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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

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). (

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.

30 May 2006: $0) = (1 - e-7 cos T) - d o ) e-7 sin 7, (3.25) 8 0 ) = e-7 sin 7 d o ) (1 - ec7 cos 7), (3.26). ... 1 2 ) and using (3.22c), we obtain. G(1) = 2M;(5) 2v$o)[l- e-T(cos 7 sin)].

1 Jun 2006: 1. In ( b ) the. A new oscillatory instability in a mushy layer 257. ... 4. 5 ) 8 = 8 B 6[-2 Sin 712 cos(kZ 60117)],.

30 May 2006: Higher normal modes of the form. W m j $oshmTax] (2.11a) T ( x , z , t ) = To cos mmax ept sin nmz ( 2. ... S ( x , x , t ) So cos m m x (2.11 c ). (2.12). and exchange of stabilities a t a thermal Rayleigh number given by.