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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.

2 Sep 2012: solution to (24) subject to the boundary conditions is(Carslaw and Jaeger, 1959). ... computations. Fig. 2. The radial distribution function R0(r) for the parameter valuesP 1, N 8 , and Q 24.

10 Feb 2023: 2.23 and comments after. Remark 2.24) Let γ1 and γ2 be future pointing causal curves from a to b and from b to c.

29 Dec 2023: Z(1.24). If we make a measurement described by an operator Ô, then the probability that wefind ourselves in the eigenstate |ϕ⟩ is given by.

2 Sep 2012: We obtain ‘p 103 m, a value consistentwith the known range [13,14,24]. ... J. Hinch, J. Fluid Mech. 75,765 (1976). [24] H. Isambert et al., J.

25 Jan 2007: N M Mt 2ik2[σ3, M ] kA(x, t)M B(x, t)M. (4.3.13). ... iqt qxx 2rq2 = 0irt rxx 2r2q = 0. (4.3.24). The reduction r = σq̄, σ = 1, yields the celebrated NLS equation (1.2).

23 Jul 2009: 1 ) , ( 2. 2 ) ; and for the case ka 4 1 and a general form of P by Batchelor (1983). ... pO as r / a c o , (3.24). and P-D;Vp = ikD,.P on r = 2a.

2 Oct 2016: KL ::::: - ( 1 /24)[2 (rods, length I), = - ( 1/3)c2 (disks, radius c). ... p = -kT[C-C'],. which gives the following when substituted into (24) and (27):.

30 Nov 2015: φ|Rn{0} = 0 , tφ|Rn{0} = 0 and lim|x|. φ = 0 , (10.24). ... From (8.43) we have that. F1[sin ka] = i2(δ(x a) δ(x a)) (10.32).

4 Mar 2016: ux0 þ rðtÞ; t uðx0; tÞþ rðtÞu xðx0; tÞ: (24). As the velocity field, the distance x and the displacement rare made dimensionless using n0x, k.