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5 Mar 2021: Γ ={(r,θ): r = R(θ) = b cos θ. c2 b2 sin2 θ. = ... 2. ξ2+ η2. )and h = a. cosh η cos ξ ,(4.33a,b).

16 Nov 2005: c) P = 2. q(. 1. q cos p). sin p and Q = log(. ... q1 = Q1 cos λ P2 sin λ , q2 = Q2 cos λ P1 sin λ.

21 Feb 2005: sin(θ ψ) sin(θ ψ) = 2 sin θ cos ψ. We have. ... ρ(B) = cos. (. π. m 1. ). 1 π2. 2m2,.

10 May 2022: µs(θ)=. l=0. µslPl(cos θ), µp(θ)=. l=0. µplPl(cos θ). (2.8a,b). In the case of our spherical particle, the slip velocity due to either one ... s0 =sb(rc)r sR3. 2r2(sz cos θ s. y sin θ sin φ).

8 Jul 2009: B′ cos KX instead of A and B, with solution. N (0) ce(. ... A′ B′. 2sin 2T. B′. (2)2cos 2T. )cos KX. (5.10). Note that the oscillating part has to be retained in the present analysis up to the secondorder.

8 Jul 2021: φ̇n = ω0 βω0 (cos χn cos χn1 ) βτ ω20 (sin χn1 sin χn ), (16). ... θ1 sin θ2 ) sin α1 sin δ cos θ1 (sin θ1 sin θ2 )]}, (46).

13 Dec 2018: b/ø (x,h) = exp{ik [xS h (x) cos 0 ] }. •ei•$[ 1 ikh(x)cos O- ( k2/2 )h2(x )cos 2 0]. (9). ... k2 •b• ø (x,zt) = - 1 ikh (x) cos 0- • cos 2 0 X [h2(x) (Zl-h) 2] )e ikxs -(za-h) •xx ikcos0.

26 Feb 2008: e1 B. (cos(βt) i sin(βt). )e2}. = eαt(a cos(βt) b sin(βt). ... J =. (0 1. (g/l) cos θ 0. ). The phase diagram has this form:.

19 Oct 2018: µs(θ)=. l=0. µslPl(cos θ), µp(θ)=. l=0. µplPl(cos θ). (2.8a,b). In the case of our spherical particle, the slip velocity due to either one ... s0 =sb(rc)r sR3. 2r2(sz cos θ s. y sin θ sin φ).

2 Jul 2024: along the a = (sin, 0, cos) axis or along the b = (sin, 0, cos) axis. ... hEPR|Az B |EPRi = sin2. 2 cos2. 2= cos. This means that if we take the four operators A2, B1, A1 and B2 to be spin operators,.

7 Feb 2012: a magnetic moment m aligned along its axis. Let there be a magnetic field B, so that. ... each molecule has a potential energy mB cos θ (θ being the angle between the axis.

27 Mar 2022: R. cos. (. ω. (. t. -. t. 0. ). ). (4.299). with. a. ,. b. ,. R. and. t. 0. integration constants. ... A. cos. (. ω. (. t. -. t. 0. ). ). and. p. =. -. m. ω. A. sin. (. ω. (.

18 Nov 2015: 2. z2h5 B. 2. θ2h5 = 0,. (curiously A α/6, but B β/80). ... t1/4. h(π) =1 cos L. t1/4. (K ((π L) cos L sin L.

5 Nov 2016: n=0. P(cos θ)Pn(cos θ). 3. Show that whenever r 6= r′, the function 1/|r r′| satisfies Laplace’s equation. ... y(x, 0) = A sin(πx/L) andy. t(x, 0) = 0. (b) If an extra transverse force F0 sin(πx/L) cos(πct/L) per unit length acts on the

9 Dec 2016: φjφk〉c =. dDq. (2π)Deiq(xjxk). r0 2βJD. µ=1 cos qµ. where qµ (π,π] is the µ-component of q.(b) Now take as the Hamiltonian (restoring the explicit ... G̃(q) =. [2. a2. Dµ=1. (1 cos(aqµ)) m2]1. [You need not repeat all the steps that are in

26 Jul 2013: With this in mind we let g2 = g1(z φ) with g1(z) = cos(z), andwe set equal for both sheets the bending stiffness B and the forcing amplitude A. ... B(576 B2. ) C, (3.7a)η2(z) = A. 12B [cos z cos(z φ)] [288 sin z (288 B2) sin(z φ)]B(576 B2).

12 Aug 2016: 2 2 ). ( 3. 2 3 ). (3.24). so that the normal derivative dominates the tangential terms; recall that our choice (3.1) has XR = b cos 0. ... B 3). - b/2R3. The magnetic boundary conditions then become. 0 6&=, = [ 3ih cos 812 - a ] ei@,. -

26 Jan 2023: The general solution is. x(t) = X R sin(!Bt ) and y(t) = Y R cos(!Bt ) (1.2). ... m!B. Y = y(t) R cos(!Bt ) = y ẋ! B= y.

25 Nov 2011: u(1, t) = ν(t) = eω2t cos ω b. ω2. (1 eω. ... p0(1, t) = ν(t) = eω2t cos ω b. ω2. (1 eω.

26 Jul 2013: us(ξ,η)= u0s. 2(1 cos ξ)( cos ξ)(1 e2) (1 e2) cos(2ξ) s(ξ,η). ... Here hwis the minimal distance of the centroid to the wall without penetration, hw =(1 e2) (1 e2) cos(2θ)/2.