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24 Aug 2009: 3.20). (3.21). (3.22). subject to the matching conditions. go = 0, U, = 0, = [(I -i)/24] a$,/& at z = 1. ... 14 0 0.04 0.08 0.12 0.16 0.20. - 0.2. - 0.4 a. -

17 Sep 2009: Penrose transform. In 1969, Penrose gave a formula for solutions to the wave equation inthe Minkowski space [20]:. ... Phys. 8 345–66[20] Penrose R 1969 Solutions of the zero-rest-mass equations J.

15 Oct 2009: mN. )Φ1. ψ2r. 2rψ2 F1 = 0, (1.20). 2mψ2. Φ1r. F2 = 0, (1.21). ... the function u in (1.4)), and(eliminating ψ2 between (1.20)-(1.21)) the function Φ1 approaches ϕ, a solution of the Newton-Schrödingersystem (1.4).

31 Jul 2009: 10. 8. 6. 4. 2. I I I I I 0 4 8 12 16 20. ... 20 - 4. 0 4 8 12 16 20. Y l D FIQIJRE 4.

8 Jul 2009: 20. 15. 10. 5. 0 1 2 3 0 2 4 6 8 10. ... 0 5 10(R – R0)/R0. 15 20 25. ES/(. mSgR. 0)E.

10 Jul 2009: and22. No pre-stretch was applied. The best fit of the adjustable parameter is L1 = 20. ... i ) (in Pa) as a function of time (in s) for the case L1 = 20.

10 Jul 2009: S0031-9007(97)05092-8]. PACS numbers: 47.11. j, 02.70.Pt, 47.20.Cq, 68.10. – m. ,. aher. al. lst,. ln,ed. gesngr-n. ffy.dutof. Inviscid pinchoff is ... With these numerical valueof A, B, and a, the three terms in the Bernoulli equa-tion (1) become in

23 Jul 2009: p E { ( T - 6,O) e(3fT cos 2 8 , 2 f ( T - 6 ) sin 20) O ( e 2 ) } , in which changes in the internal pressure have been ruled out at ... R ( 0 , z ; e ) = R,(z) [ 1 e f ( z ) cos 20 O(e2)].

13 Jul 2009: a’=20 11-022-633. E.J. Hinch / J. Non-Newtonian Fluid Mech. 54 (1994) 209-230 211. ... 14. In Fig. 20 we have plotted ,uextNb4/pR4(t), which the above estimates give as k,/12k2.

13 Jul 2009: a = 0.62 (stress Y -“.38) for 4 = 20. ... References. 1 R. Keunings, J. Non-Newtonian Fluid Mech., 20 (1986) 209-226.