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25 Feb 2008: l1 =2V 21 sin θ1 cos θ1. g=. 2eV 20 sin θ0 cos θ0g. = ... L =. n=0. enl0 =2V 20 sin θ0 cos θ0. g(1 e).

11 Aug 2008: AdaptingEqs. (4) and (5) for a rotating solution [20] gives. r r r Rr; (6). ... 15. 20. 25. 30. 35. µ. FIG. 3 (color online). Chemical potential as a function ofpumping spot size R.

1 Jul 2008: 4, 8, 14, 17, 19, 20] andsetting c = 0, b = 1 gives the dKP equation. ... Math. Phys. 273 357–78. [20] Pavlov M V 2003 Integrable hydrodynamic chains J.

26 Feb 2008: 1 Two cyclists start 40 km apart and head towards each other, each going at a steadyrate of 20 km per hour.

7 Oct 2008: 4.1features as a dashed line. 8. 0 0.05 0.1 0.15 0.20.82. ... 9. 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20.82.

26 Feb 2008: f (x1)f (x2). =f (αx1)f (αx2). 20. for any α. (If we changed our units of length from metres to centimetres, for instance, we would haveα = 100.).

31 Jul 2008: 13.1.29]. U (a, c, z) = z1cU (a 1 c, 2 c, z), (20). ... 10. 9. 8. 7. 6. 5. 4. 3. 2. K0 10 20 30 40 50.

18 Mar 2008: Eight points have beencomputed for each integral. 0 20 40 60 80 1000. ... w0 20 40 60 80 100. 0. 0.5. 1. 1.5. 2.

7 May 2008: We note that in contrast to the kinetic equations pre-sented in [19] and [20] for the one-component case, Eq. ... 20] C. Connaughton, and Y. Pomeau, C.R. Physique 5 5, 91 (2004).[21] S.

29 Jan 2008: 502, Campus Universitaire, Orsay, F-91405, France. Received 29 June 2007; accepted 15 November 2007; published online 20 December 2007. ... s1. Most experiments took place insidea tube of diameter d = 20 mm; other diameter values.