Assignment 2 solutions

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 18.085 Computational Science and Engineering I Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. 18.085 - Mathematical Methods for Engineers I Prof. Gilbert Strang Solutions - Problem Set 2 Section 1.3 7) Suppose A is rectangular (m by n) and C is symmetric (m by m) matrix i) (AT C A)T = AT C T (AT )T = AT CA � AT CA is symmetry # (AT )n×m (C )m×m (A)m×n � AT CA is (n × n) # . . �. � a1 ii) Let A = ⎭ . . . � � . . . a2 . . . . . . a3 . . . ⎢ . . .⎧ an ⎧ ⎨ . . . . . . a1 . . . ... .. . . . a2 . . . ... since C = C T (symmetric) Since aT ai = a2 → 0, we conclude that i i AT A has no negative numbers on its diagonal # � ⎬ 13 A= 32 � ⎬ �⎬ 10 1 3 = 31 0 −7 �⎬ ⎬ �⎬ � 0 13 10 1 = 01 31 0 −7 � �� � � �� � � �� � L D LT # � � AT A = � · · · a T 2 ⎭ · · · aT n �T a1 a1 �. �. . =� . � ⎭. . ... · · · aT · · · 1 ⎧� ⎧ ··· ⎧� ⎨⎭ ··· ... aT a2 2 ⎢� ⎢ . . .⎧ an ⎧ ⎨ . . . ⎢ ⎧ ⎧ ⎧ ⎧ ⎨ . aT an n 1 A= = ⎬ ⎬ ⎬ 1b bc 10 b1 � 21 A=⎭1 2 01 � 1 = ⎭ 1/2 0 � 1 = ⎭ 1/2 0 � Section 1.4 �⎬ �⎬ � 10 1 0 1b = 01 b1 0 c − b2 � �� � � �� � � �� � L D LT �⎬ 1 b 0 c − b2 � # � ⎢ 00 1 0⎨ 2/3 1 ⎢ 00 1 0⎨ 2/3 1 �� � L ⎢ 0 1 ⎨ 2 ⎢ 21 0 ⎭ 0 3/2 1/2 ⎨ 0 0 4/3 � ⎢� ⎢ 20 0 1 1/2 0 ⎭ 0 3/2 0 ⎨ ⎭ 0 1 2/3 ⎨ 0 0 4/3 00 1 � �� �� �� � D LT � 3) −u �� = � (x − 1/3) + � (x − 2/3) General Fixed-Fixed Solution � (1 − a)x , x � a u= (1 − x)a , x → a −u �� = � (x − 1/3) � (1 − 1/3)x , x � 1/3 u= (1 − x)1/3 , x → 1/3 −u �� = � (x − 2/3) � (1 − 2/3)x , x � 2/3 u= (1 − x)2/3 , x → 2/3 Combining two single-load solutions: ⎡ ⎥(1 − 1/3)x + (1 − 2/3)x , x � 1/3 ⎣ u = (1 − x)1/3 + (1 − 2/3)x , 1/3 � x � 2/3 ⎥ ⎤ (1 − x)1/3 + (1 − x)2/3 , x → 2/3 ⎡ ⎥x for x � 1/3 ⎣ = 1/3 for 1/3 � x � 2/3 ⎥ ⎤ 1 − x for x → 2/3 # 2 Second Method ⎡ ⎥Ax + B ⎣ u(x) = C x + D ⎥ ⎤ Ex + F u(0) = 0 A · 0+B =0 �B =0 0 , x � 1/3 , 1/3 � x � 2/3 , x → 2/3 # � A(1/3) +� = C (1/3) + D B A = C + 3D A−1 =C C (2/3) + D = E (2/3) + F 2C + 3D = 2E + 3F C −1 =E E (1) + F = 0 E = −F � 1 � 3 � 2 � 5 � 6 � 4 �� � 2 3 C C � + 3D − 1 = � 1 D= 3# �� � 6 4 2C + 3(1/3) = 2E + 3(−E ) 2C + 1 = −E � 7 � −2×� 7 5 1 + 3 = −3E E = −1 # C = 1 + E = 1 + (−1) � C = 0# E = −F � F = 1# A − 1 = C A = 1 + 0 � A = 1# ⎡ ⎥x ⎣ � u(x) = 1/3 ⎥ ⎤ 1−x , x � 1/3 , 1/3 � x � 2/3 , x → 2/3 3 5) Free-Free condition u(x) = −R(x − a) + C x + D u � (1) = 0 u � (0) = 0 u � (0) = 0 + C = 0 � C = 0 u � (1) = −1 + C = 0 �C =1 � There are no solutions for C and D C cannot be 0 and 1 at the same time 7) f (x) = � (x − 1/3) − � (x − 2/3) u � (1) = 0 u � (0) = 0, ⎡ ⎥Ax + B ⎣ u(x) = C x + D ⎥ ⎤ Ex + F , x � 1/3 , 1/3 � x � 2/3 , x → 2/3 # u � (0) = A = 0 # �0 A �� �(1/3) + B = C (1/3) + D 3B = C + 3D A −1 = C � C = −1 # � 2 C (2/3) + D = E (2/3) + F 2C + 3D = 2E + 3F C +1 = E E = 0# d (Ex + F ) = 0 dx E =0 From � 3 2C + 3D = 3F 3D = 3F − 2C = 3F + 2 D = F + 2/3 From � 2 3B = −1 + 3D B= −1 + 3F + 2 3 = F + 1/3 ⎩ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ � � 3 redundant, less one equation 4 F can take any value F ≤ R ⎡ , x � 1/3 ⎥F + 1/3 ⎥ ⎣ u(x) = −x + F + 2/3 , 1/3 � x � 2/3 ⎥ ⎥ ⎤ F , x → 2/3 � inﬁnitely many solutions for u(x) # ⎡ ⎣0 , x�0 ���� = � (x) 12) u , C (x) = x3 ⎤ , x→0 6 u(x) = C (x) + Ax3 + B x2 + Gx + D Given that u(1) = 0 u(−1) = 0 u �� (−1) u �� (1) = 0 =0 Cubic spline C (x) is a particular solution for u u(1) = 1 +A+B+G+D =0 6 A+B+G+D =− 1 6 � 1 � 2 u(−1) = 0 + (−A) + B − G + D = 0 A−B+G−D =0 u �� (1) = 1 + 6A(1) + 2B = 0 6A + 2B = −1 u �� (−1) = 0 − 6A + 2B = 0 � B = 3A A=− − − 1 12 1 12 1 1 B =− 4# 12 # ⎪ ⎫ 1 1 +− +G+D =− 4 6 ⎪ ⎫ 1 −− +G−D =0 4 1 6 1 6 D= 1 6# � 3 G+D = G−D =− �G=0 , � u(x) = C (x) − 1 3 12 1 x− x+ 12 4 6# 5 Section 1.5 ⎬ � 2 −1 1) K = 2 −1 1 Q= ∗ 2 � ⎬ � 1 1 10 �= 1 −1 03 � ⎬ �⎬ �⎬ 1 1 1 1 10 1 1 ∗ Q�QT = ∗ 03 1 −1 2 1 −1 2 � ⎬ �⎬ 11 3 1 1 = 1 −3 1 −1 2 � ⎬ 1 4 −2 = 4 2 −2 � ⎬ 2 −1 = 2 −1 ⎬ =K# 4) >> K = toeplitz([2 − 1 zeros(1, 3)]); >> [Q, E ] = eig(K ); >> [DST = Q � diag([−1 − 1 DST = 0.2887 0.5000 0.5000 0.5774 0.5000 0.0000 0.2887 0.5774 1 −1 1]) ⎩ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 0.5774 −0.0000 −0.5774 0.2887 −0.5000 0.5000 −0.0000 −0.5000 −0.5000 0.5000 −0.5000 0.2887 0.5000 −0.5000 −0.0000 0.5774 −0.5000 >> J K = [1 : 5]� � [1 : 5]; ans = 0.2887 0.5000 0.5774 0.5000 0.5000 >> sin(JK � pi/6)/sqrt(3) 0.5774 0.0000 0.5000 −0.5000 −0.0000 0.2887 −0.5000 0.0000 −0.5774 0.5774 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 0.5000 0.2887 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ −0.5000 −0.5000 ⎥ ⎥ ⎥ ⎥ ⎥ −0.0000 0.5774 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 0.5000 −0.5000 ⎥ ⎥ ⎥ � −0.5000 0.2887 DST = sin(JK � pi/6)/sqrt(3) 6 >> DST � ans = 0.2887 0.5000 0.5000 0.5774 0.5000 ⎩ 0.2887 ⎥ ⎥ ⎥ ⎥ ⎥ −0.5000 ⎥ ⎥ ⎥ ⎥ ⎥ 0.5774 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ −0.5000 ⎥ ⎥ ⎥ ⎥ ⎥ 0.2887 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ 0.5774 −0.0000 −0.5774 −0.0000 0.5000 −0.5000 0.0000 0.2887 −0.5000 0.5000 −0.0000 −0.5000 0.5000 0.5774 −0.5000 >> inv(DST ) ans = 0.2887 0.5000 0.5000 0.5774 DST � = inv(DST ) 0.5774 −0.0000 −0.5774 0.2887 −0.5000 2 −1 0.5774 ⎢ 0.5000 −0.0000 0.5000 −0.5000 −0.0000 7) � � −1 2 −1 0 C4 = � � 0 −1 2 −1 ⎭ −1 0 −1 2 [Q, E ] = eig(C4 ) � 1/2 ∗ 1/ 2 � 0 −1 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 0.5000 0.2887 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ −0.5000 −0.5000 ⎥ ⎥ ⎥ ⎥ ⎥ −0.0000 0.5774 ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ 0.5000 −0.5000 ⎥ ⎥ ⎥ ⎥ � −0.5000 0.2887 ⎥ ⎧ ⎧ ⎧ ⎧ ⎨ � � 1/2 0 � Q=� ∗ ⎭ 1/2 −1/ 2 1/2 0 � �1 i F =� � 1 i2 ⎭ 1 i3 � 1 1 1 i2 i4 i6 1 0 ∗ 1/ 2 0 ∗ −1/ 2 ⎢ � 1 −1/2 ⎧ 1/2 ⎧ ⎧ ⎧ −1/2 ⎨ 1/2 1 ⎢ � � E=� � ⎭ 1 ⎧ ⎧ ⎧ ⎧ ⎨ ⎢ � 0 2 0 2 0 4 ⎧ ⎧ ⎧ ⎧ ⎨ ⎢ 1 ⎧� i3 ⎧ � 1 i −1 −i ⎧=� � 1 −1 6⎧ 1 −1 i⎨ ⎭ 9 i 1 −i −1 i f1 = a 1 q1 + a 2 q2 + a 3 q3 + a 4 q4 f2 = b 1 q1 + b 2 q2 + b 3 q3 + b 4 q4 f3 = c 1 q1 + c 2 q2 + c 3 q3 + c 4 q4 f4 = d 1 q1 + d 2 q2 + d 3 q3 + d 4 q4 7 [f1 f2 f3 f4 ] = [q1 q2 q3 q4 ] � a1 b1 c1 d1 � � a 2 b2 c2 � �a b c ⎭3 3 3 a 4 b4 c4 � �� denote by fi and qi are the column i of their respective matrix F = QA Q−1 QA = Q−1 F A = Q−1 F From MATLAB � 2 0 ∗ � �0 2 A=� ∗ � 2i ⎭0 0 � 1 ⎧ d2 ⎧ ⎧ d3 ⎧ ⎨ d4 � A ⎢ 0 0 −2 0 0 ∗ 0 2 ∗ 0 − 2i 0 ⎢ ⎧ ⎧ ⎧ ⎧ ⎨ # 0 9) � � −1 1 0 �− = � � 0 −1 1 ⎭ 0 0 −1 �− T �− 1 −1 0 � 1 −1 =⎭0 0 0 � = ⎭ −1 � 2 −1 � 0 ⎢ ⎧ ⎧ ⎧ ⎧ ⎨ 4×3 � � 1 0 ⎧ � −1 0 ⎨� 1 ⎭ 1 −1 1 −1 0 −1 0 0 2 ⎢ ⎢ � 1 0 0 ⎢ ⎧ ⎧ ⎧ ⎧ ⎨ 0 −1 ⎧ 2 −1 ⎨ = K3 ⎢ # �− �− T � 1 0 � � −1 1 � =� ⎭ 0 −1 0 � ⎢ 0 0 ⎧ 1 −1 0 ⎧� ⎧ ⎧ 1 −1 0⎨ ⎧⎭ 0 1⎨ 0 0 1 −1 0 −1 0 8 � � −1 2 −1 0 =� � 0 −1 2 −1 ⎭ 0 0 −1 1 � ⎧ eig(K3 ) = ⎭ 2.0000 ⎨ 3.4142 [u s v ] = svd(�− ) � 1.8478 1.4142 � 0.5858 ⎢ � 1 −1 0 0 ⎢ ⎧ ⎧ ⎧ = B4 ⎧ ⎨ # s=⎭ ⎢ 0.7654 ⎧ ⎨ 2 �1 = 1.84782 = 3.4142 = �3 2 �2 = 1.41422 = 2 = �2 2 �3 = 0.76542 = 0.5858 = �1 � The eigenvalues of K3 are the squared singular values � 2 of �− # 9 ...
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• Math, Pallavolo Modena, Sisley Volley Treviso, B-1 Lancer, MIT OpenCourseWare http://ocw.mit.edu, Prof. Gilbert Strang, General Fixed-Fixed Solution

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