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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 Quiz 1 October 2, 2006 Professor Strang
Grading 1 2 3 Your PRINTED name is: SOLUTIONS 1) (36 pts.) (a) Suppose u(x) is linear on each side of x = 0, with slopes u � (x) = A on the left and u � (x) = B on the right: ⎡ ⎥ ⎣Ax for x � 0 u(x) = ⎥ ⎤B x for x � 0 What is the second derivative u �� (x) ? Give the answer at every x. (b) Take discrete values Un at all the whole numbers x = n: ⎡ ⎥ ⎣An for n � 0 Un = ⎥ ⎤Bn for n � 0 For each n, what is the second diﬀerence �2 Un ? Using co eﬃcients 1, −2, 1 (notice signs !) give the answer �2 Un at every n. (c) Solve the diﬀerential equation −u �� (x) = � (x) from x = −2 to x = 3 with boundary values u(−2) = 0 and u(3) = 0. (d) Approximate problem (c) by a diﬀerence equation with h = �x = 1. What is the matrix in the equation K U = F ? What is the solution U ? 1 Solutions. (a) u �� (x) = (B − A) � (x). This is not B − A at x = 0. ⎡ ⎣B − A n=0 (b) �2 un = ⎤ 0 n=0 � (c) u(x) = ⎡ ⎣ 3/5 (x + 2) ⎤ 2/5 (3 − x) � −2 � x � 0 ⎢ 0�x�3 3/5 � ⎧ � ⎧ � 6/5 ⎧ � ⎧ U =� ⎧ � 4/5 ⎧ � ⎨ 2/5 � ⎢ (d) K+++ U lies right on the graph of u(x) 2 −1 � ⎧ � ⎧ � −1 ⎧ 2 −1 � ⎧ =� ⎧ � ⎧ −1 2 −1 � ⎨ −1 2 ﬁxed ﬁxed 2 2) (24 pts.) A symmetric matrix K is “positive deﬁnite ” if uT Ku > 0 for every nonzero vector u. (a) Suppose K is p ositive deﬁnite and u is a (nonzero) eigenvector, so K u = �u. From du the deﬁnition above show that � > 0. What solution u(t) to = K u comes from dt knowing this eigenvector and eigenvalue ? (b) Our seconddiﬀerence matrix K4 has the form AT A: � ⎢ � ⎢ � ⎢ 1 � ⎧ 1 −1 2 −1 � ⎧ � −1 ⎧ ⎧� 1 � ⎧� ⎧ ⎧ � −1 1 −1 2 −1 � ⎧� ⎧=� ⎧ K4 = � −1 1 ⎧� ⎧� 1 −1 −1 2 −1 ⎧ ⎧� ⎨� ⎨ � −1 1⎨ � 1 −1 −1 2 −1 vector u. (Show me why uT AT Au � 0 and why > 0.) (c) This matrix is p ositive deﬁnite for which b ? Semideﬁnite for which b ? What are its pivots ?? S= ⎩ 2b b4 ⎦ Convince me how K4 = AT A proves that uT K4 u = uT AT Au > 0 for every nonzero 3 Solutions. (8 pts each) (a) “. . . show that � > 0” K u = �u uT Ku = �uT u “What solution u(t) . . . ” u(t) = e�t u (b) uT AT Au = (Au)T (Au) � 0 The particular matrix A in this problem has independent columns. The only solution to Au = 0 is u = 0. So uT AT Au > 0 except when u = 0. Other proofs are possible. (c) Positive deﬁnite if b2 < 8 Semideﬁnite if b2 = 8 Pivots 2 and 4 −
b2 2 uT Ku so � = T > 0 u u 4 3) (40 pts.) (a) Suppose I measure (with possible error) u1 = b1 and u2 − u1 = b2 and u3 − u2 = b3 and ﬁnally u3 = b4 . What matrix equation would I solve ��� to ﬁnd the best least squares estimate u1 , u2 , u3 ? Tell me the matrix (b) What 3 by 2 matrix A gives the spring stretching e = A u from the � and the right side in K u = f . c1 m1 c2 m2 c3 u1
� displacements u1 , u2 of the masses ? (c) Find the stiﬀness matrix K = AT CA. Assuming positive c1 , c2 , c3 show that K is invertible and positive deﬁnite. u2
� 5 Solutions. (a) (12 p oints) ⎢ b1 �⎧ ⎧ u1 ⎧ � ⎧ � b2 ⎧ ⎧�u ⎨ = � ⎧ 2 �b ⎧ ⎧ 1⎨ � 3⎨ u3 1 b4 � ⎢� ⎢ � ⎢ 2 −1 u1 � b2 − b 1 � ⎧� ⎧ � ⎧ � −1 2 −1 ⎨ � u2 ⎨ = � b3 − b2 ⎨ � −1 2 u3 � b4 + b 3 � �⎪ � � �⎪ � K f 1 � � −1 1 � � −1 � � ⎢
3pts f or numbers 3pts Au = b is � ⎢ � T T A�⎪A u = ��⎪� � �� A b K f 3pts 3pts is ⎢ 10 � ⎧ (b) (10 p oints) A = � −1 0 ⎨ −1 1 (c) (16 p oints) AT CA = ⎩ � Any of these proofs is OK: (1) det = c1 c3 + c2 c3 > 0 1 −1 −1 � 0 0 1 ⎦ � c1 c2 ⎢ ⎩ ⎦ 10 c1 + c2 + c3 −c3 ⎧� ⎧ ⎨ � −1 0 ⎨ = −c3 c3 c3 −1 1 ⎢� (2) AT CA always p ositive deﬁnite with independent columns in A (3) other ideas . . . 6 ...
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This note was uploaded on 04/12/2011 for the course MATH 18.085 taught by Professor Staff during the Fall '10 term at MIT.
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