FinalExamCHEE3334Fall2002Solutions2

FinalExamCHEE3334Fall2002Solutions2 - CHEE 3334 Fall 2002...

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CHEE 3334 Fall 2002 Michael Nikolaou Student Name FINAL EXAM Open-everything. Total points 100 (10 points per question). Please, budget your time ! Do NOT try to read the book during the exam! Write neatly! Do NOT use your programmable calculator to find results you are asked to compute in a certain way (but you may use anything you want to check numerical results)! PLEASE SHOW ALL RELEVANT WORK IN THE INDICATED SPACES. DO NOT RETURN ADDITIONAL PAGES. (You may use scratch paper for scratch work.) 1. ( 10 points ) Find the analytical solution of the differential equations / 98 198 du dt u v = + , / 99 199 dv dt u v = - - with initial conditions (0) 1 u = , (0) 0 v = . Would you characterize these equations as stiff? Eigenvalues of [98 198; -99 –199]: (x – 98)(x + 199) + 99*198 = 0 ==> x^2 + 101x + 100 = 0 ==> x1 = -1, x2 = - 100. Equations are stiff (eigevalues differ by two orders of magnitude). Eigenvectors of [98 198; -99 –199]: (-1-98)z1 – 198z2 = 0 ==> z1 = - 198.z2/99 = -2.z2 ==> z = [-2; 1] (-100 – 98)w1 – 198w2 = 0 ==> w1 = -198.w2/198 = -w2 ==> w = [-1; 1] Modal matrix: P = [z w] ==> P^(-1) = [-2 –1 ; 1 1]^(-1) = 1/(-1)[1 1; -1 –2] = [-1 -1; 1 2] Similarity transformation: y = [y1; y2] = P[u; v] ==> dy1/dt = -y1, dy2/dt = -100y2 ==> y1 = y1(0)exp(-t), y2 = y2(0)exp(-100t). Therefore: [u; v] = Sum{P(:,i)Pinverse(i,:)[u(0); v(0)] exp(eigenvalue_i*t)} = = [-2; 1] [-1 –1] [1; 0]exp(-t) + [-1; 1] [ 1 2] [1; 0] exp(-100t) = [2 exp(-t) – exp(-100t); -exp(-t) + exp(-100t) { or, step by step: [y1(0); y2(0)] = P^(-1)[u(0); v(0)] = [-1 –1 ; 1 2][1; 0] = [-1; 1] Therefore y1 = -exp(-t), y2 = exp(-100t). [u, v] = Py = [-2
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FinalExamCHEE3334Fall2002Solutions2 - CHEE 3334 Fall 2002...

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