Normal Mode Analysis for Nano- Scale Hydrodynamic Model Determination

Size: px

Start display at page:

Download "Normal Mode Analysis for Nano- Scale Hydrodynamic Model Determination"

Amberlynn Tyler
5 years ago
Views:

1 Normal Mode Analysis for Nano- Scale Hydrodynamic Model Determination Xiaoyu Wei, Department of Mathematics, HKUST (with Tiezheng Qian, Xiao-Ping Wang and Ping Sheng) HKUST-ICERM Program, Jan 05, 2017 (Best viewed in Chrome Canary) 1. 1

2 Motivation - Determining "Hydrodynamical" Boundary (Inverse Problem) Model Problem - Stokes Equation + Navier Slip Boundary Condition Analytical Results - Eigenvalues & Eigenfunctions in Periodic Channels 1. 2

3 Micro uidic Dynamics At small scales (channel diameters from around around 100μm). If Navier-Stokes still holds, ρ ( u t 2 + u u ) = p + η u + f, with Low Reynolds number, Re = ρvl < nm Dissipation dominated by viscosity effects. Fluid particles moving along smooth paths in laminar or layers. μ to 2. 1

4 Stokes Approximation, Boundary Slip Dropping inertial term (Stokes approximation), assuming zero body force and incompressibility, ρ t v v = p + (η v), = 0. Instead of usual non-slip bounadry conditions, there is a slip at the solid surface due to intermediate Knudsen number, u slip = L s u τ n 2. 2

5 Navier Slip Length L s L MFP When Kn = L MFP H small enough, non-slip boundary condition is accurate. Image from

6 Even Narrower Channels If channel width is only around 100σ, the model above fails again due to the presence of density boundary layer. Image from [Ping Sheng et al., Physical Review E, 2015] 2.4

7 Molecular Dynamics Model validation and determining model parameters directly from MD simulation. At equilibrium, the system experiences thermal uctuations (FDT). In linear response regime, longevity of a speci c mode can be determined by: Calculating its autocorrelation function ( N i=1 (t= )U(, ))( (t= +Δt)U(, )) C U (Δt) = v i t0 x i y i N i=1 v i t0 x i y i N i=1 v i t0 x i y i N i=1 v i t0 x i y i 1 τ decay [ln( C U )] ( (t= )U(, ))( (t= )U(, )). U.

8 Matching Spectral Fingerprints Take U from a oneparameter group by xing k. Plot τ decay against λ, and obtain eigenvalues as local peaks. Image from [Ping Sheng et al., Physical Review E, 2015] 2. 56

9 Even Narrower Channels Cont. A modi ed model can be derived in this way Image from [Ping Sheng et al., Physical Review E, 2015] 2. 7

10 However.. One obvious caveat is that not all families of eigenfunctions are investigated. To validate the conjectured model above, we have to obtain the full set of eigenfunctions of the model. 2. 8

11 Existing Results Taking l s = 0, the problem reduces to the one considered by Steven A. Orszag in Later, Orszag's results are applied to analyse accuracy and stability of Chorin's projection method and its variants. See, for example [Weinan E and Jianguo Liu, 1995, 1996]. 2. 9

12 A 1D System We consider the domain to be a 2D periodic channel Ω = [ l, l] [ h, h]. The generalized eigenvalue problem can be written as 2 u + λu = p, u Subject to periodic boundary conditions at x = ±l: u(x + 2l, y) = u(x, y), = 0. and slip boundary conditions at y = ±h: l s y u x p(x + 2l, y) = p(x, y), =. Key observation: pressure Poisson equation is a Laplace equation! u x 3. 1

13 Solution Procedure Solutions to the eigenvalue problem is of the form where. k π l u k (x, y) p k (x, u) General solution to the pressure Poisson equation: Then from Stokes equation, u k,x (x, y) u k,y (x, y) u k e ikx = (y), = p k (y) e ikx, p k c 1 e ky c 2 e ky (x, y) = +. = a x e ik λ y + b x e ik y ik λ + ( c 1 e ky + c 2 e ky ), = a y e ik λ y + b y e k y ik λ + ( c 1 e ky c 2 e ky ), λ λ

14 Solution Procedure (2) Now, the solution is fully determined by six constants Par = [ a x, a y, b x, b y, c 1, c 2 ] T. Using boundary conditions and incompressibility constraints, the eigenvalue problem is reduced to a linear system A(h, l, l s, λ)par = 0, where A is a 6 6 matrix. For the problem to have nontrivial solutions, the characteristic equation has to be satis ed det(a) =

15 Results When k = 0, there are two cases: Asymmetric: tan( λh) = l s λ, u 0λ Symmetric: cot( λh) = l s λ, u 0λ When k 0, also two cases ( k λ = λ k 2 ): cot( h) k coth(kh) λ = 0. k λ k λ l s = [sin( λy), 0 ] T. = [cos( λy), 0 ] T. k λ cosh(kh) ik cosh(ky) i cos( y) u kλ = e ikx sin( k λ h sinh(kh) k sinh(ky) k sin( y) [ ] k λ tan( k λ h) + k tanh(kh) + l s λ = 0 sin( h) ik sinh(ky) i sin( y) u kλ = e ikx cos( k λ h cosh(kh) k cosh(ky) k cos( y). k λ k λ cosh(kh) k λ k λ k λ [ ] k λ cos( h) k λ

16 Discussion Theorem: Each eigenvalue λ obtained is monotonically decreasing in Navier length l s. Proof: Directly calculate λ/ l s via implicit function theorem. More slip = more dissipation on ALL modes. A counterintuitive fact: Numerical tests show that the rst antisymmetric mode for and the rst mode for has close eigenvalues even when. k = 0 k = 1 l/h

17 Summary A normal mode analysis is carried out for Stokes equation subject to Navier slip boundary conditions. Link to previous studies is drawn. Application to determining hydrodynamics boudary conditions is discussed. Thanks for your attention! 5. 1

18 Wish List A generic numerical method would be nice to have to deal with Two phase ow General geometry BIM might be preferable for this task, being more friendly to shifted inverse iteration / Arnoldi iteration. 5. 2

Navier-Stokes equations

1 Navier-Stokes equations Introduction to spectral methods for the CSC Lunchbytes Seminar Series. Incompressible, hydrodynamic turbulence is described completely by the Navier-Stokes equations where t