (c) [9] (i) u1 = c1 e3t and 1 . Then 2 ab cd 1 1 = 1 ab 1 1 and =2 . The top entries give the equations a + b = 3 and 1 cd 2 2 a + 2b = 2, which imply a = 4, b = −1. The bottom entries give the equations c + d = 3, 4 −1 c + 2d = 4, which imply c = 2, d = 1. Thus A = . 21 ...
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This note was uploaded on 02/06/2009 for the course 18 18.03 taught by Professor Unknown during the Spring '09 term at MIT.
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