Search

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 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 ).

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.

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.

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: 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.