Views
5 years ago

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

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

q odd q even 180 °T

q odd q even 180 °T s~llx-all • T o • O Figure 4: Sheets of reversible subharmonics for the case IX and the secondary bifurcations of nonreversible subharmonics. Case IX : This last case is the first one where nonreversible subharmonics branch off, as secondary bifurcations from the manifolds of reversible subharmonics. The normal form and its unfolding are (6a,~2~ + elA,~4v + ~3A + a) for all q. The unfolding parameter only affects the nonreversible subharmonics bifurcation. The bifurcation equations for the reversible subharmonics are $a + e2s q -I- elAsq -2 = 0 (+ only if q is odd). The line (0, A, 0) is the only intersection with 7~. From the q conjugate parabolas, el A + e2s 2 = 0, a = 0, there are branches of nonreversible periodic orbits [in'klng the manifolds of reversible subharmonics along tubes of width proportional to ~ (Fig~lre 4). These branches axe branching off at The system they satisfy is 8 2 = ~le2e3a + h.o.t, a = 0, A = -~3a + h.o.t. a=0,~2u+~lA=0,~4v+~3A+a=0. C J

3.2 dimB--1 181 We are now in the situation where "antireversibility" dominates "reversibility". As we discussed earlier, a second parameter is very much welcomed to enrich the bifurcations. As in the last section, let consider the following model equation "~" + A~ + f(t,~, ~,£) + p = 0, (10) where the only difference with (7) is that f(-t,-x,y,-z) = f(t,x,y,z). We look at the same questions as before with the kernel of the linearisation being generated by the same vectors. If last time the mean of the solutions of (7) was in effect arbitrary, parametrizing the families, it is not so now. The role of p is to provide enough freedom for the mean, allowing bifurcating branches to occur. We should point out that the mean of each solution is still represented by the third variable, in this case b. The generic expression for ~b, after some trivial changes of coordinates, is given by : b(hoz +,.. + 1~o~ q-l) + iz(g~u + 92b 2 + e4A +...) + i~oz -q-~ +... = 0 (11) e]u + cab 2 + ealz +... = O. (12) After sca~ng~ ~ = ~I = ,I = ~] = 1. The Z~ equation is ,2b 2 + '3t' +--- = 0. It is therefore clear that 7~ = {(0,0,A,0)} and T = {(0,b2,A, b2)}. We have an illustration of our previous analysis. As a one parameter problem, the origin is an isolated reversibh solution, a turning point (generically) of a branch of nonreversib]e 1-periodic solutions. For subharmonics it is useful, if not almost necessary, to have a second parameter p. Its effect on the solution of period 1 is straightforward. We get a cylinder in A on the previous diagram. The bifurcation of the reversible subharmonics takes now place along the following curves : q=3 p=-ele3s 2+h.o.t, A=-e4~0s+... q = 4 p = -ele3 ~2 -k h.o.t, A = -q(gl -kgo)82 + ..- q>_5 #=-ele3s 2+h.o.t, A=-e4gls 2+... We suppose that every relevant coefficient is nonzero. The question around nonreversible subharmonics is more complicated. For a start we prove that there axe bifurcations for (11,12). Later on we shall see that the second parameter is not necessary for q = 3 and in some cases for q = 4. Let introduce the polar coordinates re i° for z. Separating the real and imaginary parts of (11), the system (11,12) becomes b(ho + horq-~co~(qO)) + ~or~-~i~(qO) + .... 0 (13) 91r 2 + g2b 2 + e4A - hobrq-2sin(qO) + ~orq-2cos(qO) +... = 0 (14) elr 2 + e2b 2 + eat +... = 0 (15) A simple application of the implicit function theorem shows that we can always solve that system for b, A and p near the origin and we find the following unique solutions, representing nonreversible subharmonics of (10) :

Page 1 Page 2 D B 5 6 D MOTOR Mo~, lmotowon T R T ...
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
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 ...
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
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 ...
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 ...
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 ...
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 ...
1. NC 2. L 3. NP 4. D 5. F 6. A 7. M a) b) b) 8. T 9 ... - Bangladesh Bank
1 2 (a) 2 (b) 3 4 5 6 7 8 9 10 11 - Travel Insurance
1. 2. 3. 4. 5. 6. 7. 8. Meeting A. M B. Pl Approval Awards a Public He ...
A. LISTENING. 1-5 B. VOCABULARY 6-15 C. GRAMMAR. 16-25
Case 3:13-cv-03813-B Document 6 Filed 09/20/13 Page 1 of 5 ...