Directory UMM :Data Elmu:jurnal:J-a:Journal of Computational And Applied Mathematics:Vol104.Issue2.1999:
Journal of Computational and Applied Mathematics 104 (1999) 173–191
Two-level method and some a priori estimates in unsteady
Navier–Stokes calculations
Maxim A. Olshanskii ∗
Dept. Mech. and Math., Moscow State University, Moscow 119899, Russia
Received 3 December 1997; received in revised form 3 December 1998; accepted 3 December 1998
Abstract
A two-level method proposed for quasielliptic problems is adapted in this paper to the simulation of unsteady incompressible Navier–Stokes
ows. The method requires a solution of a nonlinear problem on a coarse grid and a solution of
linear symmetric problem on a ne grid, the scaling between these two grids is superlinear. Approximation, stability, and
convergence aspects of a fully discrete scheme are considered. Stability properties of the two-level scheme are compared
c 1999 Elsevier
with those for a commonly used semi-implicit scheme, some new estimates are also proved for the latter.
Science B.V. All rights reserved.
Keywords: Navier–Stokes equation; Two-level method; A priori estimate
1. Introduction
Numerical simulation of unsteady incompressible viscous
ow is a fundamental problem both of
numerical analyses and
uid dynamics. The governing equations are the incompressible Navier–
Stokes ones:
@u
− u + (u · 3)u + 3p = f ;
@t
div u = 0;
in
× (0; T ];
(1.1)
with a given force eld f , kinematic viscosity ¿ 0;
⊂ Rn ; n = 2; 3. The velocity vector function
u and pressure scalar function p to be found are subject to some boundary conditions, which we
∗
Corresponding author. E-mail address: [email protected] (M.A. Olshanskii).
c 1999 Elsevier Science B.V. All rights reserved.
0377-0427/99/$ - see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 9 ) 0 0 0 5 6 - 4
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
assume to be Dirichlet and homogeneous for the velocity:
u = 0 on @
× [0; T ]
(1.2)
and initial condition at t = 0:
u = u0 (x)
in
:
(1.3)
R
Another common assumption is
p(x; t) dx = 0 for the pressure p(x; t) for all t ∈ (0; T ]. For
a detailed consideration of mechanical, mathematical, and computational aspects associated with
Navier–Stokes problem we refer to, among others, [9,12,13,25].
Belonging of a weak solution of (1.1) – (1.3) to a space of solenoidal functions and domination
of nonlinear convection terms in the momentum equation for moderate and high Reynolds numbers
(small and=or high velocities) are commonly considered to be the main diculties in numerical and
computational theory of Navier–Stokes equations. The objective of this paper is to apply two-level
mesh reduction method to the treatment of convection phenomenon in unsteady numerical simulations. The method is closely related to the nonlinear Galerkin method [1,18–20] and was developed
in [28,3,14 –16].
Using numerous solution schemes for (1.1) – (1.3) with u−p coupling (see, e.g., [5,11]) or operator
splitting [6], one has to choose between a fully implicit treatment of nonlinear terms in (1.1) or some
of their linearization using, e.g. extrapolation in time. In the rst case one faces the necessity of
solving nonlinear problem of the Burgers or Navier–Stokes type on each time step, otherwise linear
and even symmetric problems can be obtained on each time step. However, the latter approach
may cause stability problems, i.e. time step becomes subject to some conditions involving spatial
discretization and=or Reynolds number. We refer to [25] for theoretical considerations and [27] for
experimental comparison of various schemes.
The method presented and studied here can be roughly described as follows. For numerical solution
of (1.1) – (1.3) choose some spatial nite element or nite di erence discretization and two meshes:
the coarse one with the step H and the ne one with the step h such that h ∼ H ; ¿1: For temporal
integration on [t0 ; t0 + ] with time step ¿ 0 make fully implicit step on the coarse grid and obtain
uH ; pH for t0 + via solution of nonlinear problem on the coarse grid. Then extrapolate uH on the
ne grid. Using linearization of convective terms about uH , make semi-implicit integration step on
the ne grid and obtain uh ; ph for t0 + solving linear symmetric problem on the ne grid.
Compared to the recent studies [1,20] for unsteady problems, the primary innovations in this paper
are the treatment of fully discrete (both in time and space) case of the method. Hence the stability
results are quite important. We also avoid the use of any intermediate nite element subspaces.
Therefore well established solvers can be readily applied to the auxiliary nite element problems.
In Section 2 of the paper we introduce necessary notations and preliminary results. The algorithm to
be studied is described in Section 3. Where possible we compare results obtained for the constructed
algorithm with appropriate ones for commonly used algorithm based on a fully explicit treatment
of the nonlinearity via extrapolation in time. For this purpose some results on stability of the latest
algorithm are also proved.
In Section 4 some approximation results are proved. In particular, it follows that in the case
of linear velocity – constant pressure nite elements the scaling h ∼ H 2 gives the same order of
spatial discretization error as the usual Galerkin method with mesh size h. We note that this relation
between coarse and ne grids is somewhat less impressive than the one recovered in the framework
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
175
of Newton-type methods (see [2,28], and references cited therein). However, application of the latter
techniques for the problem considered requires highly nonsymmetrical problems to be solved on
each time step. Moreover, it causes the appearance of undesirable reactive term in the linearized
equation (see also [16]).
The stability of the schemes is studied in Section 5. It is proved that while usual semi-implicit
scheme requires time step to be small enough to guarantee the stability, the two-level scheme (at
least theoretically) requires the spatial step to be small enough to ensure stability. Moreover, if
the problem is regular enough, the use of high order nite elements weakens the condition on h.
In Section 6 we consider convergence of the two-level scheme. The appropriate convergence for
velocity is proved in two dimensions.
Throughout the paper we deal with saddle point formulations of the corresponding nite element
problems (discrete velocity is not solenoidal in general). This causes some extra complications but
corresponds to real-life situations.
2. Preliminaries
Throughout the paper we assume
to be a bounded domain in R2 or R3 with suciently smooth
boundary, or a convex polygon (polyhedron).
Later on we need the following functional spaces:
H01 ≡ {u ∈ W21 (
)2 : u = 0 on @
};
V ≡ {u ∈ H01 : div u = 0 in
}
with energy scalar product (u; C)1 = (3u; 3C); u; C ∈ H01 ;
L0 ≡ {u∈ L2 (
)2 : div
u · n = 0 on @
};
Z u = 0 in
;
p dx = 0
L2 =R ≡ p ∈ L2 (
):
with L2 -scalar product. Let H −1 be a dual, with respect to L2 -duality, space to H01 with the corresponding norm:
k f k−1 = sup
06=u∈H01
h f ; ui
;
k u k1
f ∈ H −1 :
We also use Sobolev spaces of a real exponent s: H s (
) with a norm k · ks .
The following forms are associated with the Navier–Stokes problem:
a(u; C) = (u; C)1 ; u; C ∈ H01 ;
a (u; C) = (u; C) + (u; C)1 ; u; C ∈ H01 ; ¿ 0;
b(p; u) = (p; div u); p ∈ L2 =R; u ∈ H01 ;
N (u; C; w) = 21 [((u · 3)C; w) − (u · 3)w; C)]; u; C; w ∈ H01 :
The weak formulation of (1.1) – (1.3) is to nd
u ∈ L2 (0; T ; H01 ) ∩ L∞ (0; T ; L0 ) and
p ∈ L2 (0; T ; L2 =R)
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
satisfying for given f ∈ L2 (0; T ; H −1 ) in the sense of distributions on (0; T ]
@u
; C + a(u; C) + N (u; u; C) − b(p; C) = h f ; Ci;
@t
b(q; u) = 0 ∀{C; q} ∈ H01 × L2 =R;
u(0; x) = u0 (x) in
;
or, alternatively, to nd
u ∈ L2 (0; T ; V ) ∩ L∞ (0; T ; L0 )
satisfying
d
(u; C) + a(u; C) + N (u; u; C) = h f ; Ci
dt
u(0; x) = u0 (x) in
;
∀C ∈ V ;
De Rham theorem connects both formulations (cf. [9,25]).
It is worth mentioning that a weak solution de ned above exists and for n=2 is unique; moreover,
some extra assumptions on f ; @f =@t, and u0 provide u ∈ L2 (0; T ; H 2 (
)2 ) (cf. [17]).
Further, we shall use the following estimates due to [12,25]:
b(p; u)6 k p k0 k u k1
( f ; u)6 k f k−1 k u k1
|((u · 3)C; w)|6c k u kL4 k C k1 k w kL4
|N (u; C; w)|6c k u ks k C k1 k w k1
|N (u; C; w)|6c k u k1 k C ks k w k1
|N (u; C; w)|6c k u k0 k C k1 k w k1+s
∀p ∈ L2 =R; u ∈ H01 ;
∀f ∈ H −1 ; u ∈ H01 ;
∀u; C; w ∈ H01 ;
∀u; C; w ∈ H01 ;
∀u; C; w ∈ H01 ;
∀u; C ∈ H01 ; w ∈ H01 ∩ H 1+s :
(2.1)
From now on we agree to consider s as an arbitrary number from (0; 1] for two-dimensional problem
and s ∈ [ 21 ; 1] for three-dimensional problem, if it is not stated otherwise. We also need
√
k u k2L4 6 2 k u k0 k u k1
∀u ∈ H01 ;
∈ R2 ;
(2.2)
k u k2L4 62 k u k01=2 k u k13=2 ∀u ∈ H01 ;
∈ R3 ;
s
1−s
1
∀u ∈ H0 ; s ∈ (0; 1):
k u ks 6c k u k0 k u k1
Here and later on we denote by c(
); c; c0 ; c1 ; : : : some constants independent of both spatial and
temporal discretization parameters (h and ) and , otherwise we shall use, for example, c() for a
constant depending possibly on .
So called j-inequality:
|ab|6
1 2 j 2
|a| + |b| ;
2j
2
∀a; b ∈ R; j ¿ 0;
will be also used throughout the paper.
Let us denote by h a mesh size parameter, and denote by Hh a nite element subspace of H01 and
by Qh a nite element subspace of L2 =R. Assume that for some real h0 ¿ 0; positive integers k1 ; k2 ;
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
177
and h ∈ (0; h0 ] the following hypotheses hold. Examples of such nite element spaces can be found
in [5,9,11].
(H1) Approximation hypothesis
(a) For all u ∈ H01 ∩ H d (
)n ; with integer d ¿ 0
inf k u − C k1 6chl k u kd ;
l = min(k1 ; d − 1):
C∈Hh
(b) For all p ∈ L2 =R ∩ H d (
); with integer d ¿ 0
inf k p − q k0 6chl k p kd ;
q∈Qh
l = min(k2 ; d):
(H2) Inverse hypothesis
For all u ∈ Hh
k u k1 6h−1 k u k0 :
(H3) Stability hypothesis. There exists some real constant c0 ¿ 0 independent on h such that
sup
06=u∈Hh
b(p; u)
¿c0 k p k0 ;
k u k1
∀p ∈ Qh :
The above hypotheses give the following standard result concerning Stokes problem.
Lemma 2.1. For any given {u; p} ∈ H01 × L2 =R; there exist unique uh ∈ Hh and ph ∈ Qh such that
a(u − uh ; C) − b(p − ph ; C) = 0;
b(q; u − uh ) = 0;
∀{C; q} ∈ Hh × Qh :
Moreover;
k u − uh k1 + k p − ph k0 6c(
) inf k u − C k1 + inf k p − q k0 ;
C∈Hh
q∈Qh
k u − uh k0 6c(
)h inf k u − C k1 :
C∈Hh
Denote also by Vh a subspace of discretely divergence-free functions from Hh : Vh = {uh ∈
Hh : b(q; uh ) = 0 ∀q ∈ Qh }.
3. Two algorithms for the unsteady problem
The common and e ective way of numerical treatment of time-dependent problems is separation
of spatial and temporal discretizations. For spatial discretization one can choose nite element, nite
di erence, spectral methods, while for the temporal discretization the nite di erence method is the
most natural choice. Two schemes described below utilize this idea.
First let us consider one widely used semi-implicit scheme for solving unsteady Navier–
Stokes problem (1.1) – (1.3). For given uh0 ∈ Hh nd {uhi+1 ; phi+1 } ∈ Hh × Qh for i = 0; 1; : : : from
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
the relations
a i (uhi+1 ; C) − i b(phi+1 ; C) + i N (uhi ; uhi ; C) = i (f i+1 ; C) + (uhi ; C);
b(q; uhi+1 ) = 0
(3.1)
∀ {C; q} ∈ HH × QH :
where i ¿ 0 are steps of temporal
discretization,
which we assume generally to be variable. In (3.1)
R
Pi
−1 t2
i+1
and below we set f
= i t1 f dt with t2 = k=0 k ; t1 = t2 − i .
On every time step scheme (3.1) requires solution of the generalized Stokes problem, e ective
solution methods for such problem are available [22,8]. With the above assumptions scheme (3.1)
is conditionally stable and the sucient conditions for stability from [25] are 6c(;
; f (t); u0 )hn ,
(
⊂ Rn ); moreover, the careful reading of the proof gives c = O(2 ); → 0.
The scheme being proposed works as follows. For given uh0 ∈ Hh nd {uhi+1 ; phi+1 } ∈ Hh × Qh for
i = 0; 1; : : : from the two-level algorithm:
Find {uHi+1 ; pHi+1 } ∈ HH × QH
a i (uHi+1 ; C) − i b(pHi+1 ; C) + i N (uHi+1 ; uHi+1 ; C) = i ( f i+1 ; C) + (uhi ; C);
b(uHi+1 ; q) = 0
(3.2a)
∀ {C; q} ∈ HH × QH ;
with known {uHi+1 ; pHi+1 } nd {uhi+1 ; phi+1 } ∈ Hh × Qh
a i (uhi+1 ; C) − i b(phi+1 ; C) + i N (uHi+1 ; uHi+1 ; C) = i ( f i+1 ; C) + (uhi ; C);
b(uhi+1 ; q) = 0
(3.2b)
∀ {C; q} ∈ Hh × Qh :
The above two-level algorithm can be observed simultaneously from two points of view. On
the one hand it can be considered as an improvement of scheme (3.1) by obtaining a qualitative
information (on a coarse grid) about the solution at time (t0 + i ). On the other hand, let us consider
fully implicit unconditionally stable scheme for (1.1) – (1.3) without any spatial discretization: for
given u0 ∈ H01 nd {ui+1 ; pi+1 } ∈ H01 × L2 =R for i = 0; 1; : : : from
a i (ui+1 ; C) − i b(pi+1 ; C) + i N (ui+1 ; ui+1 ; C) = i ( f i+1 ; C) + (ui ; C);
b(u
i+1
; q) = 0
∀ {C; q} ∈
H01
(3.3)
× L2 =R
then (3.2a) and (3.2b) is a straightforward application of the two-level method from [15] to the
solution of steady nonlinear problem arising on every time step of (3.3). Note that the two-level
algorithm (3:2) requires on every time step solution of the nonlinear problem of Navier–Stokes type
on the coarse grid and solution of the linear symmetric problem of Stokes type on the ne grid.
Remark 3.1. To improve the accuracy and stability of scheme (3.1) such variants of (3.1) as Crank–
Nicolson, fractional-step [21], with high-order extrapolation in time of nonlinear terms, multistep [4],
with upwinding (see, e.g. [24]) are known and used in practice. We note that all or at least most of
these improvements are quite applicable to (3.2a) and (3.2b) as well as splitting techniques leading
to a class of projection type methods [6,10].
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
4. Approximation
Consider the following problem. For given g ∈ H −1 nd {u; p} ∈ H01 × L2 =R satisfying
a (u; C) − b(p; C) + N (u; u; C) = (g; C);
(4.1)
∀ {C; q} ∈ H01 × L2 =R:
b(q; u) = 0
(4.1) exists (see, e.g. [25, Lemma √
IV.4.3]). For the uniqueness it is sucient
√ Solution of problem
2 k g k−1 6 1=2 3=2 for two-dimensional problem and 2 2 k g k−1 6 3=4 7=4 for three-dimensional
problem. If g ∈ Ln (
) then sucient conditions can be written as k g k0 6 and 2 k g k0 6 1=2 3=2 ,
respectively. The proof is quite standard, it follows from (2.1), (2.2) and a priori estimates for weak
solutions of (4.1):
2 k u k20 + k u k21 6( )−1 k g k2−1 ;
k u k20 +2 k u k21 6 k g k20
and certain relation for the di erence w = u − u˜ between two weak solutions of (4.1) that can be
obtained from (4.1) with C = w.
Two-level method for problem (4.1) means: nd sequentially {uH ; pH } ∈ HH × QH and {uh ; ph } ∈
Hh × Qh from
a (uH ; C) − b(pH ; C) + N (uH ; uH ; C) = (g; C);
b(q; uH ) = 0
(4.2a)
∀ {C; q} ∈ HH × QH
and
a (uh ; C) − b(ph ; C) + N (uH ; uH ; C) = (g; C);
b(q; uh ) = 0
(4.2b)
∀ {C; q} ∈ Hh × Qh :
Theorem 4.1. For given g ∈ H −1 let {u; p} ∈ H01 ∩ H 1+s (
)n × L2 =R; n = 2; 3 be a solution to
problem (4:1) and let {uh ; ph } ∈ Hh ×Qh be a solution to problem (4:2a); (4:2b); then the following
estimate is valid:
k uh − u
k20
+ k uh − u
k21
+ k ph − p
+
h ( ; h; ) inf k p − q
q∈Qh
2
k20
k20
+ k uH − u
6C(
; s) (h2 + ) inf k u − C k21
k20 k u
k21+s
+ k uH − u
C∈Hh
k02−2s k uH
−u
k2+2s
1
; (4.3)
with
h = min( =; 2 =h2 ).
Proof. For given {u; p} from (4.1) let {uS ; pS } ∈ Vh × Qh be a solution to the Stokes problem
(3uS ; 3C) − b(pS ; C) = (3u; 3C) − b(p; C);
b(uS ; q) = 0; ∀{C; q} ∈ Hh × Qh ;
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
then, by virtue of Lemma 2.1, the following estimates are valid:
k u − uS k1 6c(
) inf k u − C k1 + inf k p − q k0 ;
C∈Hh
q∈Qh
k u − uS k0 6c(
)h inf k u − C k1 :
(4.4)
C∈Hh
We also choose pI = arg minq∈Qh k p − q k0 :
We choose in (4.1) and (4.2b) C = uS − uh and substract (4.2b) from (4.1). Then we obtain
k uS − uh k20 + k uS − uh k21 =a (uS − u; uS − uh )
+ (b(p − pI ; uS − uh ) − N (u; u; uS − uh ) + N (uH ; uH ; uS − uh ))
= a (uS − u; uS − uh ) + (b(p − pI ; uS − uh ) − N (u − uH ; u; uS − uh )
+ N (u − uH ; u − uH ; uS − uh ) − N (u; u − uH ; uS − uh ))
1
6 21 k u − uS k20 + 12 k uS − uh k20 + k u − uS k21 + 21 k uS − uh k21
2
+ c (k p − pI k0 + k u − uH k0 k u k1+s + k u − uH k01−s k u − uH k1+s
1 ) k uS − uh k1 :
(4.5)
Using Cauchy inequality, we get
c (k p − pI k0 + k u − uH k0 k u k1+s + k u − uH k01−s k u − uH k1+s
1 ) k uS − uh k1
)
6c2 −1 (k p − pI k20 + k u − uH k20 k u k21+s + k u − uH k02−2s k u − uH k2+2s
1
+
k uS − uh k21 ;
4
(4.6a)
and
c (k p − pI k0 + k u − uH k0 k u k1+s + k u − uH k01−s k u − uH k1+s
1 ) k uS − uh k1
c
(k p − pI k + k u − uH kk u k1+s + k u − uH k01−s k u − uH k1+s
6
1 ) k uS − uh k0
h
c 2 2 2
)
6 2 (k p − pI k20 + k u − uH k20 k u k21+s + k u − uH k02−2s k u − uH k2+2s
1
h
+ 14 k uS − uh k20 :
(4.6b)
Now in order to estimate k uh − u k, we apply the triangle inequality: k uh − u k 6 k uh − uS k +
k uS − u k; estimates (4.4), (4.5), and one of (4.6a) and (4.6b). To estimate k ph − p k0 we use the
following standard arguments (see, e.g. [9]). From (4.1) and (4.2b) we get for all {C; q} ∈ Hh × Qh
b(ph − q; C) = a (u − uh ; C) + b(p − q; C) + N (u; u; C) − N (uH ; uH ; C):
Thus, similar to (4.5) we obtain for all {C; q} ∈ Hh × Qh :
b(ph − q; C)6(c + )1=2 (k uh − u k20 + k uh − u k21 )1=2 k C k1
+ c (k p − q k0 + k u − uH k0 k u k1+s + k u − uH k01−s k u − uH k1+s
1 ) k C k1 :
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
181
Dividing both sides of the last inequality by k C k1 for 0 6= C ∈ Hh , taking sup over all such C and
inf over all q ∈ Qh , and utilizing hypothesis (H3) we get
c0 inf k ph − q k0 6(c + )1=2 (k uh − u k20 + k uh − u k21 )1=2
q∈Qh
+ c inf k p − q k0 + k u − uH k0 k u k1+s + k u − uH k01−s k u − uH k1+s
:
1
q∈Qh
Now from the triangle inequality: k ph − p k0 6inf q∈Qh k ph − q k0 +inf q∈Qh k p − q k0 we get an
estimate on k ph − p k0 . Combining estimates for k uh − u k20 + k uh − u k21 and 2 k ph − p k20 ; we
prove the theorem.
Remark 4.2. Estimate (4.3) shows a loss of convergence in pressure for too small time steps .
Indeed, the same standard arguments as in the proof of Theorem 4.1 when applied to some nite
element approximation of linear (generalized) Stokes problem
−1 u − u + 3p = f ;
div u = 0;
(4.7)
u|@
= 0
give the estimate
k uh − u k20 + k uh − u k21 + 2 k ph − p k20
6C(
) (h2 + ) inf k u − C k21 +
h inf k p − q k20 :
C∈Hh
q∈Qh
(4.8)
Assumed to be optimal for pressure this estimate recovers the necessity of h2k2 ≪ or h2(k2 −1) ≪1 for
pressure nite elements and hk1 +1 ≪ for velocity nite elements to ensure the convergence of pressure
in L2 (
): These conditions on h; ; and force us to use schemes of equal order (k2 = k1 + 1¿2)
or high order spacial interpolation (k2 ¿2; k1 ¿2), and implicit schemes, which do not require to
be too small.
Remark 4.3. Further we shall need an estimate for k u −uH k0 ; where u is a solution to (4.1) and uH
is a solution to (4.2a). To obtain such an estimate assume the solution of (4.1) to be nonsingular.
With our assumptions on
, problem (4.7) is W22 -regular. Hence we use standard arguments from
[9]: duality estimates for problem (4.7) (Theorem II.1.2), Theorems IV.3.3, IV.3.5, and IV.4.2,
nally for {u; p} ∈ H01 ∩ H d (
)n × L2 =R ∩ H d−1 (
) with integer d ¿ 0 and suciently small H we
obtain
k u − uH k20 6C(
; ; ; k u k‘1 +1 ; k p k‘2 )(H 2(‘1 +1) + H 2(‘2 +1) )
(4.9)
with ‘1 = min(k1 ; d − 1); ‘2 = min(k2 ; d − 1). This estimate provides us with the choice of scaling
h = O(H 1+1=‘1 ) to obtain an optimal accuracy O(h2‘1 ) in (4.3). However, in (4.11) a dependence of
the constant C on and is implicit.
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
Corollary 4.4. Let {u; p} ∈ H01 ∩ H d (
)2 × L2 =R ∩ H d−1 (
) for some integer d ¿ 0 be a solution
to problem (4:1) and {uH ; pH ; uh ; ph } ∈ HH × QH × Hh × Qh be a solution to problem (4:2a); (4:2b)
then for suciently small H the following estimates hold:
k uh − uH k20 6C(
; ; ; k u k‘1 +1 ; k p k‘2 )(H 2(‘1 +1) + H 2(‘2 +1) )
with ‘1 = min(k1 ; d − 1); ‘2 = min(k2 ; d − 1).
Proof. The inequality
k uh − uH k20 62(k u − uh k20 + k u − uH k20 )
together with (4.9), Theorem 4.1, and assumption h ¡ H give the result.
Another estimate on uH − uh without explicitly involving solution of (4.1) is given by the following
theorem.
Theorem 4.5. For given g ∈ H −1 let {uH ; pH } ∈ HH × QH and {uh ; ph } ∈ Hh × Qh be a solution
to problem (4:2a); (4:2b); then the following estimate is valid:
k uh − uH k20 + k uh − uH k21
2
6C(
) (H + ) inf k uh − C
C∈HH
k21
+( +
H ) inf k ph − q
q∈QH
k20
:
(4.10)
Proof. For given {uh ; ph } from (4.2a) let {uS ; pS } ∈ VH × QH be a solution to the Stokes problem
(3uS ; 3C) − b(pS ; C) = (3uh ; 3C) − b(ph ; C);
b(q; uS ) = 0;
∀{C; q} ∈ HH × QH ;
then the following estimates are valid:
k uh − uS k1 6c(
) inf k uh − C k1 + inf k ph − q k0 ;
q∈QH
C∈HH
k uh − uS k0 6c(
) H inf k uh − C k1 ;
(4.11)
C∈HH
Choose also pI = arg minq∈QH k ph − q k0 :
Let us take in (4.2a) and (4.2b) C = uS − uH and substract (4.2b) from (4.2a), then we obtain
k uS − uH k20 + k uS − uH k21 = a (uS − uh ; uS − uH ) + b(ph − pI ; uS − uH )
6 k uS − uh k20 + k uS − uh k21 +2 k ph − pI k0 k uS − uH k1 :
(4.12)
Estimate the last term as
and
2 k ph − pI k0 k uS − uH k1 6 −1 k ph − pI k20 + k uS − uH k21
(4.13a)
2 k ph − pI k0 k uS − uH k1 62 2 H −2 k ph − pI k20 + 21 k uS − uH k20 :
(4.13b)
Now apply the triangle inequality, estimates (4.11), (4.12) and one of the estimates (4.13a) and
(4.13b) and get (4.10). The theorem is proved.
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
183
5. Stability
We understand the stability of schemes (3:1)–(3:3) as a validation of some a priori estimates for
nite element solutions obtained using these schemes. Week solution u ∈ L2 (0; T ; H01 ) ∩ L∞ (0; T ; L0 )
of problem (1.1) – (1.3) satis es energy estimate
k u(t) k20 +2
Z
0
t
k u(s) k2 ds6 k u0 k20 +2
Z
0
t
| ( f (s); u(s)) | ds:
Further we will deduce some conditions that provide nite element solution with nite analogue of
energy estimate (see also [25,26]).
The following theorem gives condition on i that ensure (3.1) to be stable. This condition depends
on uhi ; Corollary 5.2 shows that the condition can be strengthened and made depend only on the
given data.
Theorem 5.1. For
∈ Rn ; n = 1; 2 any natural m¿0 and any i ; i = 0; : : : ; m satisfying condition
!
hn
h2
i 6min
;
;
82 k uhi k20 22
i = 0; 1; : : : ;
(5.1)
the following estimate holds for uhi determined from (3:1):
k uhm+1 k20 +
m
m
m
X
1X
2X
i k uhi+1 k21 6 k u0 k20 +
k uhi+1 − uhi k20 +
i k f i+1 k2−1 ;
4 i=0
i=0
i=0
(5.2)
Proof. Let us take in (3.1) C = 2uhi+1 , we get the equality
k uhi+1 k20 − k uhi k20 + k uhi+1 − uhi k20 +2 i k uhi+1 k21 = −2 i N (uhi ; uhi ; uhi+1 ) + 2 i ( f i+1 ; uhi+1 ):
(5.3)
Right-hand side in (5.3) can be estimated as follows:
−2 i N (uhi ; uhi ; uhi+1 ) + 2 i ( f i+1 ; uhi+1 ) = −2 i N (uhi ; uhi ; uhi+1 − uhi ) + 2 i ( f i+1 ; uhi+1 )
2 i
62 i h−n=2 k uhi k0 k uhi k1 k uhi+1 − uhi k0 + 12 i k uhi+1 k21 +
k f i+1 k2−1
2 i
622 2i h−n k uhi k20 k uhi k21 + 21 k uhi+1 − uhi k20 + 12 i k uhi+1 k21 +
k f i+1 k2−1 :
From (5.3) and the last estimate we get
2 i
k f i+1 k2−1 :
k uhi+1 k20 + 12 k uhi+1 − uhi k20 + 23 i k uhi+1 k21 −42 2i h−n k uhi k20 k uhi k21 6 k uhi k20 +
(5.4)
By virtue of condition (5.1), for i the following estimates are valid:
3
2 i
k uhi+1 k21 −42 2i h−n k uhi k20 k uhi k21 ¿ i k uhi+1 k21 + 12 i (k uhi+1 k21 − k uhi k21 )
¿ i k uhi+1 k21 − 21 i k uhi+1 − uhi k21 ¿ i k uhi+1 k21 − 14 k uhi+1 − uhi k20 :
Now we get from (5.4)
k uhi+1 k20 + 41 k uhi+1 − uhi k20 + i k uhi+1 k21 6 k uhi k20 +
2 i
k f i+1 k2−1 :
(5.5)
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
Taking for i = 0; : : : ; m a sum of (5.5), we obtain (5.2) and prove the theorem.
Corollary 5.2. There exist some constant c = c(
; f (t); u0 ) ¿ 0 such that for any set of i ; i =
0; : : : ; m; satisfying condition
!
h2
i 6min c h ; 2 ;
2
n 2
i = 0; 1; : : : ;
(5.6)
and uhi determined from (3:1) estimate (5:2) holds with any natural m¿0.
Proof. Estimate (5.2) for any set of i satisfying (5.1) provides k uhi k20 6c1 (u0 ; f (t)) −1 . Setting in
(5.6) c = (82 c1 )−1 , we prove the corollary.
Note that Corollary 5.2 recovers asymptotically the same condition on the constant step as
Lemma 5.3 from [25]. However, if in a particular problem k uhi k0 depends on in some more
advantageous way, then the condition on can be weakened; this is the situation which we have
sometime in practice (see [23] for calculations of critical (for stability) for scheme (3.1) in the
case of one substantially nonlinear and unsteady
ow).
The following theorem sharpens the estimate on with respect to . The theorem requires that
function f satis es k f (t) kL2 (G) 6c ¡ ∞ for some T ¿ 0; G = (0; T ) ×
.
Theorem 5.3. For any T ¿ 0 there exist some constant c = c(T;
; f (t); u0 ) ¿ 0 such that for any
satisfying the condition
!
h2
6min chn ;
;
22
i = 0; 1; : : : ;
(5.7)
and for any natural m¿0 such that m 6T the following estimate holds:
k uhm+1
k20
with t = m :
m
m
X
1X
+
k uhi+1 − uhi k20 +
i k uhi+1 k21 62 exp(t) k u0 k20 +
4 i=0
i=0
Z
0
t
kf
k20
(5.8)
Proof. Fix T and choose satisfying (5.8) with
1
[2 exp(T )(k u0 k20 + k f k2L2 (G) )]−1 :
82
Let us prove by induction with respect to m the following estimate:
c=
k uhm+1
k20
(5.9)
!
m
m
m
X
X
1X
+
k f i+1 k20 :
k uhm+1 − uhm k20 +
k uhm+1 k21 6(1 − )−m k u0 k20 +
4 i=0
i=0
i=0
(5.10)
Consider the case m = 0. Indeed from (5.9) and (5.7) we have 6hn (82 k u k20 )−1 and condition
(5.1) for 0 = is ful lled. Similar to the proof of Theorem 5.1 let us estimate right-hand side of
(5.1) with i = 0. However, now we apply the inequality 2 ( f 0 ; uh1 )6 k uh1 k20 + k f 0 k20 instead of
2 ( f 0 ; uh1 )6 21 k uh1 k21 +(2=) k f 0 k2−1 .
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
185
Finally, we get
k uh1 k20 + 41 k uh1 − u0 k20 + k uh1 k21 6 k u0 k20 + k uh1 k20 + k f 0 k20
that implies
k uh1 k20 + 41 k uh1 − u0 k20 + k uh1 k21 6(1 − )−1 (k u0 k20 + k f 0 k20 ):
Thus, the “trivial” case of m = 0 is checked.
Suppose now that (5.10) is proved for all m = 0; 1; : : : ; k. Let us prove it for m = k + 1, assuming
(k + 1) 6T . From inductive hypothesis we have (5.10) for m = k, hence
k uhk+1 k20 6(1 − )−k (k u0 k20 + k f k2L2 (G) )
and condition (5.1) for k = is valid. With the same arguments as in the proof of Theorem 5.1
and another estimate for 2 ( f k ; uhk+1 ) we get
k uhk+1 k20 + 41 k uhk+1 − uhk k20 + k uhk+1 k21 6 k uhk k20 + k uhk+1 k20 + k f k k20 ;
and with inductive hypothesis
k uhk+1 k20 + 41 k uhk+1 − uhk k20 + k uhk+1 k21
6(1 − )
−1
"
(1 − )
−1
+ (1 − ) k f
−
k
−(k−1)
k20
ku
0
k20
6(1 − )
−k
+
k−1
X
i=0
ku
0
kf
k20
k
k
X
1X
i k uhk k21 :
k uhk − uhk−1 k20 −
4 i=1
i=1
+
i+1
k20
k
X
i=0
!
k
k
X
1X
−
k uhk − uhk−1 k20 −
i k uhk k21
4 i=1
i=1
kf
i+1
k20
#
!
Thus, (5.10) is proved for all m:m 6T: Estimate (5.8) follows from (5.10) and obvious relations:
(1 − )
Z
0
−m
t
= 1−
m
t
kf
k20
=
m
X
i=0
−m
¡ exp(t)(1 − )−1 62 exp t;
k f i+1 k20 :
The theorem is proved.
Note that Theorem 5.3 permits the exponential growth of k uh k; however, this is an admissible
assumption in the stability theory for sti systems (see, e.g. [7]).
Now we prove a stability result for two-level scheme (3:2) with i = ; i = 0; 1; : : : :
Theorem 5.4. Assume that H = o(h(1+s)=2 ). Then there exists some real positive constant
c1 (; ; f ; u0 ; T ) such that for any real h ¿ 0 satisfying
h6c1 (; ; f ; u0 ; T );
(5.11)
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
the following estimate holds for uhi ; i = 0; : : : ; m determined from (3:2a) and (3:2b):
k uhm+1
k20
m
m
m
X
1X
2X
i 2
i+1
2
i+1 2
+
k uh − uh k0 +
k f i+1 k2−1
k uh k1 6ct+ k u0 k0 +
4 i=0
i=0
i=0
(5.12)
with t = (m + 1) ; t6T and any integer m¿0. Moreover
m
X
i=0
k phi+1 k20 6H −s c2 (;
; f ; u0 ; s; T )
(5.13)
with some real positive constant c2 (;
; f ; u0 ; s; T ).
Proof. Let us take in (3.2b) C = 2uhi+1 . We get the equality
k uhi+1 k20 − k uhi k20 + k uhi+1 − uhi k20 +2 k uhi+1 k21
= − 2 N (uHi+1 ; uHi+1 ; uhi+1 ) + 2 ( f i+1 ; uhi+1 ):
(5.14)
The right-hand side in (5.14) can be estimated as follows:
−2 N (uHi+1 ; uHi+1 ; uhi+1 ) + 2 ( f i+1 ; uhi+1 )
= 2 N (uhi+1 ; uhi+1 − uHi+1 ; uhi+1 ) − 2 N (uhi+1 − uHi+1 ; uhi+1 − uHi+1 ; uhi+1 )
+ 2 ( f i+1 ; uhi+1 )6c(s) k uhi+1 − uHi+1 ks k uhi+1 k21
+ c(s) k uhi+1 − uHi+1 ks k uhi+1 − uHi+1 k1 k uhi+1 k1 + 12 k uhi+1 k21 +
6c(s) h−s k uhi+1 − uHi+1 k0 k uhi+1 k21
+ c(s) h−1−s k uhi+1 − uHi+1 k20 k uhi+1 k1 + 21 k uhi+1 k21 +
2
k f i+1 k2−1
(5.15)
2
k f i+1 k2−1 :
Denote by c3 (; ) a positive constant from the estimate
k u k1 + k p k0 6c3 (; ) k g k0 ;
(5.16)
for solution of (4.1) with given g ∈ L2 (
)2 . Let us consider
a = c3 (; )
"
cT +
k uh0
k20
2
+ max k f k2−1
[0;T ]
1=2
#
+ k f kL2 (G) :
Then from Corollary 4.4 (with ‘1 = ‘2 = 0) it follows that
c(s)h−s k uh1 − uH1 k0 6C(
; ; ; a; a)1=2 H
(5.17)
with C(
; ; ; · ; · ) from Corollary 4.4.
Further we prove (5.12) by induction with respect to m. Consider the case m = 0: Indeed (5.17)
and relation H = o(h(1+s)=2 ) provide us with suciently small h such that c(s)h−s k uh1 − uH1 k0 6=4
and c(s)h−2−2s k uh1 − uH1 k40 6=4. Thus,
c(s) h−s k uh1 − uH1 k0 k uh1 k21 6 41 k uh1 k21
(5.18)
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
and
c(s) h−1−s k uhi+1 − uHi+1 k20 k uh1 k1 6
c(s)
(1 + h−2−2s k uh1 − uH1 k40 k uh1 k21 )
2
6 12 (c(s) + 12 k uh1 k21 ):
(5.19)
From (5.14), (5.15) with i = 0, (5.18), and (5.19) we get (5.12) for m = 0. The case of m = 0 is
checked.
Assuming that (5.12) is valid for some m = j − 1, we prove this estimate for m = j ((m + 1) 6T )
in the same way as for m = 0: starting with (5.15) for i = j and checking with the help of inductive
hypothesis
c(s)h−s k uhj − uHj k0 6C(
; ; ; a; a)1=2 H:
(5.20)
Then in the similar way as (5.18), (5.19) for suciently small h it follows
c(s) h−s k uhj+1 − uHj+1 k0 k uhj+1 k21 6 41 k uhj+1 k21
and
c(s) h−1−s k uhj+1 − uHj+1 k20 k uhj+1 k1 6 21 (c(s) + 21 k uhj+1 k21 ):
Now we obtain from (5.14)
k uhj+1 k20 + k uhj+1 − uhj k20 + k uhj+1 k21 6c + k uhj k20 +2 k f j+1 k2−1 :
(5.21)
Taking a sum of (5.21) and (5.12) with m = j − 1 we obtain (5.12) for m = j; hence we complete
the inductive step and prove (5.12) for any m satisfying (m + 1) ¡ T:
To obtain an estimate on pressure function, let us rewrite (3.2b) as
b(phi+1 ; C) = a(uhi+1 ; C) + (uhi+1 − uhi ; C) + (fi+1 ; C) + N (uHi+1 ; uHi+1 ; C):
with arbitrary C ∈ Hh : Hence the stability hypothesis provides us with estimate
and
k phi+1 k20 6c(
; )k uhi+1
ui+1 − ui 2
h
k21 + h
+ k fi+1 k2−1 +c(s)H −s k uHi+1 k20 k uHi+1 k21
m
X
m
X
i=0
k
phi+1
k20
6c(
; )
−1
i=0
k uhi+1
k21
+
m
X
i=0
k fi+1 k2−1
!
m
m
ui+1 − ui
2
X
X
h
h
i+1 2
−s
i+1 2
k uH k 1 :
+
+ c(s)H max(k uH k0 )
i
i=0
−1
(5.22)
i=0
The rst P
term in estimate (5.22) is bounded due to (5.12), the second depends only on given data.
Estimate mi=0 k −1 (uhi+1 − uhi ) k2−1 6c H −s is proved similar to Lemma III:4:6 in [25].
To obtain estimates on uH let us write down (3.2a) with C = uHi+1 (note that (5.12) provides us
with estimate on uh but not on uH ):
(uHi+1 ; uHi+1 ) + a(uHi+1 ; uHi+1 ) = (uhi ; uHi+1 ) + (fi+1 ; uHi+1 ):
(5.23)
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
From (5.23) we readily get for i = 0; : : : ; m
k uHi+1 k20 + k uHi+1 k21 6 k uhi k20 + k f i+1 k2−1 ;
i+1 2
i+1 2
2
i
k uH k0 + k uH k1 6 k uH k0 + k f i+1 k2−1 +2(uhi − uHi ; uHi+1 );
(5.24)
in the second inequality, uHi should be replaced by uhi for i = 0.
The rst inequality inP(5.24) and (5.12) gives an estimate on K =maxi (k uHi+1 k20 )6c(
; u0 ; f ; ; T ):
To obtain a bound for mi=0 k uHi+1 k21 in (5.22) let us take a sum of the second inequality in (5.24)
for i = 0; : : : ; m:
m
X
i=0
k uHi+1 k21 6 k uh0 k20 +
m
m
X
1X
k f i+1 k2−1 +2 (uhi − uHi ; uHi+1 )
i=0
i=1
(5.25)
and act as follows. For suciently small h, using (5.20), we obtain
m
X
i=1
(uhi
−
√
uHi ; uHi+1 )6 max(k uHi+1
i
m
X
1=2
k0 )
m
X
i=1
k uhi − uHi k0
H 6c(
; u0 ; f ; ; T ):
6 KC(
; ; ; a; a)
(5.26)
i=1
Estimate (5.26) together with (5.25) gives
m
X
i=0
k uHi+1 k21 6c3 (
; u0 ; f ; ; T; s):
Thus all the terms in a right-hand side of (5.22) are estimated and the desired inequality (5.13)
follows.
Remark 5.5. The crucial point in the proof of the theorem is obtaining an estimate on k uh − uH k.
Corollary 4.4 shows that this estimate improves for equal order interpolation or high-order nite
element schemes, if we assume an extra regularity of the solution obtained. Hence, in this case the
stability conditions could be weakened. However, the lack of information about high-order norms in
our stability estimates does not permit us to state precise results on this matter.
Compare condition (5.6) for scheme (3.1) with condition (5.11) for two-level scheme (3:2).
Scheme (3.1) imposes restriction on ; while scheme (3:2) imposes restriction on H . Although
extensive calculations are needed to check whether the condition on H is restrictive indeed, the
latest fact is of certain theoretical interest.
Finally we note that scheme (3.3) is unconditionally stable.
6. Convergence
Let us x some T ¿ 0 and for a sequence of uhi ; i = 1; : : : ; m; m = T de ned from any of schemes
(3:1)–(3:3) denoted by uh = uh (t) : [0; T ] → Hh , the following function:
uh (t) = uhi
for t ∈ [(i − 1) ; i ):
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
189
For scheme (3.3) function uh (t) converges to solution u of (1.1) – (1.3) strongly in L2 (G); G =
(0; T ) ×
and weakly in L2 (0; T ; H01 ): The same is valid for scheme (3.1) with ; h satisfying
stability condition (5.6).
Here we prove
Theorem 6.1. Assume
∈ R2 : Let uhi ; i = 0; : : : ; m; from Hh recovered via two-level scheme (3:2)
then
uh → u strongly in L2 (G) for h; → 0;
uh → u weakly in L2 (0; T ; H01 ) for h; → 0;
if condition (5:11) and assumptions of Theorem 5:4 are satis ed. Here u is a unique weak solution
to problem (1:1)–(1:3).
Proof. The proof is close to those of Temam [25, Theorem III.5.4] for scheme 3:3. And for some
omitted technical details we refer to the [25].
Along with uh consider
wh (t) =
(i − 1) − t i
t − i
uh +
ui−1
(i − 1) − i
(i − 1) − i h
for t ∈ [(i − 1) ; i ];
and wh (t) = 0 for t 6∈ [0; T ]:
Since condition (5.11) is assumed to be satis ed, from estimate (5.12) we get
k uh kL∞ (0;T ; L2 (
)2 ) 6c;
k wh kL∞ (0;T ; L2 (
)2 ) 6c;
and for some subsequence still denoted by h; (→ 0)
wh → w weakly in L2 (0; T ; H01 )
uh → u weakly in L2 (0; T ; H01 );
(6.1)
with some u; w ∈ L∞ (0; T ; L2 (
)2 ) ∩ L2 (0; T ; H01 ): Moreover [25],
wh → w strongly in L2 (G):
(6.2)
Further, from equality
r
k uh − wh kL2 (G) =
m
X
k uhi+1 − uhi k20
3 i=0
!1=2
estimate (5.12), (6.1) and (6.2) we get
u = w;
uh → u strongly in L2 (G):
(6.3)
Now let us set
uH (t) = uHi
for t ∈ [(i − 1) ; i ):
From (4.10) and hypothesis (H1) we get
"
k uh − uH k2L2 (G) 6c() (H 2 + )
m
X
i=0
k uhi k21 +
m
X
i=0
k pi k20
#
(6.4a)
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
and
k uh −
uH k2L2 (0;T ; H01 ) 6c()
"
2
(H + )
m
X
i=0
k uhi k21
+
m
X
i=0
kp
i
k20
#
:
(6.4b)
Hence we imply (5.12), (5.13) with suciently small s, (6.1), (6.3) and obtain
uH → u strongly in L2 (G);
uH → u weakly in L2 (0; T ; H01 ):
(6.5)
To prove that u is a weak solution of (1.1) – (1.3) it is sucient to note that (3.2b) provides
d
(wh (t); C) + a(uh (t); C) + N (uH (t); uH (t); C) = ( fh (t); C) ∀C ∈ Vh
dt
and imply (6:1)–(6:3); (6:5) and arguments of Lemma III.5.9 from [25] in a straightforward way.
The theorem is proved.
Remark 6.2. The diculty in proving Theorem 6.1 for
⊂ R3 is caused by the impossibility of
taking suciently small s in (5.3). Dependence of constant c2 in Theorem 5.3 on is generally
unknown. Therefore pressure terms in (6.4a) and (6.4b) failed to be properly estimated.
Acknowledgements
This work was supported in part by INTAS Grant No. 93-377 EXT.
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Two-level method and some a priori estimates in unsteady
Navier–Stokes calculations
Maxim A. Olshanskii ∗
Dept. Mech. and Math., Moscow State University, Moscow 119899, Russia
Received 3 December 1997; received in revised form 3 December 1998; accepted 3 December 1998
Abstract
A two-level method proposed for quasielliptic problems is adapted in this paper to the simulation of unsteady incompressible Navier–Stokes
ows. The method requires a solution of a nonlinear problem on a coarse grid and a solution of
linear symmetric problem on a ne grid, the scaling between these two grids is superlinear. Approximation, stability, and
convergence aspects of a fully discrete scheme are considered. Stability properties of the two-level scheme are compared
c 1999 Elsevier
with those for a commonly used semi-implicit scheme, some new estimates are also proved for the latter.
Science B.V. All rights reserved.
Keywords: Navier–Stokes equation; Two-level method; A priori estimate
1. Introduction
Numerical simulation of unsteady incompressible viscous
ow is a fundamental problem both of
numerical analyses and
uid dynamics. The governing equations are the incompressible Navier–
Stokes ones:
@u
− u + (u · 3)u + 3p = f ;
@t
div u = 0;
in
× (0; T ];
(1.1)
with a given force eld f , kinematic viscosity ¿ 0;
⊂ Rn ; n = 2; 3. The velocity vector function
u and pressure scalar function p to be found are subject to some boundary conditions, which we
∗
Corresponding author. E-mail address: [email protected] (M.A. Olshanskii).
c 1999 Elsevier Science B.V. All rights reserved.
0377-0427/99/$ - see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 9 ) 0 0 0 5 6 - 4
174
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
assume to be Dirichlet and homogeneous for the velocity:
u = 0 on @
× [0; T ]
(1.2)
and initial condition at t = 0:
u = u0 (x)
in
:
(1.3)
R
Another common assumption is
p(x; t) dx = 0 for the pressure p(x; t) for all t ∈ (0; T ]. For
a detailed consideration of mechanical, mathematical, and computational aspects associated with
Navier–Stokes problem we refer to, among others, [9,12,13,25].
Belonging of a weak solution of (1.1) – (1.3) to a space of solenoidal functions and domination
of nonlinear convection terms in the momentum equation for moderate and high Reynolds numbers
(small and=or high velocities) are commonly considered to be the main diculties in numerical and
computational theory of Navier–Stokes equations. The objective of this paper is to apply two-level
mesh reduction method to the treatment of convection phenomenon in unsteady numerical simulations. The method is closely related to the nonlinear Galerkin method [1,18–20] and was developed
in [28,3,14 –16].
Using numerous solution schemes for (1.1) – (1.3) with u−p coupling (see, e.g., [5,11]) or operator
splitting [6], one has to choose between a fully implicit treatment of nonlinear terms in (1.1) or some
of their linearization using, e.g. extrapolation in time. In the rst case one faces the necessity of
solving nonlinear problem of the Burgers or Navier–Stokes type on each time step, otherwise linear
and even symmetric problems can be obtained on each time step. However, the latter approach
may cause stability problems, i.e. time step becomes subject to some conditions involving spatial
discretization and=or Reynolds number. We refer to [25] for theoretical considerations and [27] for
experimental comparison of various schemes.
The method presented and studied here can be roughly described as follows. For numerical solution
of (1.1) – (1.3) choose some spatial nite element or nite di erence discretization and two meshes:
the coarse one with the step H and the ne one with the step h such that h ∼ H ; ¿1: For temporal
integration on [t0 ; t0 + ] with time step ¿ 0 make fully implicit step on the coarse grid and obtain
uH ; pH for t0 + via solution of nonlinear problem on the coarse grid. Then extrapolate uH on the
ne grid. Using linearization of convective terms about uH , make semi-implicit integration step on
the ne grid and obtain uh ; ph for t0 + solving linear symmetric problem on the ne grid.
Compared to the recent studies [1,20] for unsteady problems, the primary innovations in this paper
are the treatment of fully discrete (both in time and space) case of the method. Hence the stability
results are quite important. We also avoid the use of any intermediate nite element subspaces.
Therefore well established solvers can be readily applied to the auxiliary nite element problems.
In Section 2 of the paper we introduce necessary notations and preliminary results. The algorithm to
be studied is described in Section 3. Where possible we compare results obtained for the constructed
algorithm with appropriate ones for commonly used algorithm based on a fully explicit treatment
of the nonlinearity via extrapolation in time. For this purpose some results on stability of the latest
algorithm are also proved.
In Section 4 some approximation results are proved. In particular, it follows that in the case
of linear velocity – constant pressure nite elements the scaling h ∼ H 2 gives the same order of
spatial discretization error as the usual Galerkin method with mesh size h. We note that this relation
between coarse and ne grids is somewhat less impressive than the one recovered in the framework
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
175
of Newton-type methods (see [2,28], and references cited therein). However, application of the latter
techniques for the problem considered requires highly nonsymmetrical problems to be solved on
each time step. Moreover, it causes the appearance of undesirable reactive term in the linearized
equation (see also [16]).
The stability of the schemes is studied in Section 5. It is proved that while usual semi-implicit
scheme requires time step to be small enough to guarantee the stability, the two-level scheme (at
least theoretically) requires the spatial step to be small enough to ensure stability. Moreover, if
the problem is regular enough, the use of high order nite elements weakens the condition on h.
In Section 6 we consider convergence of the two-level scheme. The appropriate convergence for
velocity is proved in two dimensions.
Throughout the paper we deal with saddle point formulations of the corresponding nite element
problems (discrete velocity is not solenoidal in general). This causes some extra complications but
corresponds to real-life situations.
2. Preliminaries
Throughout the paper we assume
to be a bounded domain in R2 or R3 with suciently smooth
boundary, or a convex polygon (polyhedron).
Later on we need the following functional spaces:
H01 ≡ {u ∈ W21 (
)2 : u = 0 on @
};
V ≡ {u ∈ H01 : div u = 0 in
}
with energy scalar product (u; C)1 = (3u; 3C); u; C ∈ H01 ;
L0 ≡ {u∈ L2 (
)2 : div
u · n = 0 on @
};
Z u = 0 in
;
p dx = 0
L2 =R ≡ p ∈ L2 (
):
with L2 -scalar product. Let H −1 be a dual, with respect to L2 -duality, space to H01 with the corresponding norm:
k f k−1 = sup
06=u∈H01
h f ; ui
;
k u k1
f ∈ H −1 :
We also use Sobolev spaces of a real exponent s: H s (
) with a norm k · ks .
The following forms are associated with the Navier–Stokes problem:
a(u; C) = (u; C)1 ; u; C ∈ H01 ;
a (u; C) = (u; C) + (u; C)1 ; u; C ∈ H01 ; ¿ 0;
b(p; u) = (p; div u); p ∈ L2 =R; u ∈ H01 ;
N (u; C; w) = 21 [((u · 3)C; w) − (u · 3)w; C)]; u; C; w ∈ H01 :
The weak formulation of (1.1) – (1.3) is to nd
u ∈ L2 (0; T ; H01 ) ∩ L∞ (0; T ; L0 ) and
p ∈ L2 (0; T ; L2 =R)
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
satisfying for given f ∈ L2 (0; T ; H −1 ) in the sense of distributions on (0; T ]
@u
; C + a(u; C) + N (u; u; C) − b(p; C) = h f ; Ci;
@t
b(q; u) = 0 ∀{C; q} ∈ H01 × L2 =R;
u(0; x) = u0 (x) in
;
or, alternatively, to nd
u ∈ L2 (0; T ; V ) ∩ L∞ (0; T ; L0 )
satisfying
d
(u; C) + a(u; C) + N (u; u; C) = h f ; Ci
dt
u(0; x) = u0 (x) in
;
∀C ∈ V ;
De Rham theorem connects both formulations (cf. [9,25]).
It is worth mentioning that a weak solution de ned above exists and for n=2 is unique; moreover,
some extra assumptions on f ; @f =@t, and u0 provide u ∈ L2 (0; T ; H 2 (
)2 ) (cf. [17]).
Further, we shall use the following estimates due to [12,25]:
b(p; u)6 k p k0 k u k1
( f ; u)6 k f k−1 k u k1
|((u · 3)C; w)|6c k u kL4 k C k1 k w kL4
|N (u; C; w)|6c k u ks k C k1 k w k1
|N (u; C; w)|6c k u k1 k C ks k w k1
|N (u; C; w)|6c k u k0 k C k1 k w k1+s
∀p ∈ L2 =R; u ∈ H01 ;
∀f ∈ H −1 ; u ∈ H01 ;
∀u; C; w ∈ H01 ;
∀u; C; w ∈ H01 ;
∀u; C; w ∈ H01 ;
∀u; C ∈ H01 ; w ∈ H01 ∩ H 1+s :
(2.1)
From now on we agree to consider s as an arbitrary number from (0; 1] for two-dimensional problem
and s ∈ [ 21 ; 1] for three-dimensional problem, if it is not stated otherwise. We also need
√
k u k2L4 6 2 k u k0 k u k1
∀u ∈ H01 ;
∈ R2 ;
(2.2)
k u k2L4 62 k u k01=2 k u k13=2 ∀u ∈ H01 ;
∈ R3 ;
s
1−s
1
∀u ∈ H0 ; s ∈ (0; 1):
k u ks 6c k u k0 k u k1
Here and later on we denote by c(
); c; c0 ; c1 ; : : : some constants independent of both spatial and
temporal discretization parameters (h and ) and , otherwise we shall use, for example, c() for a
constant depending possibly on .
So called j-inequality:
|ab|6
1 2 j 2
|a| + |b| ;
2j
2
∀a; b ∈ R; j ¿ 0;
will be also used throughout the paper.
Let us denote by h a mesh size parameter, and denote by Hh a nite element subspace of H01 and
by Qh a nite element subspace of L2 =R. Assume that for some real h0 ¿ 0; positive integers k1 ; k2 ;
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
177
and h ∈ (0; h0 ] the following hypotheses hold. Examples of such nite element spaces can be found
in [5,9,11].
(H1) Approximation hypothesis
(a) For all u ∈ H01 ∩ H d (
)n ; with integer d ¿ 0
inf k u − C k1 6chl k u kd ;
l = min(k1 ; d − 1):
C∈Hh
(b) For all p ∈ L2 =R ∩ H d (
); with integer d ¿ 0
inf k p − q k0 6chl k p kd ;
q∈Qh
l = min(k2 ; d):
(H2) Inverse hypothesis
For all u ∈ Hh
k u k1 6h−1 k u k0 :
(H3) Stability hypothesis. There exists some real constant c0 ¿ 0 independent on h such that
sup
06=u∈Hh
b(p; u)
¿c0 k p k0 ;
k u k1
∀p ∈ Qh :
The above hypotheses give the following standard result concerning Stokes problem.
Lemma 2.1. For any given {u; p} ∈ H01 × L2 =R; there exist unique uh ∈ Hh and ph ∈ Qh such that
a(u − uh ; C) − b(p − ph ; C) = 0;
b(q; u − uh ) = 0;
∀{C; q} ∈ Hh × Qh :
Moreover;
k u − uh k1 + k p − ph k0 6c(
) inf k u − C k1 + inf k p − q k0 ;
C∈Hh
q∈Qh
k u − uh k0 6c(
)h inf k u − C k1 :
C∈Hh
Denote also by Vh a subspace of discretely divergence-free functions from Hh : Vh = {uh ∈
Hh : b(q; uh ) = 0 ∀q ∈ Qh }.
3. Two algorithms for the unsteady problem
The common and e ective way of numerical treatment of time-dependent problems is separation
of spatial and temporal discretizations. For spatial discretization one can choose nite element, nite
di erence, spectral methods, while for the temporal discretization the nite di erence method is the
most natural choice. Two schemes described below utilize this idea.
First let us consider one widely used semi-implicit scheme for solving unsteady Navier–
Stokes problem (1.1) – (1.3). For given uh0 ∈ Hh nd {uhi+1 ; phi+1 } ∈ Hh × Qh for i = 0; 1; : : : from
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
the relations
a i (uhi+1 ; C) − i b(phi+1 ; C) + i N (uhi ; uhi ; C) = i (f i+1 ; C) + (uhi ; C);
b(q; uhi+1 ) = 0
(3.1)
∀ {C; q} ∈ HH × QH :
where i ¿ 0 are steps of temporal
discretization,
which we assume generally to be variable. In (3.1)
R
Pi
−1 t2
i+1
and below we set f
= i t1 f dt with t2 = k=0 k ; t1 = t2 − i .
On every time step scheme (3.1) requires solution of the generalized Stokes problem, e ective
solution methods for such problem are available [22,8]. With the above assumptions scheme (3.1)
is conditionally stable and the sucient conditions for stability from [25] are 6c(;
; f (t); u0 )hn ,
(
⊂ Rn ); moreover, the careful reading of the proof gives c = O(2 ); → 0.
The scheme being proposed works as follows. For given uh0 ∈ Hh nd {uhi+1 ; phi+1 } ∈ Hh × Qh for
i = 0; 1; : : : from the two-level algorithm:
Find {uHi+1 ; pHi+1 } ∈ HH × QH
a i (uHi+1 ; C) − i b(pHi+1 ; C) + i N (uHi+1 ; uHi+1 ; C) = i ( f i+1 ; C) + (uhi ; C);
b(uHi+1 ; q) = 0
(3.2a)
∀ {C; q} ∈ HH × QH ;
with known {uHi+1 ; pHi+1 } nd {uhi+1 ; phi+1 } ∈ Hh × Qh
a i (uhi+1 ; C) − i b(phi+1 ; C) + i N (uHi+1 ; uHi+1 ; C) = i ( f i+1 ; C) + (uhi ; C);
b(uhi+1 ; q) = 0
(3.2b)
∀ {C; q} ∈ Hh × Qh :
The above two-level algorithm can be observed simultaneously from two points of view. On
the one hand it can be considered as an improvement of scheme (3.1) by obtaining a qualitative
information (on a coarse grid) about the solution at time (t0 + i ). On the other hand, let us consider
fully implicit unconditionally stable scheme for (1.1) – (1.3) without any spatial discretization: for
given u0 ∈ H01 nd {ui+1 ; pi+1 } ∈ H01 × L2 =R for i = 0; 1; : : : from
a i (ui+1 ; C) − i b(pi+1 ; C) + i N (ui+1 ; ui+1 ; C) = i ( f i+1 ; C) + (ui ; C);
b(u
i+1
; q) = 0
∀ {C; q} ∈
H01
(3.3)
× L2 =R
then (3.2a) and (3.2b) is a straightforward application of the two-level method from [15] to the
solution of steady nonlinear problem arising on every time step of (3.3). Note that the two-level
algorithm (3:2) requires on every time step solution of the nonlinear problem of Navier–Stokes type
on the coarse grid and solution of the linear symmetric problem of Stokes type on the ne grid.
Remark 3.1. To improve the accuracy and stability of scheme (3.1) such variants of (3.1) as Crank–
Nicolson, fractional-step [21], with high-order extrapolation in time of nonlinear terms, multistep [4],
with upwinding (see, e.g. [24]) are known and used in practice. We note that all or at least most of
these improvements are quite applicable to (3.2a) and (3.2b) as well as splitting techniques leading
to a class of projection type methods [6,10].
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
4. Approximation
Consider the following problem. For given g ∈ H −1 nd {u; p} ∈ H01 × L2 =R satisfying
a (u; C) − b(p; C) + N (u; u; C) = (g; C);
(4.1)
∀ {C; q} ∈ H01 × L2 =R:
b(q; u) = 0
(4.1) exists (see, e.g. [25, Lemma √
IV.4.3]). For the uniqueness it is sucient
√ Solution of problem
2 k g k−1 6 1=2 3=2 for two-dimensional problem and 2 2 k g k−1 6 3=4 7=4 for three-dimensional
problem. If g ∈ Ln (
) then sucient conditions can be written as k g k0 6 and 2 k g k0 6 1=2 3=2 ,
respectively. The proof is quite standard, it follows from (2.1), (2.2) and a priori estimates for weak
solutions of (4.1):
2 k u k20 + k u k21 6( )−1 k g k2−1 ;
k u k20 +2 k u k21 6 k g k20
and certain relation for the di erence w = u − u˜ between two weak solutions of (4.1) that can be
obtained from (4.1) with C = w.
Two-level method for problem (4.1) means: nd sequentially {uH ; pH } ∈ HH × QH and {uh ; ph } ∈
Hh × Qh from
a (uH ; C) − b(pH ; C) + N (uH ; uH ; C) = (g; C);
b(q; uH ) = 0
(4.2a)
∀ {C; q} ∈ HH × QH
and
a (uh ; C) − b(ph ; C) + N (uH ; uH ; C) = (g; C);
b(q; uh ) = 0
(4.2b)
∀ {C; q} ∈ Hh × Qh :
Theorem 4.1. For given g ∈ H −1 let {u; p} ∈ H01 ∩ H 1+s (
)n × L2 =R; n = 2; 3 be a solution to
problem (4:1) and let {uh ; ph } ∈ Hh ×Qh be a solution to problem (4:2a); (4:2b); then the following
estimate is valid:
k uh − u
k20
+ k uh − u
k21
+ k ph − p
+
h ( ; h; ) inf k p − q
q∈Qh
2
k20
k20
+ k uH − u
6C(
; s) (h2 + ) inf k u − C k21
k20 k u
k21+s
+ k uH − u
C∈Hh
k02−2s k uH
−u
k2+2s
1
; (4.3)
with
h = min( =; 2 =h2 ).
Proof. For given {u; p} from (4.1) let {uS ; pS } ∈ Vh × Qh be a solution to the Stokes problem
(3uS ; 3C) − b(pS ; C) = (3u; 3C) − b(p; C);
b(uS ; q) = 0; ∀{C; q} ∈ Hh × Qh ;
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
then, by virtue of Lemma 2.1, the following estimates are valid:
k u − uS k1 6c(
) inf k u − C k1 + inf k p − q k0 ;
C∈Hh
q∈Qh
k u − uS k0 6c(
)h inf k u − C k1 :
(4.4)
C∈Hh
We also choose pI = arg minq∈Qh k p − q k0 :
We choose in (4.1) and (4.2b) C = uS − uh and substract (4.2b) from (4.1). Then we obtain
k uS − uh k20 + k uS − uh k21 =a (uS − u; uS − uh )
+ (b(p − pI ; uS − uh ) − N (u; u; uS − uh ) + N (uH ; uH ; uS − uh ))
= a (uS − u; uS − uh ) + (b(p − pI ; uS − uh ) − N (u − uH ; u; uS − uh )
+ N (u − uH ; u − uH ; uS − uh ) − N (u; u − uH ; uS − uh ))
1
6 21 k u − uS k20 + 12 k uS − uh k20 + k u − uS k21 + 21 k uS − uh k21
2
+ c (k p − pI k0 + k u − uH k0 k u k1+s + k u − uH k01−s k u − uH k1+s
1 ) k uS − uh k1 :
(4.5)
Using Cauchy inequality, we get
c (k p − pI k0 + k u − uH k0 k u k1+s + k u − uH k01−s k u − uH k1+s
1 ) k uS − uh k1
)
6c2 −1 (k p − pI k20 + k u − uH k20 k u k21+s + k u − uH k02−2s k u − uH k2+2s
1
+
k uS − uh k21 ;
4
(4.6a)
and
c (k p − pI k0 + k u − uH k0 k u k1+s + k u − uH k01−s k u − uH k1+s
1 ) k uS − uh k1
c
(k p − pI k + k u − uH kk u k1+s + k u − uH k01−s k u − uH k1+s
6
1 ) k uS − uh k0
h
c 2 2 2
)
6 2 (k p − pI k20 + k u − uH k20 k u k21+s + k u − uH k02−2s k u − uH k2+2s
1
h
+ 14 k uS − uh k20 :
(4.6b)
Now in order to estimate k uh − u k, we apply the triangle inequality: k uh − u k 6 k uh − uS k +
k uS − u k; estimates (4.4), (4.5), and one of (4.6a) and (4.6b). To estimate k ph − p k0 we use the
following standard arguments (see, e.g. [9]). From (4.1) and (4.2b) we get for all {C; q} ∈ Hh × Qh
b(ph − q; C) = a (u − uh ; C) + b(p − q; C) + N (u; u; C) − N (uH ; uH ; C):
Thus, similar to (4.5) we obtain for all {C; q} ∈ Hh × Qh :
b(ph − q; C)6(c + )1=2 (k uh − u k20 + k uh − u k21 )1=2 k C k1
+ c (k p − q k0 + k u − uH k0 k u k1+s + k u − uH k01−s k u − uH k1+s
1 ) k C k1 :
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
181
Dividing both sides of the last inequality by k C k1 for 0 6= C ∈ Hh , taking sup over all such C and
inf over all q ∈ Qh , and utilizing hypothesis (H3) we get
c0 inf k ph − q k0 6(c + )1=2 (k uh − u k20 + k uh − u k21 )1=2
q∈Qh
+ c inf k p − q k0 + k u − uH k0 k u k1+s + k u − uH k01−s k u − uH k1+s
:
1
q∈Qh
Now from the triangle inequality: k ph − p k0 6inf q∈Qh k ph − q k0 +inf q∈Qh k p − q k0 we get an
estimate on k ph − p k0 . Combining estimates for k uh − u k20 + k uh − u k21 and 2 k ph − p k20 ; we
prove the theorem.
Remark 4.2. Estimate (4.3) shows a loss of convergence in pressure for too small time steps .
Indeed, the same standard arguments as in the proof of Theorem 4.1 when applied to some nite
element approximation of linear (generalized) Stokes problem
−1 u − u + 3p = f ;
div u = 0;
(4.7)
u|@
= 0
give the estimate
k uh − u k20 + k uh − u k21 + 2 k ph − p k20
6C(
) (h2 + ) inf k u − C k21 +
h inf k p − q k20 :
C∈Hh
q∈Qh
(4.8)
Assumed to be optimal for pressure this estimate recovers the necessity of h2k2 ≪ or h2(k2 −1) ≪1 for
pressure nite elements and hk1 +1 ≪ for velocity nite elements to ensure the convergence of pressure
in L2 (
): These conditions on h; ; and force us to use schemes of equal order (k2 = k1 + 1¿2)
or high order spacial interpolation (k2 ¿2; k1 ¿2), and implicit schemes, which do not require to
be too small.
Remark 4.3. Further we shall need an estimate for k u −uH k0 ; where u is a solution to (4.1) and uH
is a solution to (4.2a). To obtain such an estimate assume the solution of (4.1) to be nonsingular.
With our assumptions on
, problem (4.7) is W22 -regular. Hence we use standard arguments from
[9]: duality estimates for problem (4.7) (Theorem II.1.2), Theorems IV.3.3, IV.3.5, and IV.4.2,
nally for {u; p} ∈ H01 ∩ H d (
)n × L2 =R ∩ H d−1 (
) with integer d ¿ 0 and suciently small H we
obtain
k u − uH k20 6C(
; ; ; k u k‘1 +1 ; k p k‘2 )(H 2(‘1 +1) + H 2(‘2 +1) )
(4.9)
with ‘1 = min(k1 ; d − 1); ‘2 = min(k2 ; d − 1). This estimate provides us with the choice of scaling
h = O(H 1+1=‘1 ) to obtain an optimal accuracy O(h2‘1 ) in (4.3). However, in (4.11) a dependence of
the constant C on and is implicit.
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
Corollary 4.4. Let {u; p} ∈ H01 ∩ H d (
)2 × L2 =R ∩ H d−1 (
) for some integer d ¿ 0 be a solution
to problem (4:1) and {uH ; pH ; uh ; ph } ∈ HH × QH × Hh × Qh be a solution to problem (4:2a); (4:2b)
then for suciently small H the following estimates hold:
k uh − uH k20 6C(
; ; ; k u k‘1 +1 ; k p k‘2 )(H 2(‘1 +1) + H 2(‘2 +1) )
with ‘1 = min(k1 ; d − 1); ‘2 = min(k2 ; d − 1).
Proof. The inequality
k uh − uH k20 62(k u − uh k20 + k u − uH k20 )
together with (4.9), Theorem 4.1, and assumption h ¡ H give the result.
Another estimate on uH − uh without explicitly involving solution of (4.1) is given by the following
theorem.
Theorem 4.5. For given g ∈ H −1 let {uH ; pH } ∈ HH × QH and {uh ; ph } ∈ Hh × Qh be a solution
to problem (4:2a); (4:2b); then the following estimate is valid:
k uh − uH k20 + k uh − uH k21
2
6C(
) (H + ) inf k uh − C
C∈HH
k21
+( +
H ) inf k ph − q
q∈QH
k20
:
(4.10)
Proof. For given {uh ; ph } from (4.2a) let {uS ; pS } ∈ VH × QH be a solution to the Stokes problem
(3uS ; 3C) − b(pS ; C) = (3uh ; 3C) − b(ph ; C);
b(q; uS ) = 0;
∀{C; q} ∈ HH × QH ;
then the following estimates are valid:
k uh − uS k1 6c(
) inf k uh − C k1 + inf k ph − q k0 ;
q∈QH
C∈HH
k uh − uS k0 6c(
) H inf k uh − C k1 ;
(4.11)
C∈HH
Choose also pI = arg minq∈QH k ph − q k0 :
Let us take in (4.2a) and (4.2b) C = uS − uH and substract (4.2b) from (4.2a), then we obtain
k uS − uH k20 + k uS − uH k21 = a (uS − uh ; uS − uH ) + b(ph − pI ; uS − uH )
6 k uS − uh k20 + k uS − uh k21 +2 k ph − pI k0 k uS − uH k1 :
(4.12)
Estimate the last term as
and
2 k ph − pI k0 k uS − uH k1 6 −1 k ph − pI k20 + k uS − uH k21
(4.13a)
2 k ph − pI k0 k uS − uH k1 62 2 H −2 k ph − pI k20 + 21 k uS − uH k20 :
(4.13b)
Now apply the triangle inequality, estimates (4.11), (4.12) and one of the estimates (4.13a) and
(4.13b) and get (4.10). The theorem is proved.
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
183
5. Stability
We understand the stability of schemes (3:1)–(3:3) as a validation of some a priori estimates for
nite element solutions obtained using these schemes. Week solution u ∈ L2 (0; T ; H01 ) ∩ L∞ (0; T ; L0 )
of problem (1.1) – (1.3) satis es energy estimate
k u(t) k20 +2
Z
0
t
k u(s) k2 ds6 k u0 k20 +2
Z
0
t
| ( f (s); u(s)) | ds:
Further we will deduce some conditions that provide nite element solution with nite analogue of
energy estimate (see also [25,26]).
The following theorem gives condition on i that ensure (3.1) to be stable. This condition depends
on uhi ; Corollary 5.2 shows that the condition can be strengthened and made depend only on the
given data.
Theorem 5.1. For
∈ Rn ; n = 1; 2 any natural m¿0 and any i ; i = 0; : : : ; m satisfying condition
!
hn
h2
i 6min
;
;
82 k uhi k20 22
i = 0; 1; : : : ;
(5.1)
the following estimate holds for uhi determined from (3:1):
k uhm+1 k20 +
m
m
m
X
1X
2X
i k uhi+1 k21 6 k u0 k20 +
k uhi+1 − uhi k20 +
i k f i+1 k2−1 ;
4 i=0
i=0
i=0
(5.2)
Proof. Let us take in (3.1) C = 2uhi+1 , we get the equality
k uhi+1 k20 − k uhi k20 + k uhi+1 − uhi k20 +2 i k uhi+1 k21 = −2 i N (uhi ; uhi ; uhi+1 ) + 2 i ( f i+1 ; uhi+1 ):
(5.3)
Right-hand side in (5.3) can be estimated as follows:
−2 i N (uhi ; uhi ; uhi+1 ) + 2 i ( f i+1 ; uhi+1 ) = −2 i N (uhi ; uhi ; uhi+1 − uhi ) + 2 i ( f i+1 ; uhi+1 )
2 i
62 i h−n=2 k uhi k0 k uhi k1 k uhi+1 − uhi k0 + 12 i k uhi+1 k21 +
k f i+1 k2−1
2 i
622 2i h−n k uhi k20 k uhi k21 + 21 k uhi+1 − uhi k20 + 12 i k uhi+1 k21 +
k f i+1 k2−1 :
From (5.3) and the last estimate we get
2 i
k f i+1 k2−1 :
k uhi+1 k20 + 12 k uhi+1 − uhi k20 + 23 i k uhi+1 k21 −42 2i h−n k uhi k20 k uhi k21 6 k uhi k20 +
(5.4)
By virtue of condition (5.1), for i the following estimates are valid:
3
2 i
k uhi+1 k21 −42 2i h−n k uhi k20 k uhi k21 ¿ i k uhi+1 k21 + 12 i (k uhi+1 k21 − k uhi k21 )
¿ i k uhi+1 k21 − 21 i k uhi+1 − uhi k21 ¿ i k uhi+1 k21 − 14 k uhi+1 − uhi k20 :
Now we get from (5.4)
k uhi+1 k20 + 41 k uhi+1 − uhi k20 + i k uhi+1 k21 6 k uhi k20 +
2 i
k f i+1 k2−1 :
(5.5)
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
Taking for i = 0; : : : ; m a sum of (5.5), we obtain (5.2) and prove the theorem.
Corollary 5.2. There exist some constant c = c(
; f (t); u0 ) ¿ 0 such that for any set of i ; i =
0; : : : ; m; satisfying condition
!
h2
i 6min c h ; 2 ;
2
n 2
i = 0; 1; : : : ;
(5.6)
and uhi determined from (3:1) estimate (5:2) holds with any natural m¿0.
Proof. Estimate (5.2) for any set of i satisfying (5.1) provides k uhi k20 6c1 (u0 ; f (t)) −1 . Setting in
(5.6) c = (82 c1 )−1 , we prove the corollary.
Note that Corollary 5.2 recovers asymptotically the same condition on the constant step as
Lemma 5.3 from [25]. However, if in a particular problem k uhi k0 depends on in some more
advantageous way, then the condition on can be weakened; this is the situation which we have
sometime in practice (see [23] for calculations of critical (for stability) for scheme (3.1) in the
case of one substantially nonlinear and unsteady
ow).
The following theorem sharpens the estimate on with respect to . The theorem requires that
function f satis es k f (t) kL2 (G) 6c ¡ ∞ for some T ¿ 0; G = (0; T ) ×
.
Theorem 5.3. For any T ¿ 0 there exist some constant c = c(T;
; f (t); u0 ) ¿ 0 such that for any
satisfying the condition
!
h2
6min chn ;
;
22
i = 0; 1; : : : ;
(5.7)
and for any natural m¿0 such that m 6T the following estimate holds:
k uhm+1
k20
with t = m :
m
m
X
1X
+
k uhi+1 − uhi k20 +
i k uhi+1 k21 62 exp(t) k u0 k20 +
4 i=0
i=0
Z
0
t
kf
k20
(5.8)
Proof. Fix T and choose satisfying (5.8) with
1
[2 exp(T )(k u0 k20 + k f k2L2 (G) )]−1 :
82
Let us prove by induction with respect to m the following estimate:
c=
k uhm+1
k20
(5.9)
!
m
m
m
X
X
1X
+
k f i+1 k20 :
k uhm+1 − uhm k20 +
k uhm+1 k21 6(1 − )−m k u0 k20 +
4 i=0
i=0
i=0
(5.10)
Consider the case m = 0. Indeed from (5.9) and (5.7) we have 6hn (82 k u k20 )−1 and condition
(5.1) for 0 = is ful lled. Similar to the proof of Theorem 5.1 let us estimate right-hand side of
(5.1) with i = 0. However, now we apply the inequality 2 ( f 0 ; uh1 )6 k uh1 k20 + k f 0 k20 instead of
2 ( f 0 ; uh1 )6 21 k uh1 k21 +(2=) k f 0 k2−1 .
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
185
Finally, we get
k uh1 k20 + 41 k uh1 − u0 k20 + k uh1 k21 6 k u0 k20 + k uh1 k20 + k f 0 k20
that implies
k uh1 k20 + 41 k uh1 − u0 k20 + k uh1 k21 6(1 − )−1 (k u0 k20 + k f 0 k20 ):
Thus, the “trivial” case of m = 0 is checked.
Suppose now that (5.10) is proved for all m = 0; 1; : : : ; k. Let us prove it for m = k + 1, assuming
(k + 1) 6T . From inductive hypothesis we have (5.10) for m = k, hence
k uhk+1 k20 6(1 − )−k (k u0 k20 + k f k2L2 (G) )
and condition (5.1) for k = is valid. With the same arguments as in the proof of Theorem 5.1
and another estimate for 2 ( f k ; uhk+1 ) we get
k uhk+1 k20 + 41 k uhk+1 − uhk k20 + k uhk+1 k21 6 k uhk k20 + k uhk+1 k20 + k f k k20 ;
and with inductive hypothesis
k uhk+1 k20 + 41 k uhk+1 − uhk k20 + k uhk+1 k21
6(1 − )
−1
"
(1 − )
−1
+ (1 − ) k f
−
k
−(k−1)
k20
ku
0
k20
6(1 − )
−k
+
k−1
X
i=0
ku
0
kf
k20
k
k
X
1X
i k uhk k21 :
k uhk − uhk−1 k20 −
4 i=1
i=1
+
i+1
k20
k
X
i=0
!
k
k
X
1X
−
k uhk − uhk−1 k20 −
i k uhk k21
4 i=1
i=1
kf
i+1
k20
#
!
Thus, (5.10) is proved for all m:m 6T: Estimate (5.8) follows from (5.10) and obvious relations:
(1 − )
Z
0
−m
t
= 1−
m
t
kf
k20
=
m
X
i=0
−m
¡ exp(t)(1 − )−1 62 exp t;
k f i+1 k20 :
The theorem is proved.
Note that Theorem 5.3 permits the exponential growth of k uh k; however, this is an admissible
assumption in the stability theory for sti systems (see, e.g. [7]).
Now we prove a stability result for two-level scheme (3:2) with i = ; i = 0; 1; : : : :
Theorem 5.4. Assume that H = o(h(1+s)=2 ). Then there exists some real positive constant
c1 (; ; f ; u0 ; T ) such that for any real h ¿ 0 satisfying
h6c1 (; ; f ; u0 ; T );
(5.11)
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
the following estimate holds for uhi ; i = 0; : : : ; m determined from (3:2a) and (3:2b):
k uhm+1
k20
m
m
m
X
1X
2X
i 2
i+1
2
i+1 2
+
k uh − uh k0 +
k f i+1 k2−1
k uh k1 6ct+ k u0 k0 +
4 i=0
i=0
i=0
(5.12)
with t = (m + 1) ; t6T and any integer m¿0. Moreover
m
X
i=0
k phi+1 k20 6H −s c2 (;
; f ; u0 ; s; T )
(5.13)
with some real positive constant c2 (;
; f ; u0 ; s; T ).
Proof. Let us take in (3.2b) C = 2uhi+1 . We get the equality
k uhi+1 k20 − k uhi k20 + k uhi+1 − uhi k20 +2 k uhi+1 k21
= − 2 N (uHi+1 ; uHi+1 ; uhi+1 ) + 2 ( f i+1 ; uhi+1 ):
(5.14)
The right-hand side in (5.14) can be estimated as follows:
−2 N (uHi+1 ; uHi+1 ; uhi+1 ) + 2 ( f i+1 ; uhi+1 )
= 2 N (uhi+1 ; uhi+1 − uHi+1 ; uhi+1 ) − 2 N (uhi+1 − uHi+1 ; uhi+1 − uHi+1 ; uhi+1 )
+ 2 ( f i+1 ; uhi+1 )6c(s) k uhi+1 − uHi+1 ks k uhi+1 k21
+ c(s) k uhi+1 − uHi+1 ks k uhi+1 − uHi+1 k1 k uhi+1 k1 + 12 k uhi+1 k21 +
6c(s) h−s k uhi+1 − uHi+1 k0 k uhi+1 k21
+ c(s) h−1−s k uhi+1 − uHi+1 k20 k uhi+1 k1 + 21 k uhi+1 k21 +
2
k f i+1 k2−1
(5.15)
2
k f i+1 k2−1 :
Denote by c3 (; ) a positive constant from the estimate
k u k1 + k p k0 6c3 (; ) k g k0 ;
(5.16)
for solution of (4.1) with given g ∈ L2 (
)2 . Let us consider
a = c3 (; )
"
cT +
k uh0
k20
2
+ max k f k2−1
[0;T ]
1=2
#
+ k f kL2 (G) :
Then from Corollary 4.4 (with ‘1 = ‘2 = 0) it follows that
c(s)h−s k uh1 − uH1 k0 6C(
; ; ; a; a)1=2 H
(5.17)
with C(
; ; ; · ; · ) from Corollary 4.4.
Further we prove (5.12) by induction with respect to m. Consider the case m = 0: Indeed (5.17)
and relation H = o(h(1+s)=2 ) provide us with suciently small h such that c(s)h−s k uh1 − uH1 k0 6=4
and c(s)h−2−2s k uh1 − uH1 k40 6=4. Thus,
c(s) h−s k uh1 − uH1 k0 k uh1 k21 6 41 k uh1 k21
(5.18)
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
and
c(s) h−1−s k uhi+1 − uHi+1 k20 k uh1 k1 6
c(s)
(1 + h−2−2s k uh1 − uH1 k40 k uh1 k21 )
2
6 12 (c(s) + 12 k uh1 k21 ):
(5.19)
From (5.14), (5.15) with i = 0, (5.18), and (5.19) we get (5.12) for m = 0. The case of m = 0 is
checked.
Assuming that (5.12) is valid for some m = j − 1, we prove this estimate for m = j ((m + 1) 6T )
in the same way as for m = 0: starting with (5.15) for i = j and checking with the help of inductive
hypothesis
c(s)h−s k uhj − uHj k0 6C(
; ; ; a; a)1=2 H:
(5.20)
Then in the similar way as (5.18), (5.19) for suciently small h it follows
c(s) h−s k uhj+1 − uHj+1 k0 k uhj+1 k21 6 41 k uhj+1 k21
and
c(s) h−1−s k uhj+1 − uHj+1 k20 k uhj+1 k1 6 21 (c(s) + 21 k uhj+1 k21 ):
Now we obtain from (5.14)
k uhj+1 k20 + k uhj+1 − uhj k20 + k uhj+1 k21 6c + k uhj k20 +2 k f j+1 k2−1 :
(5.21)
Taking a sum of (5.21) and (5.12) with m = j − 1 we obtain (5.12) for m = j; hence we complete
the inductive step and prove (5.12) for any m satisfying (m + 1) ¡ T:
To obtain an estimate on pressure function, let us rewrite (3.2b) as
b(phi+1 ; C) = a(uhi+1 ; C) + (uhi+1 − uhi ; C) + (fi+1 ; C) + N (uHi+1 ; uHi+1 ; C):
with arbitrary C ∈ Hh : Hence the stability hypothesis provides us with estimate
and
k phi+1 k20 6c(
; )k uhi+1
ui+1 − ui 2
h
k21 + h
+ k fi+1 k2−1 +c(s)H −s k uHi+1 k20 k uHi+1 k21
m
X
m
X
i=0
k
phi+1
k20
6c(
; )
−1
i=0
k uhi+1
k21
+
m
X
i=0
k fi+1 k2−1
!
m
m
ui+1 − ui
2
X
X
h
h
i+1 2
−s
i+1 2
k uH k 1 :
+
+ c(s)H max(k uH k0 )
i
i=0
−1
(5.22)
i=0
The rst P
term in estimate (5.22) is bounded due to (5.12), the second depends only on given data.
Estimate mi=0 k −1 (uhi+1 − uhi ) k2−1 6c H −s is proved similar to Lemma III:4:6 in [25].
To obtain estimates on uH let us write down (3.2a) with C = uHi+1 (note that (5.12) provides us
with estimate on uh but not on uH ):
(uHi+1 ; uHi+1 ) + a(uHi+1 ; uHi+1 ) = (uhi ; uHi+1 ) + (fi+1 ; uHi+1 ):
(5.23)
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M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
From (5.23) we readily get for i = 0; : : : ; m
k uHi+1 k20 + k uHi+1 k21 6 k uhi k20 + k f i+1 k2−1 ;
i+1 2
i+1 2
2
i
k uH k0 + k uH k1 6 k uH k0 + k f i+1 k2−1 +2(uhi − uHi ; uHi+1 );
(5.24)
in the second inequality, uHi should be replaced by uhi for i = 0.
The rst inequality inP(5.24) and (5.12) gives an estimate on K =maxi (k uHi+1 k20 )6c(
; u0 ; f ; ; T ):
To obtain a bound for mi=0 k uHi+1 k21 in (5.22) let us take a sum of the second inequality in (5.24)
for i = 0; : : : ; m:
m
X
i=0
k uHi+1 k21 6 k uh0 k20 +
m
m
X
1X
k f i+1 k2−1 +2 (uhi − uHi ; uHi+1 )
i=0
i=1
(5.25)
and act as follows. For suciently small h, using (5.20), we obtain
m
X
i=1
(uhi
−
√
uHi ; uHi+1 )6 max(k uHi+1
i
m
X
1=2
k0 )
m
X
i=1
k uhi − uHi k0
H 6c(
; u0 ; f ; ; T ):
6 KC(
; ; ; a; a)
(5.26)
i=1
Estimate (5.26) together with (5.25) gives
m
X
i=0
k uHi+1 k21 6c3 (
; u0 ; f ; ; T; s):
Thus all the terms in a right-hand side of (5.22) are estimated and the desired inequality (5.13)
follows.
Remark 5.5. The crucial point in the proof of the theorem is obtaining an estimate on k uh − uH k.
Corollary 4.4 shows that this estimate improves for equal order interpolation or high-order nite
element schemes, if we assume an extra regularity of the solution obtained. Hence, in this case the
stability conditions could be weakened. However, the lack of information about high-order norms in
our stability estimates does not permit us to state precise results on this matter.
Compare condition (5.6) for scheme (3.1) with condition (5.11) for two-level scheme (3:2).
Scheme (3.1) imposes restriction on ; while scheme (3:2) imposes restriction on H . Although
extensive calculations are needed to check whether the condition on H is restrictive indeed, the
latest fact is of certain theoretical interest.
Finally we note that scheme (3.3) is unconditionally stable.
6. Convergence
Let us x some T ¿ 0 and for a sequence of uhi ; i = 1; : : : ; m; m = T de ned from any of schemes
(3:1)–(3:3) denoted by uh = uh (t) : [0; T ] → Hh , the following function:
uh (t) = uhi
for t ∈ [(i − 1) ; i ):
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
189
For scheme (3.3) function uh (t) converges to solution u of (1.1) – (1.3) strongly in L2 (G); G =
(0; T ) ×
and weakly in L2 (0; T ; H01 ): The same is valid for scheme (3.1) with ; h satisfying
stability condition (5.6).
Here we prove
Theorem 6.1. Assume
∈ R2 : Let uhi ; i = 0; : : : ; m; from Hh recovered via two-level scheme (3:2)
then
uh → u strongly in L2 (G) for h; → 0;
uh → u weakly in L2 (0; T ; H01 ) for h; → 0;
if condition (5:11) and assumptions of Theorem 5:4 are satis ed. Here u is a unique weak solution
to problem (1:1)–(1:3).
Proof. The proof is close to those of Temam [25, Theorem III.5.4] for scheme 3:3. And for some
omitted technical details we refer to the [25].
Along with uh consider
wh (t) =
(i − 1) − t i
t − i
uh +
ui−1
(i − 1) − i
(i − 1) − i h
for t ∈ [(i − 1) ; i ];
and wh (t) = 0 for t 6∈ [0; T ]:
Since condition (5.11) is assumed to be satis ed, from estimate (5.12) we get
k uh kL∞ (0;T ; L2 (
)2 ) 6c;
k wh kL∞ (0;T ; L2 (
)2 ) 6c;
and for some subsequence still denoted by h; (→ 0)
wh → w weakly in L2 (0; T ; H01 )
uh → u weakly in L2 (0; T ; H01 );
(6.1)
with some u; w ∈ L∞ (0; T ; L2 (
)2 ) ∩ L2 (0; T ; H01 ): Moreover [25],
wh → w strongly in L2 (G):
(6.2)
Further, from equality
r
k uh − wh kL2 (G) =
m
X
k uhi+1 − uhi k20
3 i=0
!1=2
estimate (5.12), (6.1) and (6.2) we get
u = w;
uh → u strongly in L2 (G):
(6.3)
Now let us set
uH (t) = uHi
for t ∈ [(i − 1) ; i ):
From (4.10) and hypothesis (H1) we get
"
k uh − uH k2L2 (G) 6c() (H 2 + )
m
X
i=0
k uhi k21 +
m
X
i=0
k pi k20
#
(6.4a)
190
M.A. Olshanskii / Journal of Computational and Applied Mathematics 104 (1999) 173–191
and
k uh −
uH k2L2 (0;T ; H01 ) 6c()
"
2
(H + )
m
X
i=0
k uhi k21
+
m
X
i=0
kp
i
k20
#
:
(6.4b)
Hence we imply (5.12), (5.13) with suciently small s, (6.1), (6.3) and obtain
uH → u strongly in L2 (G);
uH → u weakly in L2 (0; T ; H01 ):
(6.5)
To prove that u is a weak solution of (1.1) – (1.3) it is sucient to note that (3.2b) provides
d
(wh (t); C) + a(uh (t); C) + N (uH (t); uH (t); C) = ( fh (t); C) ∀C ∈ Vh
dt
and imply (6:1)–(6:3); (6:5) and arguments of Lemma III.5.9 from [25] in a straightforward way.
The theorem is proved.
Remark 6.2. The diculty in proving Theorem 6.1 for
⊂ R3 is caused by the impossibility of
taking suciently small s in (5.3). Dependence of constant c2 in Theorem 5.3 on is generally
unknown. Therefore pressure terms in (6.4a) and (6.4b) failed to be properly estimated.
Acknowledgements
This work was supported in part by INTAS Grant No. 93-377 EXT.
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