Convergence Proof of the Velocity Field
for a Stokes Flow Immersed Boundary Method
YOICHIRO MORI
University of British Columbia
Abstract
The immersed boundary (IB) method is a computational framework for problems involving the interaction of a fluid and immersed elastic structures. It is
characterized by the use of a uniform Cartesian mesh for the fluid, a Lagrangian
curvilinear mesh on the elastic material, and discrete delta functions for communication between the two grids. We consider a simple IB problem in a twodimensional periodic fluid domain with a one-dimensional force generator. We
obtain error estimates for the velocity field of the IB solution for the stationary
Stokes problem. We use this result to prove convergence of a simple smallamplitude dynamic problem. We test our error estimates against computational
c 2007 Wiley Periodicals, Inc.
experiments. 1 Introduction
The immersed boundary method was first introduced by Peskin in [24] as a
computational tool to investigate blood flow in the presence of cardiac valves. It
has since proved to be a generally useful computational framework for solving
problems in fluid-structure interaction. The method has found many applications,
including blood flow in the heart [17], vibrations of the cochlear basilar membrane
[1, 10], blood clotting [7, 39, 42], aquatic locomotion [4, 6, 8, 14], insect flight
[18, 19], flow with suspended particles [9, 33], and other physical problems [5, 15,
16, 29]. We refer to [25] for a more extensive list of applications.
In this paper, we present a proof of convergence for a stationary immersed
boundary problem and a small-amplitude dynamic problem. We shall be concerned
solely with the velocity field. We consider a model problem that consists of a onedimensional closed filament in a two-dimensional periodic fluid domain U. This
is a standard model problem for the immersed boundary method. A Lagrangian
coordinate is placed on the one-dimensional filament so that X./ traces the
curve . A boundary force F is given as a function of . Our problem is to find the
fluid velocity field u under Stokes flow.
The problem can be formulated as an interface problem, where the Stokes equations are satisfied in the two regions demarcated by . One requires that the velocity field u be continuous at and satisfy stress jump conditions expressed in
terms of F and X. A mathematically equivalent way to formulate this problem is
Communications on Pure and Applied Mathematics, Vol. LXI, 1213–1263 (2008)
c 2007 Wiley Periodicals, Inc.
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Y. MORI
to consider Stokes flow under a singular external force field f [26]. This force field
f is supported on the curve and is expressed in terms of the boundary force F as
follows:
Z
(1.1)
f D F. /ı.x X.//d
where x is a point in the fluid domain and ı is the Dirac delta function. The immersed boundary method uses this formulation to approach the problem.
We place a uniform Cartesian mesh on the fluid domain. In order for the singular force field f to be “felt” by the Cartesian grid, we must regularize f. This is
achieved by making use of discrete delta functions, whose properties will be discussed in Section 3. In this paper, we shall employ a spectral discretization for the
periodic Stokes problem.
Our problem is thus to solve a partial differential equation with a singular
source term. Such problems arise not only in the context of the immersed boundary
method but also in the front tracking method [37] and the level set method [23, 30].
The numerical approximation of such problems has been analyzed in detail for the
one-dimensional Poisson problem in [2, 36, 38], and a general framework for a
convergence analysis of such problems is proposed in [36]. However, to the best of
the author’s knowledge, there has been no convergence analysis for such problems
for PDEs in dimension 2 or higher.
In Section 2, we discuss the model problem above and its immersed boundary
discretization. Following the framework proposed in [36], we write the exact velocity field and its immersed boundary approximation in the form of an integral
and a sum, respectively. This reduces the problem to that of an approximation of
an integral. In Section 3, we review some properties of the discrete delta functions
given in [2, 32, 36] and present them in a fashion useful for our analysis.
In Section 4, we prove some estimates on the discrete and continuous Green’s
functions. A key estimate is Proposition 4.5 on the difference between the discrete and continuous Green’s functions. We obtain this estimate by studying their
Fourier sum representations.
In Section 5, we prove pointwise error estimates valid away from the immersed
boundary . Let the fluid mesh spacing be h. Suppose we refine the Lagrangian
mesh so that the mesh spacing is proportional to h˛ , ˛ > 0. If F and X are
smooth functions of , we shall see that the following error estimate holds for
“generic” (the precise meaning of which will be discussed in Section 5) points x
away from :
(1.2)
ju.x/ uh .x/j C h.h˛ C h log.h1 // C hp
where uh denotes the immersed boundary solution and p is the number of discrete
moment conditions satisfied by the discrete delta function (see Section 3). If this
estimate is optimal, it shows that ˛ D 1 and p D 2 are the optimal choices for
these parameters.
IMMERSED BOUNDARY METHOD PROOF
1215
In Section 6, we prove a global L1 error estimate for the velocity field of the
immersed boundary solution. This estimate ensures that the velocity field converges to the true solution up to the immersed boundary. Suppose we refine proportionally to h˛ . We shall prove
(1.3)
ku uh kL1 .U/ C.h C h˛ / log.h1 /:
This shows that the immersed boundary solution converges to the true solution
everywhere as long as ˛ > 0. This also shows that should be refined proportionally to h to obtain the optimal global convergence rate.
The stationary immersed boundary problem considered above constitutes only
a part of a full immersed boundary computation, which may be described algorithmically as follows:
(1) Compute the boundary force F. / given the configuration X. /.
(2) Given F./, compute the regularized external force field f using discrete
delta functions.
(3) Compute the velocity field u.
(4) Interpolate the velocity field u at the immersed boundary points.
(5) Advect the immersed boundary points X./ using the interpolated velocities.
So far we have only dealt with steps 2 and 3 above. In Section 7, we prove
convergence of a simple dynamic immersed boundary problem. We consider an
approximate problem in which we assume that the immersed boundary displacements are small enough so that the spreading (step 2) and the interpolation (step 4)
operations may be performed at fixed locations. We take a Hookean restoring
force to the fixed locations as the force law used in step 1. The global L1 estimate
proven in Section 6 allows us to perform step 4, which involves point evaluation of
the velocity field. This estimate also gives us the requisite consistency.
In Section 8, we shall test the error estimates for the stationary problem against
a computational experiment for various choices of the discrete delta function. All
computational results are consistent with the error estimates. We shall see that the
error estimates capture the overall features of the convergence rates.
2 Model Problem
We consider the following force balance problem for Stokes flow (Figure 2.1).
The equations we solve are:
(2.1)
(2.2)
(2.3)
u D rp f C g;
r u D 0;
Z fD
F./ı.x X.//d:
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Y. MORI
Velocity Field
u.x/
Immersed Boundary
; X./
y
Fluid Domain
U
x
F IGURE 2.1. Schematic diagram describing the model problem.
The equations have already been rendered dimensionless. We shall work on a
square periodic fluid domain x D .x; y/ 2 U where x and y . The immersed boundary X./ parametrized by 2 ‚ R=2Z defines
a simple closed curve in the fluid domain. We note that is not necessarily an
angular coordinate. We require that stay away from @U, where @U consists of
the points x 2 U that satisfy x D ˙ or y D ˙. The force F./ is a given
function of the immersed boundary coordinate .
We have an additional constant force term g in the force balance equation (2.1)
to ensure that the above problem has a solution in a periodic domain:
Z
Z
.f g/d x D .u C rp/d x
(2.4)
U
U
Z
.ru C p/n dS D 0
D
@U
where n is the unit normal on @U and dS signifies integration on @U. The last
equality holds because u and p are assumed periodic. Using (2.3), we conclude
that
Z 1
(2.5)
gD
F./d:
.2/2 The force field g serves to subtract the uniform (zero wavelength) component of
the boundary force field. We must also place a constraint on the velocity field u so
IMMERSED BOUNDARY METHOD PROOF
1217
that we have a unique solution. We require that
Z
(2.6)
u d x D 0:
U
In what follows, we assume that u satisfies this constraint.
We now discretize the above problem using the immersed boundary method:
(2.7)
(2.8)
(2.9)
Lh uh D Dh ph fh C gh ;
Dh uh D 0;
fh D
M
X
F.m /ıh .x X.m //:
mD1
In the above, h is the mesh width of the uniformly discretized fluid domain and is the width of the immersed boundary mesh. We let
(2.10)
N h D 2;
(2.11)
M D 2;
m D C m:
The discrete operators Lh and Dh are approximations to the Laplacian and the
gradient/divergence operators, respectively. The function ıh .x X/ in (2.9) is a
discrete approximation to the Dirac delta function centered around X. The discrete
delta function ıh will be discussed in detail in Section 3.
We must supplement the above with an expression for g furnished here as a
simple discretization of (2.5):
(2.12)
gh D
M
1 X
F.m /:
.2/2
mD1
In accordance with (2.6), we require that uh satisfy
X
uh .x/h2 D 0
(2.13)
x2Gh
where Gh denotes the set of fluid grid points.
Following [36], we now write down an integral or sum representation of the
solution to the continuous and discrete problems.
The periodic Stokes problem is well-studied (see, e.g., [12, 27, 28]). The solution of the continuous Stokes problem can be written as follows:
Z (2.14)
u.x/ D
G.x X. //F. /d
where G is a 2 2 matrix Green’s function that corresponds to the Stokes problem
on a two-dimensional periodic domain. The function G can be expressed as a
1218
Y. MORI
lattice sum of the two-dimensional free-space Stokeslet (care must be taken so that
the sum is convergent) or can be written as the following Fourier sum:
1 X
1
exp.ik x/ 2 Pk ;
2
.2/
jkj
jkj¤0
kkT
1
l 2 kl
:
Pk D I 2 D
jkj
jkj2 kl k 2
G.x/ D
(2.15)
k
kD
;
l
where I is the 2 2 identity matrix, and the components k and l of the vector k
are integers. In the above and in what follows, we let j j of a vector denote its
Euclidean length. We shall often use the notation
Gx .y/ G.x y/ D G.y x/:
Note that the constraint in (2.6) is already built into expression (2.14) since the
Green’s function G does not have a zero-frequency component.
We would now like to write down a similar expression for the discrete problem.
Taking the discrete Fourier transform of equations (2.7)–(2.9) and taking a spectral
discretization for the Laplacian and gradient/divergence operators, we find
X
(2.16)
uh .x/ D
Gh .x y/fh .y/h2 :
y2Gh
We shall often make use of the notation Gh;x .y/ Gh .x y/ D Gh .y x/. The
function Gh is given by
(2.17)
Gh .x/ D
X
1
1
exp.ik x/ 2 Pk
2
.2/
jkj
k2Kh
where the matrix Pk is the same as in equation (2.15), and Kh is a square-shaped
subset of the k-plane,
(2.18)
Kh D fk D .k; l/T j jkj ¤ 0; jkj < = h; jlj < = hg:
We shall call Gh the discrete Green’s function. The discrete Green’s function Gh
is nothing more than a band-limited version of the continuous Green’s function G.
Now, consider the expression for fh :
(2.19)
fh .x/ D .S.X/F/.x/ M
X
F.m /ıh .x X.m //:
mD1
S.X/ can be seen as a linear operator that maps a function defined on the immersed
boundary points to a function defined on the Eulerian grid. This operator S.X/,
aside from X, depends on h, , and the choice of the discrete delta function ıh .
IMMERSED BOUNDARY METHOD PROOF
1219
We shall call this the spreading operation. Substituting this into (2.16), we find
uh .x/ D
X
Gh;x .y/.S.X/F/.y/h2
y2Gh
(2.20)
D
X
Gh;x .y/
X
M
mD1
y2Gh
D
M X
X
F.m /ıh .y X.m // h2
Gh;x .y/ıh .y X.m //h
2
F.m /:
mD1 y2Gh
For any function q on the fluid grid, define S as follows:
(2.21)
.S .X/q/.m / D
X
q.x/ıh .x X.m //h2 :
x2Gh
This linear map S .X/ interpolates the function q defined on the fluid grid at the
immersed boundary points X.m /. We shall call S .X/ the interpolation operator.
The operator S .X/ will also act on vectors and matrices, with the understanding that it acts componentwise. Note that S .X/ is the transpose of S.X/, as the
notation suggests. We can now rewrite (2.20) as
(2.22)
uh .x/ D
M
X
.S .X/Gh;x /.m /F.m /:
mD1
Our task now is to understand under what conditions (2.22) converges to (2.14) and
at what rate.
3 Discrete Delta Function
In this section, we discuss the discrete delta function and some properties of the
linear operations that make use of the discrete delta function. A mathematically
equivalent presentation of the facts to be shown below can be found in [2, 32,
36]. We also note that essentially equivalent results are obtained in [20], where
these functions are used for image reconstruction. We include these results here
nonetheless to present them in a form most conducive to our purposes.
We assume that ıh has the form
(3.1)
ıh .x/ D
1 x y h2
h
h
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Y. MORI
for some function defined on the real line. Some of the commonly imposed
conditions on are [25]:
has compact support. We let the support of .r/ satisfy jrj r .
satisfies moment conditions. When satisfies
X
(3.2)
.n r/ D 1 if p D 0;
(3.3)
X
n2Z
.n r/p .n r/ D 0 if p > 0;
n2Z
for all r 2 R, is said to satisfy the p th -order moment condition. If has
compact support, the above sums contain only finitely many terms. If satisfies moment conditions up to order p 1, we shall say, following [36],
that is of moment order p.
satisfies the following even-odd condition:
X
X
.n r/ D
.n r/
(3.4)
n even
n odd
for all r 2 R.
is continuous.
The first condition is for computational efficiency. It ensures that each immersed boundary point only communicates with a finite number of grid points
independent of the mesh spacing h. For any point X on the fluid domain, we define
the following square-shaped region:
˚
(3.5) RX D x D .x; y/T j jx Xj r h; jy Y j r h; X D .X; Y /T :
RX , which depends also on and h, is a subset of the fluid region with which an
immersed boundary point located at X may communicate.
The moment conditions, as we shall see, determine the accuracy of the operations performed with the discrete delta function. The third condition was introduced to avoid the “checkerboard” type instability that manifests itself when using
the central difference operator for the discretization of the rp and the r u terms
in the Stokes equations [25]. Here we are using a spectral scheme, so we do not
expect this condition to play an important role. We shall, however, briefly come
back to this condition in Section 8. The fourth condition will not be used in this
paper except to ensure that remains bounded if has compact support. This
condition may have important consequences especially in relation to the advection
of the immersed boundary.
We henceforth assume that has compact support and is bounded, and is of
moment order at least 1 (that is to say, satisfies the zeroth-order moment condition).
We now prove two lemmas about the interpolation operator S . The first one is
trivial.
IMMERSED BOUNDARY METHOD PROOF
1221
L EMMA 3.1 Let q.x/ be a function defined on the Cartesian fluid domain. Suppose
we are interpolating at a point X0 D X.0 /. Then
j.S .X/q/.0 /j C max jq.x/j
(3.6)
x2RX0
for some constant C that depends only on .
P ROOF : We have:
.S .X/q/.0 / D
(3.7)
X
ıh .x X0 /q.x/h2 :
x2Gh
Substitute (3.1) in the above:
(3.8)
X
.S .X/q/.0 / D
x2Gh \RX0
x X0
y Y0
q.x/
h
h
where X0 D .X0 ; Y0 /T . By taking absolute values on both sides,
j.S .X/q/.0 /j
(3.9)
max jq.x/j
x2RX0
X
x2Gh \RX0
ˇ ˇ ˇ ˇ
ˇ x X0 ˇ ˇ y Y0 ˇ
ˇ
ˇ ˇ
ˇ
ˇ
ˇˇ
ˇ
h
h
2
X
j.n r/j :
max jq.x/j max
x2RX0
r2R
n2Z
The sum in the last line is bounded from above since has compact support and
is bounded. In fact, if is positive, the above sum is equal to 1 thanks to the
zeroth-order moment condition.
Our second lemma tells us that the accuracy of the interpolation operation is
determined by the moment conditions.
L EMMA 3.2 Let q.x/ be a C n function defined on the Cartesian fluid domain, and
be of moment order n. Suppose we interpolate q at a point X0 D .X0 ; Y0 /T D
X.0 /,
(3.10)
j..S .X/q/.0 / q.X0 /j C hn max . max j@˛x @yˇ q.x/j/
˛Cˇ Dn x2RX0
ˇ
where @˛x denotes the ˛ th derivative in the x-direction and likewise for @y . The
above constant C depends only on and n.
P ROOF : For any point x D .x; y/T 2 RX0 , one can define the following function g:
(3.11)
g.t / D q.X0 C t.x X0 //:
1222
Y. MORI
Note g.0/ D q.X0 / and g.1/ D q.x/. Applying the Taylor formula to this equation, we have
q.x/ D q.X0 /
C
n1
X
X
kD1 ˛Cˇ Dk
(3.12)
r.x/ D
X
˛Cˇ Dn
1 ˛ ˇ
@ @ q.X0 /.x X0 /˛ .y Y0 /ˇ C r.x/;
˛ŠˇŠ x y
1 ˛ ˇ
@ @ q.zx /.x X0 /˛ .y Y0 /ˇ ;
˛ŠˇŠ x y
zx D X0 C .x X0 /
where is some constant 0 < < 1 and depends on x. Note that zx 2 RX0 since
zx lies on the straight line between x and X0 and RX0 is convex.
First we have
X
.S .X/.1//.0 / D
ıh .x X0 /h2
x2Gh
x X0
y Y0
D
h
h
x2Gh
X X X0
Y0
D
m
m
D1
h
h
X
(3.13)
m2Z
m2Z
where we used the zeroth-order moment condition in the last line. Note for any
.˛; ˇ/ ¤ .0; 0/, ˛ < n, and ˇ < n,
.S .X/..x X0 /˛ .y Y0 /ˇ //.0 /
X
D
ıh .x X0 /.x X0 /˛ .y Y0 /ˇ h2
x2Gh
(3.14)
x X0
y Y0
D
.x X0 /˛ .y Y0 /ˇ
h
h
x2Gh
X X0
X0 ˛
D
m
m
h
h
m2Z
X Y0
Y0 ˇ ˛Cˇ
h
m
D0
m
h
h
X
m2Z
where we used the moment conditions in the last line. Applying S to (3.12) and
using the above two calculations, we have
(3.15)
.S .X/q/.0 / D q.X0 / C .S .X/r/.0 /:
IMMERSED BOUNDARY METHOD PROOF
1223
We compute j.S .X/r/.0 /j. First,
ˇ ˇ
ˇ S .X/ @˛ @ˇ q.zx /.x X0 /˛ .y Y0 /ˇ .0 /ˇ
x y
ˇX
ˇ
ˇ
ˇ
x
X
y
Y
0
0
˛
ˇ
˛
ˇ
ˇˇ
@x @y q.zx /
.x X0 / .y Y0 / ˇˇ
h
h
(3.16)
x2Gh
max j@˛x @yˇ q.zx /jC˛ Cˇ hn max j@˛x @yˇ q.x/jC˛ Cˇ hn ;
x2RX0
C˛;ˇ D max
r2R
X
x2RX0
j.m r/jjm rj˛;ˇ
m2Z
where we used zx 2 RX0 . Thus,
X C˛ Cˇ
j.S .X/r/.0 /j max j@˛x @yˇ q.x/jhn
x2R
˛Š ˇŠ
X0
˛Cˇ Dn
X
(3.17)
C˛ Cˇ n
D
h max . max j@˛x @yˇ q.x/j/:
˛ŠˇŠ
˛Cˇ Dn x2RX0
˛Cˇ Dn
This proves the assertion.
We remark that if the function q in the above is only Lipschitz and is at least
of moment order 1, we still have the following bound, which can be readily proved:
(3.18)
j.S .X/q/.0 / q.X0 /j C h
where the constant C above depends only on the Lipschitz constant and the function . We shall use this fact later in Section 7.
4 Estimates for Green’s Function
In this section we shall state some useful estimates for the continuous and discrete Green’s function.
We first consider the continuous Green’s function. All of the results in the following lemma are implicit in classical studies of the periodic Stokes problem [12].
L EMMA 4.1 Suppose
x
(4.1)
xD
;
y
jxj 2 d; jyj 2 d; d > 0:
Then, we have the following:
(4.2)
(4.3)
jGij .x/j C log.jxj1 / C C;
C
j@˛x @yˇ Gij .x/j n C Cdn ;
jxj
where Gij is the ij th element of the Green’s function matrix, and ˛ C ˇ D n. The
constant C depends only on n and Cdn depends only on d and n.
1224
Y. MORI
P ROOF : Define the set
(4.4)
Dd D fx j jxj 2 d; jyj 2 d g:
Because our computational domain is a periodic square domain of side 2, the
Green’s function G can be written as a lattice sum of the free space two-dimensional Stokeslets G free (care must be taken so as to obtain a convergent sum).Thus,
on Dd , G can be written as follows:
1
1
xxT
free
free
(4.5)
G.x/ D G .x/ C R.x/; G .x/ D
I log
C 2
4
jxj
jxj
where R.x/ is a smooth function and I is the identity matrix. Taking any nth partial
derivative of the above,
(4.6)
@˛x @yˇ Gij .x/ D @˛x @yˇ Gijfree .x/ C @˛x @yˇ Rij .x/:
Since R is smooth up to the boundary of Dd , its nth partial derivatives are bounded
by a universal constant Cdn . The derivatives of G free can be computed explicitly,
and we see that each element is bounded by C =jxjn for a constant C that only
depends on n.
We turn to the discrete Green’s function Gh . We start with the following trivial
bound:
L EMMA 4.2
jGh;ij .x/j C log.h1 /
(4.7)
for a universal constant C , where Gh;ij denotes the ij th element of the discrete
Green’s function matrix.
P ROOF : Take (2.17):
ˇ
ˇ
X 1
ˇX
ˇ
1
ˇ
ˇ
jG
.x/j
exp.ik
x/
P
jPk;ij j:
k;ij
h;ij
(4.8)
ˇ
ˇ
jkj2
jkj2
k2Kh
k2Kh
Noting that Pk;ij is bounded in absolute value by 1, we see that
Z p2 = h
X 1
1
jGh;ij .x/j C
r dr C log.h1 /
(4.9)
2
jkj2
r
1
k2Kh
where we used k D .k; l/T , jkj < = h, and jlj < = h for k 2 Kh . The last
constant C is a universal constant (not all constants in the above may be equal, as
will be the case elsewhere in the paper).
We now prove estimates that tell us how well the discrete Green’s function
approximates the continuous Green’s function.
IMMERSED BOUNDARY METHOD PROOF
1225
Consider expressions (2.15) and (2.17) for the continuous and discrete Green’s
function. A first attempt to estimate the difference between the two might be the
following:
ˇ X
ˇ
ˇ
ˇ
1
exp.ik x/ 2 Pk;ij ˇˇ
jGij .x/ Gh;ij .x/j D ˇˇ
jkj
k¤0;k…Kh
(4.10)
X
X
1
1
jP
j
D
:
k;ij
jkj2
k2 C l 2
jkj;jlj= h
k¤0;k…Kh
The difficulty is that the last sum is not convergent. The above calculation indicates
that the Fourier sum representation (2.15) of G is nowhere absolutely convergent.
Equation (2.15) is an expression valid only in the L2 sense. This difficulty comes
from the fact that the free space two-dimensional Stokeslet (and hence the Green’s
function G) possesses a logarithmic singularity at the origin. We have thus a rather
delicate summation problem that we must understand.
We start by estimating the magnitude of the following expression:
L
D
GK
(4.11)
X
1
1
exp.ik x/ 2 Pk
2
.2/
jkj
L
k2BK
L is a square-toric shaped subset of the k-plane
where BK
(4.12)
L
D fk j K jkj L; K jlj Lg:
BK
L EMMA 4.3 Suppose
(4.13)
x D .x; y/T ¤ 0;
jxj < 2; jyj < 2:
L
The ij th element of GK
is bounded by
(4.14)
ˇ
ˇ L
ˇ
ˇG
K;ij .x/ C
;
Kw.x/
jxj
jyj
w.x/ D max sin
; sin
2
2
where the constant C is a universal constant and max.a; b/ denotes the greater
value of a and b.
P ROOF : Assume that jxj ¤ 0 and thus 0 < jxj < 2. Consider the sum
l
D
SP;Q
(4.15)
Q
X
˛ k ˇ l kl ;
kDP
˛ D exp.ix/; ˇ D exp.iy/; kl D
1
Pk;ij ;
jkj2
1226
Y. MORI
for some P and Q. Note that l is fixed in the above sum. Multiply by .1 ˛/ and
rearrange the above sum:
.1 l
˛/SP;Q
Dˇ
l
Q
X
.˛ k ˛ kC1 /kl
kDP
(4.16)
Dˇ
l
˛ P l ˛
P
QC1
Q
X
Ql C
k
˛ .kl
k1;l / :
kDP C1
The reader will recognize the above as a summation-by-parts procedure. By
assumption, 0 < jxj < 2, and thus ˛ ¤ 1. We can therefore divide both sides of
(4.16) by 1 ˛:
(4.17)
l
SP;Q
D ˇl
˛ P P l ˛ QC1 Ql
C ˇl
1˛
Q
X
kDP C1
˛k
.kl k1;l /:
1˛
We take the absolute value of both sides:
(4.18)
l
jSP;Q
j
Q
X
1
jkl k1;l j
jP l j C jQl j C
j1 ˛j
kDP C1
where we used jˇj D 1 and j˛j D 1. We have
(4.19)
jP l j 1
jkj2kDP
1
;
jP j2
jQl j 1
jkj2kDQ
1
:
jQj2
We can also check that
jkl k1;l j (4.20)
C
jkj3
where C is a universal constant. Thus,
l
j
jSP;Q
(4.21)
Q
X
C
1
1
1
:
C
C
j1 ˛j jP j2
jQj2
jkj3
kDP
L . GL
We now use this to estimate the magnitude of GK;ij
K;ij can be written as
(4.22)
L
GK;ij
D
K
X
lDL
l
SL;L
C
K1
X
l
.SL;K
lDKC1
C
l
SK;L
/
C
L
X
lDK
l
SL;L
:
IMMERSED BOUNDARY METHOD PROOF
1227
Thus,
L
jGK;ij
j
K
X
l
jSL;L
j
lDL
C
K1
X
l
l
.jSL;K
j C jSK;L
j/ C
lDKC1
(4.23)
K1
X
1
1
C
C 2
2
K
K
lDKC1
X 1 C
8
C
j1 ˛j K
jkj3
L
l
jSL;L
j
lDK
X
L C
1
1
C 2
j1 ˛j
L2
L
lDL
L
X
C
X
L
k2BK
1
jkj3
k2BK
where we used (4.21) to estimate the above sum and used L > K to get the last
line. For the last sum, we have
Z p2 L
X 1
C
1
(4.24)
C
r dr 3
3
jkj
r
K
K
L
k2BK
for some constant C . Putting this back into (4.23) and noting that j1 ˛j D
2 sin.jxj=2/, we have
L
jGK;ij
j
(4.25)
C
:
K sin.jxj=2/
If y ¤ 0, we can likewise prove the bound
L
j
jGK;ij
(4.26)
C
:
K sin.jyj=2/
Since x ¤ 0, at least one of x ¤ 0 or y ¤ 0 is true, and thus the desired conclusion
follows.
L
We can go on to prove a similar but stronger bound on GK
if x ¤ 0 and y ¤ 0.
The idea of the proof is that if x ¤ 0 and y ¤ 0, one can perform the above
summation-by-parts argument in each coordinate to obtain a better bound. We
shall relegate its proof to Appendix A.
L EMMA 4.4 Suppose
(4.27)
x D .x; y/T ;
0 < jxj < 2; 0 < jyj < 2:
1228
Y. MORI
L is bounded by
The ij th element of GK
(4.28)
L
.x/j jGK;ij
C
K 2 sin.jxj=2/ sin.jyj=2/
where the constant C is a universal constant.
With the use of the two lemmas above, we can prove the following estimate on
the difference between the discrete and continuous Green’s functions:
P ROPOSITION 4.5 Suppose
(4.29)
x D .x; y/T ¤ 0;
jxj < 2; jyj < 2:
Then Gh;ij .x/ converges to Gij .x/. Furthermore, the difference between G and
Gh satisfies the bound
jxj
Ch
jyj
; w.x/ D max sin
(4.30) jGij .x/ Gh;ij .x/j ; sin
:
w.x/
2
2
If x ¤ 0 and y ¤ 0, the difference satisfies a stronger bound:
C h2
sin.jxj=2/ sin.jyj=2/
where the constants C in the above do not depend on x or h.
(4.31)
jGij .x/ Gh;ij .x/j P ROOF : Define the square-shaped subset AK in the k-plane:
˚
(4.32)
AK D B1K D k D .k; l/T j 0 < jkj K; 0 < jlj K :
Define G K as
(4.33)
G K .x/ X
k2AK
exp.ik x/
1
Pkl :
jkj2
We now prove that
converges to G.x/. Let B D fx j x ¤ 0; jxj <
2; jyj < 2g. Take any point x0 in B and a closed disc D of radius > 0
centered at x0 so that this disc is contained in B. For any point x in D,
ˇ L
ˇ ˇ
ˇ
C
ˇG .x/ G K .x/ˇ D ˇG L
ˇ
(4.34)
.x/
;
ij
ij
KC1;ij
.K C 1/w.x/
where we have assumed L > K and used Lemma 4.3. The function w is given in
the statement of this lemma. By taking the maximum on both sides with respect to
x, we have
ˇ
ˇ
CD
(4.35)
max ˇGijL .x/ GijK .x/ˇ K C1
x2D
G K .x/
where CD is a constant that depends on the disc D. This shows that GijK is a
Cauchy sequence in the maximum norm on this disc and thus converges to some
continuous function GQ ij on D. On the other hand, it is clear from expression (4.33)
and (2.15) that GijK ! Gij in the L2 norm on the periodic unit cell (a coordinate
IMMERSED BOUNDARY METHOD PROOF
1229
square of side 2). If we let D denote the characteristic function of the disc D,
it follows that D GijK ! D Gij in the L2 norm. This shows that GQ ij should be
equal to the continuous version of the L2 function Gij on D. This proves that
GijK .x/ ! Gij .x/ on all points of D. Since the center of the disc D was arbitrary,
GijK .x/ ! Gij .x/ for all x 2 B.
Now we can take L ! 1 in (4.34) to find
ˇ
ˇ
C
ˇGij .x/ G K .x/ˇ :
(4.36)
ij
.K C 1/w.x/
Noting that Gh D G K for K < = h K C 1, we find
(4.37)
jGij .x/ Gh;ij .x/j If x ¤ 0 and y ¤ 0, we have from Lemma 4.4
ˇ ˇ L
ˇ
ˇ
.x/ˇ (4.38) ˇGijL .x/ GijK .x/ˇ D ˇGKC1;ij
Ch
:
w.x/
C
.K C
1/2 sin.jxj=2/ sin.jyj=2/
:
Taking L ! 1 and noting Gh D G K and K < = h K C 1, we have
(4.39)
jGij .x/ Gh;ij .x/j C h2
:
sin.jxj=2/ sin.jyj=2/
It is interesting to note that, although we use a spectral scheme to discretize the
fluid equations, the discrete Green’s function converges to the continuous Green’s
function only at an order 1 or order 2 rate. This is, after all, hardly surprising given
that the Green’s function represents the solution to the Stokes problem with a delta
function as the external force field. We can only expect spectral convergence if the
external force field is a smooth function.
Where does the convergence rate of 2 come from? By looking through the
above argument, we realize that this rate comes from the fact that the force field
f “gains” two derivatives when the Stokes problem is solved to find the velocity
field u. The order 1 rate seen where x D 0 or y D 0 in Proposition 4.5 may be
understood as coming from grid effects.
5 Local Error Estimates
We are now ready to prove error estimates about the immersed boundary solution. In this section, we concern ourselves with pointwise error estimates valid at
points away from the immersed boundary. We assume that the immersed boundary
curve D fx 2 U j x D X. /g is at least a distance dU > 0 away from the
boundary of U.
In the following theorem, we estimate the difference between the exact and
approximate solutions by dividing it into three parts, a quadrature error term, an
1230
Y. MORI
interpolation error term, and an error term due to the discretization of the fluid
domain (the difference between the continuous and discrete Green’s functions).
This idea is based in part on a discussion given in [36].
T HEOREM 5.1 Let be of moment order p, and F./ and X. / be C n as functions
of . For x … , the following error bound holds for sufficiently small h and :
ju.x/ uh .x/j C.h C hmin.p;n/ C . /n /:
(5.1)
If no points on share the same x- or y-coordinate as the point x, uh satisfies the
stronger error estimate
ju.x/ uh .x/j C.h2 C hmin.p;n/ C . /n /
(5.2)
for h and small enough. The constants C in the above do not depend on h or
but may depend on x.
P ROOF : We first prove (5.1). Suppose
dist.x; / minjx X. /j d > 0:
(5.3)
Consider the difference between (2.14) and (2.22):
R
u uh D I C Iı C IG ;
Z M
R
X
I D
Gx .X.//F. /d Gx .X.m //F.m /;
(5.4)
Iı D
M
X
mD1
.Gx .X.m // .S .X/Gx /.m //F.m /;
mD1
IG D
M
X
.S .X/.Gx Gh;x //.m /F.m /:
mD1
R
The term I is a quadrature error. The integrand is a periodic function, and the
quadrature method is the trapezoidal rule. Recalling that the trapezoidal rule is
spectrally accurate for periodic functions, we have
R
R
jI j C . /n
(5.5)
R
where the constant C R depends on the derivatives of Gx .X.//F. / with respect
that F./
to . We know that C is bounded by Lemma 4.1 and the assumption
R
and X./ are C n . From equation (4.3) of Lemma 4.1, we see that C depends on
x only through d .
IMMERSED BOUNDARY METHOD PROOF
We now estimate Iı . Let h be small enough so that
ı
I D .I1ı ; I2ı /T , F D .F1 ; F2 /T , and q D min.p; n/.
2
M X
X
jIiı j 1231
p
2 r h < d =2. Let
j.Gx;ij .X.m // .S .X/Gx;ij /.m //Fj .m /j
mD1 j D1
(5.6)
C
2
M X
X
mD1 j D1
C
M
X
mD1
max
max j@˛x @yˇ Gx .y/j hq jFj .m /j
˛Cˇ Dq y2RX.m /
max
y2RX.m /
C
q
C Cd hq jF.m /j
jx yjq
where we used Lemma 3.2 in the second inequality and Lemma 4.1 in the third.
Note that
p
d
(5.7)
jx yj jx X./j jy X./j d 2 r h 2
where we used (5.3) and y 2 RX./ . Thus,
jIiı j
(5.8)
C
M
X
C
q
q
C
C
jF.m /j
h
d
.d =2/q
mD1
C .; d ; max jFj; q/h :
ı
q
We now estimate IG . We take h small enough so that
r h < dU =2. Let IG D .I1G ; I2G /T ,
jIiG j
2
M X
X
p
2 r h < d =2 and
jS .X/.Gx;ij Gh;x;ij /.m /Fj .m /j
mD1 j D1
(5.9)
2
M X
X
mD1 j D1
M
X
mD1
max
y2RX.m /
max
y2RX.m /
C j.Gx;ij .y/ Gh;x;ij .y//jjFj .m /j
Ch
jF.m /j
w.x y/
where we have used Lemma 3.1 in the second inequality and Lemma 4.5 in the
third. Let x D .x0 ; y0 /T , y D .x1 ; y1 /T , z D x y D .x2 ; y2 /T , and X./ D
.X./; Y.//T . We first note that
jx2 j D jx0 x1 j jx0 X. /j C jx1 X. /j
(5.10)
2 dU C r h 2 dU
2
1232
Y. MORI
where we have used y 2 RX./ and so jx1 X. /j r h. An identical inequality
holds for jy2 j D jy0 y1 j. We also have
d
(5.11)
jzj D jx yj 2
where we can argue as in (5.7). Thus the argument z of the function w.z/ D
w.x y/ in (5.9) only ranges over the following domain Dz :
ˇ
ˇ
dU
dU
d
T ˇ
; x2 2 ; y2 2 (5.12) Dz D z D .x2 ; y2 / ˇ jzj :
2
2
2
Since w.z/ is never 0 and is a continuous function that is always positive on this
compact set Dz , w.z/ > C.dU ; d / > 0 where C is a constant that depends on dU
and d . Substituting this back into (5.9), we have
(5.13)
jIiG j C h
M
X
jF.m /j C G .dU ; d ; maxjFj/h:
mD1
Collecting (5.5), (5.8), and (5.13), we obtain (5.1).
We now turn to (5.2). Let x D .x0 ; y0 /, and suppose the lines x D x0 and
y D y0 never touch or intersect . Define
(5.14)
dx D min jx0 X./j;
dy D min jy0 Y./j;
dx D min.dx ; dy /:
A similar proof to that of inequality (5.13) yields
jIiG j C.dU ; dx ; maxjFj/h2
(5.15)
where instead of equation (4.30) in Proposition 4.5, equation (4.31) is used. Collecting (5.5), (5.8), and (5.15), we obtain the bound (5.2).
We make the following observation. Fix a positive number d and consider all
x such that dist.x; / d . Then, forR any such x the constant in (5.1) can be taken
uniformly in x, since the constants C , C ı , and C G in (5.5), (5.8), and (5.13) only
depend on d and not directly on x. In particular, if we take p D 1 and n D 1 we
have
(5.16)
ju.x/ uh .x/j C.h C / for x such that dist.x; / d
where the constant C can be taken uniformly in x.
We now prove the following generalization of (5.16), which we shall need later
in the proof of Theorem 6.3. Consider a connected subset a of the closed curve
; a would thus be a filament with endpoints. Define
Z
vD
Gx .X. //F. /d;
X. /2a
(5.17)
vh D
X
X.m /2a
.S .X/Gh;x /.m /F.m /:
IMMERSED BOUNDARY METHOD PROOF
1233
The only difference between v; vh and u; uh is that the integration or summation is
over a and not . We have the following corollary for the difference between v
and vh :
C OROLLARY 5.2 Suppose F and X are C 1 functions in , and is of moment order
at least 1. Fix a positive number d and consider x such that dist.x; a / d > 0.
The following holds:
(5.18)
jv.x/ vh .x/j C.h C /
where the constant C may be taken to be the same as the constant in (5.16). Thus,
the constant C does not depend on x, dist.x, a / d , or the choice of connected
subset a .
R
P ROOF : We estimate the difference v vRh by dividing it into three parts, Iv , Iıv ,
ı
G
and IG
v , which correspond respectively to I , I , and I in equation (5.4). Since
a is a subset of , it is clear that inequalities (5.8) and (5.13) hold for Iıv and IG
v
where we take q D 1, with the same constants C ı and C G .
The fact that is a closed filament allowed us to use the properties of the
trapezoidal rule for quadrature of periodic functions
in the proof of (5.5). Since a
R
is not a closed filament, (5.5) does not hold for Iv when n > 1, but does
still hold
R
R
for n D 1 with the same constant C . Combining the estimates for Iv , Iıv , and IG
v ,
we obtain the desired estimate.
We proved in Theorem 5.1 that at points away from the immersed boundary,
the generally applicable error estimate is (5.1), whereas if the point x0 D .x0 ; y0 /T
does not share the x- or the y-coordinate with , (5.2) holds. For points that do
not share an x- or y-coordinate, convergence is considerably faster, according to
Theorem 5.1. However, the following is true:
Suppose x D x0 and y D y0 is never tangent to . We shall call such points
x0 D .x0 ; y0 /T nontangent points. For nontangent points, we can prove the following bound, whose proof is given in Appendix B.
T HEOREM 5.3 Suppose X and F satisfy the conditions of Theorem 5.1 and, in
addition, j@X=@j ¤ 0. Nontangent points are generic: almost all points x 2 Un,
in the sense of Lebesgue measure, are nontangent points. For nontangent x, the
following error bound holds for small enough h and :
(5.19)
ju.x/ uh .x/j C h. C h log.h1 // C hmin.p;n/ C ./n
where the constant C may depend on x but not on h or .
Since (5.19) holds almost everywhere in Un, we take this as the representative
error bound that holds for points in the computational domain. Suppose we let
1234
Y. MORI
d
d
X?
y
x
x
d
F IGURE 6.1. Schematic diagram of the geometric constructs used in the
proof of Theorem 6.3. The region is the region between the innermost
and outermost curves, which are both a distance d away from . When
the point x is within the region , we consider the closest point on ,
which we denote by X? , and a circle of radius d around X? .
h˛ as h ! 0 and X and F are smooth so that n can be taken arbitrarily large.
Then the error estimate (5.19) reduces to
(5.20)
ju.x/ uh .x/j C h.h˛ C h log.h1 // C hp / :
The best convergence rate that can be obtained with the above estimate is for ˛ 1
and p 2, in which case the asymptotic convergence rate is h2 log.h1 /. We shall
see in Section 8 that this estimate seems to be optimal (modulo logarithmic factors)
except for certain types of the discrete delta functions. We shall come back to this
point at the end of Section 8.
6 Global Error Estimate
We now turn to a global error estimate valid everywhere including points at or
close to the immersed boundary. Such bounds are important not least because the
velocity field of the immersed boundary solution is interpolated at the immersed
boundary points. In order to obtain this bound, we require some geometric constructs. We assume in the rest of this section that X is at least C 2 as a function of
and that j@X=@j ¤ 0.
We introduce a local coordinate system in the vicinity of the curve (Figure 6.1):
(6.1)
D fx j dist.x; / minjx X. /j < d g:
We must specify d as well as the local coordinate on .
IMMERSED BOUNDARY METHOD PROOF
1235
Since is a simple closed curve, it separates U into two regions. Given the
tangent vector @X=@ ¤ 0 on , we let U1 be the region to the left of this tangent
Q
vector and U2 to the right. Define the signed-distance function :
8
ˆ
if x 2 U1 ;
<dist.x; /
(6.2)
.x/
Q
D 0
if x 2 ;
:̂
dist.x; / if x 2 U2 :
Denote by 0 the set of points (level set) for which Q D 0 . We note 0 D .
There is a real number d > 0 such that if j
0 j < d , 0 is a simple closed curve.
We require that d < d .
Now define the local coordinate system as a map Z from „ to :
Z./ D .Z.; /; W .; //T
(6.3)
D X./ C n./;
n./ D
1
@Y =@
;
j@X=@j @X=@
„ D f D .; / j 2 R=.2Z/; j
j < d g:
The vector n is the unit normal on . The image of D 0 by the map Z is the
curve . This coordinate map can be extended to include j
j D d . We compute
the Jacobian and require that it be positive and bounded from above:
ˇ ˇ
ˇ @X ˇ
@Z
(6.4)
det JZ ./ D ˇˇ ˇˇ . /; JZ ./ D
;
@
@
where . / is the curvature of at X. / and is therefore bounded so long as X is
a C 2 function of .
We now let
1 min j@X=@j
0
0
;
; d ; dU /; d
D
(6.5)
d D min.d
2 max ./
where dU , introduced earlier, is the distance between and the boundary @U of
the computational domain. With this definition of d , the map Z defines a local
coordinate system on . Within , may be identified with the signed-distance
function .
Q We see that
ˇ ˇ
ˇ ˇ
ˇ @X ˇ
ˇ @X ˇ
3
1
(6.6)
0 < minˇˇ ˇˇ det JZ ./ maxˇˇ ˇˇ:
2 @
2 @
We can also compute
(6.7)
ˇ ˇ
ˇ @X ˇ
3
jJZ ./j2 maxˇˇ ˇˇ C1 ;
2 @
jJZ1 ./j2
ˇ ˇ1
ˇ @X ˇ
1
C2 ;
minˇˇ ˇˇ
2 @
where j j2 denotes the matrix norm induced by the Euclidean distance. Note that
C1 and C2 are constants that do not depend on .
We start with the following observation:
1236
Y. MORI
L EMMA 6.1 Let w be as in Lemma 4.3. Then, for x 2 U,
8
d
1 p
<
if jx X. /j < d ;
1
sin.d =.2 2// jxX./j
(6.8)
1 p
w.x X.// :
if jx X. /j d :
sin.d =.2 2//
P ROOF : Let y D .x 0 ; y 0 /T D .x X. /; y Y.//T D x X. /. We first note
that
(6.9)
w.y/ D max.sin.x 0 =2/; sin.y 0 =2// sin.max.x 0 =2; y 0 =2// sin.wm /:
wm is bounded by
(6.10)
jyj
2wm max.x 0 ; y 0 / min.max.jyj cos. /; jyj sin. /// D p :
2
Combined with a trivial bound, we have
(6.11)
jyj
jyj
:
p wm 2
2 2
Suppose jyj d . Equation (6.11) implies
(6.12)
d
wm p :
2 2
By definition (6.5) of d , X./ is at least a distance d away from @U. Recalling
that U has side 2, 2wm D max.x 0 ; y 0 / D max.x X. /; y Y.// 2 d .
Combined with (6.12),
(6.13)
Thus, we have
(6.14)
d
d
:
p wm 2
2 2
d
w sin.wm / sin p :
2 2
On the other hand, suppose jyj < d . Equation (6.11) implies
(6.15)
wm <
d
<
2
2
where we used d < , as should be geometrically clear. We have
p sin.d =.2 2//
jyj
jyj:
(6.16)
w sin.wm / sin p
d
2 2
The first inequality follows from (6.15) and (6.11), whereas the second inequality
follows from the convexity of the sine function for .0; /. Inequalities (6.14) and
(6.16) produce the desired result.
IMMERSED BOUNDARY METHOD PROOF
1237
To state the next lemma, we must introduce some notation. Take any point
x D Z. 0 / D .Z.0 ; 0 /; W .0 ; 0 //T in . Denote the closest point from x on as X? D Z.0 ; 0/ (Figure 6.1). Draw an open disc of radius d centered at X? ,
which we shall call D0 . Note that D0 . Take any point X.1 / D Z. 1 / D
.Z.1 ; 0/; W .1 ; 0// 2 D0 .
L EMMA 6.2 For x, X.1 /, 0 , and 1 introduced above, we have
(6.17)
jx X.1 /j D jZ. 0 / Z. 1 /j CZ j 0 1 j
where CZ does not depend on x, X.1 /, 0 , or 1 .
P ROOF : Draw a straight line L between x and X.1 /. This line is wholly contained in D0 since a disc is convex. Thus the inverse image Z1 .L/ is
contained in „ and is a curve that connects 0 and 1 . Let the length of Z1 .L/
be ƒ and introduce an arc length coordinate :
(6.18)
./ D ../; . //T ;
.0/ D 0 ;
.ƒ/ D 1 :
Since a straight line is shorter in length than a curve, we have
(6.19)
j 0 1 j ƒ:
The line L can also be parametrized with , and its length jLj calculated in terms
of an integral in :
ˇ
ˇ
Z ƒˇ
Z ƒˇ
ˇ d Z.. // ˇ
ˇ d ˇ
ˇ
ˇJZ
ˇd D
ˇ
(6.20)
jLj D
ˇ
ˇ d ˇd
ˇ
d
0
0
where JZ D @Z=@. We can obtain the following lower bound on the integrand”:
ˇ
ˇ
ˇ d ˇ
jJZ vj
jvj
ˇJZ
ˇ
D min 1
ˇ d ˇ min
v¤0 jvj
v¤0 jJZ vj
(6.21)
1
1
D 1 C > 0
1
D
maxv¤0 jJZ vj=jvj
jJZ j2
where we used jd =d j D 1 ( is the arc length coordinate) in the first inequality
and (6.7) in the second. Substituting the above back into the integral,
(6.22)
jLj Cƒ:
Combine the above with (6.19) to obtain the desired inequality.
Now we are ready to prove the following global error estimate:
T HEOREM 6.3 Suppose F is C 1 and X is C 2 as a function of , and is of moment
order at least 1. Assume also that j@X=@j never vanishes. For sufficiently small h
and the following global error estimate is valid:
(6.23)
ku uh kL1 .U/ C.h C / log.h1 / C log../1 /
where C is a constant that does not depend on h or .
1238
Y. MORI
P ROOF : Suppose x 2 U n . Then, the distance from x to is at least d . We
may thus apply inequality (5.16):
ju.x/ uh .x/j C.h C /
(6.24)
where the constant C depends only on d , F, X, and its derivatives, and not on
x 2 U n , h, or .
Next suppose that x 2 . Without loss of generality, we assume that the coordinate that corresponds to x is D 0. That is to say, x D Z. 0 / D Z.0; 0 /
for some 0 < d . We let X? D Z.0; 0/ D X.0/, the closest point on from x.
We first write the difference between u and uh as follows: Define the following
subset of :
D D f j jX? X. /j < d g:
(6.25)
Then
u uh D Ia C Ib ;
Z
(6.26)
Ia D
Gx .X. //F./d .S .X/Gh;x /.m /F.m /;
m …D
…D
Z
Ib D
X
Gx .X.//F./d X
.S .X/Gh;x /.m /F.m /:
m 2D
D
We thus divided the difference into regions of where jX? X./j < d or not.
Denote by a the portion of that corresponds to … D. Ia can be written as
Ia D v vh ;
Z
(6.27)
vD
Gx .X. //F. /d;
X. /2a
vh D
X
.S .X/Gh;x /.m /F.m /:
X.m /2a
Any point on a is at least a distance d away from the three points Z.0; d /,
Z.0; d /, and X? D Z.0; 0/, as can be seen from the definition of the region
. Elementary
geometric arguments show that the distance between a and x is at
p
least . 3=2/d . We can thus apply Corollary 5.2 to find that
(6.28)
jIa j C.h C /
where the constant C does not depend on x 2 , h, or .
IMMERSED BOUNDARY METHOD PROOF
1239
The task that remains is to bound Ib . Define the subset of as
(6.29)
p
V D f j j j ; max.; . 2 r h=CZ / C h/g
where CZ is the same CZ that appears in Lemma 6.2. We write Ib as
Ib D Ib1 C Ib2 ;
Z
Gx .X.//F. /d Ib1 D
(6.30)
.S .X/Gh;x /.m /F.m /;
m 2DnV
DnV
Z
Ib2 D
X
Gx .X.//F./d X
.S Gh;x /.X.m //F.m /:
m 2V
V
We start by deriving bounds for Ib1 . Write Ib1 as
R
Ib1 D Ib1 C Iıb1 C IG
b1 ;
Z
R
Ib1 D
Gx .X.//F./d Iıb1 D
X
Gx .X.m //F.m /;
m 2DnV
DnV
(6.31)
X
.Gx .X.m // .S .X/Gx //.m //F.m /;
m 2DnV
IG
b1 D
X
.S .X/.Gx Gh;x //.m /F.m /:
m 2DnV
We first introduce some notation. Let D n V be characterized by
(6.32)
D n V D f j < < ; < < C g:
Consider the region < < C . We let the smallest and the largest m that belong
C to be
to this region to be P and Q , respectively. Define Qm
(6.33)
C
Qm
D C .m P /
C
C
so that Qm
m < Qm
C :
1240
Y. MORI
R
R
R
C
We start by estimating Ib1 . Denote the portion of Ib1 where > 0 as Ib1 . This
can be written as
R
Z
C
Ib1 D
C
Z
J1m D
J D
J1m ;
mDP
C
QmC1
Gx .X. //F./d Gx .X.m //F.m /;
C
Qm
Z
0
Gx .X.m //F.m /
mDP
Q
X
D J0 C
(6.34)
Q
X
Gx .X. //F. /d C
C
QQC1
Gx .X.//F. /d:
C
J0 D .J10 ; J20 /T can be bounded as follows: Suppose QQC1
< C . Then
jJi0 j
(6.35)
ˇZ C X
ˇ
2
ˇ
ˇ
ˇ
ˇ
Gx;ij .X.//Fj . /d ˇˇ
QC
QC1 j D1
max
2
X
jGx;ij .X.//Fj . /j
C
C j D1
QQC1
C
max
C
C
QQC1
log.jx X. /j1 / C 1
p
where we used Lemma 4.1 in the last line. Note that jx X. C /j . 3=2/d by
the argument preceding equation (6.28). Thus,
(6.36)
p
ˇ ˇ
ˇ @X ˇ
3
d
jx X./j max
d maxˇˇ ˇˇ :
C
2
@
2
C
QQC1
C
C j . The last inequality above holds for sufficiently small
We used jQQC1
. This shows that
(6.37)
jJi0 j C:
C
> C case follows similarly.
The QQC1
IMMERSED BOUNDARY METHOD PROOF
1241
1 ; J 1 /T .
We now turn to J1m D .Jm;1
m;2
1
jJm;i
j
ˇ
ˇX
ˇ 2 @ ˇ
ˇ
./
max
Gx;ij .X. //Fj ./ ˇˇ
ˇ
C
C
@
Qm
QmC1
2
j D1
./
(6.38)
2
2
X
j D1
ˇ
ˇ
ˇ @Gx;ij .X. // ˇ
ˇ
ˇjFj ./j
max
ˇ
ˇ
C
C
@
Q
Q
m mC1
ˇ
ˇ
ˇ @Fj ./ ˇ
ˇ
ˇ :
C jGx;ij .X. //jˇ
@ ˇ
From Lemma 4.1, we have
ˇ
ˇ
ˇ
ˇ
ˇ @X. / ˇ
ˇ @Gx;ij .X.// ˇ
1
ˇ
ˇ C.jx X. /j C 1/ˇ
ˇ
ˇ @ ˇ;
ˇ
ˇ
@
jGx;ij .X.//j C log.jx X./j1 / C 1 :
(6.39)
We thus have
1
j C./2
jJm;i
(6.40)
C./2
max
C
C
QmC1
Qm
max
C
C
QmC1
Qm
jx X./j1 C log.jx X./j1 / C 1
jx X./j1
where C does not depend on x 2 , h, or . In the last inequality above, we
discarded the logarithmic and constant terms since they can be dominated by jx X./j1 for jx X./j < d by taking the constant C in the last line sufficiently
large:
Q
X
1
jJm;i
j
C. /
2
mDP
(6.41)
Q
X
mDP
C. /
2
Q
X
mDP
C
Q
X
mDP
max
1
jx X./j
max
1
q
CZ 2 C 20
C
C
QmC1
Qm
C
C
QmC1
Qm
q
C 2
.Qm
/ C 20
where we used Lemma 6.2 in the second inequality.
1242
Y. MORI
We now estimate the last sum:
Q
X
mDP
q
C 2
.Qm
/ C 20
QP
X
mD0
C m
Z dr
C
Cr
0
D
C log
C.1 C log. 1 //
(6.42)
where we used C .Q P / C in the second inequality and (see equation (6.29)) in the last inequality.
R
C
get
Combining (6.37), (6.41), and (6.42) to estimate the i th component of Ib1 , we
(6.43)
ˇ
ˇ
Q
Q
X
X
ˇ R Cˇ ˇ 0
ˇ
1 ˇ
0
1
ˇI ˇ D ˇJ C
Jm;i ˇ jJi j C
jJm;i
j
ˇ i
b1;i
mDP
mDP
C C C.1 C log. 1 //:
R
We can obtain the same bound for the < 0 part of Ib1 , and we have the estimate
ˇR ˇ
ˇI ˇ C.1 C log. 1 //
(6.44)
b1
where C does not depend on x 2 , h, or .
We turn next to Iıb1 . First consider the difference in the sum:
(6.45)
jGx;ij .X. // .S .X/Gx;ij /. /j
max j@˛x @yˇ Gx .y/j h
max
˛Cˇ D1 y2RX./
max
y2RX. /
Ch
Ch
D
jx yj
miny2RX. / jx yj
where we used Lemma 3.2 in the first inequality and Lemma 4.1 in the second. We
obtain a lower bound on the denominator of the last expression:
p
min jx yj jx X. /j 2 r h
y2RX. /
q
p
2
2
C
2 r h
(6.46)
Z m C 0 p
2 r h
CZ jm j CZ .jm j . h// :
CZ
In the first inequality, we used the fact that RX./ is contained in a closed disc
p
centered at X./ of radius 2 r h. We used Lemma 6.2 in the second inequality.
IMMERSED BOUNDARY METHOD PROOF
1243
p
In the last inequality, we used h 2 r h=CZ as follows from the definition
of , (6.29). Note that the last expression is greater than 0 since jm j > .
ı
ı
Combining (6.45) and (6.46), we can estimate Iıb1 D .Ib1;1
; Ib1;2
/T :
ı
j
jIb1;i
(6.47)
X
2
X
jGx;ij .X.m // .S .X/Gx;ij /.m /j jFj .m /j
m 2DnV j D1
X
2
X
m 2DnV j D1
X
C hjFj .m /j
Ch
CZ .jm j . h//
m 2DnV
:
jm j . h/
Consider the m > 0 part of the last sum.
Q
X
(6.48)
mDP
Z QC1
d
C
m . h/
P . h/
.
h/
P
Z d
1
C
C
C log.h / :
h
h
. h/
We obtain the same estimate for m < 0 part of the sum. Substituting this back
into (6.47),
(6.49)
jIıb1 j C. C h log.h1 //
where C is a constant that does not depend on x 2 , h, or .
We now estimate IG
. We estimate the difference in the sum:
b1
jS .X/.Gx;ij Gh;x;ij /./j C max jGx;ij .y/ Gh;x;ij .y/j
y2RX. /
(6.50)
max
y2RX./
Ch
Ch
max
w.x y/ y2RX./ jx yj
where we used Lemma 3.1 in the first inequality, Proposition 4.5 in the second, and
Lemma 6.1 in the third. This estimate has the same form as equation (6.45) and
thus allows us to proceed in the same way as when we bounded Iıb1 :
(6.51)
1
jIG
b1 j C. C h log.h //
where C is a constant that does not depend on x 2 , h, or .
Combining (6.44), (6.49), and (6.51) we have
(6.52)
jIb1 j C. C log. 1 / C h log.h1 //
where C is a constant that does not depend on x 2 , h, or .
1244
Y. MORI
Finally, we must bound Ib2 . We separate this as follows:
R
P
Ib2 D Ib2 C Ib2 ;
Z
R
(6.53)
Ib2 D
Gx .X.//F./d;
V
P
Ib2 D R
X
.S .X/Gh;x /.m /F.m /:
m 2V
R
R
First, we bound Ib2 D .Ib2;1 ; Ib2;2 /T :
Z
ˇ R ˇ
ˇI
ˇ
b2;i
2
X
Z
(6.54)
jGx;ij .X. //j jFj . /jd
j D1
C
Z
C
log jx X./j1 C 1 d
.log.1=j j/ C 1/d C .1 C log. 1 //
where we used Lemma 4.1 in the first inequality. In the
q second inequality, we used
Lemma 6.2 from which we see that jx X. /j CZ 2 C 20 CZ j j.
P
P
P
Next we bound Ib2 D .Ib2;1 ; Ib2;2 /T . This yields
2
X X
ˇ P ˇ
ˇI
ˇ
j.S .X/Gh;x;ij /.m /j jFj .m /j
b2;i
m 2V j D1
(6.55)
X
C log.h1 /
m 2V
where we used Lemma 3.1 and Lemma 4.2 in the last inequality. Since V spans
, the number of terms in the last sum above is at most 1 C 2=.
Thus
ˇ P ˇ
ˇI
ˇ C. C 2 log.h1 //:
(6.56)
b2;i
Combining (6.55) and (6.56), we have
(6.57)
jIb2 j C C C log. 1 / C log.h1 /
where C is a constant that does not depend on x 2 , h, or .
We now only have to collect results. From estimates (6.28), (6.52), and (6.57)
we can estimate the difference (6.26):
ju.x/ uh .x/j jIa j C jIb1 j C jIb2 j
(6.58)
C.h C / 1 C log.h1 / C log.. /1 /
IMMERSED BOUNDARY METHOD PROOF
1245
where x 2 and the constant C does not depend on x, h, or . We used
C.h C / and log. 1 / C.log.h1 / C log../1 // as follows from the
definition of in (6.29). We can combine the above with the estimate (6.24) valid
for x … to obtain
(6.59)
ju.x/ uh .x/j C.h C / log.h1 / C log.. /1 /
where x 2 U and the constant C does not depend on x, h, or . We have used
the fact that a constant term is dominated by log.h1 / or log../1 / for small
enough h or . We can now take the supremum on the left-hand side of the above
to obtain the desired estimate.
We can deduce some interesting conclusions from the above theorem. Refine
proportionally to h˛ where ˛ is positive. According to the above theorem:
(6.60)
ku uh kL1 .U/ C.h C h˛ / log.h1 /:
This tells us that the immersed boundary solution converges to the true solution
everywhere for any ˛ > 0. Any ˛ < 1 would lead to suboptimal convergence.
If ˛ > 1, the asymptotic rate of convergence would be the same as if ˛ D 1, but
would be a waste of computational resources since ˛ > 1 means more immersed
boundary points compared to the ˛ D 1 case. We may combine these observations
into a corollary.
C OROLLARY 6.4 Refine proportionally to h˛ . The immersed boundary solution uh converges to the true solution u everywhere in the domain U for any ˛ > 0.
If in particular is refined proportionally to h, we have the error estimate
(6.61)
ku uh kL1 .U/ C h log.h1 /
where the constant C does not depend on h.
7 A Simple Dynamic Problem
As an application of the foregoing estimates, we consider a small-amplitude
dynamic problem. We let
u.x; t/ D rp.x; t / f.x; t / C g.x; t /; r u.x; t / D 0;
Z (7.2) f.x; t / D
F.; t/ı.x X0 . //d; F.; t / D K.X.; t / X0 . //;
Z
@X.; t/
(7.3)
D u.X0 ./; t / D u.x; t /ı.x X0 . //d x:
@t
(7.1)
U
The force law we consider is a simple restoring force to the fixed locations X0 .
Assuming that the immersed boundary points do not deviate significantly from the
locations X0 , we spread the forces at fixed locations X0 and interpolate the velocity
at the same fixed locations. This is analogous to the linear approximation made in
1246
Y. MORI
the theory of small-amplitude water waves [35]. In the form of an integral, the
above system can be written as
@X
.; t/ D .LF/.; t / u.X0 ./; t /;
@t
Z
(7.4)
u.x; t / D .QF/.x; t / G.x X0 . //F.; t /d;
F.; t/ D K.X.; t / X0 . //:
A discretization of the above using the immersed boundary method is the following:
(7.5)
Lh unh .x/ D Dh phn .x/ fnh .x/ C gnh .x/;
(7.6)
fnh .x/ D .S.X0 /Fnh /.x/;
(7.7)
D t Xnh .m /
D .S
Dh unh .x/ D 0;
Fnh .m / D K.Xnh .m / X0 .m //;
.X0 /unh /.m /;
D t Xnh .m /
.m /
Xnh .m / Xn1
h
t
:
The superscript n refers to the immersed boundary approximation at time t D nt .
S.X0 / and S .X0 / denote the spreading and interpolation operators, respectively,
performed at the points X0 .m /. We have used a backward Euler discretization for
the time derivative.
The above can be written in the following form, which parallels (7.4):
D t Xnh .m / D .Lh Fnh /.m / .S .X0 /unh /.m /;
(7.8)
unh .x/ D .Qh Fnh /.x/ X
Gh .x y/.S.X0 /Fnh /.y/;
y2Gh
Fnh .m /
D
K.Xnh .m /
X0 .m //:
For simplicity, in this section, we shall assume that is refined proportionally
to h. We start with the following lemma:
L EMMA 7.1 Suppose .QF/.x/ is a Lipschitz-continuous function in x. Then
(7.9)
maxj.LF/.m / .Lh F/.m /j C h log.h1 /
m
where C is a constant independent of h and .
IMMERSED BOUNDARY METHOD PROOF
1247
P ROOF : Let v.x/ D .QF/.x/ and vh .x/ D .Qh F/.x/. We have
j.LF/.m / .Lh F/.m /j jv.X0 .m // .S .X0 /vh /.m /j
E1 C E2 ;
(7.10)
E1 D jv.X0 .m // .S .X0 /v/.m /j;
E2 D j.S .X0 /.v.x/ vh .x///.m /j:
Given that v.x/ D .QF/.x/ is Lipschitz-continuous and is of moment order at
least 1,
E1 D jv.X0 .m // .S .X0 /v/.m /j C h
(7.11)
where we used (3.18). On the other hand,
E2 D j.S .X0 /.v.x/ vh .x///.m /j C
(7.12)
max
x2RX0 .m /
jv.x/ vh .x/j
where we used Lemma 3.1 in the inequality above. We know from Corollary 6.4
that
kv vh kL1 .U/ D k.QF/ .Qh F/kL1 .U/ C h log.h1 /:
(7.13)
Therefore,
(7.14)
E2 C
max
x2RX0 .m /
jv.x/ vh .x/j C kv vh kL1 .U/ C h log.h1 /:
Combining (7.11) and (7.14), we obtain the desired result.
We note that the velocity field u D QF should be Lipschitz-continuous in x if
F and X0 are smooth enough. After all, if u is a classical solution to the Stokes
problem, u is continuous at the immersed boundary and its derivatives satisfy jump
conditions whose values are given in terms of F and X0 [26]. Note also that the
above lemma is of independent interest. It states that the interpolated velocity field
using S converges to the true solution.
We now turn to the next lemma.
L EMMA 7.2 The matrix Lh is symmetric positive semidefinite.
P ROOF : The operator Lh acting on F can be written as
X
(7.15)
Lh F D S .X0 /
Gh .x y/.S.X0 /F/.y/:
y2Gh
Since Gh is a symmetric positive semidefinite matrix (see equation (2.17)) and
S.X0 / and S .X0 / are transposes of each other, it follows that Lh is symmetric
positive semidefinite.
Define the discrete L2 norm for functions defined on the immersed boundary
grid:
X
1=2
M
2
jX.m /j :
(7.16)
kXkL2 .‚/ D
mD1
1248
Y. MORI
We can now prove the following theorem:
T HEOREM 7.3 Suppose X is a smooth function in t and that and X0 satisfy
the conditions of Theorem 6.3. Then the difference between X and the immersed
boundary approximation Xh satisfies the following bound:
(7.17)
kX. ; T / Xnh kL2 .‚/ C T .t C h log.h1 //;
T D nt;
where the constant C does not depend on t or h.
P ROOF : The idea of the proof is that Lemma 7.1 gives us consistency and
Lemma 7.2 gives us the requisite stability.
Let Xn ./ D X.; nt/. We first observe
@X
.m ; nt / C Jnt .m /; jJnt .m /j Ct;
@t
where C does not depend on m, n, h, or t . Next, we see that
(7.18)
.D t Xn /.m / D
(7.19) .Lh K.Xn X0 //.m / D .LK.Xn X0 //.m / C Jnh .m /;
jJnh .m /j C h log.h1 /;
where we used Lemma 7.1. The constant C above does not depend on m, n, h, or
t. Noting that
@X
.m ; nt / D .LK.Xn X0 //.m /;
@t
we immediately conclude that
(7.20)
(7.21)
.D t Xn /.m / D .Lh K.Xn X0 //.m / C Jnt .m / C Jnh .m /:
Since Xnh satisfies the above discrete evolution equation exactly, we have
(7.22)
.D t En /.m / D K.Lh En /.m // C Jnt .m / C Jnh .m /;
En Xn Xnh ;
kE0 kL2 .‚/ D 0:
Define the discrete inner product:
(7.23)
.U; V/ D
M
X
.U.m /; V.m //
mD1
where . ; / denotes the Euclidean inner product. Taking the discrete inner product on both sides of equation (7.22),
(7.24)
.En ; D t En / D K.En ; Lh En / C .En ; Jnt C Jnh / :
Using Lemma 7.2, we see that the first term on the right-hand side is nonpositive.
Dropping this term and substituting the definition of D t , we find
(7.25)
2
.En ; En1 / t .En ; Jnt C Jnh / :
kEn kL
2
.‚/
IMMERSED BOUNDARY METHOD PROOF
1249
Using the Cauchy-Schwartz inequality and dividing both sides by kEn kL2 .‚/ ,
kE kL2 .‚/ kE
n
(7.26)
n1
kL2 .‚/ t kJnt
C
Jnh kL2 .‚/
t C.t C h log.h1 //:
Since kE0 kL2 .‚/ D 0, we immediately conclude that
(7.27)
kEn kL2 .‚/ T C.t C h log.h1 //:
8 Computational Demonstration
We take the stationary model problem for which we proved convergence and
perform computational experiments.
We shall examine the following six different discrete delta functions. We mainly
follow [11] in the choice of discrete delta functions used to examine convergence.
The simplest one we use is the following:
(
1
; 1 r < 1;
(8.1)
ch .r/ D 2
0 otherwise:
This function is discontinuous and never used in practice, but is nonetheless included here as an example of a discrete delta function that satisfies only one moment condition.
The next function we consider is the hat function:
(
min.r C 1; r 1/; jrj 1;
(8.2)
H .r/ D
0
otherwise:
We shall also use the wider hat function:
(
1
min.r C 2; r 2/;
(8.3)
wH .r/ D 4
0
jrj 2;
otherwise:
When the support of the delta function is 4, one can accommodate up to four moment conditions [36], and the following is one such function [21, 36, 38]:
8
1
2
3
ˆ
jrj < 1;
<1 jrj jrj C 2 jrj ;
c
11
1
2
3
(8.4)
.r/ D 1 6 jrj C jrj 6 jrj ; 1 < jrj 2;
:̂
0:
2 < jrj:
In [25], is derived systematically by imposing certain conditions to be satisfied by including the conditions we discussed in Section 3. Two functions that
satisfy these conditions to varying degrees are the functions IB and IB6 . IB
is defined as
p
81
.3
2jrj
C
1 C 4jrj 4r 2 /;
jrj 1;
ˆ
<8
p
IB
1
2
(8.5)
.r/ D 8 .5 2jrj 7 C 12jrj 4r /; 1 < jrj 2;
:̂
0;
2 < jrj:
1250
Y. MORI
IB6 first appeared in [32]:
(8.6)
IB6 .r/ D
8
ˆ
1IB6 .r/;
jrj 1;
ˆ
ˆ
ˆ
ˆ
< 21 C 7 jrj 7 jrj2 C 1 jrj3 3 IB6 .jrj 1/; 1 < jrj 2;
16
12
8
6
2 1
9
23
3
1
2
3
ˆ jrj C jrj jrj C 1 IB6 .jrj 2/; 2 < jrj 3;
ˆ
ˆ
4
12
2 1
ˆ 8 12
:̂0;
3 < jrj:
1IB6 .r/ D
2 1
3
61
11
11
112 42 jrj 56 jrj 12 jrj
p C 3363 243 C 1584jrj 748jrj2
1560jrj3
1
C 500jrj4 C 336jrj5 112jrj6 2 :
We let N D 128 2k , k D 0; 1; 2; 3. We compute the following quantities to
compute the convergence rate.
For a vector field w.x/ D .w1 .x/; w2 .x// defined on the Cartesian fluid domain U, we define the discrete Lp norm as follows:
1=p
X
2
2
p=2 2
kwkp D
(8.7)
.w1 .x/ C w2 .x// h
; 1 p < 1;
x2Gh
(8.8)
kwk1 D max.w12 .x/ C w22 .x//1=2 :
x2Gh
Let uN be the computed velocity field for the N N mesh. We define a measure
of error epN as follows:
(8.9)
epN D kuN I 2N !N u2N kp :
Here I 2N !N is an interpolation operator from the finer to the coarser grid. As an
empirical measure of global convergence rate, we use
(8.10)
rpN D log2 .epN =ep2N /:
In order to examine the local convergence rate away from the immersed boundary , we compute the following quantity at all points x on the N N mesh that
are at least two mesh widths away from the support of the discrete delta functions:
N
ju .x/ u2N .x/j
N
(8.11)
.x/ D log2
:
ju2N .x/ u4N .x/j
We note that this quantity can be arbitrarily large or small at certain locations since
there are places where the velocity field u vanishes. We list the average of this
quantity NN as well as its mean deviation N .
IMMERSED BOUNDARY METHOD PROOF
2r
ch
2
H
2
wH
4
c
4
IB
4
IB6 6
p
1
2
2
4
2
4
r1256
1.01
1.95
1.98
1.86
1.98
2.00
r2256
1.16
1.50
1.51
1.48
1.50
1.49
1251
256 N 256 256
r1
0.95 0.94 0.83
0.97 2.02 0.26
0.98 2.00 0.03
0.99 2.20 0.90
0.98 2.00 0.01
0.95 3.50 0.81
TABLE 8.1. Convergence rates for the stationary immersed boundary
method for different choices of discrete delta function. 2r denotes the
width of the support, and p the moment order. In these computations,
M D 4N so that is refined proportionally to h.
F IGURE 8.1. Velocity field computed with the immersed boundary
method for the model problem. A 32 32 mesh was used to generate
this figure.
We must specify X./ and F. /. We tested various geometries and force functions. To generate Table 8.1, we used
.6 C cos.3 // cos. /
1 C sin. /
(8.12)
X./ D
; F. / D
:
1 C cos./
12 .6 C sin.3 // sin. /
We let M D 4N and M D 2 so that is refined proportionally to h. The
velocity field as computed with this method is shown in Figure 8.1.
N , N N , and N for N D 256. The values at
In Table 8.1, we list r1N , r2N , r1
N D 128 were similar to those shown in the table. First of all, we see that the
1252
Y. MORI
2r
ch
2
H
2
wH
4
c
4
IB
4
IB6 6
p
1
2
2
4
2
4
r1256
0.80
1.95
1.98
1.86
1.99
2.00
r2256
1.17
1.50
1.51
1.48
1.50
1.49
256 N 256 256
r1
0.89 0.36 0.87
1.00 2.03 0.26
0.98 2.00 0.03
1.00 2.16 0.87
0.98 2.00 0.01
0.95 3.50 0.81
TABLE 8.2. Convergence rates for the stationary immersed boundary
method for different choices of discrete delta function. 2r denotes the
width of the support, and p denotes the moment order. In these computations, M D N 2 =32 so that is refined proportionally to h2 .
L1 convergence rate is always invariably close to 1. This is in agreement with
the global error estimate we presented in Corollary 6.4. Except for ch , whose
moment order is 1, the L1 error decays approximately at a second-order rate and
the L2 error at a rate of about 1:5.
This phenomenon can be explained as follows: We showed in Theorem 5.3 that
the local convergence rate away from the boundary is 2 at generic points if we
refine proportionally to h and if the discrete delta function is of moment order
at least 2. Thus, over most of the computational domain, the pointwise error must
be proportional to h2 . At points within distance O.h/ of the immersed boundary
, however, the error is proportional to h, as given by Corollary 6.4. Thus the
L1 error will likely behave as
(8.13)
C h2 jUj C C h.hjj/ D O.h2 /
where jUj denotes the area of the fluid domain U and jj the length of the curve
. For the L2 norm, we have
(8.14)
.C.h2 /2 jUj C C h2 .hjj//1=2 D O.h3=2 /:
The average local error N follows the trend predicted in Section 5. We see
approximately second-order convergence on average for discrete delta functions of
moment order 2 or higher.
In Table 8.2, we have listed the convergence rates for M D N 2 =32 so that is proportional to h2 . In agreement with our theory, we do not see any significant
change in the order of convergence from Table 8.1.
We make two observations that cannot be fully understood with our error analysis. The first is the behavior of IB6 . The local convergence rate away from the
immersed boundary is approximately order 3:5 in Tables 8.1 and 8.2. For different
choices X and F we have seen local convergence rates of 3:3 3:7. Although this
behavior is consistent with our error estimate, our analysis cannot explain convergence rates above order 2. This behavior cannot be due solely to the fact that IB6
IMMERSED BOUNDARY METHOD PROOF
1253
is of moment order 4, since c , which is also of moment order 4, exhibits only approximately second-order local convergence in line with our predictions. We also
note that c exhibits highly erratic behavior as is evidenced by the relatively high
-value. For different choices of X and F, N for c ranges from approximately
1:5 to 2:5.
The other observation is that there are some fine details in the convergence
properties that are different even among discrete delta functions of moment order 2.
IB and wH exhibit superior convergence properties in the sense that is close
to 0. This property was seen regardless of choice of X and F. On the other hand,
the -value for H is considerably higher.
A common feature of the discrete delta functions that exhibit better convergence properties is that they satisfy the following even-odd condition discussed
briefly in Section 3:
X
X
.n r/ D
.n r/
(8.15)
n even
n odd
for all r 2 R. We believe that this condition results in cancellation of highfrequency error components in the far field.
9 Conclusion
We proved convergence of the velocity field for a simple immersed boundary
method. We considered a stationary Stokes problem in a two-dimensional fluid domain with a closed one-dimensional immersed boundary curve. A key to proving
convergence of the velocity field was the estimate on the difference between the
discrete and continuous Green’s function. We applied this estimate to prove local
and global error bounds on the immersed boundary solution. The global error estimate ensures convergence of the immersed boundary solution up to the immersed
boundary points.
We have not dealt with the convergence of the pressure field. The pressure field
has a jump across the immersed boundary, and thus its convergence analysis may
be somewhat more delicate. We hope to extend our analysis in a future paper to
resolve this issue.
We used a spectral method to discretize the Stokes problem. In a forthcoming
paper, we plan to generalize the results obtained in this paper to other differencing
schemes.
The immersed boundary method uses a finite difference grid for the fluid domain, as was the case in this paper. There is recent work that combines some
features of the immersed boundary method with finite element discretizations of
the fluid domain [3, 13, 40, 41]. Such methods may be easier to study theoretically
given that they are constructed within a variational framework for which powerful
analytical tools are available. It would be of interest to see what bearing the present
analysis may have on such methods, or whether the tools of finite element analysis
1254
Y. MORI
may lead to a better understanding of finite-difference-based immersed boundary
methods.
We hope to generalize our analysis to three-dimensional problems. Our analysis, especially near the immersed boundary points, was facilitated by the weak
logarithmic singularity of the Stokeslet in two dimensions. In three dimensions the
singularity of the Stokeslet is much stronger than in two dimensions.
We tested our error estimates against computational experiment. The overall
features of the convergence rates are well explained by our theory. The global L1
convergence rate is order 1 regardless of the choice of discrete delta function as
long as is refined proportionally to h, as indicated by our global error estimate.
The local convergence rates are consistent with our theory, although some observations could not be fully understood. We believe that this is linked to the even-odd
condition (8.15). A full explanation will be sought in a future study.
We applied results from the stationary problem to prove convergence of the
immersed boundary solution for a small-amplitude dynamic problem. We have
only considered the simplest possible elastic forcing. Typically, the elastic forcing
involves derivatives in the Lagrangian coordinate. In relation to such problems, we
point to [31, 34], where the authors demonstrate stability of a continuous version
of such a problem for a straight filament. Such problems are yet to be studied for
their discrete analogues.
One crucial aspect of the immersed boundary method that we did not touch
upon is that the immersed boundary moves. We only proved convergence for a
dynamic problem in which the spreading and interpolation operations were performed at fixed locations. A convergence analysis for when the boundary moves is
completely open.
Convergence in the case of time-dependent Stokes or Navier-Stokes flow was
not dealt with here. We have only dealt with creeping flow, in which case we could
take full advantage of the instantaneous smoothing effects of the inverse Laplacian.
All such questions serve as directions for future investigation.
Appendix A: Proof of Lemma 4.4
Note by assumption that neither ˛ nor ˇ that appears in (4.15) is equal to 1
L
since 0 < jxj < 2 and 0 < jyj < 2. We write GK;ij
as in (4.22) and substitute
expression (4.17). First, rewrite equation (4.17) as
l
D ˇl
SP;Q
(A.1)
Q
X
˛ P P l ˛ QC1 Ql
C
1˛
kDP C1
kl D kl k1;l ;
kl D
˛k ˇl
kl ;
1˛
1
Pk;ij :
jkj2
IMMERSED BOUNDARY METHOD PROOF
1255
Substituting the above into (4.22), we find after rearrangement:
L
D IL C IK C J;
GK;ij
(A.2)
L
X
ˇ l Ll
IL D
;
1˛
lDL
K
X
ˇ l Kl
IK D
;
1˛
J D
k2BQK
lDK
Ll D ˛ L L;l ˛ LC1 Ll ;
X ˛k ˇl
kl ;
1˛
L
Kl D ˛ P C1 P;l C ˛ P P l ;
L
where the subset BQK
of the k-plane is defined as
L
D fk j K < jk 12 j < L; K jlj Lg:
BQK
(A.3)
To each sum in (A.2) we apply the same summation-by-parts procedure as in
equation (4.16), but now in the index l. We find
(A.4)
IL D
ˇ L L;l ˇ LC1 Ll
C
.1 ˛/.1 ˇ/
L
X
lDLC1
ˇ l .Ll L;l1 /
:
.1 ˛/.1 ˇ/
Thus,
(A.5)
jL;L j C jLL j
jIL j C
j.1 ˛/.1 ˇ/j
L
X
lDLC1
jLl L;l1 j
:
j.1 ˛/.1 ˇ/j
First, we have
jLL j jL;L j C jLL j (A.6)
1
;
L2
and likewise for L;L . We also have
(A.7)
jLl L;l1 j jL;l L;l1 j C jLl L;l1 j
C
C
C
C 3
3:
3
L
jkjkDL
jkjkDL
We can substitute these estimates back into (A.5):
L
X
1
1
C
4
C
C
:
(A.8) jIL j 2
3
j.1 ˛/.1 ˇ/j L
L
j.1 ˛/.1 ˇ/j L2
lDLC1
Likewise, we have
(A.9)
jIK j 1
C
:
j.1 ˛/.1 ˇ/j K 2
To estimate J in (A.2), we proceed again as in Lemma 4.3. First, define the
sum
(A.10)
k
TP;Q
Q
X
˛k ˇl
kl :
D
1˛
lDP
1256
Y. MORI
Using the summation-by-parts argument we see that
Q
X
1
k
(A.11)
jTP;Q j D
jkl k;l1 j :
jP l j C jQl j C
j.1 ˛/.1 ˇ/j
lDP
By direct calculation, one can check that
(A.12)
jP l j C
C
;
3
jP j3
jkjkDP
jQl j C
;
jQj3
jkl k;l1 j C
;
jkj4
for some constant C . Thus,
(A.13)
k
jTP;Q
j
Q
X
C
1
1
1
D
C
C
:
j.1 ˛/.1 ˇ/j jP j3
jQj3
jkj4
lDP
We now write J in terms of T and take absolute values:
K
X
jJ j kDLC1
C
L
X
k
jTL;L
j
C
j.1 ˛/.1 ˇ/j
X
L
l
l
.jTL;K
j C jTK;L
j/
kDKC1
lDKC1
(A.14)
K
X
k
jTL;L
jC
kDLC1
C
8
C
j.1 ˛/.1 ˇ/j K 2
2
C
L3
X
L
k2BK
K
X
kDKC1
1
jkj4
X 1 2
C
K3
jkj4
L
k2BK
where we used (A.13) to estimate the above sum and L > K to get the last line.
We now estimate the last sum:
Z p2 L
X 1
C
1
C
r dr 2 :
(A.15)
4
4
jkj
r
K
K
L
k2BK
Putting this back into (A.14), we have
(A.16)
jJ j C
1
:
j.1 ˛/.1 ˇ/j K 2
Collecting (A.8), (A.9), and (A.16) and substituting this back into (A.2), we have
(A.17)
L
.x/j jGK;ij
K 2 j.1
C
:
˛/.1 ˇ/j
Noting that j.1 ˛/.1 ˇ/j D 4 sin.jxj=2/ sin.jyj=2/, we have the desired result.
IMMERSED BOUNDARY METHOD PROOF
x D .x0 ; y0 /
1257
Y02 D .x2 ; y0 /
Y01 D .x1 ; y0 /
y D y0
X01 D .x0 ; y1 /
D .X.1x /; Y.1x //
X02 D .x0 ; y2 /
x D x0
F IGURE B.1. Schematic diagram for the proof of Theorem 5.3.
Appendix B: Proof of Theorem 5.3
That nontangent points are generic is a direct consequence of Sard’s theorem
[22]. Given the continuous differentiable function X. /, the values x D X. / at
which @X=@ D 0 (critical values) form a set of measure 0 on the real line. The
same is true for y D Y . /, and thus nontangent points are generic.
We next prove that for a nontangent point x0 D .x0 ; y0 /, the coordinate lines
x D x0 and y D y0 intersect at only a finite number of points. For, assume
otherwise. Suppose x D x0 intersects at X0k D .X.kx /; Y .kx // D .x0 ; yk /,
k 2 N, where xk are distinct real numbers in Œ0; 2. Since the kx form a bounded
sequence in Œ0; 2, there is a subsequence such that Qkx ! , Qkx ¤ , for some
. Then
ˇ
@X ˇˇ
x0 x0
(B.1)
D lim x
D 0:
ˇ
@ D k!1 Q k
This shows that x D x0 is tangent to at x D X. / since j@X=@j ¤ 0 by
assumption. We thus have a contradiction. The same holds for the line y D y0 .
In order to prove the error estimate (5.19), we divide the difference between u
and uh as in equation (5.4). We shall derive an estimate for IG that is different
from
the ones derived in Theorem 5.1. We shall reuse estimates (5.5) and (5.8) for
R
I and Iı .
Let X0k D .x0 ; yk /, k D 1; : : : ; nx , and Y0k D .xk ; y0 /, k D 1; : : : ; ny , be the
points at which the coordinate lines x D x0 and y D y0 intersect (Figure B.1).
1258
Y. MORI
Let
y
y
(B.2) X0k D .X.kx /; Y.kx // D .x0 ; yk /;
Y0k D .X.k /; Y .k // D .xk ; y0 /:
Take the arc length coordinate s along .
the points X0k and Y0k , we have
ˇ
ˇ
ˇ @X ˇ
ˇ
ˇ
> 0;
(B.3)
ˇ @s ˇ
XDX0
Since the coordinate lines intersect at
k
ˇ
ˇ @Y
ˇ
ˇ @s
ˇ
ˇ
ˇ
ˇ
XDY0k
> 0:
We thus have
ˇ
ˇ
ˇ
ˇ ˇ
ˇ ˇ
ˇ
ˇ
ˇ
ˇ @X ˇ
ˇ @X ˇ
ˇ @X ˇ
ˇ @s ˇ
ˇ @X ˇ
ˇ
ˇ ˇ
ˇ ˇ
ˇ
ˇ
ˇ
D ˇˇ
D ˇˇ
>0
(B.4)
ˇ @ ˇ
ˇ
ˇ
ˇ
ˇ
@s XDX0 @ D x
@s XDX0 ˇ @ ˇD x
D x
k
k
k
k
k
where we used j@X=@j Dkx ¤ 0. Likewise, we have j@Y =@jD y > 0. Since
k
j@X=@j Dkx > 0 and j@Y =@jD y , we can take a -interval in around each kx
k
y
and k so that
ˇ
ˇ
D ˇˇ j kx j < ;
ˇ
ˇ
y
y
‚k D ˇˇ j k j < ;
‚xk
(B.5)
ˇ
ˇ
ˇ @X ˇ
ˇ
ˇ > C > 0 ;
ˇ @ ˇ
ˇ ˇ
ˇ @Y ˇ
ˇ ˇ > C > 0 :
ˇ @ ˇ
y
We shall also take small enough so that the intervals ‚xk ; ‚k do not overlap. Let
‚ D
x
(B.6)
nx
[
‚ D
‚xk ;
y
kD1
ny
[
y
‚k :
kD1
Recall ‚ D R=2Z. It is clear from the definition of ‚xk that
(B.7)
2 ‚ n ‚x ;
jx0 X. /j C :
2 ‚ n ‚y ;
jy0 Y./j C :
Likewise,
(B.8)
We now estimate IG :
IG D J0 C
nx
X
Jxk C
kD1
X
J D
0
(B.9)
ny
X
y
Jk ;
kD1
.S .X/.Gx Gh;x //.m /F.m /;
m 2‚n.‚x [‚y /
Jxk D
y
Jk D
X
.S .X/.Gx Gh;x //.m /F.m /;
m 2‚x
k
X
y
m 2‚k
.S .X/.Gx Gh;x //.m /F.m /:
IMMERSED BOUNDARY METHOD PROOF
1259
First of all, we bound J0 . From Proposition 4.5, we have
2
X
X
jJi0 j j.S .X/.Gx;ij Gh;x;ij //.m /jjFj .m /j
m 2‚n.‚x [‚y / j D1
(B.10)
X
2
X
C
max
y2RX.m /
m 2‚n.‚x [‚y / j D1
h2 jFj .m /j
:
sin.jx1 x0 j=2/ sin.jy1 y0 j=2/
For y D .x1 ; y1 / 2 RX.m / , we have
(B.11)
jx1 x0 j jx0 X.m /j jx1 X.m /j C r h C
2
where the last inequality is valid for small enough h. Likewise,
(B.12)
jy1 y0 j C
2
for small enough h. It is also geometrically clear that
dU
dU
; jy1 y0 j 2 ;
2
2
for small enough h. This shows that sin.jx1 x0 j=2/ sin.jy1 y0 j=2/ is bounded
from below by a positive constant. Thus
jx1 x0 j 2 (B.13)
2
X
X
jJi0 j m 2‚n.‚x [‚y / j D1
(B.14)
2
X
X
C
max
y2RX.m /
h2 jFj .m /j
sin.jx1 x0 j=2/ sin.jy1 y0 j=2/
C h2 jFj .m /j C h2
m 2‚n.‚x [‚y / j D1
where the constant C above does not depend on h or .
Now we bound Jxk . We first introduce the set ƒ:
ƒ D f 2 ‚ j j kx j ; h C r h=C g:
(B.15)
For h small enough, ƒ ‚xk . We divide the sum Jxk into two parts:
Jx1
k D
(B.16)
Jx2
k D
X
x2
Jxk D Jx1
k C Jk ;
.S .X/.Gx Gh;x //.m /F.m /;
m 2‚x
k nƒ
X
.S .X/.Gx Gh;x //.m /F.m /:
m 2ƒ
. We have
We start with Jx1
k
(B.17)
x1
j
jJk;i
X
2
X
j D1
m 2‚x
k nƒ
C
max
y2RX.m /
h2 jFj .m /j
:
sin.jx1 x0 j=2/ sin.jy1 y0 j=2/
1260
Y. MORI
For y 2 RX.m / , m 2 ‚xk .‚ n ‚y /,
2 dU jy1 y0 j C :
2
We thus see that sin.jy1 y0 j=2/ is positive and bounded from below in the above
sum. Thus,
(B.18)
(B.19)
x1
jJk;i
j
2
X
X
C
j D1
m 2‚x
k nƒ
max
y2RX.m /
h2 jFj .m /j
:
sin.jx1 x0 j=2/
Now note, for some constant C ,
jx1 x0 j
(B.20) sin
C jx1 x0 j C.jx0 X.m /j r h/
2
C.C jm kx j r h/ C.jm kx j C h/ > 0:
We thus see that
(B.21)
x1
j
jJk;i
X
m 2‚x
k nƒ
C h2
:
jm kx j C h
We consider the part of the above sum where m > kx .
X
x
m 2‚x
k nƒ;m >k
(B.22)
C h2 m kx C h
C h2
C C h2
h
Z
C h C C h2
Z
kx C
kx C
h
kx
1
d
Ch
1
d C h. C h log.h1 //:
The same estimate holds for the m < kx part, and we thus have
x1
j C h. C h log.h1 //:
jJk;i
(B.23)
. We have
We now turn to the estimate of Jx2
k
x2
jJk;i
jD
(B.24)
2
X X
m 2ƒ j D1
2
X X
m 2ƒ j D1
We have
(B.25)
j.S .X/.Gx;ij Gh;x;ij //.m /j jFj .m /j
C
max
y2RX.m /
h
jFj .m /j:
w.x y/
jy1 y0 j
w.x y/ sin
:
2
IMMERSED BOUNDARY METHOD PROOF
1261
For y 2 ƒ, we have (B.12), and thus w.x y/ is bounded from below by a positive
constant. Thus,
X
x2
(B.26)
jJk;i
j
C h:
m 2ƒ
Since ƒ is an interval of length 2, the number of m contained in ƒ is at most
2= C 1. Thus,
x2
j C h 2 C 1 C h.h C /:
(B.27)
jJk;i
From (B.23) and (B.27) we have
jJxk j C h. C h log.h1 //:
(B.28)
y
The same estimate holds for Jk . Substituting this and (B.14) into (B.9) we have
jI j jJ j C
G
(B.29)
0
nx
X
jJxk j
kD1
nx
X
C
ny
X
y
jJk j
kD1
C h. C h log.h1 // C
C h2 C
kD1
ny
X
C h. C h log.h1 //
kD1
C h. C h log.h1 //:
Combining this with (5.5) and (5.8) from the proof of Theorem 5.1, we have the
desired result.
Acknowledgment. The author thanks Charlie Peskin and Leah Keshet for comments on the manuscript. The author would also like to thank the anonymous reviewer for pointing to references of which the author was unaware. The author was
supported by the Mathematics of Information Technology and Complex Systems
(MITACS, Canada) team grant (with Leah Keshet the team leader) and the Natural
Sciences and Engineering Research Council (NSERC) discovery grant (awarded to
Leah Keshet).
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YOICHIRO M ORI
University of British Columbia
Department of Mathematics
1984 Mathematics Road
Vancouver, BC V6T 1Z2
CANADA
E-mail: [email protected]
Received January 2007.
Revised July 2007.