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30 May 2006: Howard’s technique applied along a = 9 4 3 with n = Q & (a - J)Q and H = (cos $Y)(&--)’ yields. ... There is another neutral eigen- solution. (2.13 a, b ). J J3(a) = 2 - a2 (J2 < a < 4 3 ) (2.13 c) c = 0, 4 = (cos 9y)4(’t-J)’ (sin

26 Mar 2023: β s. [h3(. hs. cos α sin α)]. (3.11). We note that (3.11) reduces to (2.1a,b) only under the assumption |h/s| | tan α|. ... sk1N = skN vk1N t, vk1N =(. hk1n1)3. cos α. 3(skN sn1), (A3a,b).

30 May 2006: 2. 1 0 4 and. 992n(& 7) = ( - )%-l sin (o cosh < cos 7 cos t ) cos (o sinh 6 sin 11 sin t ) se2,(t) dt I 0 =. { 2.10 b ) ... 4. 1 d ). x = 8% cosh [cos@, y = & sinh %sin@,.

30 May 2006: 6 ). We note the alternative representation. sgn ($77 - t ) exp ( - ikr cos f? ... exp ( - ik< cos 8) sin (ky sin 0). (

11 Apr 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 Mar 2023: cos nθ , (6.6). so that, on r = R2 = b β cos nθ ,pr. ... r=R2. = 6μh3. dhdt. (b β cos nθ ) [n k̂ bn1 n m̂ bn1.

26 Apr 2006: dd. dy 0 at y 0 ð4aÞ. and. d! H R cos ycos y. ... Wecomplete the variable transformation by definingthe horizontal coordinate x 1 cos y so that f isa positive, monotonically decreasing function of x,where x ranges between 0 and 1.Eq.

26 Apr 2006: Fk"(2pR2)6kR2;Ph#. 0. sin h cos hAPh. h#. sin /. d3d/B dh. ... oq. j3. 6kR2:h#0. sin h cos h d3dh. :h#0. sin h cos hA:hh# sin / d3 d/B dh,.

30 May 2006: 5.13 a) = cos{Ic(y-sy))-Sjsin{k(y-sy)) (5.13 b ) = ( q 2 c2) cos {Ic(y - 9 ) tan-1 (&I)} ( 5. ... 5.16). (5.1 7 a). 00. = 2 K 2 b / (7 cos K 7 csin q)2dx, (5.17 b ) --m.

30 May 2006: A ) = 1 m(sin h f - Rh cos A f ) h(6) df nrh(lR2h2) 0 For R = 0, the second term in the integrals in ( 6. ... RBO(5) [sin (C/R) i sgn k cos (C/B)], (A 1 ) where.