NUMERICAL SIMULATION OF 2D INCOMPERSSIBLE NAVIER-STOKES FLOW
DRIVEN BY ROTLETS IN A UNIT DISK
A thesis submitted
to Kent State University in
partial fulfillment of the requirements
for the degree of Master of Science
by
MANSOOR A. ALSULAMI
August, 2017 Thesis written by
MANSOOR A. ALSULAMI
B.S., King Abdulaziz University, 2008
M.S., Kent State University, 2017
Approved by
Xiaoyu Zheng , Advisor
Andrew Tonge , Chair, Department of Mathematical
Sciences
James L. Blank , Dean, College of Arts and Sciences TABLE OF CONTENTS
LIST OF TABLES ...... v
LIST OF FIGURES ...... vi
Acknowledgement ...... vii
1 Introduction ...... 1
2 Mathematical Models ...... 4
2.1 Incompressible Navier-Stokes Equations ...... 4
2.2 Stream Function and Velocity Formulation of NSEs in 2D ...... 5
2.2.1 Vorticity Formulation in 2D ...... 5
2.2.2 Poisson Equation for Stream Function ...... 6
2.3 Boundary Conditions ...... 6
2.4 Nondimensionlization of NSEs ...... 7
2.5 Decomposition of Solutions ...... 8
3 Numerical Method ...... 11
3.1 Spatial Discretization and Fourier Spectral Method ...... 11
3.2 Time Integration ...... 14
3.2.1 Initialization ...... 15
3.2.2 Vorticity Equation ...... 15
3.2.3 Fourier Filtering for Explicit Time Stepping ...... 16
3.2.4 Poisson Equation ...... 17
3.2.5 Vorticity at Boundary and Velocity Field ...... 18
iii 4 Numerical Results ...... 20
4.1 Convergence Rate ...... 20
4.2 Numerical Experiments ...... 23
5 Conclusion ...... 29
BIBLIOGRAPHY ...... 30
iv LIST OF TABLES
1 The errors inω ˜ for Re =1...... 22
2 The error in ψ˜ for Re =1 ...... 22
3 The error inu ˜ for Re =1 ...... 22
4 The error inv ˜ for Re =1 ...... 23
v LIST OF FIGURES
1 discretization of the domain where N =8...... 12
2 Contour plots for stream function and vorticity at c = 0.2, Re = 0.1, N =
64. Top row: singular part of the solution; Middle row: combined solution;
Bottom row: regular part of the solution...... 24
3 Contour plots for stream function and vorticity at c = 0.5, Re = 0.1, N = 64. 25
4 Contour plots for stream function and vorticity at c = 0.7238, Re = 0.1, N = 64. 26
5 Contour plots for stream function and vorticity at c = 0.5, N = 64, Re = 1. . 27
6 Contour plots for stream function and vorticity at c = 0.5, N = 64, Re = 10. 28
7 Normalized traction for different values of Re, for c = 0.5, N = 64...... 28
vi Acknowledgement
I would like to express my deep appreciation to my supervisor professor Xiaoyu Zheng for her support, help and encouragement. She has been always supportive, and I am grateful to her. I wish to express my sincere gratitude to the committee members, Professor Gartland and Professor Jing Le. I would like to thank them for their time that they have given me. I would like to take this opportunity to record my thanks to the Department of Mathematics at Kent State University for giving me the chance to study and gain my Master’s degree.
I would like to thank my father and my mother who have supported me to study abroad.
Special thanks to my great family members who have been with me all the time.
vii CHAPTER 1
Introduction
In many active systems, individual self-propellers in a fluid generate the translational propulsion [1]. Recently, there are increased interest in active rotors. Individual particles rotate due to either internal torque or external torque. In experiments, particle rotations can be driven by chemicals [2], optical tweezers or light [3], magnetic [4] or electric fields [5].
The interaction between the individual rotors and the surrounding environment can lead to intriguing collective motion in the population.
In this thesis, we study the simple steady, two dimensional flow problem when viscous
fluid contained in a disk is disturbed by a fixed rotlet inside the disk. The line rotlet is simply the limiting case of the fluid generated by a rotating cylinder with radius b, angular
velocity ω. As the radius b → 0, and ωb2 is kept as a constant, one obtains the flow driven
by a line rotlet. The constant ωb2, usually denoted as Γ, which is known as strength of the
rotlet [8].
For a free line rotlet in 2D infinite medium, the stream function is given by
ψ = Γ log R,R = |~x − ~x0|, (1.0.1)
where R is the distance to the line rotlet. The stream function not only represents a Stokes
flow, but also the exact solution of Navier-Stokes equations with an appropriate pressure
field p. The velocity field and vorticity field are given by
Γ ~u = θ,ˆ (1.0.2) R ~ω = 0. (1.0.3)
1 The stream lines form concentric circles around the rotlet. The velocity is tangential to
the circles, which blows up near the rotlet and decays as 1/R as one moves away from the
rotlet. The vorticity is zero everywhere. The line rotlet exerts a torque per unit length with
magnitude of 4πΓµ to the surrounding fluid whose dynamic viscosity is µ.
When solid boundaries present, the imposed boundary conditions will alter the velocity
field and break the symmetry of the flow, and might cause flow separation and eddies for- mation in the flow field [8]. A separation occurs either on the boundary or in the interior of the domain. A separation point on the boundary is where one or more streamlines meet the boundary. An interior separation point is where two or more streamlines meet. The streamlines which intersect either type of the separation point are called dividing stream lines. If P is a separation point on a smooth boundary, then the vorticity, or equivalently, the tangential stress vanishes at P . If Q is an interior separation point, then the velocity
vanishes at Q.
There are two kinds of eddies : attached eddy and free eddy [8]. An attached eddy
is an open, simply connected region of flow, containing closed stream lines, and no flow
singularities or rigid boundaries, and bounded by one or more rigid boundaries and one or
more dividing streamlines. A free eddy differs from the attached eddy in that a free eddy is
bounded by one or more dividing streamlines.
In the Stokes limit, this problem was solved analytically by Ranger [6] . He showed that
if a line rotlet is contained in a fixed disk, an attached eddy is formed when the rotlet is
within a certain distance to the boundary of the disk. If a line rotlet is inside a rotating
disk, then free eddies can be found for certain relative positions of the rotlet.
In this thesis, we are interested in the problem where the inertial term cannot be ne-
glected, and the goal is to examine the effect of the nonlinear convection term on the flow
2 field. The thesis is organized as follows. In Chapter 2, we introduce the mathematical mod- els which govern the motion of the fluid induced by a line rotlet in 2D disk. In Chapter 3, a combination of finite difference and Fourier collocation methods is developed to solve the equations. In Chapter 4, we demonstrate the numerical results with convergence rate and various numerical experiments. We conclude in Chapter 5.
3 CHAPTER 2
Mathematical Models
In this chapter, we present the mathematical models that govern the dynamics of the flow
field driven by a line rotlet in a 2D disk. We first write down the Navier-Stokes equations in stream function-vorticity formulation with appropriate boundary conditions. We then nondimensionalize the equations, and finally decompose the solutions into two parts, with a known singular part and a regular part, which needs to be solved.
2.1 Incompressible Navier-Stokes Equations
We consider the Navier-Stokes equations (NSEs) in two dimensions. The general form of the incompressible NSEs is given by
∂~u p + (~u · ∇)~u = −∇ + ν∇2~u, on Ω, (2.1.1) ∂t ρ
∇ · ~u = 0, on Ω. (2.1.2)
where, ~u(~x,t) is the velocity field of the fluid, p(~x,t) is the pressure field, ρ is density of the
fluid, and ν = µ/ρ is the kinematic viscosity of the fluid. The first equation is the consequence
of the conservation of momentum, and the second equation describes the incompressibility of
the fluid. We are going to consider the simplest Dirichlet boundary condition in this thesis,
which is given by
~u = ~0 on ∂Ω. (2.1.3)
In 2D, we can rewrite the NSEs in terms of the stream function-vorticity formulation. The
advantage of the stream function-vorticity formulation is that we can eliminate the pressure
term, which in general is difficult to handle both analytically and numerically.
4 2.2 Stream Function and Velocity Formulation of NSEs in 2D
In this section, we provide the details on deriving the stream function-vorcitity formula-
tion of the Navier-Stokes equations in 2D.
2.2.1 Vorticity Formulation in 2D
By taking the curl of Eq. (2.1.1), we have
∂~u p ∇ × + ∇ × ((~u · ∇)~u) = −∇ × ∇ + ∇ × (ν∇2~u). (2.2.1) ∂t ρ
Since the pressure p/ρ is a scalar function, so ∇ × ∇(p/ρ) = 0, and using the identity
1 ∇ × ((~u · ∇)~u) = ∇ × ( 2 ∇(~u · ~u) − ~u × (∇ × ~u)), we get ∂ 1 ∇ × ~u + ∇ × ∇(~u · ~u) − ~u × (∇ × ~u) = ν∇2(∇ × ~u). (2.2.2) ∂t 2
The vorticity vector, ~ω, is defined at a point as
~ω = ∇ × ~u, (2.2.3) which in 2D can be reduced to
∂u ∂u ~ω = ωzˆ = 0, 0, y − x . (2.2.4) ∂x ∂y
Once we substitute the expression of ~ω in Eq. (2.2.3) in Eq. (2.2.2), we have
∂~ω 1 + ∇ × ∇(~u · ~u) − ∇ × (~u × ~ω) = ν∇2~ω. (2.2.5) ∂t 2
1 Since 2 ∇ × ∇(~u · ~u) = 0, and ∇ × (~u × ~ω) = ~u(∇ · ~ω) − ~ω(∇ · ~u) + (~ω · ∇)~u − (~u · ∇)~ω, we have ∂~ω + −[~u(∇ · ~ω) − ~ω(∇ · ~u) + (~ω · ∇)~u − (~u · ∇)~ω] = ν∇2~ω. (2.2.6) ∂t Because the only non-zero component of vorticity ~ω is along the z direction, it is easy to show that ~u(∇ · ~ω) = 0, and (~ω · ∇)~u = 0, and the incompressibility gives ~ω(∇ · ~u) = 0. The incompressible Navier Stokes equations can be rewritten as
∂~ω + (~u · ∇)~ω = ν∇2~ω, on Ω. (2.2.7) ∂t 5 2.2.2 Poisson Equation for Stream Function
Since ∇ · ~u = 0 , the velocity can be represented as the curl of a vector potential. In 2D,
it reduces to
~u = −∇ × (ψzˆ), (2.2.8) where ψ is known as the stream function. Therefore, from Eqs. (2.2.3) and (2.2.8), we obtain
~ω = ∇ × ~u = −∇ × ∇ × ψzˆ = −[∇(∇ · ψzˆ) − ∇2ψzˆ], (2.2.9)
Since ψ only depends on x and y, so ∇(∇·ψzˆ) = 0, and this leads us to the Poisson equation for ψ, which is
∇2ψ = ω. (2.2.10)
This is the kinematic equation connecting the stream function ψ and the vorticity ~ω.
The original incompressible Navier-Stokes equations (2.1.1) and (2.1.2) have now an alternative formulation as:
∂~ω + (~u · ∇)~ω = ν∇2~ω, on Ω, (2.2.11) ∂t ∇2ψ = ω, on Ω. (2.2.12)
In our thesis, we are interested in Navier Stoke’s equation in a disk, so it is natural to write
the equations in polar coordinates. The Laplacian operator in polar coordinate is given by
∂2 1 ∂ 1 ∂2 ∇2 = + + . (2.2.13) ∂r2 r ∂r r2 ∂θ2
The gradient operator is give by
∂ 1 ∂ ∇ = , . (2.2.14) ∂r r ∂θ
2.3 Boundary Conditions
The velocity in terms of stream function in polar coordinates can be written as
1 ∂ψ ∂ψ ~u = −∇ × (ψzˆ) = − , . (2.3.1) r ∂θ ∂r
6 Thus the boundary condition, ~u = urˆ + vθˆ = ~0, can be rewritten as
1 ∂ψ u = − = 0, on ∂Ω, (2.3.2) r ∂θ ∂ψ v = = 0, on ∂Ω. (2.3.3) ∂r
From Eqs. (2.3.2) and (2.3.3), it can be shown that ψ = c on ∂Ω while c is a constant. Since
ψ appears only in differentiated form, so the constant can be chosen as zero. Therefore, we will have the following conditions
ψ = 0, on ∂Ω, (2.3.4) ∂ψ = 0, on ∂Ω. (2.3.5) ∂r
Thus the original zero velocity boundary conditions are expressed as one Dirichlet and one
Neumann boundary condition for stream function.
In our study, there is a rotlet in the interior of the domain. The stream function in the
vicinity of the rotlet satisfies
ψ ∼ Γ log R, as R → 0, (2.3.6)
where R is the distance to the location of the line rotlet. This is the interior boundary
condition near the rotlet.
2.4 Nondimensionlization of NSEs
To nondimensionlize Navier-Stokes equations, we introduce a characteristic length scale
rc = a, which is the radius of the disk, and a characteristic speed Uc. By introducing the
dimensionless variables,
r a t Uc r = , tc = , t¯= , ωc = , ψc = aUc, (2.4.1) a Uc tc a ~u ω ψ ~u∗ = , ω∗ = , ψ∗ = . (2.4.2) Uc ωc ψc
7 The governing equations after nondimensionalization become
1 ω∗ + (u∗ · ∇)ω∗ = (∇2ω∗), 0 ≤ r¯ < 1, 0 ≤ θ < 2π, (2.4.3) t¯ Re ∇2ψ∗ = ω∗, 0 ≤ r¯ < 1, 0 ≤ θ < 2π, (2.4.4)
∗ ψ |r¯=1 = 0, 0 ≤ θ < 2π, (2.4.5) ∂ψ∗ | = 0, 0 ≤ θ < 2π, (2.4.6) ∂r¯ r¯=1 ψ∗ ∼ β log R∗, as R∗ → 0, (2.4.7)
∗ 2 2 1 where R = (¯r +c −2¯rc cos θ) 2 and the rotlet is located in the positive horizontal axis with
aUc distance c (0 < c < 1) to the disk center. The symbol Re = ν is the Reynolds number. When one is concerned with situations where the fluid is extremely viscous, namely ν is
large, or the motion of the body is very slow, then 0 < Re 1, and the inertial terms
on the left hand side can be neglected, and one has the Stokes equation. β = Γ is the Uca nondimensionalized strength of the rotlet. For a fixed stength Γ, we can always set β = 1 by choosing an appropriate value for the characteristic speed Uc. For simplicity, we drop the bars and superscript ∗ for the rest of the thesis.
2.5 Decomposition of Solutions
In order to solve the above equations, we observe that both the velocity and stream function will blow up near the rotlet. This is a big challenge in numerical simulations. To avoid dealing with huge numbers in numerical solutions, we write the solutions as
˜ ψ = ψ + ψs, (2.5.1)
ω =ω ˜ + ωs, (2.5.2)
~u = ~u˜ + ~us, (2.5.3) where ψ,˜ ω,˜ ~u˜ are the regular parts of the stream function, vorticity, and velocity field, which are expected to behave nicely, and ψs, ωs, ~us are the singular part of the solutions. We note
8 that the singular and regular part of ω and ~u can be obtained through their relations through
their corresponding part of stream function ψ, as
2 ˜ 2 ω˜ = ∇ ψ, ωs = ∇ ψs, (2.5.4)
˜ ~u˜ = −∇ × ψz,ˆ ~us = −∇ × ψsz.ˆ (2.5.5)
The advantage of this splitting scheme is that an analytical form of singular part solution
exists. In the work of Ranger [6], he obtained the exact solution for Stokes equation in a
disk, where the flow is induced by a fixed line rotlet. By solving the following biharmonic
equation of stream function, which is satisfied in stokes limit,
4 ∇ ψs = 0, 0 ≤ r < 1, 0 ≤ θ < 2π, (2.5.6)
ψs|r=1 = 0, 0 ≤ θ < 2π, (2.5.7) ∂ψ s | = 0, 0 ≤ θ < 2π, (2.5.8) ∂r r=1
ψs ∼ log(R), as R → 0, (2.5.9) the singular part of the stream function is given by
1 1 1 r 1 r cos θ − 1 1 ψ = log r2 + c2 − 2rc cos θ− log c2 r2 + − 2 cos θ + r2 − 1 c + . s 2 2 1 r 2 2 c c c r + c2 − 2 c cos θ 2 (2.5.10)
We now rewrite Eq. (2.4.3) in term of singular part and regular part of the vorticity and
velocity, such as, 1 (˜ω + ω ) + (~u · ∇)(˜ω + ω ) = ∇2(˜ω + ω ). (2.5.11) s t s Re s
2 4 From Eq. (2.5.10) and (ωs)t = 0, ∇ ωs = ∇ ψs = 0, we have
1 ω˜ + (~u · ∇)(˜ω + ω ) = ∇2ω,˜ (2.5.12) t s Re
where the term ~u = ~u˜ + ~us. We perform similar calculations for the Poisson equation for the stream function and boundary conditions, and finally we have the governing equations for
9 the regular part of solution as,
1 ω˜ + ((~u˜ + ~u ) · ∇)(˜ω + ω ) = ∇2ω,˜ 0 ≤ r < 1, 0 ≤ θ < 2π, (2.5.13) t s s Re ∇2ψ˜ =ω, ˜ 0 ≤ r < 1, 0 ≤ θ < 2π (2.5.14) ∂ψ˜ ψ˜| = | = 0, (2.5.15) r=1 ∂r r=1
Equations (2.5.13)–(2.5.15) are the main equations that we need to solve. We note that the
boundary conditions in the governing equation are only for stream function ψ˜, and there is no boundary condition for vorticityω ˜.
In particular, if the rotlet is located in the center of the disk, then the singular part of
the solutions ψs, ωs, and ~us are given by the following simple form
1 1 ψ = − r2 + log r, (2.5.16) s 2 2 −1 ∂ ∂ 1 ~u = −∇ × ψ z = ψ , ψ = 0, −r + , (2.5.17) s sb r ∂θ s ∂r s r
ωs = −2. (2.5.18)
We remark that the velocity field is tangential, and the speed blows up near the rotlet and
decays to zero at the disk boundary. It is interesting to note that the vorticity is a constant
everywhere in this case. In this case, the regular solutions ψ˜,ω ˜ andu ˜ will be identically zero
due to the symmetry. In other words, the solutions for steady state Navier-Stokes equations
are the same as those of Stokes equation. Thus, in our numerical simulations, we are going
to explore only the situation when the rotlet is off the center of the disk.
10 CHAPTER 3
Numerical Method
In this chapter, we describe the numerical method developed to solve the regular part of the solutions. Since the domain under consideration is a disk, it is natural to use the Fourier collocation method along θ direction. We will use finite difference scheme to approximate the derivatives along the r direction. Simple first order forward Euler, fourth order Runge-Kutta
method and first order backward Euler method will be implemented for time integration. Our
problem has a strong convection term since the velocity blows up near the rotlet. In addition,
the diffusion term in polar coordinate is more stiff than that in Cartesian coordinate. If the
diffusion term is treated explicitly, we use the Fourier filtering technique near the disk center
to make sure that the time step constraint is mainly from the strong convection term. Since
we are only interested in the steady state solution, the implicit backward Euler method on
diffusion term enables us to use larger time steps for the solution to converge to steady state.
Details are addressed in this chapter.
3.1 Spatial Discretization and Fourier Spectral Method
First, we note that our equation is singular at r = 0. This is not due to the singularity in
the solutions, but the coordinate system we work with, see Eq. (2.2.13). To avoid working
directly at the center of the disk, we will use the following radial grid points [7]
1 r = (i − )dr, i = 1,...,N, (3.1.1) i 2
1 where dr = 1 , such that r1 = dr/2, r2 = r1 + dr, . . . , and rN = 1. This way we make sure N− 2 that we shift a half mesh from the origin.
11 Figure 1: discretization of the domain where N = 8
One of the advantages of this mesh is that the Laplacian can be approximated without any difficulty at each grid point. For example, for a smooth scalar function f(r, θ),
∂2 1 ∂ 1 ∂2 ∇2f = f(r, θ) + f(r, θ) + f(r, θ). (3.1.2) ∂r2 r ∂r r2 ∂θ2
Expand this smooth function in a discrete Fourier expansion in θ as
M/2 X ˆ Ikθj f(r, θ, t) = fk(r, t)e (3.1.3) k=−M/2+1
12 Take the discrete Fourier transform of equation (3.1.2), we get
∂2 1 ∂ k2 2 ˆ ˆ ˆ ∇df(r)k = fk(r) + fk(r) − fk(r), k = −M/2 + 1,...,M/2. (3.1.4) ∂r2 r ∂r r2
The centered finite difference approximation reads
fˆ (r ) − 2fˆ (r ) + fˆ (r ) 1 fˆ (r ) − fˆ (r ) k2 2 k i+1 k i k i−1 k i+1 k i−1 ˆ ∇df(ri)k ≈ 2 + − 2 fk(ri), i = 1,...,N. dr ri 2dr ri (3.1.5) ˆ In particular, we note that for i = 1, the terms involving f(r0) cancel out automatically, and it simply becomes 2fˆ (r ) − 2fˆ (r ) k2 2 k 2 k 1 ˆ ∇df(r1) = 2 − 2 fk(ri). (3.1.6) dr r1 ˆ We also note that for i = N, we would need the values of fk at rN+1, which is outside the
domain, or called the ghost point. This will be given through Neumann boundary conditions.
We remark that one can use the regular uniform grids in the disk, when the first grid is
put right at the center. However, one has to impose an appropriate boundary condition for
the solution has to satisfy at the origin. Using the shifted grid, we don’t need to impose any
boundary condition at the origin.
Uniform grid points along the θ direction will be used,
θj = jdθ, j = 1,...,M, (3.1.7)
2π where dθ = M .
13 We next expand our solutions in a discrete Fourier expansion in θ as
M/2 ˜ X ˆ Ikθj ψ(ri, θj, t) ≈ ψk(ri, t)e , (3.1.8) k=−M/2+1 M/2 X Ikθj ω˜(ri, θj, t) ≈ ωˆk(ri, t)e , (3.1.9) k=−M/2+1 M/2 X Ikθj u˜(ri, θj, t) ≈ uˆk(ri, t)e , (3.1.10) k=−M/2+1 M/2 X Ikθj v˜(ri, θj, t) ≈ vˆk(ri, t)e , (3.1.11) k=−M/2+1
ˆ for i = 1, . . . , N, j = 1,...,M. Here I denotes the pure imaginary number, and ψk,ω ˆk ,ˆuk
, andv ˆk, k = −M/2 + 1,...,M/2 are complex Fourier coefficients of the regular part of
solutions, which are given by discrete Fourier transform,
M 1 X ψˆ (r , t) = ψ˜(r , θ , t)e−Ikθj , i = 1,...,N, (3.1.12) k i M i j j=1
The expressions forω ˆk, uˆk, vˆk are similar. The main goal now is to find the Fourier coeffi- cients of all the unknowns.
3.2 Time Integration
In this section, we describe the details of solving Eqs.(2.5.13)–(2.5.15). We are going to find the steady state solutions by time integration until solutions reach equilibrium. We consider two explicit time-stepping methods: forward Euler and 4th-order Runge-Kutta methods, and one implicit time-stepping method: backward Euler method. Since we are only interested in the equilibrium solutions, we anticipate all three methods produce the same results.
The simple forward Euler method can be summarized as follows:
ω˜n+1 − ω˜n 1 + (~un · ∇)(˜ωn + ω ) = ∇2ω˜n, ∇2ψ˜n+1 =ω ˜n+1, ψ˜n+1| = 0. (3.2.1) dt s Re r=1
14 where the superscript on a variable represents the time step index, and dt is the time step size.
The 4th-order Runge-Kutta (RK4) method can be summarized as [7]:
ω˜ − ω˜n 1 1 + (~un · ∇)(˜ωn + ω ) = ∇2ω˜n, ∇2ψ˜ =ω ˜ , ψ˜ | = 0, (3.2.2) dt/2 s Re 1 1 1 r=1 ω˜ − ω˜n 1 2 + (~u · ∇)(˜ω + ω ) = ∇2ω˜ , ∇2ψ˜ =ω ˜ , ψ˜ | = 0, (3.2.3) dt/2 1 1 s Re 1 2 2 2 r=1 ω˜ − ω˜n 1 3 + (~u · ∇)(˜ω + ω ) = ∇2ω˜ , ∇2ψ˜ =ω ˜ , ψ˜ | = 0, (3.2.4) dt 2 2 s Re 2 3 3 3 r=1 1 K4 = −(~u · ∇)(˜ω + ω ) + ∇2ω˜ , (3.2.5) 3 3 s Re 3 1 dt ω˜n+1 = (−ω˜n +ω ˜ + 2˜ω +ω ˜ ) + K4, ∇2ψ˜n+1 =ω ˜n+1, ψ˜n+1| = 0, (3.2.6) 3 1 2 3 6 r=1
The implicit method we adoped here is to only treat the diffusion term implicitly as
ω˜n+1 − ω˜n 1 + (~un · ∇)(˜ωn + ω ) = ∇2ω˜n+1, ∇2ψ˜n+1 =ω ˜n+1, ψ˜n+1| = 0. (3.2.7) dt s Re r=1
We remark that instead of solving the above equations directly, we work on the Fourier
transform of them. Below we provide details in solving the vorticity equation and Poisson
equation in each time step that are required in both forward Euler and RK4 methods.
3.2.1 Initialization
We start our scheme with an initial value of ψ˜, for simplicity, we choose
˜ ψ(ri, θj) = 0, i = 1, . . . , N, j = 1,...,M. (3.2.8)
This will set all variables such asω, ˜ u,˜ v˜ to zero.
3.2.2 Vorticity Equation
Consider the vorticity equation
1 ω˜ + (~u · ∇)ω = ∇2ω,˜ 0 ≤ r < 1, 0 ≤ θ < 2π, (3.2.9) t Re
where ~u = ~u˜+~us, and ω =ω ˜ +ωs. From this equation, we update the interior values ofω ˜(ri) only, i.e., for i = 1,...,N − 1. The boundary value forω ˜ will be updated after we update
15 ψ˜. Since we want to find the Fourier coefficients of unknowns, we write down the discrete
Fourier transform of the equation as
d 1 2 −M M ωˆk + ((~ud· ∇)ω)k = ∇dω˜k, k = + 1,..., . (3.2.10) dt Re 2 2
Let us first consider the convection term (~u · ∇)ω. Explicitly, it reads
∂ ∂ 1 ∂ 1 ∂ (~u · ∇)ω = (˜u + u )( ω˜n + ω ) + (˜v + v )( ω˜n + ω ). (3.2.11) s ∂r ∂r s s r ∂θ r ∂θ s
In this term, all terms involving singular part of the solution are evaluated analytically.
n ∂ n Given the discrete Fourier coefficientsω ˆk , we can approximate ∂r ω˜ (ri, θj) in the interior of the domain as
∂ ωˆn(r ) − ωˆn(r ) ω˜n(r , θ ) ≈ F −1( k i+1 k i−1 ), i = 2,...,N − 1. (3.2.12) ∂r i j 2dr
Near the center i = 1, a one-sided finite difference scheme is employed to attain second order accuracy, ∂ −3ˆω (r ) + 4ˆω (r ) − ωˆ (r ) ω˜n(r , θ ) = F −1( k 1 k 2 k 3 ). (3.2.13) ∂r 1 j 2dr The derivative of theω ˜ in θ direction is obtained by
∂ ω˜n(r , θ ) = F −1((Ik)ˆω (r )), i = 1,...,N − 1, j = 1,...,M. (3.2.14) ∂θ i j k i
We remark that we have to have ~u and ∇~ω in physical space, perform the product, and then
2 transform the product to Fourier space. The diffusion term ∇dω˜k is treated as explained in
the beginning of this chapter.
3.2.3 Fourier Filtering for Explicit Time Stepping
Our form of equation puts a severe stability constraint on the size of time step if one
chooses a explicit time stepping method. This is because of the factor 1/r2 before the second
derivative in azimuthal direction. When r is close to the center, the coefficient is large, thus making the term very stiff. We can also see this by observing that the uniform lattice near the
16 center of the disk is much more dense than that near the boundary. For stability purposes,
we want to make sure that we can use as large time step as we can without sacrificing the
accuracy. In Fourier space, the term becomes k2/r2, so the idea is to get rid of large wave
numbers once we get closer to the origin. This is called Fourier filtering [7]. In practice, we
set the coefficients for the large wave numbers near the center zero. To illustrate that, we
will divide the domain into core region which is the half of the disk, i.e. r < 0.5, and the
outer region. In the first layer in core region, r = dr/2, we keep only the wave numbers
k = 0, 1, −1 and remove the rest. On the second layer, r = 3dr/2, we keep the modes with wave numbers k = 0, 1, −1, 2, −2, and so on. The outer region will take all modes with all
wave numbers. This is equivalent to make the lattice spacing coarser near the center of the
disk.
3.2.4 Poisson Equation
Now let’s consider the Poisson equation, which relates the steam function and vorticity,
∇2ψ˜n+1 =ω ˜n, 0 ≤ r < 1, 0 ≤ θ < 2π, (3.2.15)
˜n+1 ψ |r=1 = 0. (3.2.16)
Applying Fourier transform to these equations, we get
2 2 d ˆn+1 1 d ˆn+1 k ˆ n+1 2 ψk (r) + ψk (r) − 2 ψ(r)k =ω ˆk(r), (3.2.17) dr ri dr ri ˆn+1 ψk |r=1 = 0, k = −M/2 + 1,...,M/2. (3.2.18) A second order finite difference approximation to all derivatives leads to the following tridi- agonal system
2 −2 − k 2 0 ··· 0 dr2 2 dr2 ˆn+1 n+1 r1 ψ (r1) ωˆ (r1) k k . . . . . . . . . . . 0 0 . . . . 1 1 −2 k2 1 1 0 − − + 0 ψˆn+1(r ) = ωˆn+1(r ) , (3.2.19) dr2 2ridr dr2 r2 dr2 2ridr k i k i i . . . . . . . . . . . . . . . 0 0 1 1 −2 k2 ˆn+1 n+1 0 0 0 − − ψ (rN−1) ωˆ (rN−1) dr2 2r dr dr2 r2 k k N−1 N−1 for k = −M/2 + 1,...,M/2. This linear system can be solved easily by Thomas algorithm.
17 3.2.5 Vorticity at Boundary and Velocity Field
At the end of each time step, we need to find the vorticity at the domain boundary and
velocity throughout the domain for preparation of next time step. There are two boundary
conditions for ψ. The Dirichlet boundary condition is used to solve the Poisson equation for
ψ. The other Neumann boundary condition is going to help find the vorticity at the domain boundary.
First, the Poisson equation, when extend to the boundary, reads
ˆ ˆ ˆ ˆ ˆ 2 ψk(rN+1) − 2ψk(rN ) + ψk(rN−1) 1 ψk(rN+1) − ψk(rN−1) k ˆ 2 + − 2 ψk(rN ) =ω ˆk(rN ). (3.2.20) dr rN 2dr rN ˆ The value ψk(rN+1) is evaluated outside the domain, which needs to be specified. We notice
d ˜ d ˆ that to satisfy the Neumann boundary condition dr ψ|r=1 = 0, or dr ψk|r=1 = 0, we can approximate that as
ψˆ (r ) − ψˆ (r ) k N+1 k N−1 = 0, k = −M/2 + 1,...,M/2, (3.2.21) 2dr
which gives at once
ˆ ˆ ψk(rN+1) = ψk(rN−1), k = −M/2 + 1,...,M/2. (3.2.22)
ˆ From Eq. (3.2.22) and ψk(rN ) = 0, Eq. (3.2.20) becomes
2ψˆ (r ) ωˆ (r ) = k N−1 , k = −M/2 + 1,...,M/2. (3.2.23) k N dr2
This is the vorticity boundary condition.
~ ~ 1 ∂ψ˜ ∂ψ˜ To obtain the values of velocity u˜, we use the relation u˜ = (˜u, v˜) = (− r ∂θ , ∂r ). By applying the discrete Fourier transform, we immediately have
1 ˆ uˆk(ri) = − (Ik)ψk(ri), i = 1,...,N − 1, (3.2.24) ri ψˆ (r ) − ψˆ (r ) vˆ (r ) = k i+1 k i−1 , i = 2,...,N − 1, (3.2.25) k i 2dr
18 −M M for k = 2 + 1,..., 2 . Special care is needed forv ˆk(r1), since we do not have the value ˆ ψk(r0). Here we adopt the one-sided finite-difference approximation forv ˆk(r1) as follows
−3ψˆ (r ) + 4ψˆ (r ) − ψˆ (r ) −M M vˆ (r ) = k 1 k 2 k 3 , k = + 1,..., . (3.2.26) k 1 2dr 2 2
We note thatu ˆk andv ˆk are zero at the boundary due to the zero boundary condition for velocity.
19 CHAPTER 4
Numerical Results
In this chapter, we first show the rate of convergence for our numerical algorithm. Second,
we demonstrate how the numerical results depend on Reynolds numbers and locations of the
rotlet in the disk. We finally calculate the traction exerted by the flow on the domain
boundary.
4.1 Convergence Rate
In this section, we are going to show the convergence rate for all of our variables including
the regular part of vorticity, stream function, and two components of the velocity field.
Reynolds number was chosen as Re = 1 and c = 0.5 for this test. We have chosen N =
256, 128, 64, 32, 16 in radial directions, and M = 4N in azimuthal direction. We start the
algorithm with initial value of ψ˜ as zero, and the iterations run until the solutions between two consecutive time steps change within a tolerance 10−8. The time step that was chosen as
dt = Redr2/5 for forward Euler method and dt = Redr2/3 for RK4, and dt = 1 for backward
Euler method.
Since there are no exact solutions to our problem, the errors have to be calculated by
comparing steady state solutions on two consecutive grid points. That is, we compare the
solutions for N = 256 and N = 128, and so on. The convergence rate is obtained by
computing the ratio of differences. To explain this, let us consider the regular part of stream
function ψ˜, the convergence rate for ψ˜ can be found by,
ψ˜ − ψ˜ log dr dr/2 , (4.1.1) 2 ˜ ˜ ψdr/2 − ψdr/4
20 ˜ ˜ where ψdr means the value of ψ obtained when the mesh size along radial direction as dr, ˜ ˜ and ψdr/2 means the value of ψ using the half of mesh size and so on.
We remark here that there is more effort involved in calculating the error, since the mesh points on a coarser grid do not conincide with the mesh points on the finer grid. Thus to compare the results at different mesh sizes, a cubic spline interpolation of the finer grid solution to the coarser mesh points is implemented. The error is then calculated between the interpolated solution from the finer grid and the solution directly from the coarser grid at the same coarser mesh points.
Tables 1–4 shows the l1, l2, and l∞ errors for different numbers of grid points. In addition to compare the errors inω ˜, we also list the error for ψ˜, andu, ˜ v˜. A first order accuracy is shown for all variables for all different norms, in Tables 1-4. In fact, for different values
Re and c, the rate of convergence would be different, and mostly less than 1. The errors in ω and ψ concentrate in a local region near the rotlet. Throughout the algorithm, we have used second order accurate finite difference schemes for radial derivatives, the lower convergence rate might be attributed to the singular coefficients in the convection term. It is well known that finite difference scheme to solve stream function-vorticity equations poses some numerical challenge on the order of accuracy. The nonslip boundary conditions for velocity become two boundary conditions for stream function, and which seems redundant since there is only one Poisson equation for stream function. When dealing with the boundary conditions numerically, the Dirichlet tangential velocity boundary condition was used to determine the boundary value for vorticity, thus the tangential velocity near the boundary is not as accurate as expected. We suspect this might also attribute to the lower order of convergence of the numerics.
21 Table 1: The errors inω ˜ for Re = 1
N l1 rate l2 rate l∞ rate
256-128 1.16E-2 1.15 1.01E-2 1.15 5.95E-2 0.94
128-64 2.61E-2 0.979 2.27E-2 0.98 1.14E-1 0.72
64-32 5.15E-2 0.73 4.49E-2 0.73 1.89E-1 0.41
32-16 8.55E-2 — 7.48E-2 – 2.52E-1 —
Table 2: The error in ψ˜ for Re = 1
N l1 rate l2 rate l∞ rate
256-128 2E-3 1.16 1.4E-3 1.16 1.71E-3 1.16
128-64 4.48E-3 0.98 3.13E-3 0.98 3.83E-3 0.98
64-32 8.87E-3 0.74 6.21E-3 0.75 7.61E-3 0.75
32-16 1.49E-2 — 1.04E-2 — 1.28E-2 —
Table 3: The error inu ˜ for Re = 1
N l1 rate l2 rate l∞ rate
256-128 1.12E-3 1.15 9.22E-4 1.15 1.41E-3 1.16
128-64 2.5E-3 0.98 2.06E-3 0.98 3.17E-3 0.98
64-32 8.87E-3 0.74 6.21E-3 0.75 7.61E-3 0.75
32-16 1.49E-2 — 1.04E-2 — 1.28E-2 —
22 Table 4: The error inv ˜ for Re = 1
N l1 rate l2 rate l∞ rate
256-128 5.63E-3 1.149 3.43E-3 1.15 3.68E-3 1.15
128-64 1.25E-2 0.95 7.64E-3 0.97 8.22E-3 0.98
64-32 2.43E-2 0.69 1.49E-2 0.71 1.62E-2 0.73
32-16 3.92E-2 — 2.46E-2 — 2.71E-2 —
4.2 Numerical Experiments
In this section, we demonstrate the numerical results for various values of Reynolds
number and different positions of the rotlet. In each of the figures, we show the contour
plots of singular (top), combined (middle), regular part (bottom) of stream functions (left)
and vorticity (right).
In Figs.1–3, results are shown with Re = 0.1, with the line rotlet locates at the horizontal
axis with distance to the origin c = 0.2, 0.5, and 0.7238 respectively. The distortion sin the stream function ψ and vorticity ω are hard to observe due to the small Re number.
Nonetherless, the bottom row of the Figs.1–3 aims to show that the small distortions in the contours from the regular ψ˜ andω ˜. The maximum of absolute values inω ˜ and ψ˜ is not at the rotlet center, but approximately symmetrically located at above and below the rotlet.
In Figs.4–5, results are shown for a fixed line rotlet c = 0.5 with two different values
of Reynolds number, Re = 1 and 10. For Stokes flow, the attached eddy is formed when √ c > 2 − 1 [6]. That can be observed from the contour plot of singular part of stream
function ψs. Near the rotlet, the contours are more or less concentric circles centered at the
rotlet. There exists a dividing stream line, across which the directions of velocities reverse,
and the attached eddy forms, see tau row of Figs. 2 and 3. For Navier-Stokes flow, the
attached eddy forms for even smaller values of c, and the region of the eddy enlarges with
23 increasing Re. The stream lines become asymmetric with respect to the horizontal line, and the contours of vorticities become distorted, more pronounced near the rotlet, as shown in the middle row of Figs.4 and 5.
Figure 2: Contour plots for stream function and vorticity at c = 0.2, Re = 0.1, N = 64. Top row: singular part of the solution; Middle row: combined solution; Bottom row: regular part of the solution.
Another physical quantity that is ready to be evaluated is the traction exerted by the
flow on the disk boundary. The traction on the surface of the circular cylinder, which we
assume having the height H, is given in dimensional form by,
Z 2π T = Ha τrθ|r=adθ, (4.2.1) 0 where τrθ is the rθ components of the deviatoric stress tensor,
τ = 2µD, (4.2.2)
24 Figure 3: Contour plots for stream function and vorticity at c = 0.5, Re = 0.1, N = 64. where µ is dynamic viscosity and D is the rate of strain tensor, which, in polar coordinates, is given by, ∂u 1 ∂v v 1 ∂u 1 ( − + ) D = (∇~u + (∇~u)T ) = ∂r 2 ∂r r r ∂θ . (4.2.3) 2 1 ∂v v 1 ∂u 1 ∂v u 2 ( ∂r − r + r ∂θ ) r ∂θ + r Thus
Z 2π Z 2π 1 ∂u ∂v v T (r = 1) = Ha τrθ|r=adθ = µaH + − dθ, (4.2.4) 0 0 r ∂θ ∂r r r=a Z 2π ∂v = µaH dθ, (4.2.5) 0 ∂r r=a or in dimensionless form T 1 Z 2π ∂v∗ 2 = dθ, (4.2.6) ρaHUc Re 0 ∂r r¯=1 in which the radial derivative of tangential velocity can be numerically approximated by one-sided finite difference scheme along the boundary.
25 Figure 4: Contour plots for stream function and vorticity at c = 0.7238, Re = 0.1, N = 64.
Figure 6 shows that the numerical calculated normalized traction decreases with increas- ing Re when c = 0.5. This indicates that the more viscous the flow is, the more traction the
flow will exert on the boundary, as one might expect.
26 Figure 5: Contour plots for stream function and vorticity at c = 0.5, N = 64, Re = 1.
27 Figure 6: Contour plots for stream function and vorticity at c = 0.5, N = 64, Re = 10.
Figure 7: Normalized traction for different values of Re, for c = 0.5, N = 64.
28 CHAPTER 5
Conclusion
In this thesis, we numerically investigate solutions of the Navier-Stokes equations in a
2D disk where the flow is driven by a single line rotlet. To model the problem, we first write down the Navier-Stokes equations in vorticity-stream function formulation with appropriate boundary conditions both at the disk boundary and near the rolet. We further decompose the solutions into the singular part and regular part. The singular part of the solution is analytically known, so we only need to solve for the regular part of the solutions.
We use the finite difference method in the r direction and Fourier spectral method in the
θ direction to numerically solve the equations for the regular part of the solutions. We choose uniform radial grid points with shifting a half mesh point from the origin, and uniform regular grid points along azimuthal direction. We have implemented both forward Euler and fourth order Runge-Kutta method for time integration. The algorithm stops when the solutions reach equilibrium. Fourier filtering is used to allow for a larger time step.
The convergence rates are shown to be first order for vorticity, and less than first order for stream function and velocity. This can be attributed to the singular nature of the solution, i.e., the velocity field blows up near the rotlet. The numerical experiments for different
Reynolds numbers show that the higher the Reynolds number, or unsteady the flow is, the more asymmetry of the flow pattern appears near the rotlet. We also demonstrate results for various positions of rotlet inside the domain. Our results imply that the inertial term helps to develop eddies in the flow field as expected. The calculation for the traction on the domain boundary shows that the more viscous the flow is, the more traction the flow exerts on the boundary.
29 BIBLIOGRAPHY
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