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30 Nov 2015: φ|Rn{0} = 0 , tφ|Rn{0} = 0 and lim|x|. φ = 0 , (10.24).

26 Oct 2015: and hence that the volume element is. dV = dx dy dz = r2 sin θ dr dθ dφ (3.24).

9 Nov 2015: shape. y(x) =µg. 2Tx(x L). (7.24). that obeys y(0) = y(L) = 0. –

18 Nov 2015: f, f) =1. 2π(f̃, f̃) (8.24). so that the L2-norm of a Fourier transform agrees with the L2-norm the original function,.

31 Dec 2015: 2p 1)" 2p1. L. x. 2+. 2q1Ly. 2# ez, (24). while the sensitivity on the vertical walls Hvert(y) at x = 0 and x = Lx can be obtained by symmetry.The ... Rev. Lett. 99, 048102 (2007).24 R. Golestanian, T. B. Liverpool, and A.

4 May 2015: Charney [24], Aleksandr Mikhailovich Obukhov [25],. and Ernst Kleinschmidt [26]. 1.4 Cyclone and jet structure. ... dent pioneering work of Charney [24] and Obukhov [25] and are described.

9 Nov 2015: We have (P 1)K = 24 equa-tions for K(K 1)/2 = 28 unknowns. ... Dj,k = 0 for |k j| 6. There remain 25 unknowns for the 24 equations.

4 Oct 2015: This new vibrational modes are called surface phononsdue to their localisation at the surface [24–26]. ... m & N µ(A1W1) s log(N), (24). in order to guaranty successful recovery with high probability, where µ(B) = max1i,jN |Bi,j|2.

4 Oct 2015: This new vibrational modes are called surface phononsdue to their localisation at the surface [24–26]. ... m & N µ(A1W1) s log(N), (24). in order to guaranty successful recovery with high probability, where µ(B) = max1i,jN |Bi,j|2.

13 Jan 2015: Hence1. ǫ0Q =. S=V. E dS. (24). The right-hand side is the flux of E out of V. ... The statement (24) is Gauss’s Law. It is ofpractical use.Ex. Consider a point charge q at rest at O, and let V be the sphere of radius r centred