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j1 =12[sin(Φ) dΘ sin(Θ) cos(Θ) cos(Φ) dΦ],. j2 =12[cos(Φ) dΘ sin(Θ) cos(Θ) sin(Φ) dΦ],. ... with (recall that a2 a b2 = 0, see (3.4)). J′1 =12sin(Θ) cos(Θ) (a dΦ b dΦ),.

2) isgiven by B(?): = @cos(2c:/#) 8@6sin(2c:/#) @for the periodic axis, while for the reflective axis wehave B(A)0 = 0 or B. ... A): (=, =0) =. #1 5 (==01)/26(=)6(=0)/[f cos(c:/#)] with6(<) =. 5 sin(c:</#) sin[c:(< 1)/# ] when. : >

bottom topography considered is described by:. b(x) =. . 0.25 cos(10π(x 1.5)) 1)) m if 1.4 m x 1.6 m,0 otherwise. ... 0. 0.5. 1. 1.5. x (m). (b. b U)/. b ref.

stabilising short-range interaction, Π̃2(h̃) = B/h̃6. Here h̃ is the dimensional film thickness,. ... Using cylindrical polar coordinates (r, θ, z), such that y1 = r cos θ, y2 = r sin θ and.

1/2Pmn1(x) , (5). where x = cos θ, we can rewrite (4) as. ... Bs =n=1. nm=0. (reqr. )n2[gmn cos mφ h. mn sin mφ].

a(x, y) = 1. m=1. αm cos(2πβ1mx1) cos(2πβ. 2mx2)ym,. where β1m = mkm(km 1)/2, β2m = km β1m and km = b1/2 (1/4 2m)1/2c ... φij = 2 cos(iπx1) cos(jπx2), νij =1. 4exp(π(i2 j2)l2). We choose the correlation length l = 0.65 and rewrite the sum

For example, Fig. 1 shows two block designs with v = 15, b = 7 and. ... Thus current is defined on( B) (B ) and voltage is defined on T B.

For (DBC) another straightforward, although a little longer computation gives thatbifurcation values are of the form α = 1/|cos b|, where b solves. ... Cb tan b = 0. (16). We denote by b0 (π/2,π) the smallest positive solution, set α0 = 1/|cos b0|,

Wκ(α) =κ. 2. Q. {|sα|2 η cos(2α). }dVol, (1.4). where η is a function which depends only on the geometry of the torus. ... Examples. 1. (γ [0, 1] v(γ)). Let v(x,y) := (cos(y), sin(y)), θ̄ (0, 2π), and γ : [0, θ̄] R2, γ(t) := (0, t)

trjt = r. jt. (rjt )2 1. (rjt )2 2rjt cos(ϑ. ... jTt). (rjt )2 2rjt cos(ϑ. jt ξ. jTt) 1. (23). (where we have suppressed the dependence on z to ease notation).By observing that the right hand side of (22)