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MIT18_06S10_pset8_s10_soln

MIT18_06S10_pset8_s10_soln - 18.06 PSET 8 SOLUTIONS Problem...

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summationdisplay summationdisplay summationdisplay bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg 18.06 PSET 8 SOLUTIONS APRIL 15, 2010 Problem 1. ( § 6.3, #14) The matrix in this question is skew-symmetric ( A T = A ): 0 c b u = cu 2 bu 3 1 d u = c 0 a u or u 2 = au 3 cu 1 dt b a 0 u = bu 1 au 2 3 (a) The derivative of bardbl u ( t ) bardbl 2 = u 1 2 + u 2 2 + u 3 3 is 2 u 1 u 1 + 2 u 2 u 2 + 2 u 3 u 3 . Substitute u 1 , u 2 , u 3 to get zero . Then bardbl u ( t ) bardbl 2 stays equal to bardbl u (0) bardbl 2 . (b) When A is skew-symmetric, Q = e At is orthogonal. Prove Q T = e At from the series for Q = e At . Then Q T Q = I . Solution. (4 points) (a) 2 u 1 u 1 + 2 u 2 u 2 + 2 u 3 u = 2 u 1 ( cu 2 bu 3 ) + 2 u 2 ( au 3 cu 1 ) + 2 u 3 ( bu 1 au 2 ) = 0 . 3 (b) The important points are that ( A n ) T = ( A T ) n = ( A ) n , and that we can take transpose termwise in a sum: parenleftBigg parenrightBigg T Q T = A n t n = ( A n ) T t n = ( A ) n t n = e At . n ! n ! n ! n =0 n =0 n =0 Then, Q T Q At At 0 = e e = e = I because A and A commute (but I don’t think the problem intended for you to have to actually check this!). Problem 2. ( § 6.3, #24) Write A = 1 3 as S Λ S 1 . Multiply Se Λ t S 1 to find the matrix exponential 0 0 e At . Check e At and the derivative of e At when t = 0. Solution. (4 points) 2 2 Λ = 1 0 and S = 1 1 . 0 3 0 1 Then, 3 t t t e e Se Λ t S 1 = e 2 3 t 2 0 e This is the identity matrix when t = 0, as it should be. The derivative matrix is e t 3 / 2 e 3 t 1 / 2 e t 0 3 e 3 t which is equal to A when t = 0, as it should be. 1
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bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg ′′ Problem 3. ( § 6.3, #28) Centering y = y in Example 3 will produce Y n +1 2 Y n + Y n 1 = t ) 2 Y n . This can be written as a one-step difference equation for U = ( Y, Z ): Y n +1 = Y n + Δ tZ n bracketleftbigg 1 0 bracketrightbiggbracketleftbigg Y n +1 bracketrightbigg bracketleftbigg 1 Δ t bracketrightbiggbracketleftbigg Y n bracketrightbigg = . Z n +1 = Z n Δ tY n +1 Δ t 1 Z n +1 0 1 Z n Invert the matrix on the left side to write this as U n +1 = A U n . Show that det A = 1. Choose the large time step Δ t = 1 and find the eigenvalues λ 1 and λ 2 = λ 1 of A : A = 1 1 1 has | λ 1 | = | λ 2 | = 1. Show that A 6 is exactly I . 0 After 6 steps to t = 6, U 6 equals U 0 . The exact y = cos t returns to 1 at t = 2 π . Solution. (12 points) We have bracketleftbigg bracketrightbigg 1 bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbiggbracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg 1 0 1 0 1 0 1 Δ t 1 Δ t Δ t 1 = Δ t 1 and so A = Δ t 1 0 1 = Δ t 1 t ) 2 Clearly det A = 1: it is the product of two matrices that are triangular with ones on the diagonal, and so each have determinant 1.
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