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NANOSYSTEMS: PHYSICS, CHEMISTRY, MATHEMATICS, 2015, 6 (1), P. 140-145
ON THE STOKES FLOW COMPUTATION ALGORITHM BASED ON WOODBURY FORMULA
The Stokes approximation is used for the description of flow in nanostructures. An algorithm for Stokes flow computation in cases when there is great variation in the viscosity over a small spatial region is described. This method allows us to overcome computational difficulties of the finite-difference method. The background of the approach is using the Woodbury formula - a discrete analog of the Krein resolvent formula. The particular example of a rectangular domain is considered in detail. The inversion of the discrete Stokes operator is made in analytic form for the case of constant viscosity.
Keywords: nanotube, Stokes flow, finite-difference method.
Received: 1 December 2014
Revised: 4 January 2015
1. Introduction
Flow in nanostructures have many peculiarities which are not explained by the conventional theory of fluid flow [1,2]. At present, there is no general theory of nanoflows. We can only mention a few specific models [3-5]. Experiments show that nano-sized confinement leads to strong variation of liquid viscosity within the nanostructure [6-8]. One observes a similar situation when dealing with flow with nanoparticles or with singularities (see, e.g., [9,10]). Additionally, there is a fraction separation effect [11]. This is an indirect confirmation of the crystalline model of nanoflow which is based on assumption that nano-liquid contains solid-like inclusions. To describe such flow in the framework of the model it is necessary to solve the Stokes equation with strong variation in viscosity. This creates difficulties for computations, particularly for the finite difference scheme [12-15]. To overcome these obstacles an approach based on the Woodbury formula was suggested [16]. The formula reduces the inversion of the discrete Stokes operator with small inclusions having strong viscosity contrast to the inversion of the operator with constant (or slowly varying) viscosity and the inversion of small rank operator. In the present paper, we consider in detail the first step.
In more detail, the governing equations in d-dimensional (d=2,3) case (i.e. the Stokes and continuity equations for velocity v and pressure p under the action of force F) can be written as follows:
A.I. Popov1, I. S. Lobanov1, I.Yu. Popov1, T. V. Gerya2
1 ITMO University, Kronverkskiy 49, 197101, St. Petersburg, Russia
2 Institute of Geophysics, Department of Earth Sciences, Swiss Federal Institute of Technology Zurich (ETH), 5 Sonnegstrasse, CH-8092 Zurich, Switzerland
popov239@gmail.com, taras.gerya@erdw.ethz.ch
2
PACS 47.10.ad, 47.11.Fg
DOI 10.17586/2220-8054-2015-6-1-140-145
We assume that the viscosity n has strong variation in small spatial domain and is constant outside of it. We use finite-difference discretization on non-staggered grid:
h"1Aj(n(h"1vjVi + h"1viVj)) - h"1viP = Fh
= 0,
(2)
where h is a mesh size, Ai is a forward difference, Vi is a backward difference,
Ajf (x1,... ,xd) = f (x1,... ,Xi + hi,... ,xd) - f (x1,... ,xi,... ,xd),
Vjf(X1,... ,Xd) = f(X1,... ,Xi,... ,Xd) - f(X1,... ,Xi - hi,... ,Xd). The Woodbury formula allows us to improve the procedure of solving equations (2) in the case of strong viscosity variation.
2. Woodbury formula
Let us briefly describe the finite-difference approach based on the Woodbury formula (as for details, see [16]). Consider the discrete Stokes operator S on the lattice:
SVi = aj (n(vj Vi + viVj)),
where V is the liquid velocity, n is the viscosity. Symbols Ap and vp are used, correspondingly, for the forward and the backward finite differences for coordinate p where we assume for simplicity (it is not a limitation) that the lattice has a unit step. We use the following operator representation:
S — Sa + Sp , Sa :— nS(
P
0,
where S0 is the Stokes operator with the viscosity equals one:
SoVi = ^(aj vj Vi + aj viVj), j
and:
SpT Vi = (aj n)Tj (vj Vi + viVj). Here Ti is the translation operator:
Tif (X1,... ,Xd) = f(X1,... ,Xi + hi,... ,Xd).
Note that the operator Sp has small rank due to the property of the viscosity n, i.e. the corresponding matrix has many zero entries. We would like to solve the following equation:
SV = G,
where G is preassigned vector. To find a solution in our case it is convenient to use the Woodbury formula:
Vi = S-1Gi = (nSo + Sp)-1G = [1 - (nSo)"1 (1 + SpT(nSo)-1)"1SpT](nSo)"1G
One can see that we have real computational improvement (due to small rank of Sj) if we can find the inverse operator (nSo)-1 without high computational complexity. It is really so. We can find the inverse operator for the case of constant viscosity, e.g., by using of the discrete Fourier transform. Namely, let:
X = (nSo)-1G = So"1n"1G.
To find X one should solve the equation:
nSoX = G.
After the Fourier transformation we obtain:
SOX = ^ X = (So)-V^
These operations are not associated with high complexity.
In the next section we consider in detail how to use the discrete Fourier transform in a particular case.
3. Discrete Stokes operator inversion for the case of constant viscosity in a rectangle
To find the inverse matrix we use the discrete Fourier transform. To introduce it we need the corresponding system of eigenfunctions. Let us choose the system.
Consider the discrete Stokes equations coupled with the discrete continuity equation for the case of constant viscosity n =1:
-2A1v1V1 - A2(v2VI + v1V2) + viP = Fi, -2A2v2V2 - A1(v1V2 + v2V1) + v2P = F2, (3)
A1V1 + A2V2 = 0.
We deal with vector fields V and F and scalar field P, determined on a two-dimensional lattice with N = L2 nodes. The nodes are numbered by vectors with integer terms x = (x1,x2), xk g d = [0, L] h Z. Symbols Ap and vp are used, correspondingly, for the forward and the backward finite differences for coordinate p:
Af (x) = f (X1 + 1,X2) - f (X), v1/(x) = f (x) - f (X1 - 1,X2), A2/(x) = f (X1,X2 + 1) - f (x), v2f (x) = f (x) - f (X1,X2 - 1), where we assume for simplicity (it is not a limitation) that the lattice has a unit step. We will solve the system (3) with the free slip boundary condition:
'Vkl =0
k\xk =±L (4)
Vj Vk|xj=±L = 0, j = k
Consider a one-dimensional lattice d and the family of functions n = 1..L - 1 determined on the lattice by the following expressions:
... . nnt &n(t) = sin —-—, t g d. L
One can verify that the functions &n form a basis in the space of functions on the lattice d, satisfying the conditions:
& (0) = & (L) = 0. (5)
Consider the operator:
Avf (t) = vf (t +1) - vf (t) = f (t + 1) - 2f (t) + f (t - 1). It is simple to verify that Av = vA). Direct calculation gives us:
Av&(t) = Av sin j = sin (t + 1) - 2 sin j + sin (t~ 1) = 2 sin j(cos ^ - 1), L L L L L L
hence,
Av& = vA& = a &, a = 2 ^cos - . (6)
We introduce the notation:
Oj = A&. (7)
Then,
Wj = vA0j = aj . (8)
Particularly,
wj (d) = a (d) = 0, Wj (0) = 0. (9)
One can verify that 6j is also the eigenvector of the operator of second symmetric finite difference corresponding to the same eigenvalue as . Actually,
Av6j = AvA^j = Aaj = aj 6j. (10)
We additionally determine 6o = 1 and 0o = 0, which satisfy all the above-described properties.
Consider the family of functions , where the vector index j = (j\, j2) has coordinates jp = 0..L - 1:
(x) = 6ji (X1)6j2 (X2). Let us seek the solution of equation (3) in the form:
Vk(x) = E Vjvk(x),k = 1, 2, P(x) = E Pj(x), (11)
jj
for some real coefficients Vj, Pj. One can see that the expression of such type satisfies the boundary conditions (4). One can show that any vector field V satisfying the conditions (4) can be represented in the form of (11). But arbitrary pressure field P, generally speaking, can not be presented in such form. Now we consider the component related to the velocity field only.
Substituting (11) into the Stokes equation from (3), one obtains:
- E Ap i'vp E Vj vk + vk E Vj vpj + vfc E Pj = E Fj vk , k =1, 2, p=1 V j j ' j j
where Fj is the projection of Fk on . Changing the order of summation, one gets:
E
N
pjvk- Fjvk- J] (Vfc Apvpvk- VpjApvkvp^j)
p=1
0.
Using the commutativity Anvm = vmAn and the property Apvp^j = ajp , One finds that
2
E
Pj - Fk - £ ajp (Vj - Vj
p=1
Vk = 0.
Due to linear independence of the vectors vk, the Stokes equation is equivalent to a set of systems:
2
'k - Vp J = Fj,
p=1
Pj -E ajp (Vj - Vj = Fj, k = 1, 2, vj.
Substituting (11) into the continuity equation from (3), we come to the equation:
2
E Ap E Vpj vp(x) = 0. p=1 j
After simplification one gets:
2
EE j Vpj (x) = 0. j p=1
Due to linear independence of , the system is equivalent to the following one:
2
E jVpj = 0 vj. p=1
Finally, the problem (3) with the boundary conditions (4) is reduced to the set of systems:
'Pj - £j=1 j №' - Vj) = Fj k = 1, 2, vj,
Ep=1 j Vj = 0 vj.
Using the second equation, one can simplify the first one:
Pj - Vj Ej=1 j = Fj k = 1, 2, vj, Ep=1 j Vj = 0 vj.
Multiplying the first equation by and making a summation by k, one gets:
Pj (E - (E j Vjj (E "j)} = (E j j j
Due to the second equation of the system, one has:
pj = EP=1 ajP Fj vj.
EP=1 ajp
Hence, the first equation gives us the velocity field:
Pj Fj
Vj = -- k =1, 2, vj.
2^p=1 ajp
Thus, for this particular case we obtained the solution in an explicit form. As for more complicated domain, one can follow the suggested procedure if the system of eigenfunctions is known. In this case the approach described allows one to obtain essential acceleration and guaranteed convergence of the computations for strongly varying viscosity.
Acknowledgements
This work was partially financially supported by the Government of the Russian Federation (grant 074-U01), by the Ministry of Science and Education of the Russian Federation (GOSZADANIE 2014/190, Project 14.Z50.31.0031), by grant MK-5001.2015.1 of the President of the Russian Federation.
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