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19 Dec 2006: 24, 0.41, 0.56 and 1.,where T is the total duration of the spreading, and for a restitution e = 0.5 and a inter-grain friction µ = 1.

24 Nov 2006: ẏ4 = 1. Σ = 1σ( rr r )σ( r) = 24 nr rr r ẏ1 = 12 y2y3ẏ2 = y1y4. ... 24 A. ISERLES and G. R. W. QUISPEL and P. S.

23 Nov 2006: 16. 13. 13. 16. Part III: Numerical Solution of Differential Equations 24.

11 Nov 2006: 2.6.2 Three dimensional space. 23. 2.6.3 Higher dimensional spaces. 24. 2.7 Components. ... 24. 2.7.1 The Cartesian or standard basis in 3D. 24. 2.7.2 Direction cosines.

28 Oct 2006: 1.7.13 Aide Memoire. 24. 2 Partial Differential Equations 25. 2.0 Why Study This? ... You should be able to do example sheets 1/2/3/4 after lectures 6/12/18/24 respectively.

28 Oct 2006: 1.7.13 Aide Memoire. 24. 2 Partial Differential Equations 25. 2.0 Why Study This? ... You should be able to do example sheets 1/2/3/4 after lectures 6/12/18/24 respectively.

20 Oct 2006: From Eq. 22 we have. U1 =E1/y0p1/y0. =1. 2y0, 24. as expected.For large y0 an accurate approximation to the solution.

7 Sep 2006: 1 Λ2. (27). In a compact form the equations (24) can be written as. ... 24). The solid lines show three families of solutionswith α1 = α2 = 0.5 and Λ.

2 Sep 2006: the literature (see [11, 16, 23, 24, 27]and references therein) which have been specifically designed for different familiesof problems. ... x/xj )2 1). (2.24). This completes the proof. 2. 9. 2.4 Accuracy.

31 Aug 2006: We deduce from (56), (57), (58), (22) and the first part of (24)that. ... Numer. Anal., 25(2005), pp. 1–24. R.K. Beatson, M.J.D. Powell and A.M.