Views
5 years ago

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

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

72 held fixed and the

72 held fixed and the speed of the inner cylinder is increased quasistatically. A schematic of his results is shown in Fig. 3. There are two interesting features of Fig. 3. First, there is a range of speeds for which no laminar-like flow exists; secondly, there are two stable half-branches correspond to a regular mode and an anomalous mode, related by a shift of one vortex along the axis of the cylinder. Cliffe obtains similar results numerically, as follows. Following Schaeffer [1980] he includes a homotopy parameter v in the boundary conditions at the ends of the cylinder, such that the boundary conditions are Neumann for "~ = 0 and physically realistic for "~ = 1. When "v = 0 Cliffe finds a pitchfork bifurcation which, as "V is turned on, breaks to a perturbed pitchfork as in Fig. 3. The discussion in §2 leads to the following observations. The 'q = 0 model with NBC should indeed lead to a pitchfork, via the introduction of PBC and O(2)-symmetry. Moroever, the two half-branches of the pitchfork are related by the translation (1.7). That is, one expects to find one half-branch of regular modes and one half-branch of anomalous modes. Note that this conclusion can only be reached by making the extension to PBC. Note also that the only genuine axial symmetry in the apparatus, reflection in the midplane, acts trivially in this pitchfork bifurcation, because the number of vortices is even. Thus it is not surprising that when NBC are violated (v ~ 0) the pitchfork becomes imperfect, as in Fig. 3. In this instance the experiments of Mullin are in good agreement with the numerics of Cliffe, and they both agree for perturbed NBC as in Schaeffer's approach. It should be remembered, however, that there are numerous fluid states - such as wavy vortices and spirals - that are not consistent with NBC, but are admitted by PBC. The work cited previously leads to equally good predictions concerning these other PBC- based patterns, several of which have been verified by experiment. The situation appears to be that neither NBC nor PBC provides a fully adequate model, but that each works surprisingly well for an appropriate range of flow patterns. 4 Rayleigh-Brnard Convection Next we discuss Rayleigh-Brnard convection in a box. Consider the onset of the convective instability in a 2-dimensional problem in {0 < x < re, 0 < y < n} with (x,y) denoting horizontal and vertical directions respectively. The problem is more complex than reaction-diffusion equations on a line because boundary conditions must be imposed in both x and y. We consider here the case in which the boundary conditions on the horizontal surfaces y = 0, ~ are homogeneous and distinct. For example, a Robin-type boundary condition applies to the temperature at the top if the top surface radiates heat according to Newton's law of cooling. In this case there are no symmetries associated with the boundary conditions in y, and there is no modal structure in y. In the absence of the vertical sidewalls the equations of motion are invariant under translation x ~ x+£ and reflections x ~ x0-x. When sidewalls are present we may take the boundary conditions to be or ~v 20 u(x,y) = ~-x (x,y) = ~xx (x,y)= 0 on x = 0, ~ (4.1a)

73 00 u(x,y) = v(x,y) = ~-x (x,y) = 0 on x = 0, n, (4.1b) where (u,v) are the (x,y)-components of the velocity, and 0 is the temperature departure from pure conduction. In both cases the boundary conditions at the sides are identical, and the problem therefore has Z 2 symmetry x: x orc-x. The boundary conditions (4.1a) describe free-slip perfectly insulating boundaries, and extend to PBC on -n < x < n with u(-x,y) = -u(x,y) v(-x,y) = v(x,y) (4.2) 0(-x,y) = 0(x,y). Consequently there is a well-defined mode number m. Consider now the action of the reflection x. Since (u,v) are components of a vector we know that x(u,v) = (-u,v). (4.3) Therefore x acts on mode m by X(Um,Vm,0m) = (-(-1)m+lum , (-1)mvm,(-1)m0m) = (-1)m(um,Vm,0m), (4.4) and the reflection symmetry acts nontrivially on the odd modes and trivially on the even modes. The odd modes therefore automatically undergo a pitchfork bifurcation. The even modes also undergo a pitchfork, but only because the horizontal translation (1.7) acts by -I on both even and odd modes. Thus the pitchfork bifurcation in the even modes is a consequence of the translation symmetry of PBC. Case (4.1b) corresponds to no-slip, thermally insulating boundaries. Since u(x,y) = v(x,y) = 0 on x = 0, n one might try to extend the solution to -n ~ x < n by u(-x,y) = -u(x,y) v(-x,y) = -v(x,y) (4.5) as in the scalar case (2.1). But since this violates (4.3) this problem cannot be extended to PBC on -~ < x < n, and hence there is no mode structure of the form (1.5). Indeed, explicit calculation shows that the eigenfunctions are sums of trigonometric and hyperbolic functions, Drazin [1975]. These, nonetheless, divide into two classes, odd and even with respect to x. The odd eigenfunctions break z and bifurcate in pitchforks. Since there is no translational symmetry we do not expect the even modes to bifurcate in pitchforks. An additional reflectional symmetry ~ is present if the boundary conditions on top and bottom are identical. In the special case 0u ~-z (x,y) = v(x,y) = 0(x,y) = 0 on y = 0, r~ (4.6) the boundary conditions extend to PBC on -~ < y < ~ under u(x,-y) = u(x,y) v(x,-y) = -v(x,y) (4.7) 0(x,-y) = -0(x,y) and a mode structure exists in the vertical direction. Since ~:y ~ n-y acts by ~(u,v) = (u,-v), ~(0) = -0. (4.8) it acts on mode n by "~(Un,Vn,0n) = (-1) n (Un, Vn, On). (4.9)

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 ...
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 ...
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
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
9 2 3 7 5 4 8 B M A 1 6 - phfactor.net
August 6 & 7, 2011 19 Sunday in Ordinary Time Writing Team B 5 ...
Midterm 1 KEY MC Answers 1. B 2. B 3. A 4. D 5. A 6. D 7. C 8. A 9 ...