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The boundary element method for three-dimensional stokes flows exterior to an open surface
Math1 Comput.
Modelling
Printed in Great Britain.
Vol. 15, No. 6, pp. 19-41,
0895-7177191
1991 Copyright@
All rights reserved
$3.00 + 0.00
1991 Pergamon
Press plc
THE BOUNDARY ELEMENT METHOD FOR THREE-DIMENSIONAL STOKES FLOWS EXTERIOR TO AN OPEN SURFACE WENDLAND
W. L. Mathematics
Institute A, University
Stuttgart,
Federal Republic J.
Department Chongqing
Abstract-The
purpose
ZHU for Mathematics
Institute of Architecture Chongqing,
of Stuttgart
of Germany
and Engineering
Sichuan, P.R. China
of this paper is to analyse the velocity field of an incompressible
flow exterior to au open bounded
surface in three dimensions,
which is modelled
viscous
by a system
of
integral equations of the first kind on the open surface. The existence and uniqueness of the solution of the integral equations can be proved either by using the coercive variational formulation theory of strongly elliptic pseudo-differential The singular behaviour the Wiener-Hopf corresponding
pseudo-differential
we introduce
we use augmented
a Lagrangian elements,
approximation
element method
of the matrix-valued
by using
principal symbol
of the
operator.
of a boundary
boundary
admit the geometrical of the boundary
near the edge of the open surface is analysed
technique based on the factorization
For the construction equations,
of the solution
or by the
operators.
element approximation
with Gale&in
multiplier in order to incorporate
schemes for the integral
constraint conditions.
Then
which simulate the singular behaviour near the edge and also of the surface and its edge. The asymptotic
as well as those for the approximate
convergence
rates
velocity field are presented.
1. INTRODUCTION Boundary Element Methods based on different boundary integral formulations have been applied to the numerical computation of viscous incompressible flows. Here, the schemes for solving the linear stationary Stokes problem play an important part, since the nonlinear or non-stationary problems can be reduced to linear and stationary ones by means of perturbation and time-stepping procedures [l-4]. For twodimensional exterior Stokes flows around smoothly bounded obstacles, the boundary element methods are well established and provide excellent computational procedures [4-71. For three-dimensional exterior Stokes problems, where the obstacles have closed boundary surfaces, Fischer, Hsiao and Wendland [8], Fischer [2] and Fischer and Rosenberger [9], Hebeker [lo] and Zhu [11,12] have proposed different approaches by using boundary integral equation methods. Theoretical analysis and numerical experiments confirm that the boundary element methods are very well suitable for solving these problems in unbounded domains. In this paper we consider the Stokes flow exterior to an open obstacle which is an arbitrarily shaped open surface in R3. This kind of problem presents some difficulties due to the singularities at the edge or the boundary of the open surface. Similar problems in elasticity and wave propagation such as crack and screen problems have been investigated by Costabel and Stephan [13], Stephan [14-161. These investigations are here applied to the Stokes problem. We use boundary integral equations of the first kind and show the solvability of the boundary integral equations defined on the open surface. We extend the open surface to an arbitrary smooth This work was supported
by the German
a guest professor at the University Education
Research Foundation
of Stuttgart,
(DFG
We-659/8-l)
who also wishes to acknowledge
while the second author was a grant from the Chinese State
Commission. Typeset 19
by A@-TEX
W.L.
20
WENDLAND,
J. ZHU
closed surface. Then uniqueness and existence of the solution to the integral equations can be analysed by using the coercive variational formulations in suitable Sobolov spaces or by using the theory of pseudo-differential operators. Since the boundary integral equations are strongly elliptic pseudo-differential operators of order -1, the uniqueness of the solution implies the existence. However, the boundary integral equations are subject to some additional constraint, therefore we meet the same difficulties when incorporating the constraint into the three-dimensional corresponding numerical boundary element method. In the process of numerical approximation, we introduce a Lagrangian multipler to replace this constraint. The explicit singular edge behaviour of the unknown vector of the boundary integral equations is analysed by applying the Wiener-Hopf technique developed by Eskin [17]. It seems to us that this is the appropriate method for the analysis of the singularity of the open surface problem, since it is far simpler than the methods of analysis for the singular behaviour near an edge or a conical point [18-221. We use the factorization of the 3 x 3 matrix-valued principal symbol of the integral equations, and decompose the solution of the integral equations into a regular part and a singular part. In order to incorporate the singular behaviour into the approximation, we augment the boundary elements with special singular functions. For the boundary element approximation we use curved elements which approximate the surface elements by polynomial interpolation of order lc. The boundary functions are approximated by piecewise polynomials of order m in the parameter domain generating corresponding composed elements on the boundary. 2. BOUNDARY
INTEGRAL
EQUATIONS
Let the open obstacle l? be a bounded simply connected orientable smooth surface in R3, with a smooth non-self intersecting boundary y. The Stokes problem for the open surface reads as: Find the velocity field E and the pressure field p such that
-vht~+TJp=$
in SZr =
R3\r,
div E = 0,
in 52r, and,
,uP=g,
on r,
(2.1)
(Thereafter, the spaces of vector-valued where v is the given constant viscosity of the fluid. functions or distributions will be underlined by -.) We begin with the integral representation of the solution to (2.1). We extend l? to an arbitrary smooth, simply connected closed surface dR enclosing a bounded domain R. Let R’ denote the complement of d = Sl U ds1. Then our obstacle I’ forms a piece of aR. The following analysis is based on the only assumption that the velocity field ; has local finite energy.
Let us consider
the weighted
Sobolev
space
&
Au E L2(0)}
,
where T = 1x1, i = 1,2,3 and where 0 denotes any open domain. We require ,u E yl’(A,CJr) := (lV’(A, The Green
formula
corresponding
((-,~Au-tVp),~)n _
Rr))3
where fir = R3\1\.
to the Stokes equations
= ~(V~,v_v)n--
,
(p, div :)n
in R reads as - _/a n . (r- (5, P)) : ds,
(2.2)
21
Boundary element method whereas
in the exterior
((-v
it takes the form,
A: + YP), _~)s-v= v (Vz, V;)W
for all 2 E y”(n,). superscript
-(+)
Here, the unit denotes
is the stress tensor
solution
2 = (nr,n2,ns)
the limit on the boundary
;6 (2.3) J .(c+(g,p)) Bn 24
to T points
towards
Q’, and the
dR from R (from R’, respectively).
(U(lJ, P))ij
=
be th e solution
-_D
=
6ij
(2.4)
i, j = 1,2,3.
+
(2.1) and CJk(z, Y), Ph(z, Y) be the fundamental
of the Stokes problem
of the Stokes equations,
pk(x,
The application Q’, respectively,
leads to the Faxen
and, by taking
la E(I,
=J
an
I+, Y)[I(
the limits
2-(x> + z+(x) 2
(2.2) and (2.3) to (i,p) of (~,p)
(2.5)
and (,v = uk, g = pk) in Cl and
in terms
d% -
la aE(? Y> k(Y)]
ds, -
J
an
>’
k,i = 1,2,3.
representation
Y) [Z(Y)]
Yk) (Xi - Yi)
lx - Y13
XL - Yk 47r 12 - y1s ’
of the Green formulas
g(x) = P(X)
Y) =
(Xk-
(&+
uki(x7Y)=&
of boundary
integrals,
dSyj for 2 E RUCl’,
aP(xc, Y) MY)Idsyy
1, aax!
= 1, E(X, Y) [:(Y)] dSy-
the matrix-valued
qx,
y)
=
@1(x,
Y), Vz(x,
P(x,
y)
=
(P1(x,
Y),
fundamental
solution,
= pk(X, y) ni(y) +
Y) k!(Y)]
dSyr
for
x E aq
=
3 apk(W) ay
2x
j=l I:(y)1
isthejump
=
of the stress
[!j .42,P)l
P2(x,
Y),
V3(x,
Y))!
Y), P3(x1 Y)),
with
2 (au:f’y)+au~~‘y’> nj(Y), and j=l
(af%Y>)k
(2.6)
onto 6% from both sides,
where
denotes
u(z,p)
given by
Uij
Let (;,p)
normal
- (P, div :)CP +
3
i
n.(y)
f
j
=2
vector 14* ~(z,P),
* qj,P)
and
- 2. u+&,P),
for Y E
ac2,
(2.7)
W.L.
22
WENDLAND,
[E(Y)1= is the jump Hence,
of the velocity
2 across dfl.
~(2) can be expressed
+> due to (2.6) and (2.7).
We denote
to the representation
on bR.
dSl!?
for E E R3,
(2.8)
for c E I.
(2.9)
we have
en
E(z, Y) [T(Y)] dsy,
formula
(2.9),
z( x ) can be extended
continuously
onto dR.
by i, satisfying
Since the field of normal
J E(x,
vectors
an
Y>MY)I
ds,
for
7
x E an.
(2.10)
n on dR defines the null space of the integral
operator
J E(xc,
PI, the orthogonality
defined
(2.11)
Y) . C(Y) ds, = 0,
an
i satisfies
YEa%
[i] = 0, since u- = E+ everywhere
Y) L:(Y)1
Bn
$4 =
by (2.10)
Actually
J E(x,
=
In particular,
this extension
,u+,
by
g(x) = / According
z- -
J. ZHU
condition
Jan-G(X) . c(x) ds, = J,,J,,E(x, Y>. [:(Y)I
- +> 4 dsz= 0.
(2.12)
It is well known [23-251, that the following respective interior and exterior Stokes problems with the closed smooth boundary 80 have unique solutions. 1s the condition (2.12), we can find an unique solution When j E H*(&2) * g’iven satisfying (c,p)
E (g’(Q)
x L2(sZ)/R)
in fl, or (u,p)
E (Wl(sZ’)
x L2(s2’)) in Cl’, respectively,
satisfying
in RUQ’,
-vA~+~p=_O, div E = 0,
in R U R’, and,
I
:an = !'
on
(2.13)
an.
Note that, for the interior problem, p is determined only up to an arbitrary additive constant A, so there might exist a jump of p through Kl. Because of 2~E yl(A, Qr), we have @-/an = @+/an
on Xl\I’,
whereas
on I, the jump
of &/an
does not vanish.
We denote
by t(y)
the
jump L(Y) =
where G(Y)
=
[tl,t2,t31,
across
vc[nj(y) (Jy
r
at Y,
+y)]
.
(2.14)
j=l
Now, the jump
of the stress
vector
[z(y)] on dS2 can be expressed
by
for y E I’, and, (2.15) for y E as2\l?
with X E R.
Boundary element method
(2.6) and (2.7), we obtain
In view of (2.11), and combining the problem (2.1) in terms of i,
g(“) = J P(X) = In particular,
ax, Y) *L(Y) ds!J
Thus we arrive at the following Let (2,~)
THEOREM 2.1.
g E @(I’).
boundary
to
(2.16)
for x E I’, we find i+>
given
of the solution
for x E Rr.
Y) . i(Y) 44,
r
the expression
for c E R3, and
9
J P(z, r
23
Then
in tegraJ equation
J,E(x,Y>i(Y)
4.
(2.17)
conclusion:
E (w’(%) (2,~)
=
x L’(i&-))
a dmi ts an integral
b e a variational representation
solution (2.16),
of Problem where i(y)
(2.1) with satisfied
the
(2.17).
Before we discuss the mapping properties of the boundary integral operator defined by (2.17), we introduce the following Sobolev spaces, according to Hijrmander [26] and Lions, Magenes [27]. For s 1 0 let Hd(dSZ) be the usual trace space of H”+;(n) or lV”+~(a’). Then H’(r) is defined by the restriction of H’(afl) to r,
w(r) = (14IF; 21E w(xq, equipped
with the norm IIGf~(r)
:=
r c an),
inf Ilfllffqm). f/,=u
We also define
R*(r)= where No denotes
the non-negative
g(r)= {uI u H,&+‘(I’)
E
&(I’),
fors#p++,
withpENo,
&o(r)>
fors=p+i,
withpENo,
integers
and
Ha(r),u = O}, vial= p E No},
= {u I u E Ho”+ +(r),
where p = d(x,r) is the distance from x E f to the boundary y of r. It is well known that the prolongation by zero to dSJ?is a continuous er :
For negative
indices,
I?$(I’) -
H’(an)
mapping
for s > 0.
we take the dual spaces
H-S(r):=(P(r))', (s > O), with respect
to the Lz(l?) duality,
and then define the subspace
R--b(r) = {UE H--b(dil), which
fi-*(r)
of H-a(dL?)
supp u c F},
is the completion of Cr(r) = {u I u E P(r), ulr= is a subspace of Hmb(I’). We have, aa in [28]: +(I’)
0) inthenormofH-'(r).
Actually,
H-*(r):=@P(r))'3(Hyr))'=: i?(r) MCM 15:6-c
by
for s > 0,
W.L.
24
since, according to the definition the extension operator
WENDLAND, J. ZHU
of fi’(I’)
for s > 0, kS C H’(r).
er : H’(r) and the restriction
-
the above definitions,
W(&?),
operator Py : IP(Lm)
are continuous
With
for any real s. We further
-
w(r),
denote for s 2 0.
THEOREM
2.2.
onto Hi(r), Find
The boundary
which admits
t E g$(I’)
integral
equation
the following
(2.17)
variational
defines
an isomorphism
from &$(I’)
formulation:
such that
(2.18) where
J,l, (-qw)e(d)+wYdS~~
b(L, i’) : =
(_9,t’) : = J, +I
PROOF.
If 21 is solution
of problem
. t’(4
(2.19) (2.20)
d%.
(2.1) with given g E g+(r),
then
the Green
formulas
(2.2)
and (2.3) yield:
v(Vz,VtJn
V$$y
I@,
V_v E v(Qr)
Consequently,
it is a linear
(2.21)
J an”
r+ . v ds,
(2.22)
-
: = (2 E yl(&)
1 div z
=o}.
of (2.21) and (2.22) are continuous
continuous
we extend
Any given 4 E @(I’)
. z ds,
r-
= -
For any given 2 E II, the left members y.
Ja-i”
=
functional
v .pds
functionals
on
on
by x, hence z r= 4. If 2 satisfies
continuously
J
linear
the condition
= 0,
(2.23)
851”
then,
by the theorems
divergence
free field y(n)
in [23,25],
there
exists
a continuous
U y(Sl’), i.e., we can find a (%),
(%J) E y(n)
u g-q,
mapping
% from
such that
@_v)Ian= 21 where J!
Ir=+
HFO)(r)
to the
25
Boundary element method
Then,
we define a linear
continuous
U6) = v L(4) depends
continuously
(2.“21) and (2.22),
functional
on f E g*(I)
J v_u*
Vpb)
dx,
for all q!JE @(I).
nr
on 4 via the trace 2 of (%),
we have
by
and the restriction
of;
onto I. Adding
_
[I].-uds:=uJ VpV(zJhJdx. J t Q$ds=J r”
Therefore,
an
-
we define t as an element
If z, the extension
in the dual space of H*(T),
of 4 onto ds2, does not satisfy
J the extension
getting
the continuous
mapping
(2.23), i.e.,
v.cds=a#O,
m-
then we modify
(2.24)
G-
of 4 by (Y
v*=v,., W which satisfies
mes(dQ)
J
V,*
.
on Xl,
”
Eds = 0.
Then i,!+ds-J
t *(v* + a mes(aQ) an”
Jt
.sds=
an”
n) ds. -
Jt
.zds=O,
(2.25)
r”
we still have
which defines the duality the above mapping For the inverse
between
H*(I)
-
mapping,
@(I’)
e;!(I)
and &f(I).
Find ,u E y such that
The bilinear
form
that
is continuous.
if i E $(I’)
Y
H ence, in all cases we have proved
VEV_vdx= J SZr
is g iven, we consider
Jt r”
._vds,
I the following
v_vq.
variational
problem:
(2.26)
W.L.
26
is coercive
on w’(Qr),
WENDLAND, J. ZHU
since here the semi-norm
is equivalent
to the norm
in y’(&+)
[29], i.e.,
(2.27) where c > 0 is independent to the Lax-Milgram
theorem,
the variational
the trace of 2~ and the restriction Actually,
Jr t e; ds is a continuous
of ,u. Moreover,
problem
linear
functional
(2.26) has a unique
solution
on y. Due in y.
Hence,
of this trace on I? gives t E H*(I).
we have also proved
that
the isomorphism
between
g E ta(I’)
and i E @&!(I’)
is
expressed by the integral equation (2.17) in a similar way as in [30]: F’u-rally, the unique solvability of the variational problem (2.18) results also from the Lax-Milgram theorem, since substituting (2.27) into (2.26) and (2.18), we find
due to the already
shown isomorphism
REMARK 1. Theorem 2.2 can also differential operators. This method and screen problems in W3. Later need this approach. Therefore, we
Since for any i E c(o$(I’),
between
dR is a smooth closed surface provides the principal symbol
by (2.26).
be proved by using the properties of strongly elliptic pseudois used by Costabel and Stephan in [13-161 for crack problems on, for the application of Eskin’s technique along y, we will sketch the corresponding arguments.
th e extension
Ar :(4 =
E and i defined
J,E(Z, y)
i by zero onto K?\r . t(y)
ds, =
belongs
J,,E(t, Y) . i(y)
and the use of a partition ug(Ar)(<) of th e integral
of unity operator
= -& pi3
Q(4)(l)
--Cl<2
Thus, Ar is a strongly continuous mapping from
elliptic
pseudodifferential
&$r) c g+(m) Moreover,
see that
e,,!(r)
the mapping
only into IJto,(&2). c,?(I)
into Hi(r),
The form (2.30) of the principial in the energy other
hand,
According
0
(2.30)
.
0
ItI2I
operator
of order
-1.
--4(I’), space II(,)
into IJ+(aR).
we can prove that to the Fredholm
symbol
By continuous defined
by the integral
that Ar is a Fredholm
the field of normal
alternative,
restriction
shows that the operator
which means
vectors
we conclude Vri
Ar is a
Therefore
E(z, y)k(y) ds, ds, =0, >.E(x) JEra (/a-l
we have
thus Ar maps
(2.29)
0
IE12+sf
0
[
d%.
and the Fourier transformation Ar given by [2],
l<12+E22-4152 1
to @(o;(i %I), we have
= 9,
that
of IJ*(an)
operator Ar satisfies
to
g+(r),
we
(2.17), is continuous. G&ding’s
inequality
operator
of index zero. On the
14on r generates
the null space of Ar.
the boundary
integral
equation (2.31)
Boundary element method
27
then has a unique solution L which satisfied the condition
It r-
.Eds=O.
Again, we find the isomorphism between g(,,) --* ( r ) and @(I’). the constraint
Th e uniqueness is ensured from
condition (2.25). 3. SINGULARITY
OF THE
SOLUTION
Because of the special geometry of I’, the solution of the Stokes problem, aa well as the solution of the integral equations derived from the boundary value problem, have singularities at the edge 7 of the open obstacle I?. Our aim is to obtain the local behaviour and a decomposition of the solution of the integral equations into a regular part and a singular part. The regularity of the solution and the singular behaviour at corners of the twcFdimensiona1 Stokes problem in a polygonal domain has been thoroughly studied by Kellogg and Osborn [31,32] Grisvard [33], Costabel, Stephan and Wendland [34] and Dauge [18]. They obtained the singular behaviour of the solution near the corner points by using Kondratiev’s procedure [35], which is based on an eigenvalue problem associated with the Stokes problem. For the three-dimensional case, Maz’ya and Plamenevskii [20,21] studied the general dependence of properties of the solution on the geometry of the boundary by the same technique. Our case of an open surface r with the boundary edge y could be considered as a limit case with an edge angle 2n. This particular case, however, is not included in the general analysis of the three-dimensional edge and would require complicated additional analysis. Therefore, here we prefer to use Eskin’s procedure [17] which was further developed by Costabel and Stephan in dealing with three-dimensional screen and crack problems [13-161. Since our main concern is the singular behaviour of the solution of the boundary integral equations of the first kind, the expansion of this solution can be obtained by applying WienerHopf technique directly to integral equations, without returning to the original boundary value problem. The following theorem gives a decomposition of the solution of the integral equation (2.17) near the boundary edge 7 of the surface l?. Let s be the parameter of arc length of the smooth closed curve 7 and p(z) be the distance from 2 E l? to y. Let x(p) be a Coo cut-off function with x =_ 1 for small p. THEOREM 3.1.
Let g E H++& be given,
with 151 < i.
Then the solution i E e,,!(F)
of the
integral equation (2.17) has the form t =
P(s)P-+X(P)+ to,
(3.1)
with f(s)
=
to =
(Pl,P2,p3)
(t1,t2,t3)
E @+‘(T), - i+s’(r), E g
and for any 5 < 6’.
PROOF. We begin with the local representation of the integral equation (2.17). Using the technique of localization and a partition of unity, the integral equation (2.17) can be transformed into a finite sum of integral equations, each of them defined on a local chart, and the principal part can be represented with the local mapping by collecting compact perturbations on the right. Then (2.17) takes the following form,
P+@+ =
s7
on R:.
(3.2)
28
WENDLAND,
W.L.
J. ZHU
P+ denotes the projection operator of restriction to rS: and A is the pseudo-differential operator with principal symbol a(A)(t) given in (2.30). As the result of the Coo-diffeomorphism in every chart, the smooth surface &I is mapped into the plane 2s = 0. l?, one piece of dQ, is mapped into Fp$ given by xs = 0 and x2 > 0. Then the edge 7 of r is locally mapped into t2 = x3 = 0, tr E R. Finally, we apply the Wiener-Hopf technique to (3.34) in R2 and the halfspace R$. Following the ideas of Costabel and Stephan in [13], we factorize the 3 x 3 matrix u(A)([) as
4W)
$+(E)A,(E),
=
4 =
(&,<2,0),
(3.3)
with 2iK1)
3<2 A-(t) =
(52 -
-it2
--tl
i It1 I)-”
(E2 -
-
i ICI I) sign (1
0
i
0
IGI
0
,
0 t2 -
(3.4)
16I i
i
and 2<2+2iK11
l&l
-f
A+(E) = (t2 + i I& I)-+
Then we calculate
the corresponding
inverse
= (<2 - i I&])-~
4
and
352 AT'(t) = (E2+ i lt11)-$
matrix
(i (2
Iti I
It11
0
<1
F2+il&I
(3.5) 1
(t2 -
0
+ ltl I) sign (1
0
,
0
$il&l)sign& t2 -
i
0
it1
(t2 + ii I& I) sign t1
0
0
0
E2+iISll
of t = 0, according
to Eskin
$i I&
I
(3.7)
[17, p. 911, in
and 111I by 1+ It1I.
(3.3)
by n(a)(c),
fl@)(<)= &A-((1 + 15~l)sh&,b,O)A+((l+ lt~I)sign<1,&,0) = & ~--cr> a+(t). Now, we consider
(3.6)
I&I )
l&l - ft2
+
Since a(A)(t) is unbounded in a neighbourhood a(A)([) we replace
I&I) sign
matrices
0
i
We denote
ii
0
t2 - i AI1(O
(t2 +
0
i
0
(it2-2Ki])sign&
the pseudo-differential
equations
P+&+ = 9,
(3.9)
in the halfspace, on W:,
(3.10)
where the pseudo-differential operator a has the principal symbol c(A)(t). Actually, a is the inverse Fourier transform of the operator of multiplication by o(a)(e), which is a bounded operator from 2 into e’+‘(Wi), g$ = {t E Hs(R2), suppi C W+} for any real s, according
to Eskin
[17].
Boundary
By applying the Wiener-Hopf method, for s = -3 i- 6, (61 < 3 in the form ^ i+(t)
where
The projection
we obtain
29
any Fourier
transformed
solution
of (3.10)
A
= 41/&l
12 E g ‘+t ( R2 ) is an arbitrary
transform.
element method
^ of g from R$ to R2 and lg denotes
extension
(into the upper
(3.11)
II+ Pr,(t),
complex
halfspace)
operator
II+
defined
its Fourier by
can be extended by continuity to a bounded linear operator from H6(R2) into H6(R”+) for any 161 < 3, by Theorem 5.1 of [17]. Th en, the solution of (3.10) is given via the inverse Fourier transform of (3.11)) namely ^ (3.13) i+(c) = F&L i+K). The expression (3.11) implies the regularity result: nor a given z E w”+6(R$), 161 < i, we have i+ E IIJ-“+~(R:). result desired, decomposition require
since we want of the singular
additional
regularity
But this is not the only
to model the singularity at the edge y of I’ as a tensor-product form of Z+(X). This needs more regularity of t+. Therefore, we g E H$f6.
Applying
(5.36) in [17], we obtain
where A+ =
cc2 +
I& I>>,
i (1+
(3.15)
II’ @I) = & /_m Q&,02)d172. 00
The operator
II’ maps
Now we can modify
The second
Hb(R2)
into H~-~(W)
for s > +.
(3.11) by decomposing
term %2(E) of i+(t)
is regular,
i+
as follows,
since for s E E”+~(W$),
Setting i(h)
we have E(&) E 54@(W)
=
(cl(h),
cZ(tl),
for g E H++6(W$).
C3(&))
=
Tl lerefore,
in’kl
itr(<)
(3.18)
f!(E),
can be written
in the form
&(r) = As1lip c(t) (3.19) = (Fs + i(1+
IrlI))-*
$I(&)
+ ((2 + i (1+
M))-3
&I),
W.L.
30
WENDLAND,
J. ZHU
with i (d5)
-
i cz(&))
I I Hence, we &(&)
=
C2
(t1)sign
,
&
1
$ (1+
62(e)
=
Kll)
(QiCl(
+ C2Kl))
I&l)signtl
g(cl(rl)-4ic2(~1))(1+
0
[
have fi~ E H*+6(R),
,& E H-*+~(FP),
since ~(0 E &+6(~>.
1
(3.20)
.
Now, the solution
of (3.10) can be obtained by applying the inverse Fourier transform stepby step, first, forthevariable (2 --f 22, then for the variable & + q. Since F&(F2
i(1+
+
1<11>>-+ = cz;”
o+(z2)
(3.16)
to
e-“a(1+1q
(3.21)
where
1, 0,
@+(x2)=
x2
0
>
x2 < 0 ’
we have the first term :1(%x2)
=
P(~lr~2)2;~
@+(x2)
(3.22)
+:3(x1,22),
with P(x1,
i3(%
Note that is E kJ++6(Ri),
x2)
=
F$+ol
{ ,,-+(1+lEll)
jl(&)
x2)
=
F&,
(@
.
since the pseudodifferential
fo r any real s. Therefore,
into II”+’ of t in (3.1).
Obviously,
Lj2(&))
the singular
terms
operator
are included
* @+(x2)
=
(3.23)
with symbol
Ai4
fs
+
maps Ha(W)
to the regular
in the first term of il.
in x1 and x2, so we rewrite
22) xi
,
(3.24)
this part is of ir will contribute
term is not of the form of a tensor product L P(&,
@+(x2)}
part tc
However,
this
it as (3.25)
i4,
with n x2)
=
ct2 -*
0+(22)&(El),
gt1, z2)
=
CE2-+
@+(x2)
&(Cl,
The regularity
of !,(
x2)
~2) was studied =
FE;&
Applying the inverse Fourier The singular term is given by
(e-“a(l+‘Cll)
by Stephan
f4(Fl,
x2)
transform,
$(%X2)
(3.26)
with every 5’ < 6.
E H++6’,
A(R)
(3.27)
1) &(&).
in [14-161; we use his result
we obtain
=
-
the desired
22-3
decomposition
(3.28) (3.1),
locally.
(3.29)
0+(x2),
with PI(X) = cF& whereas
the regular
term is given by the remainders iO=i2+!3+:4
Patching
together
PI(&),
the local results,
we finally
Eg’
obtain
’ +6’
,
cs <
the desired
6.
decomposition
(3.1).
I
Boundary
4. APPROXIMATION
31
element method
WITH
BOUNDARY
ELEMENTS
For the numerical approximation of the variational problem (2.18), we use a boundary element subspace of g;ct (I’). Assume that the smooth surface I’ is given by a regular parameter represetation CC= 4(t), < = (<1,t2) E D in R 2. By the same smooth mapping 4, the boundary 62, of lJ is mapped onto the edge y of I’. With the regular triangulation Dr of 2) and the bijective mapping 4, we can establish a family of regular finite element spaces [36] on the surface I, satisfying with m + 1 > 12 0, that means that the components
of the vectors
degree greater or equal to m belonging to @‘(I’).
ih
E
(4.1)
are piecewise polynomials
g+“‘(r)
of
Moreover, Lh satisties the constraint
P
1$,-Eds=o.
(4.2)
I?
The approximate variational problem of (2.18) reads as follows: such that for all $
Find th E $+l”,
a(ih,$,)
:=
E $T+“‘,
JJ r
r
E(? Y)ih(Y) $h$ dS~ds, =
J r”
g(z) $(zc) dss := (2,$A
(4.3)
For (4.3), we obtain solvability and convergence results by applying the standard Galerkin finite element analysis, as discussed in [15]. Due to the lack of regularity of i at the edge y, with p -4 E H-“(F), by the finite elements, viz. (2.18), we can obtain only low convergence rates. Therefore, the best possible estimate for quasiuniform mesh refinements is of the form
(4.4) where h is the maximal mesh size of the family of partitions of I’, 0 < h < 1. However, the convergence can be improved by using extended finite element spaces which are augmented by singular elements near the edge y. To this end, we define
m+l>l>O,
(4.5)
where
i:+“‘(r) ={iOh and --*
E
s:+“‘(r) (ioh
=
o On
y}
,
1 - e < 3 + 6, 0 < c < 3, 161< 3. Therefore, we have ?;(I’)
c c,,!(I).
The
improved Galerkin scheme on the augmented spaces reads as: such that the variational equations (4.3) hold for all Find ih = ,f(+J-+ X(P) + $,, E &i(r), test functions {h
E %“,(r>.
(4.6)
32
W.L.
WENDLAND,
J. ZHU
With standard arguments for energy norm estimates and the approximation properties of the augmented boundary element space, as shown in [13-151, we obtain the higher rate of convergence,
where E > 0 is any positive number. Unfortunately, the implementation of the above approximation is not an easy task. When l? is an arbitrarily shaped surface in R3, the exact computation of Moreover, b($&) and (_g,$) is rather difficult and costly or, in some cases, even impossible. when we construct constraint additional
the finite
element
space $+“‘(I’)
condition (4.2). It is difficult singular term /Jh(s) p-4 x(p).
to treat
or ::(I’), this condition
we must
satisfy
exactly,
much
the additional more
with
the
To overcome these difficulties, we prefer another approach for the approximation. First, we approximate with the help of curved finite elements, the surface I by rh [29,37,38]. Then we construct the boundary element spaces on rh, by using the method of Lagrangian multipliers. Since we have supposed a smooth mapping between r and the plane domain D, we can easily define an appropriate family of triangulations 2)~ of D. We denote by IT any of these geometrical boundary elements. On each element I< we define an interpolation 4h of 4, such that the mapping 4h of V into w3 is continuous. Then the image of I< by the mapping 4h constitutes one piece rjh of the surface In which we take as an approximation of I’. When I#Jhis a polynomial interpolation of degree k to 4, rh will be defined by a connected parametric surface of degree k. The interpolation is given by curvilinear elements with nodes on I. We denote these geometrical elements of Ij, by rib (i = 1,2, . . . . N), then I,, = brih.
(4.3)
i=l
Under the mappings 4 and c#J;‘, the edge y of I will be approximated which has the same order of interpolation as Ij, to I. Then, for approximating the vector functions i belonging to E-t(I), nomial functions space s;+“‘(rh) not included
we use piecewise
poly-
of degree m on each element which is a finite dimensional
in @-+(I’),
of Ih, that means we construct a finite element subspace of e-*(rh). Although s;+i”(rh) is __r we will take it as the approximation of g ‘(I’),
g+“‘(rh) c Analogously,
by oh, the edge of rh,
we define $+“‘(rh)
In order to simulate the singular Of s:+“‘(rh), namely
&-*(rh),
(4.9)
m+l>l>O.
by using on y the grid points belonging behaviour
at the approximate
to the partition
of rh.
edge Th, we use an augmentation
(4.10) where
m+I>l>O,
and-i
On the other hand, instead of incorporating space, we introduce a Lagrangian multiplier
the constraint condition into the boundary by defining the bilinear form
element
(4.11)
Boundary
where the Vector
ch
is defined
33
by e,, = p
which is the image of the normal approximation
element method
4
0
4;’ := 34 0
0
c to r belonging
‘p,
to the mapping
of c one could also use its Lagrangian
$,>
+
dh($,
A) =
(ijh>
t6),
404;‘. (For a higher order
Q =
interpolation,
constraint (4.2) creates serious difficulties.) Now, we replace the approximate variational problem, as follow: Find (ih,A) E $(Th) x R such that bh(ih,
(4.12)
as in [38]. However,
the variational
vi6
then the
formulation
(4.3) by
E $(rh),
(4.13) dh(thr
where bh(Lj,,ii)
is the bilinear
bh(th,
A’)
=
VA’
0,
form on Th, defined
$,)
:=
E R,
by
JJ E(z,
Y) th(!/)
rh
$,(+Shy
(4.14)
dSh=,
rh
and (sh,$,)
where gh is an approximation
:=
J,,
zh@)
of g. Of course,
J
* $dc)
the second
(4.15)
dshr,
equation
of (4.13) is equivalent
to
$, * ,$, dsh = 0.
rh
Before we analyse the solvability in [39], let us discuss the properties
of (4.13) by using the abstract of the bilinear forms bh(;h,$)
First
of all, we need to compare
and
Lh defined
framework
given by Brezzi
d&,X).
on Th with its counterpart
i defined
on I?. One
could think of using the mapping $ defined by Nedelec in [29], where for z E Th, the vector z - $J(c) is normal to I’ at the point $J(z) E r. With this mapping, one obtains convergence of one order more for the Dirichlet problem of the Laplacian than here, where we use the mapping Q-’ = f$h o 4-l [38]. 1n our case, neither ih 0 $J nor ih 0 Q-l satisfies the constraint condition (4.2) in general. mapping
Therefore,
from $(Th)
both
of them
into @(of (I’). Therefore, ?’ih
where have
do not belong
4) =
J(D
J(D 4) and J(13 C$h) are the Jacobians
J r
rih.zds=
c KeD7
=
we define the mapping 0 dh
o
4-l J(D
corresponding
J (r th
.2)
0
0
dh),
to 4 and 4h, respectively.
45(0 4)) * (z 0 4) dt
J t,, = *l$dSh
0,
we need a
T by
4J(D 4) dt
(~h”4h)‘(~ho4h)J(D4h)dt
r”
Nevertheless,
K
(1’Lh
= =
ih
to #(,!(I’).
Then
we
W.L. WENDLAND,J. ZHU
34
if
ih
E zi(rh)
f
(-
q < l), satisfying
5
the L2 norms,
is easy to check that mapping r. Thus
the constraint
defined
cl b~hb(I’)
where
cl, c2 are positive
bijective
mapping
Hq(I’) =
r$,;
5
constants.
condition
on rh, then r Lh E I&,! (I’). It
on rh and r, respectively,
/~hb(I’,)
We denote
5
c2
are equivalent
under
the
(4.16)
Ilr:hb(r),
by Hq the image
of $(rh)
belonging
to the
r vector function
defined
on I?,
rib
=
:h
04h
v+,
th
E gQh(rh)
(
,
(4.17)
>
then
gr) where the space gq is equipped
with the norm
Il”(flhPh’X +iOh)llfqr),
:= lb
c ~-3(r),
+
(4.18)
_thllZq w IlrPhllH’(r) W W
The index q < 1 is limited
by Theorem
+
OSq
IbtOhlIiyq,
3.1, where q = g + 6, ISI < 3. We note that
Pq = B’(r)
for q < 0. LEMMA 4.1.
We have the inverse ll’~hllZ”
provided
0 < E 5 3.
PROOF.
On each element
inequality <
O
E go, then
Ilr~hll&
5
5
On the other
CIl(r:h)X+
(4.20)
=
by w lb= wk, and we have
(rih)(rihW)
c
(4.21)
;*
5
ciir~hllz-e
Ib’:hWII_Zea
(4.22)
l-
jK(I_h t
o~hWk12+ID(~ho~hWk)12).J(D~h)d~
5
j+hll;o.
we have
the last inequality
LEMMA 4.2.
5
hand,
lb% wll;l = & By interpolating,
cIDwk(~)l
(r, ih)W E H’ with w defined
s
Substitute
h
(4.20) implies lwk(t>l
If rib
(4.19)
K of DT, we define a function
Wk = & dist(<, aK>, Clearly,
E Ho,
Vrth
ch-‘Ilr~hllZ-er
into (4.22) to obtain
For the fundamental
solution
I
(4.19).
E(z, y) of the Stokes equations,
we have the following
estimate
P(~,Y) - JW’-‘(~,@-~(Y))I
5 ch”&,
(4.23)
Boundary element method
where 0-l = q$, ofj- ’ is a continuous of q5. The mesh size of the boundary
mapping elements
35
from Ih to I’ and f$h is the k-degree is denoted by h.
interpolation
For 12 - yj 2 c h, we have
PROOF.
I@(Z)
- o-l(y)
- (Z - y)I 5 lo-‘@!)
- II + IWl(y)
- yI 5 cP+l
5 chkle
according to the interpolation property. In the remaining case lz - yI < c h, the derivate D(O-’ -1) of the mapping 9-l -1 by c hk [29] and the mapping G-l - I is Lipschitzian with the constant c h”. Hence, I’p-l(z)
-O-‘(y)
- (Z - y)I = I(@-+)
- yl,
is bounded
- y)I 5 chkl?: - yl,
- z) - (O-‘(y)
and therefore
~I~-‘(4
- @-?Y)12
- Ix - Y121
=12(W(2)
- Q-‘(y))
- (x - Yh (x - y>) + (W4
- WY)
- (x - Y)121
5 cl hk Iz - yj2 + c2 h2” Iz - y12 5 c hk 12 - y12. Obviously Cl p-l(X)
Now we can estimate
@-l(Y)1
the two terms
I It - Yl I
I@-‘(4
c2
- WY)l.
in the left hand side of (4.23) separately
1
-I
-
b--Y1
l~-l(A-l(Y)l
1
I I@-‘(x) - WY>I” lz-yllip-‘(2)-0-l(y)ll@-1(z)-0-~(y)+;c-yl
=
- 12 - Y121 schk&
Since we have JW(4
- Q-l(Y)13
= l(@-‘(2)
-
@-l(Y))
- Ix -
Y131)
- (3 - Y)21 I(@-+)
p-l(z)
- W-‘(Y)>2 + (Q-W
- (P-l(y)
- Q-‘(Y)9
- Y) + (z -
Yj21
+ z - yl 5 c hkJ+ - yj3,
and
I(zi - Yi>(z:j-Yj)l 5 Cl2 -Yj2, I(% - Yi> (q - Yj) - (Q-‘(Q) - a--‘(Yi)) (W-‘(q) - W’(Yj))l - @-l(yj))l
I
(Xi -
yj)
(a;‘(x) -
(Xj - Yj)
Ix-Y13
-
< (Xi-
@rl(Y))
10-l(x)
(a?l(z) - W’(y)/3
Yi) (Xj - Yj) (IQ-‘(x)
(Xi - Yi) +I I
This completes
the proof.
chk
lx - yl3 10-l(2) (Xj -
Yj)
-
I
@-‘(Y)13 - @-l(y)13
-
(@,:1(X) p-1(2!)
$q.
ipr'(Y))
-
@f’(y)) O-l(y)13
- lx - Y13> (@y’(X)
-
@Tl(J/))
W.L.
36
LEMMA 4.3.
For all th,t’h
WENDLAND,J. ZHIJ
there holds
E Zht(I’h)
PROOF. a(l-ihlrt,‘h)
-
ah(ihyfh)
=c c/J KiE’D7KjEV7
Ki
(r~ho~(q))(rtjh~~(~))E(xo~(~),Yo~(~)) Kj
(th 0
x J(D
4(v))
J(D
4h(d)
=c c/J Kiev7
KiEVI
x
Ki
Kj
d(F))
dv dF
(t_‘h o 4h(‘t))
J(W(d)
J(ME))
E(x
’ +h(t),
Y o tih(v))
dl.ld<
(rt,04(v)> (rl_lh 04(O)
[E(X0 4(t), Y 0 d(‘?)>- E(x 0 dh(‘$ Y O h(d)] x J(W(v)) J(WO) dvdt X
= Using Lemma
r t (y)rti(x)[E(x,y)
JJ r
4.2 and 4.1 we have
Ib(rih,rch)
Here the inequality Laplacian
-
results
is continuous
LEMMA 4.4.
E(~-‘(z),~-‘(~))lds,ds=.
-
r “h
bh(ih,t_lh)l
<
from the fact that
on @-*(I?)
JJ
ch”
x cS3(l?)
r
b’Lh(Y)I
lx - YI
r
the bilinear = 2-3
bfdz)l
ds, ds,
form in the screen
x z-4
[16], and therefore
and k 1 2 1371, th ere exists a positive constant
For h small enough,
bh($ih)
1
problem
for the
on go x go.
1
CYsuch that (4.25)
Q Ib+thll;-+y
for all :h
E $(rh)
PROOF.
Prom the result
Since Lh
e g;(rh),
of the mapping space g,,f
=
{:h
of Lemma
dd$&,
E g&h),
0,
VAER}.
4.3, we have
We have Jr, I,, * c,, dSh = 0. Then
r. We know from Theorem
(I’), therefore
A) =
2.2 that
r th E c;!(r) the billinear
follows from the definition form b(., .) is coercive
on the
37
Boundary element method
which yields
Hence, for k >_ 2 and h sufficiently small, (4.25) follows. THEOREM 4.1.
I
The variational problem (4.13) has a unique solution.
According to Brezzi [39], we need to verify just the following two assumptions: i) There exists a constant cr* > 0, such that
PROOF.
ii) There exists a constant /J* > 0, such that
(4.27)
The first assumption is satisfied because of the result of Lemma 4.4, since I jr ih 1I,_ 4 (r) is a norm equivalent to I Ir ih Ilz_+. ..,h
For the second assumption, take ih = (sign A)?$ to obtain
dh(th,X)
th;;-* s ,h
IAl &,
llihll
’
Iphi
dSh
llnhll _
2
P*
I43
where p* = ]]ph]] 2 c > 0.
I 5. ERROR
ESTIMATES
The errors result mainly from two sources, the approximation of T by Ih and the approximation of the boundary function 1 by th. The errors can be estimated along the ideas of non-conforming finite elements, since the augmented finite element spaces gi have approximation properties. Here we follow Nedelec [38] and Wendland [40]. LEMMA 5.1.
Let sh be the orthogonalprojection
from h2(I’) onto gq(I’).
Then, for 0 _< q 5 l-c,
c>O,t~Z~,wehave
11:- sh$-+,
2
chq++ Iltllzq(r).
(5.1)
The Lemma can be proved by using basic interpolation theory related to the regular finite element partitions and the following estimate [14,15],
II’(PhP-‘X)llZq(r) 5 CIb?hIIz'+~(r)> _ _
E > 0.
-&
(5.2)
We omit the details. THEOREM 5.2.
Let i be the solution of the boundary
solution of the approximate estimate
II! -
T:hll-&,r
problem 5
(4.13).
ch ‘+*
IItIIq,r
integral equation
(2.17), and let th be the
Then for k > 2, q 5 1 - 6, 0 < c 5 3, we have the +
hk-*
Ilrt,lhllO,I’+
II_s -
rzhll+,r.
(5.3)
W.L. WENDLAND,J. ZHU
38
PROOF.
We write
bh($ iA11
=
Ibh(ih,:h
=
i(gh,th
IA21 = I& l&l
=
According
-gh) -
-
r (ih
to Lemma
and the triangle ]I! -
t_lh)
-
-
b(i,r(ih
-fh))i
(g&h
-
t_lh))i
-
5
C@
‘.
Ib(rfh,
o
-
-t_l/a)=A~+Az+As,
-t_lh,ih
t_lh))l
fh))
-
5 -
CIb
F.
-
Vhll$,r w
“t_lhll-+,r
bh(fh,r(:h
b’(Lh
b’(Lh
-ch)l
<
-
-t,lh)l&’
t_‘h)ll-$,r
chk-‘llr~hllO,r
Ib(ih
-t,‘h)ll-+,P
4.4, we have
IIr(ih
-t,‘h)llt;,r
I
inequality
yields
r:hll-+,r
5
Iii
5
C
-
bh(ih
-
“t,‘hll-;,r
II!
-
t_lh,ih
+
-t_lh)
Ik(i
‘$hll-j,F
-
+
5
IAll
+ IA21 + lA31,
t,lh)k-&,r
hk-’
Ik~hIb,r
+
.
Il_S -yhli+,r >
{ The
desired
result
using the result
is obtained
of Lemma t,i$,, _h _
The estimates
in E-‘(I)
by taking
on the right hand side the infimum
and
5.1,
IIL -
‘
II: -
I
can be obtained
Sh
ill-;,r
5 chq+* Ilill0
by using the Aubin-Nitsche
I(: IIL -
and with the integral
for all ch E zi
operator
rihll-l,r
=
(5.4)
trick [41]. Indeed,
-r:hjl)l
sup L+(r)
Il[lll,r
(5.5)
’
Ar defined by (2.17) we have II&
[llo,r
(5.6)
I c Il[llm
yielding I(&(:
II: -
rihll-l,r
2
- r$),&‘f)] (5.7)
sup {@l(r)
IIAF’fllo,r
’
Since
(A&
- Pih), A;‘!)
= (Ar@ - r ih), A;‘[
- Sh(Ar’[)) (5-g)
+ (Al-(: according to the Lemma 5.1, the continuity I(Ar(i
- r$),A;‘l-
&(&[))I
- ‘$),
of Ar implies,
Sh(AFlf))> for the first term of (5.81, the estimate
IClli - rihll-+,r llAF1i
11~- rihll-t,r
Sh(Ar[)ll-&,r
llA~‘fllo,r (5.9)
schq+’
Ilillq,r
+ h” Ilr~hllo,r
+ h+ lbw - rghllA,r w
II&‘fllo,r.
Boundary element method
For the second I(&(:
-
39
term of (5.8), we have
r$),Sh(A~‘f))l = I(&:,
S(Ai;‘f))
= Ib(:, sh(A,li)
- (Ar(r$GWG1f))l - (_gh, r-l(sh(AF’f)))r,
+bh(th,r-l(S~(A,‘f))-b(rt~,S~(A~lf))l (5.10) < c I(s - r_Qh, sh(AF1{))l + Ibh($,,r-l(Sh(AF1f)) I Inserting
c
t
119 ,w- rghllo,r e.
(5.9), (5.10) into (5.7) yields the desired
IItW- r _thll-l,r
I c {
h” IlJllo,r
Now, we are in the position and the approximate
solution
+ hk IIr$Ilo,r
(~h,ph)
the error between
J Jqx, J P(X>
Ph(X) =
rh
THEOREM 5.3. For x E Sir with dist(x,I’) k~2,O~q~l--~,O
I
+,,
q
h” Iltllo,r
+ h3 IIf - rzhII+,r
the solution
- Ph(2)l
5
c (1+ 4x9 r)) (4x9
r)Y
(5.11)
.
I given by (2.16)
(5.12)
Y) pb(Y)
hy,
= d(x, I’) 2 6 > 0, the following
estimate
holds for
+ hq+’ Il~llq,r + Il_s - r_shllo,r
b(z)
(z,p)
1
bw,
Y) $(Y)
rh
- 2Ld4l
IK1fllw
given by
ya(x) =
I:(4
>
estimate,
+ hq+’ IlLllq,r + 119- rpllo,r
to estimate
- b(rthrSdAF1f))l
h” IlLllo,r
+ hi
Il_s- rzhll+,r
,
(5.13)
+ h3
Il_s- r_shll+,r .
(5.14)
+ hq+l Ilillq,r +
119 shllo,r _ - rw
>
PROOF.
V(x) - Uh(X)I= J, E(x, Y) (L(Y)
5
lIW~,Y)lli,r
JA(Y)) ds, + r[E(r,Y)- E(z,4-‘(~))1r $4~)4, J + IIWx:,Y) - E(c,Q-‘b))llo,r Ilr$llo,r. II: - rihll-l,r -
(5.15) Because
of d(x,l?)
2 6 > 0 we have IIWz,Y)lli,r
IIWX, d - E(x, @W)llo,r MCM 15:6-D
L dtx: rj 1
-_
(5.16)
I &.
(5.17)
40
W.L.
WENDLAND,
Hence, the result (5.13) follows by collecting similar arguments if we use the estimate,
REMARK
J. ZHU
(5.15)-(5.17),
(5.11).
The result (5.14)
Since q is limited by 1 - 6, (0 < E 5 $), the estimate
2.
O(P The errors of velocity,
(5.13)
follows with
has the order
with k 2 2.
+ P-f),
pressure and their successive
derivates
have the same order of convergence.
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T.M.
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G.C.
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Acotistics Vienna,
J. of Integral Equationa
Modelling
Printed in Great Britain.
Vol. 15, No. 6, pp. 19-41,
0895-7177191
1991 Copyright@
All rights reserved
$3.00 + 0.00
1991 Pergamon
Press plc
THE BOUNDARY ELEMENT METHOD FOR THREE-DIMENSIONAL STOKES FLOWS EXTERIOR TO AN OPEN SURFACE WENDLAND
W. L. Mathematics
Institute A, University
Stuttgart,
Federal Republic J.
Department Chongqing
Abstract-The
purpose
ZHU for Mathematics
Institute of Architecture Chongqing,
of Stuttgart
of Germany
and Engineering
Sichuan, P.R. China
of this paper is to analyse the velocity field of an incompressible
flow exterior to au open bounded
surface in three dimensions,
which is modelled
viscous
by a system
of
integral equations of the first kind on the open surface. The existence and uniqueness of the solution of the integral equations can be proved either by using the coercive variational formulation theory of strongly elliptic pseudo-differential The singular behaviour the Wiener-Hopf corresponding
pseudo-differential
we introduce
we use augmented
a Lagrangian elements,
approximation
element method
of the matrix-valued
by using
principal symbol
of the
operator.
of a boundary
boundary
admit the geometrical of the boundary
near the edge of the open surface is analysed
technique based on the factorization
For the construction equations,
of the solution
or by the
operators.
element approximation
with Gale&in
multiplier in order to incorporate
schemes for the integral
constraint conditions.
Then
which simulate the singular behaviour near the edge and also of the surface and its edge. The asymptotic
as well as those for the approximate
convergence
rates
velocity field are presented.
1. INTRODUCTION Boundary Element Methods based on different boundary integral formulations have been applied to the numerical computation of viscous incompressible flows. Here, the schemes for solving the linear stationary Stokes problem play an important part, since the nonlinear or non-stationary problems can be reduced to linear and stationary ones by means of perturbation and time-stepping procedures [l-4]. For twodimensional exterior Stokes flows around smoothly bounded obstacles, the boundary element methods are well established and provide excellent computational procedures [4-71. For three-dimensional exterior Stokes problems, where the obstacles have closed boundary surfaces, Fischer, Hsiao and Wendland [8], Fischer [2] and Fischer and Rosenberger [9], Hebeker [lo] and Zhu [11,12] have proposed different approaches by using boundary integral equation methods. Theoretical analysis and numerical experiments confirm that the boundary element methods are very well suitable for solving these problems in unbounded domains. In this paper we consider the Stokes flow exterior to an open obstacle which is an arbitrarily shaped open surface in R3. This kind of problem presents some difficulties due to the singularities at the edge or the boundary of the open surface. Similar problems in elasticity and wave propagation such as crack and screen problems have been investigated by Costabel and Stephan [13], Stephan [14-161. These investigations are here applied to the Stokes problem. We use boundary integral equations of the first kind and show the solvability of the boundary integral equations defined on the open surface. We extend the open surface to an arbitrary smooth This work was supported
by the German
a guest professor at the University Education
Research Foundation
of Stuttgart,
(DFG
We-659/8-l)
who also wishes to acknowledge
while the second author was a grant from the Chinese State
Commission. Typeset 19
by A@-TEX
W.L.
20
WENDLAND,
J. ZHU
closed surface. Then uniqueness and existence of the solution to the integral equations can be analysed by using the coercive variational formulations in suitable Sobolov spaces or by using the theory of pseudo-differential operators. Since the boundary integral equations are strongly elliptic pseudo-differential operators of order -1, the uniqueness of the solution implies the existence. However, the boundary integral equations are subject to some additional constraint, therefore we meet the same difficulties when incorporating the constraint into the three-dimensional corresponding numerical boundary element method. In the process of numerical approximation, we introduce a Lagrangian multipler to replace this constraint. The explicit singular edge behaviour of the unknown vector of the boundary integral equations is analysed by applying the Wiener-Hopf technique developed by Eskin [17]. It seems to us that this is the appropriate method for the analysis of the singularity of the open surface problem, since it is far simpler than the methods of analysis for the singular behaviour near an edge or a conical point [18-221. We use the factorization of the 3 x 3 matrix-valued principal symbol of the integral equations, and decompose the solution of the integral equations into a regular part and a singular part. In order to incorporate the singular behaviour into the approximation, we augment the boundary elements with special singular functions. For the boundary element approximation we use curved elements which approximate the surface elements by polynomial interpolation of order lc. The boundary functions are approximated by piecewise polynomials of order m in the parameter domain generating corresponding composed elements on the boundary. 2. BOUNDARY
INTEGRAL
EQUATIONS
Let the open obstacle l? be a bounded simply connected orientable smooth surface in R3, with a smooth non-self intersecting boundary y. The Stokes problem for the open surface reads as: Find the velocity field E and the pressure field p such that
-vht~+TJp=$
in SZr =
R3\r,
div E = 0,
in 52r, and,
,uP=g,
on r,
(2.1)
(Thereafter, the spaces of vector-valued where v is the given constant viscosity of the fluid. functions or distributions will be underlined by -.) We begin with the integral representation of the solution to (2.1). We extend l? to an arbitrary smooth, simply connected closed surface dR enclosing a bounded domain R. Let R’ denote the complement of d = Sl U ds1. Then our obstacle I’ forms a piece of aR. The following analysis is based on the only assumption that the velocity field ; has local finite energy.
Let us consider
the weighted
Sobolev
space
&
Au E L2(0)}
,
where T = 1x1, i = 1,2,3 and where 0 denotes any open domain. We require ,u E yl’(A,CJr) := (lV’(A, The Green
formula
corresponding
((-,~Au-tVp),~)n _
Rr))3
where fir = R3\1\.
to the Stokes equations
= ~(V~,v_v)n--
,
(p, div :)n
in R reads as - _/a n . (r- (5, P)) : ds,
(2.2)
21
Boundary element method whereas
in the exterior
((-v
it takes the form,
A: + YP), _~)s-v= v (Vz, V;)W
for all 2 E y”(n,). superscript
-(+)
Here, the unit denotes
is the stress tensor
solution
2 = (nr,n2,ns)
the limit on the boundary
;6 (2.3) J .(c+(g,p)) Bn 24
to T points
towards
Q’, and the
dR from R (from R’, respectively).
(U(lJ, P))ij
=
be th e solution
-_D
=
6ij
(2.4)
i, j = 1,2,3.
+
(2.1) and CJk(z, Y), Ph(z, Y) be the fundamental
of the Stokes problem
of the Stokes equations,
pk(x,
The application Q’, respectively,
leads to the Faxen
and, by taking
la E(I,
=J
an
I+, Y)[I(
the limits
2-(x> + z+(x) 2
(2.2) and (2.3) to (i,p) of (~,p)
(2.5)
and (,v = uk, g = pk) in Cl and
in terms
d% -
la aE(? Y> k(Y)]
ds, -
J
an
>’
k,i = 1,2,3.
representation
Y) [Z(Y)]
Yk) (Xi - Yi)
lx - Y13
XL - Yk 47r 12 - y1s ’
of the Green formulas
g(x) = P(X)
Y) =
(Xk-
(&+
uki(x7Y)=&
of boundary
integrals,
dSyj for 2 E RUCl’,
aP(xc, Y) MY)Idsyy
1, aax!
= 1, E(X, Y) [:(Y)] dSy-
the matrix-valued
qx,
y)
=
@1(x,
Y), Vz(x,
P(x,
y)
=
(P1(x,
Y),
fundamental
solution,
= pk(X, y) ni(y) +
Y) k!(Y)]
dSyr
for
x E aq
=
3 apk(W) ay
2x
j=l I:(y)1
isthejump
=
of the stress
[!j .42,P)l
P2(x,
Y),
V3(x,
Y))!
Y), P3(x1 Y)),
with
2 (au:f’y)+au~~‘y’> nj(Y), and j=l
(af%Y>)k
(2.6)
onto 6% from both sides,
where
denotes
u(z,p)
given by
Uij
Let (;,p)
normal
- (P, div :)CP +
3
i
n.(y)
f
j
=2
vector 14* ~(z,P),
* qj,P)
and
- 2. u+&,P),
for Y E
ac2,
(2.7)
W.L.
22
WENDLAND,
[E(Y)1= is the jump Hence,
of the velocity
2 across dfl.
~(2) can be expressed
+> due to (2.6) and (2.7).
We denote
to the representation
on bR.
dSl!?
for E E R3,
(2.8)
for c E I.
(2.9)
we have
en
E(z, Y) [T(Y)] dsy,
formula
(2.9),
z( x ) can be extended
continuously
onto dR.
by i, satisfying
Since the field of normal
J E(x,
vectors
an
Y>MY)I
ds,
for
7
x E an.
(2.10)
n on dR defines the null space of the integral
operator
J E(xc,
PI, the orthogonality
defined
(2.11)
Y) . C(Y) ds, = 0,
an
i satisfies
YEa%
[i] = 0, since u- = E+ everywhere
Y) L:(Y)1
Bn
$4 =
by (2.10)
Actually
J E(x,
=
In particular,
this extension
,u+,
by
g(x) = / According
z- -
J. ZHU
condition
Jan-G(X) . c(x) ds, = J,,J,,E(x, Y>. [:(Y)I
- +> 4 dsz= 0.
(2.12)
It is well known [23-251, that the following respective interior and exterior Stokes problems with the closed smooth boundary 80 have unique solutions. 1s the condition (2.12), we can find an unique solution When j E H*(&2) * g’iven satisfying (c,p)
E (g’(Q)
x L2(sZ)/R)
in fl, or (u,p)
E (Wl(sZ’)
x L2(s2’)) in Cl’, respectively,
satisfying
in RUQ’,
-vA~+~p=_O, div E = 0,
in R U R’, and,
I
:an = !'
on
(2.13)
an.
Note that, for the interior problem, p is determined only up to an arbitrary additive constant A, so there might exist a jump of p through Kl. Because of 2~E yl(A, Qr), we have @-/an = @+/an
on Xl\I’,
whereas
on I, the jump
of &/an
does not vanish.
We denote
by t(y)
the
jump L(Y) =
where G(Y)
=
[tl,t2,t31,
across
vc[nj(y) (Jy
r
at Y,
+y)]
.
(2.14)
j=l
Now, the jump
of the stress
vector
[z(y)] on dS2 can be expressed
by
for y E I’, and, (2.15) for y E as2\l?
with X E R.
Boundary element method
(2.6) and (2.7), we obtain
In view of (2.11), and combining the problem (2.1) in terms of i,
g(“) = J P(X) = In particular,
ax, Y) *L(Y) ds!J
Thus we arrive at the following Let (2,~)
THEOREM 2.1.
g E @(I’).
boundary
to
(2.16)
for x E I’, we find i+>
given
of the solution
for x E Rr.
Y) . i(Y) 44,
r
the expression
for c E R3, and
9
J P(z, r
23
Then
in tegraJ equation
J,E(x,Y>i(Y)
4.
(2.17)
conclusion:
E (w’(%) (2,~)
=
x L’(i&-))
a dmi ts an integral
b e a variational representation
solution (2.16),
of Problem where i(y)
(2.1) with satisfied
the
(2.17).
Before we discuss the mapping properties of the boundary integral operator defined by (2.17), we introduce the following Sobolev spaces, according to Hijrmander [26] and Lions, Magenes [27]. For s 1 0 let Hd(dSZ) be the usual trace space of H”+;(n) or lV”+~(a’). Then H’(r) is defined by the restriction of H’(afl) to r,
w(r) = (14IF; 21E w(xq, equipped
with the norm IIGf~(r)
:=
r c an),
inf Ilfllffqm). f/,=u
We also define
R*(r)= where No denotes
the non-negative
g(r)= {uI u H,&+‘(I’)
E
&(I’),
fors#p++,
withpENo,
&o(r)>
fors=p+i,
withpENo,
integers
and
Ha(r),u = O}, vial= p E No},
= {u I u E Ho”+ +(r),
where p = d(x,r) is the distance from x E f to the boundary y of r. It is well known that the prolongation by zero to dSJ?is a continuous er :
For negative
indices,
I?$(I’) -
H’(an)
mapping
for s > 0.
we take the dual spaces
H-S(r):=(P(r))', (s > O), with respect
to the Lz(l?) duality,
and then define the subspace
R--b(r) = {UE H--b(dil), which
fi-*(r)
of H-a(dL?)
supp u c F},
is the completion of Cr(r) = {u I u E P(r), ulr= is a subspace of Hmb(I’). We have, aa in [28]: +(I’)
0) inthenormofH-'(r).
Actually,
H-*(r):=@P(r))'3(Hyr))'=: i?(r) MCM 15:6-c
by
for s > 0,
W.L.
24
since, according to the definition the extension operator
WENDLAND, J. ZHU
of fi’(I’)
for s > 0, kS C H’(r).
er : H’(r) and the restriction
-
the above definitions,
W(&?),
operator Py : IP(Lm)
are continuous
With
for any real s. We further
-
w(r),
denote for s 2 0.
THEOREM
2.2.
onto Hi(r), Find
The boundary
which admits
t E g$(I’)
integral
equation
the following
(2.17)
variational
defines
an isomorphism
from &$(I’)
formulation:
such that
(2.18) where
J,l, (-qw)e(d)+wYdS~~
b(L, i’) : =
(_9,t’) : = J, +I
PROOF.
If 21 is solution
of problem
. t’(4
(2.19) (2.20)
d%.
(2.1) with given g E g+(r),
then
the Green
formulas
(2.2)
and (2.3) yield:
v(Vz,VtJn
V$$y
I@,
V_v E v(Qr)
Consequently,
it is a linear
(2.21)
J an”
r+ . v ds,
(2.22)
-
: = (2 E yl(&)
1 div z
=o}.
of (2.21) and (2.22) are continuous
continuous
we extend
Any given 4 E @(I’)
. z ds,
r-
= -
For any given 2 E II, the left members y.
Ja-i”
=
functional
v .pds
functionals
on
on
by x, hence z r= 4. If 2 satisfies
continuously
J
linear
the condition
= 0,
(2.23)
851”
then,
by the theorems
divergence
free field y(n)
in [23,25],
there
exists
a continuous
U y(Sl’), i.e., we can find a (%),
(%J) E y(n)
u g-q,
mapping
% from
such that
@_v)Ian= 21 where J!
Ir=+
HFO)(r)
to the
25
Boundary element method
Then,
we define a linear
continuous
U6) = v L(4) depends
continuously
(2.“21) and (2.22),
functional
on f E g*(I)
J v_u*
Vpb)
dx,
for all q!JE @(I).
nr
on 4 via the trace 2 of (%),
we have
by
and the restriction
of;
onto I. Adding
_
[I].-uds:=uJ VpV(zJhJdx. J t Q$ds=J r”
Therefore,
an
-
we define t as an element
If z, the extension
in the dual space of H*(T),
of 4 onto ds2, does not satisfy
J the extension
getting
the continuous
mapping
(2.23), i.e.,
v.cds=a#O,
m-
then we modify
(2.24)
G-
of 4 by (Y
v*=v,., W which satisfies
mes(dQ)
J
V,*
.
on Xl,
”
Eds = 0.
Then i,!+ds-J
t *(v* + a mes(aQ) an”
Jt
.sds=
an”
n) ds. -
Jt
.zds=O,
(2.25)
r”
we still have
which defines the duality the above mapping For the inverse
between
H*(I)
-
mapping,
@(I’)
e;!(I)
and &f(I).
Find ,u E y such that
The bilinear
form
that
is continuous.
if i E $(I’)
Y
H ence, in all cases we have proved
VEV_vdx= J SZr
is g iven, we consider
Jt r”
._vds,
I the following
v_vq.
variational
problem:
(2.26)
W.L.
26
is coercive
on w’(Qr),
WENDLAND, J. ZHU
since here the semi-norm
is equivalent
to the norm
in y’(&+)
[29], i.e.,
(2.27) where c > 0 is independent to the Lax-Milgram
theorem,
the variational
the trace of 2~ and the restriction Actually,
Jr t e; ds is a continuous
of ,u. Moreover,
problem
linear
functional
(2.26) has a unique
solution
on y. Due in y.
Hence,
of this trace on I? gives t E H*(I).
we have also proved
that
the isomorphism
between
g E ta(I’)
and i E @&!(I’)
is
expressed by the integral equation (2.17) in a similar way as in [30]: F’u-rally, the unique solvability of the variational problem (2.18) results also from the Lax-Milgram theorem, since substituting (2.27) into (2.26) and (2.18), we find
due to the already
shown isomorphism
REMARK 1. Theorem 2.2 can also differential operators. This method and screen problems in W3. Later need this approach. Therefore, we
Since for any i E c(o$(I’),
between
dR is a smooth closed surface provides the principal symbol
by (2.26).
be proved by using the properties of strongly elliptic pseudois used by Costabel and Stephan in [13-161 for crack problems on, for the application of Eskin’s technique along y, we will sketch the corresponding arguments.
th e extension
Ar :(4 =
E and i defined
J,E(Z, y)
i by zero onto K?\r . t(y)
ds, =
belongs
J,,E(t, Y) . i(y)
and the use of a partition ug(Ar)(<) of th e integral
of unity operator
= -& pi3
Q(4)(l)
--Cl<2
Thus, Ar is a strongly continuous mapping from
elliptic
pseudodifferential
&$r) c g+(m) Moreover,
see that
e,,!(r)
the mapping
only into IJto,(&2). c,?(I)
into Hi(r),
The form (2.30) of the principial in the energy other
hand,
According
0
(2.30)
.
0
ItI2I
operator
of order
-1.
--4(I’), space II(,)
into IJ+(aR).
we can prove that to the Fredholm
symbol
By continuous defined
by the integral
that Ar is a Fredholm
the field of normal
alternative,
restriction
shows that the operator
which means
vectors
we conclude Vri
Ar is a
Therefore
E(z, y)k(y) ds, ds, =0, >.E(x) JEra (/a-l
we have
thus Ar maps
(2.29)
0
IE12+sf
0
[
d%.
and the Fourier transformation Ar given by [2],
l<12+E22-4152 1
to @(o;(i %I), we have
= 9,
that
of IJ*(an)
operator Ar satisfies
to
g+(r),
we
(2.17), is continuous. G&ding’s
inequality
operator
of index zero. On the
14on r generates
the null space of Ar.
the boundary
integral
equation (2.31)
Boundary element method
27
then has a unique solution L which satisfied the condition
It r-
.Eds=O.
Again, we find the isomorphism between g(,,) --* ( r ) and @(I’). the constraint
Th e uniqueness is ensured from
condition (2.25). 3. SINGULARITY
OF THE
SOLUTION
Because of the special geometry of I’, the solution of the Stokes problem, aa well as the solution of the integral equations derived from the boundary value problem, have singularities at the edge 7 of the open obstacle I?. Our aim is to obtain the local behaviour and a decomposition of the solution of the integral equations into a regular part and a singular part. The regularity of the solution and the singular behaviour at corners of the twcFdimensiona1 Stokes problem in a polygonal domain has been thoroughly studied by Kellogg and Osborn [31,32] Grisvard [33], Costabel, Stephan and Wendland [34] and Dauge [18]. They obtained the singular behaviour of the solution near the corner points by using Kondratiev’s procedure [35], which is based on an eigenvalue problem associated with the Stokes problem. For the three-dimensional case, Maz’ya and Plamenevskii [20,21] studied the general dependence of properties of the solution on the geometry of the boundary by the same technique. Our case of an open surface r with the boundary edge y could be considered as a limit case with an edge angle 2n. This particular case, however, is not included in the general analysis of the three-dimensional edge and would require complicated additional analysis. Therefore, here we prefer to use Eskin’s procedure [17] which was further developed by Costabel and Stephan in dealing with three-dimensional screen and crack problems [13-161. Since our main concern is the singular behaviour of the solution of the boundary integral equations of the first kind, the expansion of this solution can be obtained by applying WienerHopf technique directly to integral equations, without returning to the original boundary value problem. The following theorem gives a decomposition of the solution of the integral equation (2.17) near the boundary edge 7 of the surface l?. Let s be the parameter of arc length of the smooth closed curve 7 and p(z) be the distance from 2 E l? to y. Let x(p) be a Coo cut-off function with x =_ 1 for small p. THEOREM 3.1.
Let g E H++& be given,
with 151 < i.
Then the solution i E e,,!(F)
of the
integral equation (2.17) has the form t =
P(s)P-+X(P)+ to,
(3.1)
with f(s)
=
to =
(Pl,P2,p3)
(t1,t2,t3)
E @+‘(T), - i+s’(r), E g
and for any 5 < 6’.
PROOF. We begin with the local representation of the integral equation (2.17). Using the technique of localization and a partition of unity, the integral equation (2.17) can be transformed into a finite sum of integral equations, each of them defined on a local chart, and the principal part can be represented with the local mapping by collecting compact perturbations on the right. Then (2.17) takes the following form,
P+@+ =
s7
on R:.
(3.2)
28
WENDLAND,
W.L.
J. ZHU
P+ denotes the projection operator of restriction to rS: and A is the pseudo-differential operator with principal symbol a(A)(t) given in (2.30). As the result of the Coo-diffeomorphism in every chart, the smooth surface &I is mapped into the plane 2s = 0. l?, one piece of dQ, is mapped into Fp$ given by xs = 0 and x2 > 0. Then the edge 7 of r is locally mapped into t2 = x3 = 0, tr E R. Finally, we apply the Wiener-Hopf technique to (3.34) in R2 and the halfspace R$. Following the ideas of Costabel and Stephan in [13], we factorize the 3 x 3 matrix u(A)([) as
4W)
$+(E)A,(E),
=
4 =
(&,<2,0),
(3.3)
with 2iK1)
3<2 A-(t) =
(52 -
-it2
--tl
i It1 I)-”
(E2 -
-
i ICI I) sign (1
0
i
0
IGI
0
,
0 t2 -
(3.4)
16I i
i
and 2<2+2iK11
l&l
-f
A+(E) = (t2 + i I& I)-+
Then we calculate
the corresponding
inverse
= (<2 - i I&])-~
4
and
352 AT'(t) = (E2+ i lt11)-$
matrix
(i (2
Iti I
It11
0
<1
F2+il&I
(3.5) 1
(t2 -
0
+ ltl I) sign (1
0
,
0
$il&l)sign& t2 -
i
0
it1
(t2 + ii I& I) sign t1
0
0
0
E2+iISll
of t = 0, according
to Eskin
$i I&
I
(3.7)
[17, p. 911, in
and 111I by 1+ It1I.
(3.3)
by n(a)(c),
fl@)(<)= &A-((1 + 15~l)sh&,b,O)A+((l+ lt~I)sign<1,&,0) = & ~--cr> a+(t). Now, we consider
(3.6)
I&I )
l&l - ft2
+
Since a(A)(t) is unbounded in a neighbourhood a(A)([) we replace
I&I) sign
matrices
0
i
We denote
ii
0
t2 - i AI1(O
(t2 +
0
i
0
(it2-2Ki])sign&
the pseudo-differential
equations
P+&+ = 9,
(3.9)
in the halfspace, on W:,
(3.10)
where the pseudo-differential operator a has the principal symbol c(A)(t). Actually, a is the inverse Fourier transform of the operator of multiplication by o(a)(e), which is a bounded operator from 2 into e’+‘(Wi), g$ = {t E Hs(R2), suppi C W+} for any real s, according
to Eskin
[17].
Boundary
By applying the Wiener-Hopf method, for s = -3 i- 6, (61 < 3 in the form ^ i+(t)
where
The projection
we obtain
29
any Fourier
transformed
solution
of (3.10)
A
= 41/&l
12 E g ‘+t ( R2 ) is an arbitrary
transform.
element method
^ of g from R$ to R2 and lg denotes
extension
(into the upper
(3.11)
II+ Pr,(t),
complex
halfspace)
operator
II+
defined
its Fourier by
can be extended by continuity to a bounded linear operator from H6(R2) into H6(R”+) for any 161 < 3, by Theorem 5.1 of [17]. Th en, the solution of (3.10) is given via the inverse Fourier transform of (3.11)) namely ^ (3.13) i+(c) = F&L i+K). The expression (3.11) implies the regularity result: nor a given z E w”+6(R$), 161 < i, we have i+ E IIJ-“+~(R:). result desired, decomposition require
since we want of the singular
additional
regularity
But this is not the only
to model the singularity at the edge y of I’ as a tensor-product form of Z+(X). This needs more regularity of t+. Therefore, we g E H$f6.
Applying
(5.36) in [17], we obtain
where A+ =
cc2 +
I& I>>,
i (1+
(3.15)
II’ @I) = & /_m Q&,02)d172. 00
The operator
II’ maps
Now we can modify
The second
Hb(R2)
into H~-~(W)
for s > +.
(3.11) by decomposing
term %2(E) of i+(t)
is regular,
i+
as follows,
since for s E E”+~(W$),
Setting i(h)
we have E(&) E 54@(W)
=
(cl(h),
cZ(tl),
for g E H++6(W$).
C3(&))
=
Tl lerefore,
in’kl
itr(<)
(3.18)
f!(E),
can be written
in the form
&(r) = As1lip c(t) (3.19) = (Fs + i(1+
IrlI))-*
$I(&)
+ ((2 + i (1+
M))-3
&I),
W.L.
30
WENDLAND,
J. ZHU
with i (d5)
-
i cz(&))
I I Hence, we &(&)
=
C2
(t1)sign
,
&
1
$ (1+
62(e)
=
Kll)
(QiCl(
+ C2Kl))
I&l)signtl
g(cl(rl)-4ic2(~1))(1+
0
[
have fi~ E H*+6(R),
,& E H-*+~(FP),
since ~(0 E &+6(~>.
1
(3.20)
.
Now, the solution
of (3.10) can be obtained by applying the inverse Fourier transform stepby step, first, forthevariable (2 --f 22, then for the variable & + q. Since F&(F2
i(1+
+
1<11>>-+ = cz;”
o+(z2)
(3.16)
to
e-“a(1+1q
(3.21)
where
1, 0,
@+(x2)=
x2
0
>
x2 < 0 ’
we have the first term :1(%x2)
=
P(~lr~2)2;~
@+(x2)
(3.22)
+:3(x1,22),
with P(x1,
i3(%
Note that is E kJ++6(Ri),
x2)
=
F$+ol
{ ,,-+(1+lEll)
jl(&)
x2)
=
F&,
(@
.
since the pseudodifferential
fo r any real s. Therefore,
into II”+’ of t in (3.1).
Obviously,
Lj2(&))
the singular
terms
operator
are included
* @+(x2)
=
(3.23)
with symbol
Ai4
fs
+
maps Ha(W)
to the regular
in the first term of il.
in x1 and x2, so we rewrite
22) xi
,
(3.24)
this part is of ir will contribute
term is not of the form of a tensor product L P(&,
@+(x2)}
part tc
However,
this
it as (3.25)
i4,
with n x2)
=
ct2 -*
0+(22)&(El),
gt1, z2)
=
CE2-+
@+(x2)
&(Cl,
The regularity
of !,(
x2)
~2) was studied =
FE;&
Applying the inverse Fourier The singular term is given by
(e-“a(l+‘Cll)
by Stephan
f4(Fl,
x2)
transform,
$(%X2)
(3.26)
with every 5’ < 6.
E H++6’,
A(R)
(3.27)
1) &(&).
in [14-161; we use his result
we obtain
=
-
the desired
22-3
decomposition
(3.28) (3.1),
locally.
(3.29)
0+(x2),
with PI(X) = cF& whereas
the regular
term is given by the remainders iO=i2+!3+:4
Patching
together
PI(&),
the local results,
we finally
Eg’
obtain
’ +6’
,
cs <
the desired
6.
decomposition
(3.1).
I
Boundary
4. APPROXIMATION
31
element method
WITH
BOUNDARY
ELEMENTS
For the numerical approximation of the variational problem (2.18), we use a boundary element subspace of g;ct (I’). Assume that the smooth surface I’ is given by a regular parameter represetation CC= 4(t), < = (<1,t2) E D in R 2. By the same smooth mapping 4, the boundary 62, of lJ is mapped onto the edge y of I’. With the regular triangulation Dr of 2) and the bijective mapping 4, we can establish a family of regular finite element spaces [36] on the surface I, satisfying with m + 1 > 12 0, that means that the components
of the vectors
degree greater or equal to m belonging to @‘(I’).
ih
E
(4.1)
are piecewise polynomials
g+“‘(r)
of
Moreover, Lh satisties the constraint
P
1$,-Eds=o.
(4.2)
I?
The approximate variational problem of (2.18) reads as follows: such that for all $
Find th E $+l”,
a(ih,$,)
:=
E $T+“‘,
JJ r
r
E(? Y)ih(Y) $h$ dS~ds, =
J r”
g(z) $(zc) dss := (2,$A
(4.3)
For (4.3), we obtain solvability and convergence results by applying the standard Galerkin finite element analysis, as discussed in [15]. Due to the lack of regularity of i at the edge y, with p -4 E H-“(F), by the finite elements, viz. (2.18), we can obtain only low convergence rates. Therefore, the best possible estimate for quasiuniform mesh refinements is of the form
(4.4) where h is the maximal mesh size of the family of partitions of I’, 0 < h < 1. However, the convergence can be improved by using extended finite element spaces which are augmented by singular elements near the edge y. To this end, we define
m+l>l>O,
(4.5)
where
i:+“‘(r) ={iOh and --*
E
s:+“‘(r) (ioh
=
o On
y}
,
1 - e < 3 + 6, 0 < c < 3, 161< 3. Therefore, we have ?;(I’)
c c,,!(I).
The
improved Galerkin scheme on the augmented spaces reads as: such that the variational equations (4.3) hold for all Find ih = ,f(+J-+ X(P) + $,, E &i(r), test functions {h
E %“,(r>.
(4.6)
32
W.L.
WENDLAND,
J. ZHU
With standard arguments for energy norm estimates and the approximation properties of the augmented boundary element space, as shown in [13-151, we obtain the higher rate of convergence,
where E > 0 is any positive number. Unfortunately, the implementation of the above approximation is not an easy task. When l? is an arbitrarily shaped surface in R3, the exact computation of Moreover, b($&) and (_g,$) is rather difficult and costly or, in some cases, even impossible. when we construct constraint additional
the finite
element
space $+“‘(I’)
condition (4.2). It is difficult singular term /Jh(s) p-4 x(p).
to treat
or ::(I’), this condition
we must
satisfy
exactly,
much
the additional more
with
the
To overcome these difficulties, we prefer another approach for the approximation. First, we approximate with the help of curved finite elements, the surface I by rh [29,37,38]. Then we construct the boundary element spaces on rh, by using the method of Lagrangian multipliers. Since we have supposed a smooth mapping between r and the plane domain D, we can easily define an appropriate family of triangulations 2)~ of D. We denote by IT any of these geometrical boundary elements. On each element I< we define an interpolation 4h of 4, such that the mapping 4h of V into w3 is continuous. Then the image of I< by the mapping 4h constitutes one piece rjh of the surface In which we take as an approximation of I’. When I#Jhis a polynomial interpolation of degree k to 4, rh will be defined by a connected parametric surface of degree k. The interpolation is given by curvilinear elements with nodes on I. We denote these geometrical elements of Ij, by rib (i = 1,2, . . . . N), then I,, = brih.
(4.3)
i=l
Under the mappings 4 and c#J;‘, the edge y of I will be approximated which has the same order of interpolation as Ij, to I. Then, for approximating the vector functions i belonging to E-t(I), nomial functions space s;+“‘(rh) not included
we use piecewise
poly-
of degree m on each element which is a finite dimensional
in @-+(I’),
of Ih, that means we construct a finite element subspace of e-*(rh). Although s;+i”(rh) is __r we will take it as the approximation of g ‘(I’),
g+“‘(rh) c Analogously,
by oh, the edge of rh,
we define $+“‘(rh)
In order to simulate the singular Of s:+“‘(rh), namely
&-*(rh),
(4.9)
m+l>l>O.
by using on y the grid points belonging behaviour
at the approximate
to the partition
of rh.
edge Th, we use an augmentation
(4.10) where
m+I>l>O,
and-i
On the other hand, instead of incorporating space, we introduce a Lagrangian multiplier
the constraint condition into the boundary by defining the bilinear form
element
(4.11)
Boundary
where the Vector
ch
is defined
33
by e,, = p
which is the image of the normal approximation
element method
4
0
4;’ := 34 0
0
c to r belonging
‘p,
to the mapping
of c one could also use its Lagrangian
$,>
+
dh($,
A) =
(ijh>
t6),
404;‘. (For a higher order
Q =
interpolation,
constraint (4.2) creates serious difficulties.) Now, we replace the approximate variational problem, as follow: Find (ih,A) E $(Th) x R such that bh(ih,
(4.12)
as in [38]. However,
the variational
vi6
then the
formulation
(4.3) by
E $(rh),
(4.13) dh(thr
where bh(Lj,,ii)
is the bilinear
bh(th,
A’)
=
VA’
0,
form on Th, defined
$,)
:=
E R,
by
JJ E(z,
Y) th(!/)
rh
$,(+Shy
(4.14)
dSh=,
rh
and (sh,$,)
where gh is an approximation
:=
J,,
zh@)
of g. Of course,
J
* $dc)
the second
(4.15)
dshr,
equation
of (4.13) is equivalent
to
$, * ,$, dsh = 0.
rh
Before we analyse the solvability in [39], let us discuss the properties
of (4.13) by using the abstract of the bilinear forms bh(;h,$)
First
of all, we need to compare
and
Lh defined
framework
given by Brezzi
d&,X).
on Th with its counterpart
i defined
on I?. One
could think of using the mapping $ defined by Nedelec in [29], where for z E Th, the vector z - $J(c) is normal to I’ at the point $J(z) E r. With this mapping, one obtains convergence of one order more for the Dirichlet problem of the Laplacian than here, where we use the mapping Q-’ = f$h o 4-l [38]. 1n our case, neither ih 0 $J nor ih 0 Q-l satisfies the constraint condition (4.2) in general. mapping
Therefore,
from $(Th)
both
of them
into @(of (I’). Therefore, ?’ih
where have
do not belong
4) =
J(D
J(D 4) and J(13 C$h) are the Jacobians
J r
rih.zds=
c KeD7
=
we define the mapping 0 dh
o
4-l J(D
corresponding
J (r th
.2)
0
0
dh),
to 4 and 4h, respectively.
45(0 4)) * (z 0 4) dt
J t,, = *l$dSh
0,
we need a
T by
4J(D 4) dt
(~h”4h)‘(~ho4h)J(D4h)dt
r”
Nevertheless,
K
(1’Lh
= =
ih
to #(,!(I’).
Then
we
W.L. WENDLAND,J. ZHU
34
if
ih
E zi(rh)
f
(-
q < l), satisfying
5
the L2 norms,
is easy to check that mapping r. Thus
the constraint
defined
cl b~hb(I’)
where
cl, c2 are positive
bijective
mapping
Hq(I’) =
r$,;
5
constants.
condition
on rh, then r Lh E I&,! (I’). It
on rh and r, respectively,
/~hb(I’,)
We denote
5
c2
are equivalent
under
the
(4.16)
Ilr:hb(r),
by Hq the image
of $(rh)
belonging
to the
r vector function
defined
on I?,
rib
=
:h
04h
v+,
th
E gQh(rh)
(
,
(4.17)
>
then
gr) where the space gq is equipped
with the norm
Il”(flhPh’X +iOh)llfqr),
:= lb
c ~-3(r),
+
(4.18)
_thllZq w IlrPhllH’(r) W W
The index q < 1 is limited
by Theorem
+
OSq
IbtOhlIiyq,
3.1, where q = g + 6, ISI < 3. We note that
Pq = B’(r)
for q < 0. LEMMA 4.1.
We have the inverse ll’~hllZ”
provided
0 < E 5 3.
PROOF.
On each element
inequality <
O
E go, then
Ilr~hll&
5
5
On the other
CIl(r:h)X+
(4.20)
=
by w lb= wk, and we have
(rih)(rihW)
c
(4.21)
;*
5
ciir~hllz-e
Ib’:hWII_Zea
(4.22)
l-
jK(I_h t
o~hWk12+ID(~ho~hWk)12).J(D~h)d~
5
j+hll;o.
we have
the last inequality
LEMMA 4.2.
5
hand,
lb% wll;l = & By interpolating,
cIDwk(~)l
(r, ih)W E H’ with w defined
s
Substitute
h
(4.20) implies lwk(t>l
If rib
(4.19)
K of DT, we define a function
Wk = & dist(<, aK>, Clearly,
E Ho,
Vrth
ch-‘Ilr~hllZ-er
into (4.22) to obtain
For the fundamental
solution
I
(4.19).
E(z, y) of the Stokes equations,
we have the following
estimate
P(~,Y) - JW’-‘(~,@-~(Y))I
5 ch”&,
(4.23)
Boundary element method
where 0-l = q$, ofj- ’ is a continuous of q5. The mesh size of the boundary
mapping elements
35
from Ih to I’ and f$h is the k-degree is denoted by h.
interpolation
For 12 - yj 2 c h, we have
PROOF.
I@(Z)
- o-l(y)
- (Z - y)I 5 lo-‘@!)
- II + IWl(y)
- yI 5 cP+l
5 chkle
according to the interpolation property. In the remaining case lz - yI < c h, the derivate D(O-’ -1) of the mapping 9-l -1 by c hk [29] and the mapping G-l - I is Lipschitzian with the constant c h”. Hence, I’p-l(z)
-O-‘(y)
- (Z - y)I = I(@-+)
- yl,
is bounded
- y)I 5 chkl?: - yl,
- z) - (O-‘(y)
and therefore
~I~-‘(4
- @-?Y)12
- Ix - Y121
=12(W(2)
- Q-‘(y))
- (x - Yh (x - y>) + (W4
- WY)
- (x - Y)121
5 cl hk Iz - yj2 + c2 h2” Iz - y12 5 c hk 12 - y12. Obviously Cl p-l(X)
Now we can estimate
@-l(Y)1
the two terms
I It - Yl I
I@-‘(4
c2
- WY)l.
in the left hand side of (4.23) separately
1
-I
-
b--Y1
l~-l(A-l(Y)l
1
I I@-‘(x) - WY>I” lz-yllip-‘(2)-0-l(y)ll@-1(z)-0-~(y)+;c-yl
=
- 12 - Y121 schk&
Since we have JW(4
- Q-l(Y)13
= l(@-‘(2)
-
@-l(Y))
- Ix -
Y131)
- (3 - Y)21 I(@-+)
p-l(z)
- W-‘(Y)>2 + (Q-W
- (P-l(y)
- Q-‘(Y)9
- Y) + (z -
Yj21
+ z - yl 5 c hkJ+ - yj3,
and
I(zi - Yi>(z:j-Yj)l 5 Cl2 -Yj2, I(% - Yi> (q - Yj) - (Q-‘(Q) - a--‘(Yi)) (W-‘(q) - W’(Yj))l - @-l(yj))l
I
(Xi -
yj)
(a;‘(x) -
(Xj - Yj)
Ix-Y13
-
< (Xi-
@rl(Y))
10-l(x)
(a?l(z) - W’(y)/3
Yi) (Xj - Yj) (IQ-‘(x)
(Xi - Yi) +I I
This completes
the proof.
chk
lx - yl3 10-l(2) (Xj -
Yj)
-
I
@-‘(Y)13 - @-l(y)13
-
(@,:1(X) p-1(2!)
$q.
ipr'(Y))
-
@f’(y)) O-l(y)13
- lx - Y13> (@y’(X)
-
@Tl(J/))
W.L.
36
LEMMA 4.3.
For all th,t’h
WENDLAND,J. ZHIJ
there holds
E Zht(I’h)
PROOF. a(l-ihlrt,‘h)
-
ah(ihyfh)
=c c/J KiE’D7KjEV7
Ki
(r~ho~(q))(rtjh~~(~))E(xo~(~),Yo~(~)) Kj
(th 0
x J(D
4(v))
J(D
4h(d)
=c c/J Kiev7
KiEVI
x
Ki
Kj
d(F))
dv dF
(t_‘h o 4h(‘t))
J(W(d)
J(ME))
E(x
’ +h(t),
Y o tih(v))
dl.ld<
(rt,04(v)> (rl_lh 04(O)
[E(X0 4(t), Y 0 d(‘?)>- E(x 0 dh(‘$ Y O h(d)] x J(W(v)) J(WO) dvdt X
= Using Lemma
r t (y)rti(x)[E(x,y)
JJ r
4.2 and 4.1 we have
Ib(rih,rch)
Here the inequality Laplacian
-
results
is continuous
LEMMA 4.4.
E(~-‘(z),~-‘(~))lds,ds=.
-
r “h
bh(ih,t_lh)l
<
from the fact that
on @-*(I?)
JJ
ch”
x cS3(l?)
r
b’Lh(Y)I
lx - YI
r
the bilinear = 2-3
bfdz)l
ds, ds,
form in the screen
x z-4
[16], and therefore
and k 1 2 1371, th ere exists a positive constant
For h small enough,
bh($ih)
1
problem
for the
on go x go.
1
CYsuch that (4.25)
Q Ib+thll;-+y
for all :h
E $(rh)
PROOF.
Prom the result
Since Lh
e g;(rh),
of the mapping space g,,f
=
{:h
of Lemma
dd$&,
E g&h),
0,
VAER}.
4.3, we have
We have Jr, I,, * c,, dSh = 0. Then
r. We know from Theorem
(I’), therefore
A) =
2.2 that
r th E c;!(r) the billinear
follows from the definition form b(., .) is coercive
on the
37
Boundary element method
which yields
Hence, for k >_ 2 and h sufficiently small, (4.25) follows. THEOREM 4.1.
I
The variational problem (4.13) has a unique solution.
According to Brezzi [39], we need to verify just the following two assumptions: i) There exists a constant cr* > 0, such that
PROOF.
ii) There exists a constant /J* > 0, such that
(4.27)
The first assumption is satisfied because of the result of Lemma 4.4, since I jr ih 1I,_ 4 (r) is a norm equivalent to I Ir ih Ilz_+. ..,h
For the second assumption, take ih = (sign A)?$ to obtain
dh(th,X)
th;;-* s ,h
IAl &,
llihll
’
Iphi
dSh
llnhll _
2
P*
I43
where p* = ]]ph]] 2 c > 0.
I 5. ERROR
ESTIMATES
The errors result mainly from two sources, the approximation of T by Ih and the approximation of the boundary function 1 by th. The errors can be estimated along the ideas of non-conforming finite elements, since the augmented finite element spaces gi have approximation properties. Here we follow Nedelec [38] and Wendland [40]. LEMMA 5.1.
Let sh be the orthogonalprojection
from h2(I’) onto gq(I’).
Then, for 0 _< q 5 l-c,
c>O,t~Z~,wehave
11:- sh$-+,
2
chq++ Iltllzq(r).
(5.1)
The Lemma can be proved by using basic interpolation theory related to the regular finite element partitions and the following estimate [14,15],
II’(PhP-‘X)llZq(r) 5 CIb?hIIz'+~(r)> _ _
E > 0.
-&
(5.2)
We omit the details. THEOREM 5.2.
Let i be the solution of the boundary
solution of the approximate estimate
II! -
T:hll-&,r
problem 5
(4.13).
ch ‘+*
IItIIq,r
integral equation
(2.17), and let th be the
Then for k > 2, q 5 1 - 6, 0 < c 5 3, we have the +
hk-*
Ilrt,lhllO,I’+
II_s -
rzhll+,r.
(5.3)
W.L. WENDLAND,J. ZHU
38
PROOF.
We write
bh($ iA11
=
Ibh(ih,:h
=
i(gh,th
IA21 = I& l&l
=
According
-gh) -
-
r (ih
to Lemma
and the triangle ]I! -
t_lh)
-
-
b(i,r(ih
-fh))i
(g&h
-
t_lh))i
-
5
C@
‘.
Ib(rfh,
o
-
-t_l/a)=A~+Az+As,
-t_lh,ih
t_lh))l
fh))
-
5 -
CIb
F.
-
Vhll$,r w
“t_lhll-+,r
bh(fh,r(:h
b’(Lh
b’(Lh
-ch)l
<
-
-t,lh)l&’
t_‘h)ll-$,r
chk-‘llr~hllO,r
Ib(ih
-t,‘h)ll-+,P
4.4, we have
IIr(ih
-t,‘h)llt;,r
I
inequality
yields
r:hll-+,r
5
Iii
5
C
-
bh(ih
-
“t,‘hll-;,r
II!
-
t_lh,ih
+
-t_lh)
Ik(i
‘$hll-j,F
-
+
5
IAll
+ IA21 + lA31,
t,lh)k-&,r
hk-’
Ik~hIb,r
+
.
Il_S -yhli+,r >
{ The
desired
result
using the result
is obtained
of Lemma t,i$,, _h _
The estimates
in E-‘(I)
by taking
on the right hand side the infimum
and
5.1,
IIL -
‘
II: -
I
can be obtained
Sh
ill-;,r
5 chq+* Ilill0
by using the Aubin-Nitsche
I(: IIL -
and with the integral
for all ch E zi
operator
rihll-l,r
=
(5.4)
trick [41]. Indeed,
-r:hjl)l
sup L+(r)
Il[lll,r
(5.5)
’
Ar defined by (2.17) we have II&
[llo,r
(5.6)
I c Il[llm
yielding I(&(:
II: -
rihll-l,r
2
- r$),&‘f)] (5.7)
sup {@l(r)
IIAF’fllo,r
’
Since
(A&
- Pih), A;‘!)
= (Ar@ - r ih), A;‘[
- Sh(Ar’[)) (5-g)
+ (Al-(: according to the Lemma 5.1, the continuity I(Ar(i
- r$),A;‘l-
&(&[))I
- ‘$),
of Ar implies,
Sh(AFlf))> for the first term of (5.81, the estimate
IClli - rihll-+,r llAF1i
11~- rihll-t,r
Sh(Ar[)ll-&,r
llA~‘fllo,r (5.9)
schq+’
Ilillq,r
+ h” Ilr~hllo,r
+ h+ lbw - rghllA,r w
II&‘fllo,r.
Boundary element method
For the second I(&(:
-
39
term of (5.8), we have
r$),Sh(A~‘f))l = I(&:,
S(Ai;‘f))
= Ib(:, sh(A,li)
- (Ar(r$GWG1f))l - (_gh, r-l(sh(AF’f)))r,
+bh(th,r-l(S~(A,‘f))-b(rt~,S~(A~lf))l (5.10) < c I(s - r_Qh, sh(AF1{))l + Ibh($,,r-l(Sh(AF1f)) I Inserting
c
t
119 ,w- rghllo,r e.
(5.9), (5.10) into (5.7) yields the desired
IItW- r _thll-l,r
I c {
h” IlJllo,r
Now, we are in the position and the approximate
solution
+ hk IIr$Ilo,r
(~h,ph)
the error between
J Jqx, J P(X>
Ph(X) =
rh
THEOREM 5.3. For x E Sir with dist(x,I’) k~2,O~q~l--~,O
I
+,,
q
h” Iltllo,r
+ h3 IIf - rzhII+,r
the solution
- Ph(2)l
5
c (1+ 4x9 r)) (4x9
r)Y
(5.11)
.
I given by (2.16)
(5.12)
Y) pb(Y)
hy,
= d(x, I’) 2 6 > 0, the following
estimate
holds for
+ hq+’ Il~llq,r + Il_s - r_shllo,r
b(z)
(z,p)
1
bw,
Y) $(Y)
rh
- 2Ld4l
IK1fllw
given by
ya(x) =
I:(4
>
estimate,
+ hq+’ IlLllq,r + 119- rpllo,r
to estimate
- b(rthrSdAF1f))l
h” IlLllo,r
+ hi
Il_s- rzhll+,r
,
(5.13)
+ h3
Il_s- r_shll+,r .
(5.14)
+ hq+l Ilillq,r +
119 shllo,r _ - rw
>
PROOF.
V(x) - Uh(X)I= J, E(x, Y) (L(Y)
5
lIW~,Y)lli,r
JA(Y)) ds, + r[E(r,Y)- E(z,4-‘(~))1r $4~)4, J + IIWx:,Y) - E(c,Q-‘b))llo,r Ilr$llo,r. II: - rihll-l,r -
(5.15) Because
of d(x,l?)
2 6 > 0 we have IIWz,Y)lli,r
IIWX, d - E(x, @W)llo,r MCM 15:6-D
L dtx: rj 1
-_
(5.16)
I &.
(5.17)
40
W.L.
WENDLAND,
Hence, the result (5.13) follows by collecting similar arguments if we use the estimate,
REMARK
J. ZHU
(5.15)-(5.17),
(5.11).
The result (5.14)
Since q is limited by 1 - 6, (0 < E 5 $), the estimate
2.
O(P The errors of velocity,
(5.13)
follows with
has the order
with k 2 2.
+ P-f),
pressure and their successive
derivates
have the same order of convergence.
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