Views
5 years ago

SN~ (~6) lff 2It 3k_~ , 5 ",,x_J {b)

SN~ (~6) lff 2It 3k_~ , 5 ",,x_J {b)

66 (b) Next we discuss

66 (b) Next we discuss the expected type of bifurcation in the PBC case. When m > 0, the effective action of 0(2) on the bifurcation is by O(2)/E' = O(2)/Z m ~- 0(2). Hence, genetically we expect a pitchfork of revolution. When m = 0 and Z'= O(2), the group O(2)/Z' is trivial and genetically we expect a limit point bifurcation (in the branch of constant solutions). On the other hand when Z' = SO(2) we have O(2)/Z' ~ Z 2, and generically we expect a pitchfork bifurcation. The solutions to P(u) --- 0 satisfying NBC are found in FiX(BN) = {z ~ ker d$ > : Rz = z}. Thus, when m > 0, NBC picks out those solutions in the pitchfork of revolution that are invariant under the reflection Rx = -x, rather than a general reflection x ~ x0-x. These solutions form a pitchfork, for the following reason. Let T(x) = x + 2rc/2m. (1.7) The translation T lies in the normalizer of the isotropy subgroup D m =

U T(u) 67 m=2 -~ ................ 0 ........ ,,°,'~"JJJ" -:¢ 0 Fig. 2 As for Fig. 1 but with m even. Now only T acts by -I.. Assume that 9(u) depends on two parameters and that when these parameters are set to zero the linearized NBC problem has a double zero eigenvalue and no other eigenvalues on the imaginary axis. Then 0 lies in the intersection of two curves in parameter space along which Lu = 0 has a simple eigenvalue. Let m and n be the mode numbers along these curves, guaranteed by Proposition 1.2. Corollary 1.3 Suppose that m ~ n. Then dim ker L = 2. Remark Without the effect of 0(2) symmetry, we might have expected a Takens-Bogdanov type singularity at ~. = 0. The distinct mode numbers rule this out. Proof Extend to PBC and let = Mz + ... (1.8) be the O(2)-equivariant vector field on IR 4 -~ IR2xlR 2 obtained by centre manifold reduction. Since m ¢ n, 0(2) acts by distinct irreducible (indeed absolutely irreducible) representations on each copy of gt 2. Hence (,,° 1 M = 0 c2I 2 " The assumption of a double zero eigenvalue in NBC implies that c 1 = c 2 = 0. Thus M - 0. Restricting (1.8) to NBC proves the result. []

Page 1 Page 2 D B 5 6 D MOTOR Mo~, lmotowon T R T ...
P P P NP NP 12 3 4 5 6 7 8 W 9 10 11 12 13 14 W 15 16 0 17 A C B ...
B E R L I N C H A U S S E E S T R A S S E 5 7 - 6 1 B ... - Glen Leddy
DVAA: Flat Galaxies (by R.A.) A B C D E F G H I J K L 1 2 3 4 5 6 7 8 ...
Page 1 7 6 5 | 4 L 3 | 2 | 1 CK T ll ll ll )l KLÖ'ZOJ T ll ll ll ll klözo) A B ...
1 5 . T T O - F O R U M 1 2 . T T O - F O R U M O c t o b e r 1 6 ...
D e se m b e r 0 5, V olum e 0 6, Issue 11 w w w .ta n go n o tica s.co m
H O T E L - R E S TA U R A N T- B A R S e e s t r a s s e 2 2 5 6 3 ...
Reading grade 6 2.A.5.b - mdk12
a 1 2 a b b a 3 4 a b b a 5 6 a b b a 7 8 a b b a 9 10 ... - CodeMath.com
8 * 5 * 4 * 3 * 2 * 1 A 8 A 7 A 6 A 5 A 4 A 3 A 2 B 8 B 7 B 6 B 5 B 4 B 3 ...
Lösung 6 Aufgabe 1 a) b) c) Aufgabe 2 Aufgabe 3 a) 5 4 2 6 4 20 9 2 ...
5 b 5 6 10 c 12 a 5 8 A a 13 b Classroom Exercises - flip@mrflip.com
EAS 540 - STABLE ISOTOPE GEOCHEMISTRY Thrs., BUS B-5; 6 ...
Panel B 4:45-5:30 Round 2 Panel A 5:30-6:15 | Panel B 6:15-7 ...
TEST ANSWERS Version A 1. B 2. B 3. E 4. E 5. D 6. B 7. C 8. B 9. A ...
Solutions to Quiz 5 Sample B - Loyola University Chicago
¡' &) (10 , 2 354 6&7 98@ ) AC B$ E DG FH I 6 &Q ... - Université Lille 1
(0 12 3¡) 4(5 6 3 798 9 %@4a7(6 b cedgfhfhiqps rti u)vxwtwty
V 5 1 5 B 6 L 4 X P T S F
1. Nyelvismereti feladatsor 1. C 2. B 3. D 4. D 5. A 6. B 7. B 8. C ... - Itk
1 A 2 B 3 A 4 B 5 6 7 8 A 9 B 10B½ 11 A 12 13 14 B 15 A 16 B 17 A ...
Friday Warm-up A 5:00-5:30 PM, Warm-up B 5:30-6:00 PM Starts at ...
Midterm 1 KEY MC Answers 1. B 2. B 3. A 4. D 5. A 6. D 7. C 8. A 9 ...
1. Nyelvismereti feladatsor 1. A 2. B 3. A 4. A 5. D 6. D 7. B 8. C 9 ... - Itk
1 d 2 b 3 d 4 a 5 a 6 7 c 8 b 9 a 10 c - Qcpages.qc.edu
9 6030 EC 10 6030 B 11 6030 R 12 6030 N 5 6031 EC 6 6031 B 7 ...