This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 3 by 3
1
“element matrix” c1 E1 D Œ 1 0 0 T c1 Œ 1 0 0 has c1 in the top left corner. 6 Multiply AT C1 A1
1 7 For 5 springs and 4 masses, the 5 by 4 A has two nonzero diagonals: all ai i D 1
and ai C1;i D 1. With C D diag.c1 ; c2 ; c3 ; c4 ; c5 / we get K D AT CA, symmetric tridiagonal with diagonal entries Ki i D ci C ci C1 and offdiagonals Ki C1;i D ci C1 .
With C D I this K is the 1; 2; 1 matrix and K.2; 3; 3; 2/ D .1; 1; 1; 1/ solves K u D
ones.4; 1/. (K 1 will solve K u D ones.4/.)
3
1
u00 D 1 with u.0/ D u.1/ D 0 is u.x/ D 2 .x x 2 /. At x D 1 ; 2 ; 5 ; 4
55
5
2
this gives u D 2; 3; 3; 2 (discrete solution in Problem 7) times .x/ D 1=25. 8 The solution to
9 u 00 D mg has complete solution u.x/ D A C Bx 1 mgx 2 . From u.0/ D 0 we
2
get A D 0. From u 0 .1/ D 0 we get B D mg . Then u.x/ D 1 mg.2x x 2 / at
2
12
x D 3 ; 3 ; 3 equals mg=6; 4mg=9; mg=2. This u.x/ is not proportional to the discrete
3
u D .3mg; 5mg; 6mg/ at the meshpoints. This imperfection is because the discrete
problem uses a 1sided difference, less accurate at the free end. Perfect accuracy is
recovered by a centered difference (discussed on page 21 of my CSE textbook). 10 (added in later printing, changing 1011 below into 1112). The solution in this ﬁxed ﬁxed case is .2:25; 2:50; 1:75/ so the second mass moves furthest.
11 The two graphs of 100 points are “discrete parabolas” starting at .0; 0/: symmetric around 50 in the ﬁxedﬁxed case, ending with slope zero in the ﬁxedfree case.
12 Forward/backward/centered for du=dx has a big effect because that term has the large
coefﬁcient. MATLAB: E D diag(ones.6; 1/; 1/; K D 64 .2 eye.7/ E E 0 /;
D D 80 .E eye.7//; .K C D/nones.7; 1/; % forward; .K D 0 /nones.7; 1/;
% backward; .K C D=2 D 0 =2/nones.7; 1/I % centered is usually the best: more accurate Problem Set 8.2, page 428
1 AD " 1
1
0 1
0
1 #
"#"#
0
c
1
1 ; nullspace contains c ; 0 is not orthogonal to that nullspace.
1
c
0 2 AT y D 0 for y D .1; 1; 1/; current along edge 1, edge 3, back on edge 2 (full loop). Solutions to Exercises 3 Elimination on b1 Œ A " 83 b " D 1
1
0 1 0 b1
0 1 b2
1 1 b3 # leads to Œ U c D #
1
1 0 b1
0
1 1 b2 b1
. The nonzero rows of U come from edges 1 and 3
0
0 0 b3 b2 C b1
in a tree. The zero row comes from the loop (all 3 edges).
4 For the matrix in Problem 3, Ax D b is solvable for b D .1; 1; 0/ and not solvable for b D .1; 0; 0/. For solvable b (in the column space), b must be orthogonal to
y D .1; 1; 1/; that combination of rows is the zero row, and b1 b2 C b3 D 0 is the
third equation after elimination. 5 Kirchhoff’s Current Law AT y D f is solvable for f D .1; 1; 0/ and not solvable for f D .1; 0; 0/; f must be orthogonal to .1; 1; 1/ in the nullspace: f1 C f2 C f3 D 0.
"
#
"#
" # "#
2
1
1
3
1
c
T
1
2
1 xD
3 D f produces x D
1 C c ; potentials
6 A Ax D
1
1
2
0
0
c
x D 1; 1; 0 and currents Ax D 2, 1, 1; f sends 3 units from node 2 into node 1.
#
"
#
"#
"
#
"#
"...
View
Full
Document
This note was uploaded on 09/25/2012 for the course PHY 103 taught by Professor Minki during the Spring '12 term at Korea Advanced Institute of Science and Technology.
 Spring '12
 Minki
 Mass

Click to edit the document details