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26 Jul 2013: Received 24 May 2002; accepted 17 December 2002; published online 30 June 2003. ... ¤ and compute a matrix F with components fij = cij =( i ¤j ).The minimal-energy stabilizing feedback controller is then given by u = KÀ , whereK = B ¤ F ¡1.

26 Jul 2013: 9500 Gillman Drive, La Jolla California 92093-0411, USA(Received 15 January 2008; published 24 July 2008). ... f 2= is an intrinsic viscocapillary length scale (theOhnesorge length), (b) L < ‘m, where ‘m 2=KA isthe intrinsic membrane-viscous length

3 Feb 2013: Photosynthetically Active Radiation (Fluora; OSRAM) and 10 h in dark at 24 C.Cells were harvested during the exponential growth phase. ... Phys Rev Lett 105(16):168101. 17. Guasto JS, Johnson KA, Gollub JP (2010) Oscillatory flows induced by

7 Jan 2013: Photosynthetically Active Radiation (Fluora; OSRAM) and 10 h in dark at 24 C.Cells were harvested during the exponential growth phase. ... Phys Rev Lett 105(16):168101. 17. Guasto JS, Johnson KA, Gollub JP (2010) Oscillatory flows induced by

26 Jul 2013: 4,5,24–30]. The individual locomotion of spermatozoa was first studiedby G. I. ... Bottom: Variation of the ratio of the minimum to themaximum of the rate of energy dissipation with ka (symbols).

19 Oct 2013: GRAVITY WAVE GENERATION BY VORTICAL FLOWS. IN A ROTATING FRAME. Rupert Ford. Churchill College Cambridge. A dissertation sublnitted for the degree of Doctor of Philosophy. at the University of Cambridge. October, 1993. To my grandparents. Margaret

19 Oct 2013: GRAVITY WAVE GENERATION BY VORTICAL FLOWS. IN A ROTATING FRAME. Rupert Ford. Churchill College Cambridge. A dissertation sublnitted for the degree of Doctor of Philosophy. at the University of Cambridge. October, 1993. To my grandparents. Margaret

19 Oct 2013: Another importantwave activity measure is given by the pseudomomentum vector,. p = kA =h′ u′. ... Because of the ray-invariance of absolute frequency we have that |kA| = |kB | asympto-tically.

4 Oct 2013: n2π2 1. (24). On the other hand we had in a previous example the expression constructed from homo-geneous solutions:. ... Its FT is easy to calculate:. f̃(k) =. aaeikx dx =. aa. cos(kx) dx =2 sin(ka). k. (41). so by the Fourier inversion theorem we get.

4 Oct 2013: Xn(x) = Bn sinnπx. Lλn =. n2π2. L2. (24). Note that the first set eq. ... if ψ = 0at r = a then Rnk(a) = Jn(ka) = 0 so for each n, ka = kni with i = 1, 2,.