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10 Oct 2017: 1 q cos p. ). 6. Show that the following transformation is canonical, for any constant, λ :. q1 = Q1 cos λ P2 sin λ , q2 = Q2 cos λ P1 sin λ ,. p1 = ... Q2 sin λ P1 cos λ , p2 = Q1 sin λ P2 cos λ.

1 Nov 2017: 3ρ̄r. (b) Consider linear perturbations about the homogeneous solution. Show that the density. ... t = B (1 cos θ). where you should determine A, B and θ.

13 Dec 2017: real a=2.,b=1.;. border Gamma1(t=0,2pi) { x = a cos(t); y = bsin(t); }. border Gamma2(t=0,2pi) { x = 0.5a cos(t); y = 0.5bsin(t); }. // ... border Gamma(t=0,2pi) { x = cos(t); y = sin(t); }. mesh Th=buildmesh(Gamma(20));.

13 Dec 2017: real a=2.,b=1.;. border Gamma1(t=0,2pi) { x = a cos(t); y = bsin(t); }. border Gamma2(t=0,2pi) { x = 0.5a cos(t); y = 0.5bsin(t); }. // ... border Gamma(t=0,2pi) { x = cos(t); y = sin(t); }. mesh Th=buildmesh(Gamma(20));.

3 Dec 2017: a). 100-. a sin0. (m/s ). /L2. L2. 00 a cos () (m/s'). ... For kinematic viscosity v assumed small andfor a sinusoidal forcing f(t) =a cos(2cot), the width ishk =(8&2/3)vk m '&p where p=(a —a, )/a,.

7 Nov 2017: u u u1, 1, 0, 1,3. 2sin 1 cos , 1r q q q q b q= = = f q( ) ( ) ( ) ( ) ( ). where θ is the azimuthal angle from the squirmer orientation e and ... From equation (1) we have that the tangential velocity on the squirmer is given byu 3 2 sin 1 cos ,

14 Jun 2017: cos. (ωλ ,1(ck,t). )sin. (ελ ,1(ck,t). )sin. (ωλ ,1(ck,t). ), (29). ... 0. + cos. (ωλ ,1(ck,t). )(t ck).  t ck(t ck)21.

22 Nov 2017: rA A O ‘( cos u1, sin u1) ð4:1Þ. and. rB B O ‘( cos u1 þ cos u2, sin u1 þ sin u2), ð4:2Þ. ... 0 its magnitude andt̂ ( cos u2, sin u2) the unit tangent vector joining A and B.Under the assumption of creeping flow, the drag forces areFA 2 zvA

15 Feb 2017: cos (λ,1(ck, t)) rλ,1(ck, t)sλ,1(ck, t). ... r′λ(t) = (1 q(t)λ) sin(θλ(t)) cos(θλ(t))rλ(t). and yields the seminal result:.

7 Nov 2017: b β(b b‖) , (43). respectively, where. P (θ ) = (1 U0)b b‖ cos θ (1 U0)(b b‖) cos2 θ, (44). ... uγ̇ th =. sin2 θ 2 b‖b. (cos θ q)2, (45).

26 Jan 2017: ûc(λ,t) =. 0. cos(λx)u(x,t)dx, û0c(λ) =. 0. cos(λx)u0(x)dx. (c) Equation (3.14) immediately implies that u(x,t) satisfies the heatequation. ... Example Consider the heat equation on the half line with. u(x, 0) = ea2x, u(0, t) = cos(bt), a,b real

22 Nov 2017: rA A O ‘( cos u1, sin u1) ð4:1Þ. and. rB B O ‘( cos u1 þ cos u2, sin u1 þ sin u2), ð4:2Þ. ... 0 its magnitude andt̂ ( cos u2, sin u2) the unit tangent vector joining A and B.Under the assumption of creeping flow, the drag forces areFA 2 zvA

11 Dec 2017: eat cos (If a is a negative quantity, this will indicate an oscillation the amplitude. ... a 5 b S+ B p/,2QS 4 cos aQ 2= L >. (8)These equations apply to a tube of given section throughout. I f the.

13 Mar 2017: cosh ξ cos η 1cosh ξ cos η ez. (8b). We take the torus S to be the surface ξ = ξ0, with aspect ratio s0 = cosh ξ0 = b/a, then the ... A(η)d. sinh ξ0. s0 t =. n=0. ′(An cos nη Bn sin nη).

13 Feb 2017: 2/2 cos(x2 x) in the ‘window’ [λ,λ], where λ = 15. ... Hm(y) (2b(m 1)/2c)!(b(m 1)/2c)! ey2/2 cos. ((2m 1)1/2y πm. 2. ), m 1.

20 Jan 2017: g(y) =. 2(1 q cos ψi) u2i E[. 12. cos1(. y2 u2i2. ... 2(1 q cos ψi) u2iF. [12. cos1(. y2 u2i2. q cos ψi).

3 May 2017: vr|r=a =. n=0An(t)Pn(cos θ) and vθ|r=a =. n=1. 2n(n1)Bn(t)P. 1n(cos θ), (1.1a,b). ... Thus, ψ|r=1 = 0 and [ψ n]r=1 = vs sin θ = 3 sin2 θ(1 β cos θ)/2.

30 Apr 2017: I P One can non-dimensionalize this equation. a,2a+(ls’cos(rr))a = O (3 b ). w E T ’ P9. where r = o o t , r = - , E ’ = w0 1+-K2. ... on the detuning. (i) if cos (v-$) > 0, which corresponds to v > - h a / P , the transition is super- critical.

9 Oct 2017: qτmcE). cosh(qτmcE). 0 00 0 cos. (qτmB). sin(qτmB). 0 0 sin(qτmB). ... ẏ(0) cos. (qτmB)ż(0). ]z(τ) =. m. qB. [cos(qτmB)ẏ(0) sin.

4 Apr 2017: VOL.2 3 7. B. 641. (Price 8s.) 5 [Published I 4Augucit I 952. ... b ) T h a t if A = A' ei$ and b = a$, where A', a, P, 41 are real, then ,%?Aebt= A' eat cos (pt 4).