q1_sol_18085_f06

# q1_sol_18085_f06 - MIT OpenCourseWare http://ocw.mit.edu...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 second-diﬀerence 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 ...
View Full Document

## 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.

Ask a homework question - tutors are online