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19 Jan 2013: Find the time-averaged energy flux vector 〈I〉 for (i) a plane harmonic S-wave with u =B cos(kx ωt), (ii) a plane harmonic P-wave with u = Ak̂ ... An SV-wave with displacement. u = B(cos θ, 0,sin θ)eik(x sin θz cos θ)iωt.

25 Nov 2013: v2 =g sin2 θ. cos θ. where cos θ = d/. ... the particle and deduce that the particleleaves the surface of the sphere when cos θ = 2/3.

26 Jul 2013: cosh cos ,. it is found that. θ. (a) (b) (c). ... cos cos cosh cos. 4 sin2. cosh cos 2 sin sin.

26 Jul 2013: b) Two dimers. We now consider the case of two dimersa and b. ... 482 2d3, 3a. V = ̄ab2 b1b1b2. ̄ba2 a1. a1a2 f a f b.

26 Jul 2013: evaluated at =. In Fourier notation and given that t= cos t, they become. ... 0. = 0. 2. Pressure and stress field. We have. ̃20 =. 2De1 De21 iDe1. B. 2 cos 2 sin 22, 43. as well as zero pressure p2 = 0, so that the average

26 Jul 2013: With this in mind we let g2 = g1(z φ) with g1(z) = cos(z), andwe set equal for both sheets the bending stiffness B and the forcing amplitude A. ... B(576 B2. ) C, (3.7a)η2(z) = A. 12B [cos z cos(z φ)] [288 sin z (288 B2) sin(z φ)]B(576 B2).

4 Oct 2013: cos A cos B =1. 2[cos(AB) cos(A B)]. sin A sin B =1. ... 2[cos(AB) cos(A B)]. sin A cos B =1. 2[sin(A B) sin(AB)].

26 Jul 2013: 1(a). Our new result is illustrated in Fig. 1(c), where we showthe dynamics of swimmers undergoing reciprocal motionwith velocity Uswim UðtÞe and UðtÞ U cos!t withU 5 ... a) No swimming, (b) steady swimming at speed U 5 m=s, (c) reciprocalswimming at

4 Oct 2013: TA sin θATA cos θA. µgδxT. (3). Now. TB sin θBTB cos θB. ... T(t) = cos jmnct and sin jmnct. Putting all this together i.e.

28 Nov 2013: X = 0 is given byF′ G = c0, (cos X σ0) G sin XF = 0, (4.4a,b). ... Θ = Ae(1κ2)τ cos κX Z, W = Ae(1κ2)τ cos κX, (5.1a,b)for any amplitude A.

26 Jul 2013: mal a or b at given Pe). Starting from a random initial condi-. ... For t 1, the swimming stroke,. b cos t b1 þ sin t bk; (44).

19 Oct 2013: exponential counterpart, cos(µ) = 12 (e. ... Then φ̂ = φ̂(x, z) exp(kz) cos(kx),equivalent to a single pair of progressive deep-water waves travelling in the x directions.

26 Jul 2013: 3) leads to 2p = 0. A solutionof the harmonic equation for the pressure is, at the first order inkb, p = p1(r,θ) (A cos s B sin s) with s = ... xi = x0 b cos β sin (s φ) , (34)yi = y0 b sin β sin (s φ).

19 Oct 2013: expected. (ii) all A’s, B’s zero except A1:. φ = A1r cos θ = A1z , u = φ = A1ez. ... z. (iii) all A’s, B’s zero except B1:. φ =B1 cos θ.

4 Oct 2013: B(cos(1 ξ)) A cos ξ = 1. Solving these two equations for A and B gives the Green’s function to be:. ... Then. I. λ=. 0. cos(λx)eαxdx = <(. 0. e(αλi)xdx. ),. = <[e(αλi)x.

4 Nov 2013: r1 = [2 cos(φ1)]cos(φ1 θ ). d. 2, (3). r2 = [2. ... 2 cos(φ2)]cos(φ2 θ ). d. 2. (4). In drainage, θ = θr and the drainage capillary entrypressure corresponds to the minimum radius of curvature r1:rdr = minφ1 r1(φ1; θr ).

26 Jul 2013: uy=y1 = 1 ṠR cos y=y1 , 11a. vy=y1 = ṠR sin y=y1 , 11b. ... uy=y2 = 1 ṠR cos y=y2 U , 11c. vy=y2 = ṠR sin y=y2.

26 Jul 2013: B]lab = B0[cos(ωt ), si n(ωt ), 0]T. (2). Experimentally this magnetic field provides an external torque to the head of the artificial flagellumbut no external force. ... ξ̄‖ =ξ‖. µ= 2π. ln(2 cos θ/γ ) 1/2,. ξ̄ =ξ.

19 Oct 2013: be similar to that of the spherical harmonicP4(cos ), and so we adopt the estimate L 4:5. ... Then equations(1) and (2) imply T=T r2c 2i =gL and juj 2g T=2N2T , where we have takensin cos 21=2.

26 Jul 2013: a) The no-shear regions are transverse to the flow direction. (b) Theno-shear regions are parallel to the flow direction. ... Thesolution to this problem was derived by conformal mapping in Philip (1972a) and isrecalled in Appendix B.

26 Jul 2013: 23). This equation can be integrated in t as. e(j )i (t, τ) = a(j )i (τ) cos(t) b(j )i (τ) sin(t). ... A(τ) cos(2t) B(τ) sin(2t). C(τ) cos(t) D(τ) sin(t) E(τ), (28).

26 Jul 2013: 4 m1=1. m2=1. 1. m12 m221 m12. 2m12 m2. 2cos 2 b m1x cos 2 b m2y , 18from which we note that large contributions to Eqs. ... cos m2w W. m1=1. 1m Cm 12m1 12 ln S sdwds , 20b.

26 Jul 2013: x=0, 59. with boundary conditions. y10, t = 0,y1x. 0, t = cos t,2y1x2. ... 1 cos 12 102 Re1010.66. When the swimmer does not have a body, R FU = 0, we obtainthe “elastic scallop,” the flexible version of the two-armswimmer discussed by Purcell

26 Jul 2013: The scale bar is 10 mm. (b)Speed-frequency characteristics of flexible nanowire motors. ... ment as. x(s, t) b(s)cos[t f(s)], (29). y(s, t) b(s)sin[t f(s)], (30).

4 Nov 2013: vf (x, t) = kf. μ. ρg(h b). b= φγ kf. ... dp. dx. cos mπ 1cosh mπh/d. (B 2). correcting a factor of 2 from Happel & Brenner (1991).

26 Jul 2013: 1x, = U1ey , D4a. 1x,0 = b cos x tex a sinx t ey , D4b. ... UN =12 b. 2 2ab cos a2. D17. Allowing tangential motion does therefore reduce the swim-ming speed by the same amount, and the final results of thepapers, Eq.

19 Oct 2013: where θ is the polar angle and where the outward normal n̂ now has Cartesiancomponents (cos θ, sin θ). ... Then, noting that the Cartesian components of U atr = b are ( sin θ, cos θ)Γ /(2bπ), we can easily evaluate the integral.

19 Oct 2013: Further detailsof the method used can be found in appendix B and inN88. ... 4, and (b) that when judged by comparison withthe reinitialized primitive equation integration, the nor-.

26 Jul 2013: xm(x, t) =x 'a cos(x Re ω(t)t φ)and ym(x, t) = 'b sin(x Re ω(t)t),. ... xy|y=0 b sinx. 2ψ1. y2|y=0, (3.15). ψ2,x |(x,0) = a cos(x φ)2ψ1.

5 Oct 2013: As is evident, the quantity cos (θn,n) is exponentially small in n. ... 100. 106. 1010. (a) (b) (c). Figure 1: (a): cos(θn,n) against n.

5 Oct 2013: As is evident, the quantity cos (θn,n) is exponentially small in n. ... 100. 106. 1010. (a) (b) (c). Figure 1: (a): cos(θn,n) against n.

24 Jun 2013: FIG. 2 (color online). Steady-state circulation in highly concentrated B. subtilis droplets. ... 2(e)]. The averagevalue b4m is slightly smaller than the length ‘5mof B.

21 Mar 2013: from the start of the alphabet (e.g, a,b,c,. ) take values 1,. ... dx , as before (we note here that A, B, C and D, defined by (2.9).

26 Jul 2013: ̃1 =ir2. 4!cos 2" 1" ,! 16a". ũ1r = ir. 2sin 2" ,! 16b". ũ1" =ir. 2!1 cos 2""! 16c". B. Second-order solution. 1. ... 4-̃4!0" = 0. A4 sin 2" B4 cos 2" C4 sin 4"r2. ,!

26 Jul 2013: us(ξ,η)= u0s. 2(1 cos ξ)( cos ξ)(1 e2) (1 e2) cos(2ξ) s(ξ,η). ... Here hwis the minimal distance of the centroid to the wall without penetration, hw =(1 e2) (1 e2) cos(2θ)/2.

19 Oct 2013: determining the velocity field everywhere given only the mass under each isentrope (B-surface),. ... the potential vorticity distribution on each isentrope, and the B distribution on the boundary.

19 Oct 2013: determining the velocity field everywhere given only the mass under each isentrope (B-surface),. ... the potential vorticity distribution on each isentrope, and the B distribution on the boundary.