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MATHEMATICS OF COMPUTATION Volume 72, Number 241, Pages 307–333 S 0025-5718(02)01423-0 Article electronically published on May 3, 2002
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA LUNG-AN YING
Abstract. Using the fundamental solution of the heat equation, we give an expression of the solutions to two-dimensional initial-boundary value problems of the Navier-Stokes equations, where the vorticity is expressed in terms of a Poisson integral, a Newtonian potential, and a single layer potential. The density of the single layer potential is the solution to an integral equation of Volterra type along the boundary. We prove there is a unique solution to the integral equation. One fractional time step approximation is given, based on this expression. Error estimates are obtained for linear and nonlinear problems. The order of convergence is 14 for the Navier-Stokes equations. The result is in the direction of justifying the Chorin-Marsden formula for vortex methods. It is shown that the density of the vortex sheet is twice the tangential velocity for the half plane, while in general the density differs from it by one additional term.
1. Introduction The convergence and stability problems for the vortex method have been studied by many authors. Usually a fractional steps approach is applied in the vortex method for viscous flows; therefore viscous splitting is a basic problem in the mathematical theory. The convergence problem of viscous splitting was first studied in [2], and since then there have been quite a number of works on this subject (see [10] and references therein). Vorticity is the main variable in the computation of the vortex method. However, a velocity boundary condition is applied on a solid boundary and the boundary condition for vorticity is indirect. To approximate the velocity boundary condition, it was proposed that the velocity field be extended oddly after each convection step; then one vortex sheet was created, which resulted in an approximate zero velocity on the boundary (see [4]). This scheme was formulated as (1)
ωi = (Hk ◦ φ ◦ Ek )i ω0 ,
where k > 0 is the length of temporal steps, ωi is the vorticity at ti = ik, ω0 is the initial data, Ek is the solver of the Euler equation, Hk is the solver of the heat equation on the entire space, and φ is the “vorticity creation operator”. Stability and convergence of (1) for linear equations was proved in [5]. Convergence of the Chorin-Marsden formula for a half plane domain was proved in [3], with a slight modification, where the tangential component of velocity was Received by the editor June 5, 2000 and, in revised form, January 3, 2001. 2000 Mathematics Subject Classification. Primary 65M99; Secondary 35Q30, 76D05, 76M25. Key words and phrases. Vortex method, Navier-Stokes equation, fractional step method. c
2002 American Mathematical Society
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LUNG-AN YING
extended oddly and the normal component was extended evenly. An error estimate 1 with upper bound O(k 4 ) was obtained. Partial analysis for convex domains was gicen in [9], also with some modification, where the extended velocity field was multiplied by a Jacobian determinant. It was observed (see [9]) that both odd extension and even extension of the normal component of velocity yielded convergent results for the half plane domain. We notice that the way to create a vortex sheet is not unique, but all of them should be close enough so that the differences between them approach zero as k → 0. The aim of this paper is to study the problem for general bounded domains. The idea is to derive an expression of the density of vortex sheets for exact solutions, then make a time discretization to obtain an approximate one. In our scheme the derivation of vortex sheet is indirect, since it satisfies an integral equation of Volterra type along the boundary. A direct one may be possible. However, since we have the exact expression, the approximate one is required to be consistent with it. In [5], [3], [9] the density of the vortex sheet equals twice the tangential velocity after each convection step. We will make a remark in the final section, indicating that the density of the vortex sheet of our scheme is the same if the domain is a half plane. However, in the general case they are different by one additional term, which does not vanish in general. We hope this additional term approaches zero as k → 0 for some cases. This will be the subject of our further study. In order to state our scheme and convergence result, let us write down the equations first. Let Ω be a bounded domain in R2 , and ∂Ω its boundary. The initialboundary value problems of the Navier-Stokes equations for viscous incompressible flow are ∂u (2) + (u · ∇)u + ∇p = 4u, ∂t ∇ · u = 0,
(3) (4)
u|x∈∂Ω = 0,
u|t=0 = u0 ,
where u is the velocity and p is the pressure. Without loss of generality we set the ∂ ∂ density and viscosity equal to one. Let ∇∧ = ∂x2 , − ∂x1 , and let the vorticity ω = −∇ ∧ u; then the vorticity formulation of the Navier-Stokes equation is ∂ω + u · ∇ω = 4ω, ∂t
(5)
− 4 ψ = ω,
(6)
u = ∇ ∧ ψ,
(7)
ψ|x∈∂Ω = 0,
(8)
u|x∈∂Ω = 0,
(9)
ω|t=0 = ω0 = −∇ ∧ u0 .
Let ρ(ξ, t), ξ ∈ ∂Ω, t ∈ [0, T ] be the density of the vortex sheet of an exact solution; then the contribution of ρ to the solution is Z tZ (10) K(x − ξ, t − τ )ρ(ξ, τ ) dsξ dτ, 0
∂Ω
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where K is the fundamental solution of the heat equation: 1 − |x|2 e 4t . 4πt
K(x, t) =
Let F = u·∇ω, and we extend ω0 smoothly to R2 so that it is compactly supported; then we will prove that the solution to (5)-(9) can be expressed in terms of the sum of a Poisson integral, a Newtonian potential, and (10): Z tZ Z K(x − ξ, t)ω0 (ξ) dξ − K(x − ξ, t − τ )F (ξ, τ ) dξdτ ω= R2 Ω 0 (11) Z tZ K(x − ξ, t − τ )ρ(ξ, τ ) dsξ dτ. + 0
∂Ω
For exact solutions the velocity vanishes on the boundary, so ρ cannot be expressed in terms of velocity. Instead we will prove that ρ satisfies Z tZ (12) L(η, ξ, t − τ )ρ(ξ, τ ) dsξ dτ + f (η, t) = 0, ∂Ω
0
where
Z
Z Ω Z t
(13)
K(x − ξ, t)ω0 (ξ) dξ Z K(x − ξ, t − τ )F (ξ, τ ) dξdτ dx,
M (x, η)
f (η, t) = − 0
R2
Ω
and the kernels L, M will be given later. The following scheme is essentially the same as (1), except the method of vorticity creation is different. By induction we assume that the approximate solution ωk (t) is already known for t < ti , ti = ik, i = 0, 1, · · · ; then ωk (t) is obtained in [ti , ti+1 ) by the following fractional time step scheme. Convection step. Solve the inviscid problem ∂ω ˜k +u ˜k · ∇˜ ωk = 0, ∂t
(14) (15)
˜k , − 4 ψ˜k = ω
x ∈ Ω,
u ˜k = ∇ ∧ ψ˜k ,
t ∈ [ti , ti+1 ),
x ∈ Ω,
(16)
ψ˜k |x∈∂Ω = 0,
(17)
ω ˜ k (ti ) = ωk (ti − 0),
t ∈ [ti , ti+1 ),
where ωk (−0) = ω0 . Vorticity creation step. Solve Z tZ (18) L(η, ξ, t − τ )ρk (ξ, τ ) dsξ dτ + fk (η, t) = 0, 0
∂Ω
where ρk (ξ, t) is obtained for t ∈ [0, ti ) by the previous steps and fk is a linear interpolation (19)
fk (η, τ ) =
1 ((ti+1 − τ )fk i + (τ − ti )fk i+1 ), k
τ ∈ [ti , ti+1 ],
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and fk i , fk i+1 is given by
Z
Z fk i (η) =
M (x, η)
Ω
K(x − ξ, ti )ω0 (ξ) dξ
R2
(20) −
Z
i−1 Z X j=0
K(x − ξ, ti − tj )
tj+1
F (ξ, τ ) dξdτ
Ω
tj
dx,
ωk . where F = u˜k · ∇˜ Diffusion step. Solve the heat equation ∂ωk = 4ωk , ∂t
(21)
x ∈ R2 ,
t ∈ [ti , ti+1 ),
with initial data ωk (ti ) = ωk (ti − 0) + ω10 + ω20 ,
(22) where ω10 (x)
ω ˜ k (ti+1 − 0) − ωk (ti − 0), x ∈ Ω, 0, x 6∈ Ω,
=
and ω20 (ξ + sn nξ ) =
Z
ti+1
ρk (ξ, t) dtδ(sn ),
∀ξ ∈ ∂Ω, −∞ < sn < +∞,
ti
where nξ is the outward unit normal vector at ξ, and δ is the Dirac δ-function. More precisely, Z Z ti+1 0 ρk (ξ, t) dt ϕ(ξ) dsξ , ∀ϕ ∈ C0∞ (R2 ). hω2 , ϕi = ∂Ω
ti
We are now ready to state our convergence and error estimate results. For simplicity we assume that the domain Ω is simply connected, ∂Ω is sufficiently smooth, and the solution to (2)-(4) is sufficiently smooth. We will use the usual notations of Sobolev spaces, H s (Ω), and norms, k · ks , where s is a real number. Theorem 1.1. If 1 < σ < 32 , then (23)
k(ω − ωk )(ti − 0)kσ−1 ≤ C ∗ k 4
1
for 0 ≤ i ≤ [ Tk ], provided k ≤ k0 , where k0 > 0 is a fixed constant and C ∗ is a constant independent of k. We remark that the rate of 14 is the same as that in [3], but the norm is different. We will derive the expression (11) and the equation (12) for vortex sheets in the next section. Using those expressions, we will study the approximate scheme (14)(22) in Section 3. We will estimate the upper bound and error for linear problems in Section 4 and Section 5, then study nonlinear problems in Section 6. Finally we will make a remark on the vortex sheet in Section 7. Throughtout this paper we will always denote by C a generic positive constant, and C0 , C1 , · · · , M0 , M1 , · · · , will be different positive constants.
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2. An expression of the exact solutions To derive an equivalent formulation of (5)-(9) we prove Lemma 2.1. Let ω ∈ L2 (Ω) and ψ ∈ H01 (Ω) ∩ H 2 (Ω) satisfy − 4 ψ = ω, Then ∇ψ|∂Ω = 0 if and only if
R Ω
x ∈ Ω.
ωh dx = 0 for all harmonic functions h ∈ H 1 (Ω).
Proof. We denote by n the outward unit normal vector on ∂Ω, and by ν the anticlockwise unit tangential vector on ∂Ω. We have Z Z Z Z ∂ψ h ds ωh dx = − 4ψh dx = ∇ψ · ∇h dx − Ω Ω Ω ∂Ω ∂n Z (24) ∂ψ h ds. =− ∂Ω ∂n R Since h|∂Ω ∈ H 1/2 (∂Ω) is arbitrary, ∂ψ ∂n |∂Ω = 0 is equivalent to Ω ωh dx = 0, and ∂ψ ∂ν |∂Ω = 0 is obvious. We set H(Ω) = {h ∈ H 1 (Ω); 4h = 0}; then (5)-(9) is equivalent to (5)-(7), (9) and Z (25) ωh dx = 0, ∀h ∈ H(Ω). Ω
Let us express a harmonic function in terms of a double layer potential. We define M (x, η) =
x−η · nη , 2π|x − η|2
Then h ∈ H(Ω) can be expressed as Z (26) h(x) =
x ∈ Ω, η ∈ ∂Ω.
M (x, η)ϕ(η) dsη ,
∂Ω
where ϕ is the density. We express the solution as (11); then (25)-(26) implies Z
Z Z
Z tZ K(x − ξ, t)ω0 (ξ) dξ −
M (x, η) Ω
R2
∂Ω
Z tZ + 0
K(x − ξ, t − τ )F (ξ, τ ) dξdτ K(x − ξ, t − τ )ρ(ξ, τ ) dsξ dτ ϕ(η) dsη dx = 0, 0
Ω
∂Ω
Let
Z M (x, η)K(x − ξ, t) dx.
L(η, ξ, t) = Ω
312
Then
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Z
Z
Z tZ L(η, ξ, t − τ )ρ(ξ, τ )ϕ(η) dsξ dτ dsη +
∂Ω
∂Ω
0
f (η, t)ϕ(η) dsη = 0, ∂Ω
where f is given by (13). Since ϕ is arbitrary, we obtain (12), an integral equation for ρ. There is a variational form of (12). We use the conventional notations of Sobolev s s spaces H s, 2 (Ω × (0, T )) or H s, 2 (∂Ω × (0, T )) related to parabolic equations, s
s
H s, 2 (Ω × (0, T )) = L2 (0, T ; H s (Ω)) ∩ H 2 (0, T ; L2(Ω)), equipped with the norm 12 kf ks, s2 = kf k2L2 (0,T ;H s (Ω)) + kf k2H s2 (0,T ;L2 (Ω)) . s, s
As usual, H0 2 (Ω× (0, T )) is the closure of C0∞ (Ω× (0, T ]) with respect to the norm s s, s k · ks, s2 , and H −s,− 2 (Ω × (0, T )) is defined as the dual space of H0 2 (Ω × (0, T )). Let us define Z T Z Z tZ L(η, ξ, t − τ )ρ(ξ, τ )ϕ(η, t) dsξ dτ dsη dt. a(ρ, ϕ) = 0
∂Ω
0
∂Ω s−2, s2 −1
Then (12) can be written as: find ρ ∈ H0 (27)
a(ρ, ϕ) + hf, ϕi = 0,
, satisfying
∀ϕ ∈ H −s,− 2 . s
Since (5)-(9) is well-posed, and (27) is equivalent to it, (27) is also well-posed. However we will use (27) to define approximate solutions, the existence and uniqueness of which is unknown, so we will prove the exixtence and uniqueness of a solution to (27) in the next section. 3. Approximate solutions We will use the generalized Lax-Milgram theorem (see [1]) to study (27). For the reader’s convenience we recall it here. Proposition 3.1. If H1 and H2 are two real Hilbert spaces, and a(u, v) is a bilinear form on H1 × H2 such that |a(u, v)| ≤ CkukH1 kvkH2 , (28)
|a(u, v)| ≥ C0 , u∈H1 ,u6=0 v∈H2 ,v6=0 kukH1 kvkH2 sup
inf
sup |a(u, v)| > 0,
v 6= 0,
u∈H1
and if f ∈ H20 , then there exists a unique solution to the equation a(u, v) = f (v),
∀v ∈ H2 ,
such that kukH1 ≤ We prove the following fact.
1 kf kH20 . C0
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
Lemma 3.1. If s > s−2, s2 −1
solution ρ ∈ H0
3 2
313
s, s
and f ∈ H0 2 , then the problem (27) admits a unique
, and kρks−2, s2 −1 ≤ Ckf ks, s2 .
(29)
s−2, s −1
we construct a single Proof. Let us verify (28). For an arbitrary ρ ∈ H0 2 layer potential Z tZ (30) K(x − ξ, t − τ )ρ(ξ, τ ) dsξ dτ. ω= 0
∂Ω
Then it is known that kωks− 12 , s2 − 14 ≤ Ckρks−2, 2s −1 .
(31)
Let a function u be solved from (6), (7). Then ∇ · u = 0 and ω = −∇ ∧ u. From the equation ∂ω = 4ω ∂t we get ∇∧
∂u − 4u ∂t
= 0,
so there exists a function p such that ∂u − 4u = ∇p, ∂t
(32)
and hence u is the solution to the Stokes equation. Let x → ∂Ω in (26); then we get a Fredholm integral equation for ϕ: Z 1 M (x, η)ϕ(η) dsη , x ∈ ∂Ω. h(x) = − ϕ(x) + 2 ∂Ω Owing to potential theory the mapping from h to ϕ is an isomorphism in H s , where s is a real number. By (24) we have Z Z ωh dx = u · νh ds. Ω
We take h ∈ H (33)
−s,− 2s
∂Ω
such that Z TZ 1 ωh dx dt ≥ ku · νks, 2s khk−s,− 2s 2 Ω 0 ≥ Cku · νks, s2 kϕk−s,− s2 .
The estimate for the solutions to the Stokes equation yields (34)
kωks− 12 , s2 − 14 ≤ Ckuks, s2 ,∂Ω×(0,T ) = Cku · νks, 2s ,∂Ω×(0,T ) .
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LUNG-AN YING
There is a one-to-one correspondence between ρ and
∂ω ∂n |∂Ω
in H s−2, 2 −1 , so s
kρks−2, s2 −1 ≤ Ckωks− 12 , s2 − 14
(35)
for s > 2. (35) is also true for s ∈ ( 32 , 2]. Let us prove this assertion. Let ϕ ∈ C0∞ (∂Ω × (0, T ); then we extend it and define v so that v|∂Ω×(0,T ) = ϕ, ¯ × (0, T )) and v ∈ C0∞ (Ω kvk−s+ 52 ,− s2 + 54 ≤ Ckϕk−s+2,− s2 +1 . We have Z
T
0= 0
Z =
0
T
∂ω − 4ω v dx dt ∂t Ω Z TZ Z ∂ω ∂ω v + ∇ω · ∇v dx dt − ϕ dsx dt, ∂t Ω ∂Ω ∂n 0 Z
and Z Z T ∇ω · ∇v dx dt ≤ Ckvk−s+ 52 ,− s2 + 54 kωks− 12 , 2s − 14 , 0 Ω Z Z T ∂ω ∂ω v dx dt ≤ Ckvk − s2 + 54 k k s 5 H (0,T ;L2 (Ω)) ∂t H 2 − 4 (0,T ;L2 (Ω)) 0 Ω ∂t ≤ Ckvk−s+ 52 ,− s2 + 54 kωks− 12 , s2 − 14 . Hence k
∂ω ks−2, s2 −1,∂Ω×(0,T ) ≤ Ckωks− 12 , s2 − 14 , ∂n
which yields (35). It follows from (33)-(35) that Z 0
T
Z Ω
ωh dx dt ≥ Ckρks−2, s2 −1 kϕk−s,− 2s .
It is easy to see that Z
T
Z ωh dx dt,
a(ρ, ϕ) = 0
Ω
so (28) is verified. To verify the other conditions of Proposition 3.1 is straightforward.
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315
Using the fundamental solution K, the solution ωk of the scheme (14)-(22) can be expressed as Z K(x − ξ, ti )ω0 (ξ) dξ ωk (x, ti − 0) = R2 i−1 Z X
−
(36)
j=0
+
Z K(x − ξ, ti − tj )
Ω
F (ξ, τ ) dξdτ tj
Z
i−1 Z X j=0
tj+1
K(x − ξ, ti − tj )
∂Ω
tj+1
ρk (ξ, τ ) dτ dsξ , tj
ωk and ρk is the solution to (18). where F = u˜k · ∇˜
4. Upper bound estimate for linear problems To analyse the above scheme for the Navier-Stokes equations, we study the same scheme for linear problems first. We replace (5) by ∂ω + F (x, t) = 4ω, ∂t
(37)
where F is a known function. This makes (37) (6)-(9) a linear problem. We use the same scheme to solve it, and estimate the approximate solution in this and the next section. Likewise, we assume that the exact solution is sufficiently smooth. To estimate the approximate solution, let us first define some operators. We define A = − 4 +I, where I is the identity. The semigroup of operators generated 1 from A in L2 (R2 ) is denoted by e−At . We define B : H s (Ω) → H s+ 2 (∂Ω) by B : ω → g, Z M (x, η)ω(x) dx, g(η) = Ω
and the restriction operator from H s (R2 ) to H s (Ω) is denoted by R. Let us estimate fk first. The first term in (20) is sufficiently regular, so we need to estimate the second term only. Let i0 = [ Tk ], and hi =
i−1 X
BRe
j=0
−(A−I)(ti −tj )
Z
tj+1
F (τ ) dτ, tj
where we understand F (τ ) to be the zero extension of F to the space R2 . Lemma 4.1. If 2 ≤ s < 3, then i0 X i=1
where
1 2
> β > s − 52 .
khi k2s,∂Ω k ≤ CkF k2L2 (0,T ;H β (Ω)) ,
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Proof. By the properties of analytic semigroups of operators (see [6]) we have
2
Z tj+1 i0 i0 X i−1 X X
−(A−I)(ti −tj )
khi k2s,∂Ω k ≤ C e F (τ ) dτ k
tj
1 i=1 i=1 j=0 s− 2
2
Z i0 X X
i−1 s − 1 −(A−I)(t −t ) tj+1 i j
2 4 A e F (τ ) dτ ≤C
k
tj
i=1 j=0 0 2 Z i i−1 0 X X tj+1 s−β β 1 (ti − tj )− 2 + 4 kA 2 F (τ )k0 dτ k ≤C i=1
tj
j=0
2
i0 i−1 Z X X ≤C i=1
≤C
j=0
·
tj+1
1 − s−β 2 +4
(ti − tj )
i−1 Z X
j=0
tj+1
iX 0 −1 Z tj+1 j=0
kF (τ )kβ,Ω dτ k
tj+1
dτ
1 − s−β 2 +4
(ti − tj )
kF (τ )k2β,Ω dτ
tj
(ti − tj )−
s−β 1 2 +4
k
kF (τ )k2β,Ω dτ k
tj
i=1 j=0
=C
(ti − tj )
tj
j=0
i0 X i−1 Z X
1 − s−β 2 +4
tj
i0 X i−1 Z X i=1
≤C
tj+1
kF (τ )k2β,Ω dτ
tj
i0 X
(ti − tj )−
s−β 1 2 +4
k
i=j+1
≤ CkF k2L2 (0,T ;H β (Ω)) .
To estimate kfk kH 2s (0,T ;L2 (∂Ω)) we prove one auxiliary lemma first. Lemma 4.2. If β ∈ (0, 12 ) and g : [0, ti0 ] → R is a piecewise constant function, g(t) = gi , t ∈ (ti , ti+1 ), then Z ti0 Z ti0 X (gi − gj )2 (g(t) − g(τ ))2 dt dτ ≤ C k2 . 1+2β 1+2β |t − τ | |t − t | i j 0 0 i6=j
Proof. The left hand side can be divided into two parts: iX 0 −1 Z ti Z ti+1 k 1−2β X (gi−1 − gi )2 dt dτ ≤ (gi+1 − gi )2 , 1+2β |t − τ | 2β(1 − 2β) t t i−1 i i=1 i X Z |i−j|≥2
ti+1
ti
Z
tj+1
tj
Then the conclusion follows.
X (gi − gj )2 (gi − gj )2 dt dτ ≤ C k2 . 1+2β |t − τ | |ti − tj |1+2β |i−j|≥2
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
By (19),
∂fk ∂τ
317
is a piecewise constant function, 1 ∂fk = (fk i+1 − fk i ), ∂τ k
τ ∈ (ti , ti+1 ).
We define (38)
gi =
Lemma 4.3. If
5 2
1 (hi+1 − hi ). k
< s < 3, then
X kgi − gj k2 0 2 k ≤ CkF k2 s −1 , 1 H 2 (0,T ;H − 2 +ε (Ω)) |ti − tj |s−1 i6=j
where ε > 0. Proof. The left hand side is equal to 2
X kgi − gj k2 0 2 k s−1 |t − t | i j j
=2
X j
1 |ti − tj |s−1
i
X
BRe−(A−I)(ti+1 −tl )
l=0
−
i−1 X
BRe
−(A−I)(ti −tl )
l=0 j X
−
BRe−(A−I)(tj+1 −tl ) −
l=0
j−1 X
BRe−(A−I)(tj −tl )
!Z
F (τ ) dτ
!Z
tl tl+1
tl
l=0
tl+1
2
F (τ ) dτ .
0
We extend F by zero for τ < 0; then the right hand side is equal to
2
X j
1 |ti − tj |s−1
X Z tl+1
j −(A−I)(tj+1 −tl )
BRe (F (τ + ti − tj ) − F (τ )) dτ
tl
l=j−i −
j−1 X
BRe−(A−I)(tj −tl )
X j
1 |ti − tj |s−1
X
j−1
BR(e−(A−I)(tj+1 −tl )
l=j−i
X j
1 |ti − tj |s−1
≡ a1 + a2 .
2
(F (τ + ti − tj ) − F (τ )) dτ
0
2
−(A−I)(tj −tl ) −e ) (F (τ + ti − tj ) − F (τ )) dτ
tl
0
Z tj+1
2
(F (τ + ti − tj ) − F (τ )) dτ
BRe−(A−I)k
tj Z
+4
tl+1
tl
l=j−i
≤4
Z
tl+1
0
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Due to the properties of analytic semigroups we have
BR(e−(A−I)(tj+1 −tl ) − e−(A−I)(tj −tl ) )(F (τ + ti − tj ) − F (τ )) 0,∂Ω
3
4 −(A−I)(tj −tl )
≤ Ck A e (F (τ + ti − tj ) − F (τ )) 0
≤ Ck(tj − tl )−1+ 2 kF (τ + ti − tj ) − F (τ )k− 12 +ε ; ε
then, following the same lines of the proof of Lemma 4.1, we get j−1 Z tl+1 X X k2 ε (tj − tl )−1+ 2 kF (τ + ti − tj ) − F (τ )k2− 1 +ε dτ. a1 ≤ C s−1 2 |t − t | i j tl jl
0 i0 −1−i X
i0 =1
l=−i0
≤ CkF k2
Z
tl+1
kF (τ + i0 k) − F (τ )k2− 1 +ε 2
(i0 k)s−1
tl
s
1
H 2 −1 (0,T ;H − 2 +ε (Ω))
dτ
.
For the second term,
!2 Z tj+1 1 kF (τ + ti − tj ) − F (τ )k− 12 +ε dτ a2 ≤ C |t − tj |s−1 tj j 0. Proof. We get from (11) and (36) that
ω(x, ti ) − ωk (x, ti − 0) i−1 Z tj+1 Z X (K(x − ξ, ti − τ ) − K(x − ξ, ti − tj ))F (ξ, τ ) dξdτ =−
+ =−
j=0
tj
j=0
tj
i−1 Z X
i−1 Z X
i−1 Z X j=0
+
j=0
tj+1
Z (K(x − ξ, ti − τ ) − K(x − ξ, ti − tj ))F (ξ, τ ) dξdτ Ω
tj+1
Z (K(x − ξ, ti − τ ) − K(x − ξ, ti − tj ))ρk (ξ, τ ) dsξ dτ
tj
i−1 Z X
(K(x − ξ, ti − τ )ρ(ξ, τ ) − K(x − ξ, ti − tj )ρk (ξ, τ )) dsξ dτ
∂Ω
tj
j=0
+
Ω
tj+1 Z
∂Ω
tj+1
Z
tj
K(x − ξ, ti − τ )(ρ(ξ, τ ) − ρk (ξ, τ )) dsξ dτ ∂Ω
≡ a 1 + a2 + a3 . The sum in a1 , a2 can be divided into two parts: j = 0, · · · , i − 2, and j = i − 1. For j = i − 1 we have
Z
ti
−(A−I)(ti −τ ) −(A−I)k −e e F (τ ) dτ
ti−1 σ−2 Z ti ≤C kF (τ )kβ dτ ≤ Ck, ti−1
and for j = 0, · · · , i − 2 we have
X Z
i−2 tj+1 −(A−I)(t −τ ) −(A−I)(ti −tj ) i
−e e F (τ ) dτ
j=0 tj σ−2
i−2 Z tj+1 Z ti −τ
X
≤C e−(A−I)ζ dζF (τ ) dτ
tj ti −tj j=0
≤ Ck
i−2 Z X j=0
σ
tj+1
tj
(ti − τ )−
σ−β 2
kF (τ )kβ dτ ≤ Ck.
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We get for the first part of a2 that
Z
ti 1
RA 2 e−(A−I)(ti −τ ) − e−(A−I)k χk (τ ) dτ ka2 1 kσ−2 =
ti−1
Z ti
R e−(A−I)(ti −τ ) − e−(A−I)k ∇ · χk (τ ) dτ +
ti−1 σ−2 Z ti
σ2 − 12 −(A−I)(ti −τ ) ≤C − e−(A−I)k χk (τ ) dτ e
A 0
ti−1 Z ti
σ
2 −1 −(A−I)(ti −τ ) − e−(A−I)k ∇ · χk (τ ) dτ e
A
+C Z
0
ti−1 ti
(ti − τ )
≤C
ti−1 Z ti
(ti − τ )
+C Z ≤C
s−σ 3 ε 2 −4−2
s−σ 1 2 −4
ti−1 ti
(ti − τ )
kχk (τ )ks− 32 dτ
s−σ 3 ε 2 −4−2
dτ sup kχk (τ )ks− 52 −ε τ
ti−1
Z
kχk (τ )ks− 52 −ε dτ
! 12
ti
s−σ− 12
(ti − τ )
+C
Z
dτ
ti−1
We take s = σ +
3 2
ti
ti−1
! 12 kχk (τ )k2s− 3 2
dτ
.
and notice (41), then get ε
ka2 1 kσ−2 ≤ Ck 1− 2 . For the second part of a2 we can repeat the proof of Theorem 4.1 and get an upper bound of Ck. The details are omitted here. Finally, by potential theory and (47) we have 7
ka3 k1+ε, 12 + ε2 ≤ Ck 16 +
3β ε 8 −2
.
Then by the embedding theorem we get 7
ka3 k0,Ω ≤ Ck 16 +
3β ε 8 −2
.
Combining the above results, we obtain (48). 6. Estimates for the Navier-Stokes equation For the Navier-Stokes equation we denote F0 = u · ∇ω, where u and ω are exact solutions; then we regard F0 as a known function. We denote by f ∗ , ρ∗ and ω ∗ the approximate solutions to ∂ω + F0 = 4ω ∂t and (6)-(9). Then we define Z (49) ω ˜ ∗ (τ ) = −
τ
ti
F0 dτ + ω ∗ (ti − 0),
τ ∈ [ti , ti+1 ).
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
325
The estimates in Section 4 and Section 5 are valid for ω ∗ and ω ˜ ∗ , namely, we have (50)
ω ∗ (t)kσ ≤ C1 , kω ∗ (ti − 0)kσ , k˜
(51)
kω(ti ) − ω ∗ (ti − 0)kσ−2,Ω ≤ C2 k 16 + 7
3β ε 8 −2
,
for 0 ≤ i ≤ i0 and 0 ≤ t ≤ T . Certainly the constants C1 , C2 depend on the exact solution. Using this auxiliary solution we estimate ω ∗ − ωk ; then we obtain the estimates for ω − ωk . We need one fact about Sobolev spaces. For the reader’s convenience we recall it here: Lemma 6.1. If h, g ∈ H r (Ω) with 1 < r < 2, then khgkr ≤ Ckhkr kgkr ,
(52)
and if h ∈ H 1+ε (Ω), g ∈ H r (Ω) with ε > 0, 0 ≤ r ≤ 1, then khgkr ≤ Ckhk1+ε kgkr .
(53)
To estimate the upper bound and error for non-linear problems, let us prove the following lemmas first. Lemma 6.2. Let gi be defined by (38). Then i0 X
(54)
kgi k20,∂Ω k ≤ CkF k2 2
1
L (0,T ;H − 2 +ε (Ω))
i=1
,
where ε > 0. Proof. We notice that i−1 Z tj+1 1X BR e−(A−I)(ti+1 −tj ) − e−(A−I)(ti −tj ) F (τ ) dτ k j=0 tj Z ti+1 1 −(A−I)k F (τ ) dτ + BRe k ti Z tj+1 Z i−1 X 1 ti+1 −tj −(A−I)ζ BR (A − I)e dζ F (τ ) dτ =− k ti −tj tj j=0 Z 1 ti+1 F (τ ) dτ. + BRe−(A−I)k k ti
gi =
Following the same lines as the proof of Lemma 4.1, we get (54). For the Navier-Stokes equation, F = u ˜k ·∇˜ ωk . We apply Lemma 4.1 and Lemma 6.2 to obtain (55)
kfk k2,1 ≤ CkF k
1
+ C(ω0 );
1
+ C(ω0 ).
L2 (0,T ;H − 2 +ε (Ω))
then Lemma 3.1 gives (56)
kρk k0,0 ≤ CkF k
L2 (0,T ;H − 2 +ε (Ω))
Lemma 6.3. ωk k− 12 +ε ≤ Ck˜ ωk k21 +ε . k˜ uk · ∇˜ 2
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LUNG-AN YING
Proof. We have ωk = −∇ ∧ ((˜ uk · ∇)˜ uk ) = −∇ ∧ (∇(˜ uk ⊗ u ˜k )). u ˜k · ∇˜ Lemma 6.1 implies ωk k− 12 +ε = k˜ uk ⊗ u ˜k k 32 +ε ≤ Ck˜ uk k23 +ε ≤ Ck˜ ωk k21 +ε . k˜ uk · ∇˜ 2
Lemma 6.4. Assume that 1 < σ < 32 , 0 < ε ≤
and
ωk kσ−1−ε ≤ M0 . k˜ ωk kL2 (0,T ;H σ−ε (Ω)) + k˜
(57) Then (58)
σ−1 2 ,
2
i X
! 12 kωk (tl −
0)k2σ k
+ kωk (ti − 0)kσ−1 + kkωk (ti − 0)kσ+1 ≤ M1 ,
l=0
where M1 depends on M0 . Proof. We apply (36) and (42) to estimate ωk (ti − 0). We take β ∈ (σ − 1, 12 ), then get
X Z i−1 Z tj+1 X
i−1 tj+1 1 −(A−I)(t −t ) 1+σ−β i j
2 A e χk (τ ) dτ ≤ C (ti − tj )− 2 kχk (τ )kβ dτ.
j=0 tj j=0 tj σ
Then
2
i−1 Z tj+1 X
X 1
2 e−(A−I)(ti −tj ) χ (τ ) dτ k A k
i j=0 tj σ Z T kχk (τ )k2β dτ = Ckχk k2L2 (0,T ;H β (Ω)) . ≤C 0
The other terms in (36) and (42) can be estimated in the same way. By Lemma 4.4, kχk kL2 (0,T ;H β (Ω)) ≤ Ckρk kβ− 1 , β − 1 .
(59)
2
Now (56) can be applied. We obtain X kωk (tl − 0)k2σ k ≤ CkF k2 2
2
4
1
+ C(ω0 ).
1
+ C(ω0 ).
L (0,T ;H − 2 +ε (Ω))
l
Due to Lemma 6.3, X kωk (tl − 0)k2σ k ≤ Ck˜ ωk k 4 4
L (0,T ;H 2 +ε (Ω))
l
We apply the interpolation inequality to obtain Z T Z T 4 k˜ ωk k 1 +ε dt ≤ k˜ ωk k6−4σ+8ε k˜ ωk k4σ−2−8ε σ−ε σ−1−ε dt. 0
2
0
We notice that 6− 4σ + 8ε ≤ 2, so the assumption (57) implies the right hand side is bounded by a constant depending on M0 . Following the proof of Theorem 4.1, the ωk (ti − 0)kσ+1 can also be obtained. upper bounds for k˜ ωk (ti − 0)kσ−1 and kk˜
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
327
Lemma 6.5. If (58) holds for i ≤ i1 , then there exists a constant k0 > 0, depending on M1 , such that if k ≤ k0 , then k˜ ωk (τ )kr ≤ C3 kωk (ti − 0)kr ,
(60)
ti ≤ τ < ti+1 ,
for 0 ≤ r ≤ 2, where C3 depends on M1 . Proof. It can be proved (see [10]) that for the Euler equation ∂u + (u · ∇)u + ∇p = 0, ∂t ∇ · u = 0, there are constants C4 , C5 , such that if t ≤
1 C4 ku(0)k3 ,
then ku(t)kr ≤ C5 ku(0)kr
for 0 ≤ r ≤ 3. By (58) and interpolation we have kωk (ti − 0)k2 ≤ M1 k σ−3 kuk (ti − 0)k3 ≤ C(M1 )k 2 . Thus we take 2 σ−1 1 . k0 = C4 C(M1 )
σ−3 2
; hence
If k ≤ k0 , then (60) holds. Lemma 6.6. Under the assumptions of Lemma 6.5, ! 12 iX 1 +1 5 ∗ 2 (61) k(ω − ωk )(ti − 0)kσ k ≤ C6 k 8 (σ−1)−2ε , i=0
where ε > 0, k ≤ k0 , and C6 depends on M1 . ωk . We obtain by Lemma 6.1 and interpolation that Proof. Let F˜ = u · ∇ω − u ˜k · ∇˜ ˜k ) ⊗ u + u ⊗ (u − u˜k ) − (u − u ˜k ) ⊗ (u − u ˜k )))k− 12 +ε kF˜ k− 12 +ε = k∇ ∧ (∇((u − u ≤ Cku − u ˜k k 23 +ε + Cku − u˜k k23 +ε 2
≤ Ckω − ω ˜ k k 12 +ε + Ckω − ω ˜ k k21 +ε 2
≤ Ckω − ω ˜ k k 12 +ε + Ckω − ω ˜ k kσ−1 kω − ω ˜ k k2−σ+2ε . Then we notice (58) and (60), to get for τ < ti1 +1 that ˜ k k2−σ+2ε kF˜ k− 12 +ε ≤ Ckω − ω
≤ Ck˜ ω∗ − ω ˜ k k2−σ+2ε + Ckω − ω ˜ ∗ k2−σ+2ε .
Since ω is sufficiently smooth, (49) and (51) imply that kω(τ ) − ω ˜ ∗ (τ )kσ−2 ≤ Ck 16 + 7
3β ε 8 −2
,
τ < ti1 +1 .
Because β ∈ (0, 12 ) and ε > 0 are arbitrary, we take β close to the interpolation between it and (50) gives
1 2
kω(τ ) − ω ˜ ∗ (τ )k2−σ+2ε ≤ Ck 8 (σ−1)−2ε , 5
and consequently (62)
ω∗ − ω ˜ k k2−σ+2ε + Ck 8 (σ−1)−2ε . kF˜ k− 12 +ε ≤ Ck˜ 5
and ε small; then
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LUNG-AN YING
It follows from (36) that for i ≤ i1 + 1 Z
i−1 Z X
∗
(ω − ωk )(x, ti − 0) = −
Ω
j=0
Z K(x − ξ, ti − tj )
∂Ω
j=0
F˜ (ξ, τ ) dξdτ
tj
i−1 Z X
+
tj+1
K(x − ξ, ti − tj )
tj+1
(ρ∗ − ρk )(ξ, τ ) dsξ dτ.
tj
Then we get i−1 Z X
k(ω ∗ − ωk )(ti − 0)kσ ≤ C
tj
j=0
(63)
+C
tj+1
i−1 Z X
σ 1 ε (ti − tj )− 2 − 4 + 2 kF˜ k− 12 +ε dτ
tj+1
(ti − tj )−
β 1+σ 2 +2
kχ∗ − χk kβ dτ,
tj
j=0
where σ − 1 < β < 12 . Analogously to (55), (56) and (59), we have kχ∗ − χk kβ, β ,Ω×(0,ti ) ≤ CkF˜ k
(64)
2
1
L2 (0,ti ;H − 2 +ε (Ω))
,
where we notice that the term corresponding to ω0 is cancelled out. Let i2 be an index to be determined. We take the sum i2 X
k(ω ∗ − ωk )(ti − 0)k2σ k
i=0
≤C
i2 X i=0
+C
2
i−1 Z X
i2 X
j=0
i=0
tj+1
i−1 Z X
·
i−1 Z X j=0
+C ·
i2 X
2 tj+1
β − 1+σ 2 +2
(ti − tj )
tj+1
(ti − tj )
1 ε −σ 2 −4+2
(ti − tj )
i−1 Z X
tj+1
dτ
tj
j=0
i−1 Z X
kχ∗ − χk kβ dτ k
1 ε −σ 2 −4+2
kF˜ k2− 1 +ε dτ k 2
tj
i=0
j=0
≤ Ct
tj+1
kF˜ k− 12 +ε dτ k
tj
j=0
j=0
(ti − tj )
tj
i2 i−1 Z X X ≤C i=0
1 ε −σ 2 −4+2
tj+1
β − 1+σ 2 +2
(ti − tj )
tj
dτ
β − 1+σ 2 +2
(ti − tj )
kχ∗ − χk k2β dτ k
tj
3 2 −σ+ε i2
kF˜ k2 2
1
L (0,ti2 ;H − 2 +ε (Ω))
+ Ct1−σ+β kχ∗ − χk k2L2 (0,ti i2
2 ;H
β (Ω))
.
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
329
Then by (64) (65)
i2 X
˜ 2 k(ω ∗ − ωk )(ti − 0)k2σ k ≤ Ctα i2 kF k 2
1
L (0,ti2 ;H − 2 +ε (Ω))
i=0
,
where α = min( 32 − σ + ε, 1 − σ + β) > 0. Let us estimate the right hand side of (65) through (62). For ti ≤ τ < ti+1 we have Z τ ˜ k )(τ )k2−σ+2ε ≤ kF˜ k2−σ+2ε dτ + k(ω ∗ − ωk )(ti − 0)k2−σ+2ε . k(˜ ω∗ − ω ti
Then applying Lemma 6.1 we obtain ˜ k )(τ )k2−σ+2ε k(˜ ω∗ − ω Z τ ≤ (C + Ck˜ uk k1+ε k˜ ωk k3−σ+2ε ) dτ + k(ω ∗ − ωk )(ti − 0)k2−σ+2ε . ti
The inequalities (60) and (58) imply (66)
˜ k )(τ )k2−σ+2ε ≤ Ck σ−1−2ε + k(ω ∗ − ωk )(ti − 0)k2−σ+2ε . k(˜ ω∗ − ω
We substitute (66) and (62) into (65), and obtain i2 X
k(ω ∗ − ωk )(ti − 0)k2σ k
i=0
≤
Ctα i2
(i −1 2 X
) ∗
0)k22−σ+2ε k
∗
0)k2σ k
k(ω − ωk )(ti −
+k
5 4 (σ−1)−4ε
i=0
≤
Ctα i2
(i −1 2 X
) k(ω − ωk )(ti −
+k
5 4 (σ−1)−4ε
.
i=0 1 We take ti2 small enough so that Ctα i2 ≤ 2 , then get i2 X
k(ω ∗ − ωk )(ti − 0)k2σ k ≤ Ck 4 (σ−1)−4ε . 5
i=0
We notice that ti2 is independent of k; thus we can carry out an induction in a finite number, T /ti2 , of steps. Suppose (61) is true for i ≤ N i2 ; then X
(N +1)i2
k(ω ∗ − ωk )(ti − 0)k2σ k
i=N i2 +1
X
(N +1)i2
≤C
i=N i2 +1
X
(N +1)i2
+C
i=N i2 +1
i−1 Z X j=0
2 tj+1
1 ε −σ 2 −4+2
(ti − tj )
tj
i−1 Z X j=0
kF˜ k− 12 +ε dτ k 2
tj+1
tj
β − 1+σ 2 +2
(ti − tj )
kχ∗ − χk kβ dτ k.
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LUNG-AN YING
By (62),(64) and the estimate for i ≤ N i2 we get X
(N +1)i2
k(ω ∗ − ωk )(ti − 0)k2σ k
i=N i2 +1
X
(N +1)i2
≤C
i=N i2 +1
j=N i2 +1
X
i−1 X
(N +1)i2
+C
i=N i2 +1
+ Ck
2
Z
i−1 X
tj+1
(ti − tj )
tj
kF˜ k− 12 +ε dτ k 2
Z
j=N i2 +1
5 4 (σ−1)−4ε
1 ε −σ 2 −4+2
tj+1
(ti − tj )−
β 1+σ 2 +2
kχ∗ − χk kβ dτ k
tj
.
Following the same lines, we get the estimate for i ≤ (N + 1)i2 . After a finite number of steps we get the estimate up to i ≤ i1 + 1. Lemma 6.7. Under the assumptions of Lemma 6.5, k(ω ∗ − ωk )(ti − 0)kσ−1,Ω ≤ C7 k 8 (σ−1)−2ε , 5
(67) for i ≤ i1 + 1.
Proof. Analogously to (63), we have ∗
k(ω − ωk )(ti − 0)kσ−1 ≤ C
i−1 Z X j=0
+C
tj+1
tj
i−1 Z X j=0
1 σ ε (ti − tj ) 4 − 2 + 2 kF˜ k− 12 +ε dτ
tj+1
(ti − tj )− 2 + 2 kχ∗ − χk kβ dτ, β
σ
tj
and then k(ω ∗ − ωk )(ti − 0)kσ−1 12 i−1 Z tj+1 i−1 Z X X 1 ≤C (ti − tj ) 2 −σ+ε dτ j=0
tj
j=0
i−1 Z X +C j=0
tj+1
12 (ti − tj )−σ+β dτ
tj
tj+1
j=0
kF˜ k2− 1 +ε dτ 2
tj
i−1 Z X
12
tj+1
12 kχ∗ − χk k2β dτ .
tj
Then we use (62),(64) to get the desired estimate. Theorem 6.1. If 1 < σ < 32 , then (68)
i X
! 12 k(ω ∗ − ωk )(tl − 0)k2σ k
+ k(ω ∗ − ωk )(ti − 0)kσ−1 ≤ Ck 4 1
l=0
for 0 ≤ i ≤ [ Tk ], provided k ≤ k0 , where k0 > 0 is a fixed constant and C is independent of k.
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
331
Proof. We prove (58),(67) and (69)
i X
! 12 k(ω ∗ − ωk )(tl − 0)k2σ k
5
≤ C8 k 8 (σ−1)−2ε
l=0
by induction. They are obviously satisfied for i = 0. Let us fix the constant M0 in (57) first. By (50) it is easy to see that ω ∗ kσ−1−ε ≤ C9 . k˜ ω ∗ kL2 (0,T ;H σ−ε (Ω)) + k˜ We take M0 = 2C9 . We assume that (58), (67), and (69) are satisfied for i ≤ i1 . Analogously to (66), we have ˜ k )(τ )kσ−1 ≤ Ck 2 + k(ω ∗ − ωk )(ti − 0)kσ−1 k(˜ ω∗ − ω 1
for τ ∈ [ti , ti+1 ), i ≤ i1 . We substitute (67) in it and obtain ˜ k )(τ )kσ−1 ≤ Ck 8 (σ−1)−2ε . k(˜ ω∗ − ω 5
(50),(60) and (58) imply ˜ k )(τ )kσ ≤ C. k(˜ ω∗ − ω Then by interpolation we obtain ˜ k )(τ )kσ−ε ≤ Ck ε( 8 (σ−1)−2ε) . k(˜ ω∗ − ω 5
We take k0 small enough so that (57) holds for k ≤ k0 and τ < ti1 +1 . Then from Lemma 6.4, Lemma 6.6, and Lemma 6.7 we conclude that (58),(67),(69) are satisfied for i ≤ i1 + 1. Thus the induction is complete. We take σ close to 32 and ε small enough. Then the exponent is greater than 1 4. Finally let us prove our main result, Theorem 1.1. Proof of Theorem 1.1. We take β close to
1 2
and ε small enough in (48). Then
kω(ti ) − ω ∗ (ti − 0)kσ−2,Ω ≤ Ck 2 . 1
Theorem 4.1 implies kω ∗ (ti − 0)kσ,Ω ≤ C. Then by interpolation we get kω(ti ) − ω ∗ (ti − 0)kσ−1,Ω ≤ Ck 4 , 1
which in conjunction with (68) yields (23). 7. A remark on the vortex sheets
Rt We have shown in (36) that the density of vortex sheets is tjj+1 ρk (ξ, τ ) dτ . Let us make a heuristic analysis here to compare our result with the others in the literature. The vorticity in (36) is divided into three parts: by initial data, by convection, and the vortex sheet, ω = ω1 +ω2 +ω3 . Suppose the velocity generated by ω1 +ω2 is u = un n+uν ν on the boundary ∂Ω. Then un = 0, due to the boundary condition of the Euler equation. The tangential component of velocity generated by the vortex sheets should be −uν , to keep the no-slip boundary condition.
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LUNG-AN YING
On the other hand, potential theory tells us that for x0 ∈ ∂Ω, Z tZ ∂ K(x − ξ, t − τ )ρk (ξ, τ ) dsξ dτ lim x→x0 ∂n x ∂Ω x∈Ω 0 (70) Z tZ ∂ 1 K(x0 − ξ, t − τ )ρk (ξ, τ ) dsξ dτ. = ρk (x0 , t) + 2 ∂Ω ∂nx 0 The left hand side is
¯ν ∂ ∂u ¯n ∂ u ∂ω3 + = − , ∂n ∂Ω ∂n ∂ν ∂n
u ¯n u ¯ν ¯ν ) is generated by ω3 . Since ∂∂n + ∂∂ν = 0, we have where u ¯ = (¯ un , u ∂ 2 u¯ν ∂2u ¯ν ∂ω3 = + = 4¯ uν . 2 ∂n ∂Ω ∂ν ∂n2
¯ satisfies the Stokes equation, but since ω3 satisfies the heat equation on R2 . u ∇·u ¯ = 0 at t = 0, u¯ also satisfies the heat equation and the pressure equals to zero. We get ∂u ¯ν ∂uν ∂ω3 = = − . ∂n ∂Ω ∂t ∂Ω ∂t ∂Ω Let us omit the second term in (70), to get
∂ω3 ∂n |∂Ω
∂uν . ρk = −2 ∂t ∂Ω
= 12 ρk . Then
When we set a vortex sheet at t = tj+1 , we assume that the no-slip condition is satisfied at t = tj . Therefore Z tj+1 Z tj+1 ∂uν dt = −2uν (ξ, tj+1 ). ρk dτ = −2 ∂t tj tj We notice that uν (ξ, tj+1 ) is the tangential component of velocity after the convection step. This is exactly the strength of vortex sheets given in some works. By the above analysis the jump of velocity across the boundary is Z tj+1 ∂ω3 ∂ω3 − lim dτ lim c [uν ] = x→x0 ,x∈Ω ∂n x→x0 ,x∈Ω ∂n tj Z tj+1 ρk (ξ, t) dt = −2uν , =− tj − where Ωc is the exterior of Ω. Therefore it is assumed that u+ ν |∂Ω = −uν |∂Ω , where “ + ” and “ − ” indicate the exterior and the interior. If Ω is a half plane, the second term of (70) is exactly zero. This observation coincides with that in [3]. However, in general, this term does not vanish.
Acknowledgment The author thanks the referees for their valuable comments.
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References [1] A.K.Aziz (ed.), The Mathematical Foundations of the Finite Element Method with Applications to Partial Differential Equations, Academic Press, New York and London, 1972. MR 49:11824 [2] J.T.Beale, and A.Majda, Rates of convergence for viscous splitting of the Navier-Stokes equations, Math. Comp., 37, 243-259, 1981. MR 82i:65056 [3] G.Benfatto, and M.Pulvirenti, Convergence of Chorin-Marsden product formula in the halfplane, Comm. Math. Phys., 106, 427-458, 1986. MR 88a:35186 [4] A.J.Chorin, Numerical study of slightly viscous flow, J. Fluid Mech., 57, 785-796, 1973. MR 52:16280 [5] A.J.Chorin, T.J.R.Hughes, M.F.McCracken, and J.E.Marsden, Product formulas and numerical algorithms, Comm. Pure Appl. Math., 31, 205-256, 1978. MR 58:8230 [6] A.Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations, Springer-Verlag, 1983. MR 85g:47061 [7] R.Temam, On the Euler equations of incompressible perfect fluids, J. Funct. Anal., 20, 32-43, 1975. MR 55:3573 [8] R.Temam, Navier-Stokes Equations, Theory and Numerical Analysis, 3rd ed., North Holland, 1984. MR 86m:76003 [9] L.-A.Ying, Convergence of Chorin-Marsden formula for the Navier-Stokes equations on convex domains, J. Comp. Math., 17, 73-88, 1999. MR 2000d:65162 [10] L.-A.Ying, and P.Zhang, Vortex Methods, Science Press, Beijing/New York, and Kluwer Academic Publishers, Dordrecht/Boston/London, 1997. MR 2000f:76093 School of Mathematical Sciences, Peking University 100871, People’s Republic of China E-mail address: [email protected]
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA LUNG-AN YING
Abstract. Using the fundamental solution of the heat equation, we give an expression of the solutions to two-dimensional initial-boundary value problems of the Navier-Stokes equations, where the vorticity is expressed in terms of a Poisson integral, a Newtonian potential, and a single layer potential. The density of the single layer potential is the solution to an integral equation of Volterra type along the boundary. We prove there is a unique solution to the integral equation. One fractional time step approximation is given, based on this expression. Error estimates are obtained for linear and nonlinear problems. The order of convergence is 14 for the Navier-Stokes equations. The result is in the direction of justifying the Chorin-Marsden formula for vortex methods. It is shown that the density of the vortex sheet is twice the tangential velocity for the half plane, while in general the density differs from it by one additional term.
1. Introduction The convergence and stability problems for the vortex method have been studied by many authors. Usually a fractional steps approach is applied in the vortex method for viscous flows; therefore viscous splitting is a basic problem in the mathematical theory. The convergence problem of viscous splitting was first studied in [2], and since then there have been quite a number of works on this subject (see [10] and references therein). Vorticity is the main variable in the computation of the vortex method. However, a velocity boundary condition is applied on a solid boundary and the boundary condition for vorticity is indirect. To approximate the velocity boundary condition, it was proposed that the velocity field be extended oddly after each convection step; then one vortex sheet was created, which resulted in an approximate zero velocity on the boundary (see [4]). This scheme was formulated as (1)
ωi = (Hk ◦ φ ◦ Ek )i ω0 ,
where k > 0 is the length of temporal steps, ωi is the vorticity at ti = ik, ω0 is the initial data, Ek is the solver of the Euler equation, Hk is the solver of the heat equation on the entire space, and φ is the “vorticity creation operator”. Stability and convergence of (1) for linear equations was proved in [5]. Convergence of the Chorin-Marsden formula for a half plane domain was proved in [3], with a slight modification, where the tangential component of velocity was Received by the editor June 5, 2000 and, in revised form, January 3, 2001. 2000 Mathematics Subject Classification. Primary 65M99; Secondary 35Q30, 76D05, 76M25. Key words and phrases. Vortex method, Navier-Stokes equation, fractional step method. c
2002 American Mathematical Society
307
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LUNG-AN YING
extended oddly and the normal component was extended evenly. An error estimate 1 with upper bound O(k 4 ) was obtained. Partial analysis for convex domains was gicen in [9], also with some modification, where the extended velocity field was multiplied by a Jacobian determinant. It was observed (see [9]) that both odd extension and even extension of the normal component of velocity yielded convergent results for the half plane domain. We notice that the way to create a vortex sheet is not unique, but all of them should be close enough so that the differences between them approach zero as k → 0. The aim of this paper is to study the problem for general bounded domains. The idea is to derive an expression of the density of vortex sheets for exact solutions, then make a time discretization to obtain an approximate one. In our scheme the derivation of vortex sheet is indirect, since it satisfies an integral equation of Volterra type along the boundary. A direct one may be possible. However, since we have the exact expression, the approximate one is required to be consistent with it. In [5], [3], [9] the density of the vortex sheet equals twice the tangential velocity after each convection step. We will make a remark in the final section, indicating that the density of the vortex sheet of our scheme is the same if the domain is a half plane. However, in the general case they are different by one additional term, which does not vanish in general. We hope this additional term approaches zero as k → 0 for some cases. This will be the subject of our further study. In order to state our scheme and convergence result, let us write down the equations first. Let Ω be a bounded domain in R2 , and ∂Ω its boundary. The initialboundary value problems of the Navier-Stokes equations for viscous incompressible flow are ∂u (2) + (u · ∇)u + ∇p = 4u, ∂t ∇ · u = 0,
(3) (4)
u|x∈∂Ω = 0,
u|t=0 = u0 ,
where u is the velocity and p is the pressure. Without loss of generality we set the ∂ ∂ density and viscosity equal to one. Let ∇∧ = ∂x2 , − ∂x1 , and let the vorticity ω = −∇ ∧ u; then the vorticity formulation of the Navier-Stokes equation is ∂ω + u · ∇ω = 4ω, ∂t
(5)
− 4 ψ = ω,
(6)
u = ∇ ∧ ψ,
(7)
ψ|x∈∂Ω = 0,
(8)
u|x∈∂Ω = 0,
(9)
ω|t=0 = ω0 = −∇ ∧ u0 .
Let ρ(ξ, t), ξ ∈ ∂Ω, t ∈ [0, T ] be the density of the vortex sheet of an exact solution; then the contribution of ρ to the solution is Z tZ (10) K(x − ξ, t − τ )ρ(ξ, τ ) dsξ dτ, 0
∂Ω
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
309
where K is the fundamental solution of the heat equation: 1 − |x|2 e 4t . 4πt
K(x, t) =
Let F = u·∇ω, and we extend ω0 smoothly to R2 so that it is compactly supported; then we will prove that the solution to (5)-(9) can be expressed in terms of the sum of a Poisson integral, a Newtonian potential, and (10): Z tZ Z K(x − ξ, t)ω0 (ξ) dξ − K(x − ξ, t − τ )F (ξ, τ ) dξdτ ω= R2 Ω 0 (11) Z tZ K(x − ξ, t − τ )ρ(ξ, τ ) dsξ dτ. + 0
∂Ω
For exact solutions the velocity vanishes on the boundary, so ρ cannot be expressed in terms of velocity. Instead we will prove that ρ satisfies Z tZ (12) L(η, ξ, t − τ )ρ(ξ, τ ) dsξ dτ + f (η, t) = 0, ∂Ω
0
where
Z
Z Ω Z t
(13)
K(x − ξ, t)ω0 (ξ) dξ Z K(x − ξ, t − τ )F (ξ, τ ) dξdτ dx,
M (x, η)
f (η, t) = − 0
R2
Ω
and the kernels L, M will be given later. The following scheme is essentially the same as (1), except the method of vorticity creation is different. By induction we assume that the approximate solution ωk (t) is already known for t < ti , ti = ik, i = 0, 1, · · · ; then ωk (t) is obtained in [ti , ti+1 ) by the following fractional time step scheme. Convection step. Solve the inviscid problem ∂ω ˜k +u ˜k · ∇˜ ωk = 0, ∂t
(14) (15)
˜k , − 4 ψ˜k = ω
x ∈ Ω,
u ˜k = ∇ ∧ ψ˜k ,
t ∈ [ti , ti+1 ),
x ∈ Ω,
(16)
ψ˜k |x∈∂Ω = 0,
(17)
ω ˜ k (ti ) = ωk (ti − 0),
t ∈ [ti , ti+1 ),
where ωk (−0) = ω0 . Vorticity creation step. Solve Z tZ (18) L(η, ξ, t − τ )ρk (ξ, τ ) dsξ dτ + fk (η, t) = 0, 0
∂Ω
where ρk (ξ, t) is obtained for t ∈ [0, ti ) by the previous steps and fk is a linear interpolation (19)
fk (η, τ ) =
1 ((ti+1 − τ )fk i + (τ − ti )fk i+1 ), k
τ ∈ [ti , ti+1 ],
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LUNG-AN YING
and fk i , fk i+1 is given by
Z
Z fk i (η) =
M (x, η)
Ω
K(x − ξ, ti )ω0 (ξ) dξ
R2
(20) −
Z
i−1 Z X j=0
K(x − ξ, ti − tj )
tj+1
F (ξ, τ ) dξdτ
Ω
tj
dx,
ωk . where F = u˜k · ∇˜ Diffusion step. Solve the heat equation ∂ωk = 4ωk , ∂t
(21)
x ∈ R2 ,
t ∈ [ti , ti+1 ),
with initial data ωk (ti ) = ωk (ti − 0) + ω10 + ω20 ,
(22) where ω10 (x)
ω ˜ k (ti+1 − 0) − ωk (ti − 0), x ∈ Ω, 0, x 6∈ Ω,
=
and ω20 (ξ + sn nξ ) =
Z
ti+1
ρk (ξ, t) dtδ(sn ),
∀ξ ∈ ∂Ω, −∞ < sn < +∞,
ti
where nξ is the outward unit normal vector at ξ, and δ is the Dirac δ-function. More precisely, Z Z ti+1 0 ρk (ξ, t) dt ϕ(ξ) dsξ , ∀ϕ ∈ C0∞ (R2 ). hω2 , ϕi = ∂Ω
ti
We are now ready to state our convergence and error estimate results. For simplicity we assume that the domain Ω is simply connected, ∂Ω is sufficiently smooth, and the solution to (2)-(4) is sufficiently smooth. We will use the usual notations of Sobolev spaces, H s (Ω), and norms, k · ks , where s is a real number. Theorem 1.1. If 1 < σ < 32 , then (23)
k(ω − ωk )(ti − 0)kσ−1 ≤ C ∗ k 4
1
for 0 ≤ i ≤ [ Tk ], provided k ≤ k0 , where k0 > 0 is a fixed constant and C ∗ is a constant independent of k. We remark that the rate of 14 is the same as that in [3], but the norm is different. We will derive the expression (11) and the equation (12) for vortex sheets in the next section. Using those expressions, we will study the approximate scheme (14)(22) in Section 3. We will estimate the upper bound and error for linear problems in Section 4 and Section 5, then study nonlinear problems in Section 6. Finally we will make a remark on the vortex sheet in Section 7. Throughtout this paper we will always denote by C a generic positive constant, and C0 , C1 , · · · , M0 , M1 , · · · , will be different positive constants.
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
311
2. An expression of the exact solutions To derive an equivalent formulation of (5)-(9) we prove Lemma 2.1. Let ω ∈ L2 (Ω) and ψ ∈ H01 (Ω) ∩ H 2 (Ω) satisfy − 4 ψ = ω, Then ∇ψ|∂Ω = 0 if and only if
R Ω
x ∈ Ω.
ωh dx = 0 for all harmonic functions h ∈ H 1 (Ω).
Proof. We denote by n the outward unit normal vector on ∂Ω, and by ν the anticlockwise unit tangential vector on ∂Ω. We have Z Z Z Z ∂ψ h ds ωh dx = − 4ψh dx = ∇ψ · ∇h dx − Ω Ω Ω ∂Ω ∂n Z (24) ∂ψ h ds. =− ∂Ω ∂n R Since h|∂Ω ∈ H 1/2 (∂Ω) is arbitrary, ∂ψ ∂n |∂Ω = 0 is equivalent to Ω ωh dx = 0, and ∂ψ ∂ν |∂Ω = 0 is obvious. We set H(Ω) = {h ∈ H 1 (Ω); 4h = 0}; then (5)-(9) is equivalent to (5)-(7), (9) and Z (25) ωh dx = 0, ∀h ∈ H(Ω). Ω
Let us express a harmonic function in terms of a double layer potential. We define M (x, η) =
x−η · nη , 2π|x − η|2
Then h ∈ H(Ω) can be expressed as Z (26) h(x) =
x ∈ Ω, η ∈ ∂Ω.
M (x, η)ϕ(η) dsη ,
∂Ω
where ϕ is the density. We express the solution as (11); then (25)-(26) implies Z
Z Z
Z tZ K(x − ξ, t)ω0 (ξ) dξ −
M (x, η) Ω
R2
∂Ω
Z tZ + 0
K(x − ξ, t − τ )F (ξ, τ ) dξdτ K(x − ξ, t − τ )ρ(ξ, τ ) dsξ dτ ϕ(η) dsη dx = 0, 0
Ω
∂Ω
Let
Z M (x, η)K(x − ξ, t) dx.
L(η, ξ, t) = Ω
312
Then
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Z
Z
Z tZ L(η, ξ, t − τ )ρ(ξ, τ )ϕ(η) dsξ dτ dsη +
∂Ω
∂Ω
0
f (η, t)ϕ(η) dsη = 0, ∂Ω
where f is given by (13). Since ϕ is arbitrary, we obtain (12), an integral equation for ρ. There is a variational form of (12). We use the conventional notations of Sobolev s s spaces H s, 2 (Ω × (0, T )) or H s, 2 (∂Ω × (0, T )) related to parabolic equations, s
s
H s, 2 (Ω × (0, T )) = L2 (0, T ; H s (Ω)) ∩ H 2 (0, T ; L2(Ω)), equipped with the norm 12 kf ks, s2 = kf k2L2 (0,T ;H s (Ω)) + kf k2H s2 (0,T ;L2 (Ω)) . s, s
As usual, H0 2 (Ω× (0, T )) is the closure of C0∞ (Ω× (0, T ]) with respect to the norm s s, s k · ks, s2 , and H −s,− 2 (Ω × (0, T )) is defined as the dual space of H0 2 (Ω × (0, T )). Let us define Z T Z Z tZ L(η, ξ, t − τ )ρ(ξ, τ )ϕ(η, t) dsξ dτ dsη dt. a(ρ, ϕ) = 0
∂Ω
0
∂Ω s−2, s2 −1
Then (12) can be written as: find ρ ∈ H0 (27)
a(ρ, ϕ) + hf, ϕi = 0,
, satisfying
∀ϕ ∈ H −s,− 2 . s
Since (5)-(9) is well-posed, and (27) is equivalent to it, (27) is also well-posed. However we will use (27) to define approximate solutions, the existence and uniqueness of which is unknown, so we will prove the exixtence and uniqueness of a solution to (27) in the next section. 3. Approximate solutions We will use the generalized Lax-Milgram theorem (see [1]) to study (27). For the reader’s convenience we recall it here. Proposition 3.1. If H1 and H2 are two real Hilbert spaces, and a(u, v) is a bilinear form on H1 × H2 such that |a(u, v)| ≤ CkukH1 kvkH2 , (28)
|a(u, v)| ≥ C0 , u∈H1 ,u6=0 v∈H2 ,v6=0 kukH1 kvkH2 sup
inf
sup |a(u, v)| > 0,
v 6= 0,
u∈H1
and if f ∈ H20 , then there exists a unique solution to the equation a(u, v) = f (v),
∀v ∈ H2 ,
such that kukH1 ≤ We prove the following fact.
1 kf kH20 . C0
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
Lemma 3.1. If s > s−2, s2 −1
solution ρ ∈ H0
3 2
313
s, s
and f ∈ H0 2 , then the problem (27) admits a unique
, and kρks−2, s2 −1 ≤ Ckf ks, s2 .
(29)
s−2, s −1
we construct a single Proof. Let us verify (28). For an arbitrary ρ ∈ H0 2 layer potential Z tZ (30) K(x − ξ, t − τ )ρ(ξ, τ ) dsξ dτ. ω= 0
∂Ω
Then it is known that kωks− 12 , s2 − 14 ≤ Ckρks−2, 2s −1 .
(31)
Let a function u be solved from (6), (7). Then ∇ · u = 0 and ω = −∇ ∧ u. From the equation ∂ω = 4ω ∂t we get ∇∧
∂u − 4u ∂t
= 0,
so there exists a function p such that ∂u − 4u = ∇p, ∂t
(32)
and hence u is the solution to the Stokes equation. Let x → ∂Ω in (26); then we get a Fredholm integral equation for ϕ: Z 1 M (x, η)ϕ(η) dsη , x ∈ ∂Ω. h(x) = − ϕ(x) + 2 ∂Ω Owing to potential theory the mapping from h to ϕ is an isomorphism in H s , where s is a real number. By (24) we have Z Z ωh dx = u · νh ds. Ω
We take h ∈ H (33)
−s,− 2s
∂Ω
such that Z TZ 1 ωh dx dt ≥ ku · νks, 2s khk−s,− 2s 2 Ω 0 ≥ Cku · νks, s2 kϕk−s,− s2 .
The estimate for the solutions to the Stokes equation yields (34)
kωks− 12 , s2 − 14 ≤ Ckuks, s2 ,∂Ω×(0,T ) = Cku · νks, 2s ,∂Ω×(0,T ) .
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LUNG-AN YING
There is a one-to-one correspondence between ρ and
∂ω ∂n |∂Ω
in H s−2, 2 −1 , so s
kρks−2, s2 −1 ≤ Ckωks− 12 , s2 − 14
(35)
for s > 2. (35) is also true for s ∈ ( 32 , 2]. Let us prove this assertion. Let ϕ ∈ C0∞ (∂Ω × (0, T ); then we extend it and define v so that v|∂Ω×(0,T ) = ϕ, ¯ × (0, T )) and v ∈ C0∞ (Ω kvk−s+ 52 ,− s2 + 54 ≤ Ckϕk−s+2,− s2 +1 . We have Z
T
0= 0
Z =
0
T
∂ω − 4ω v dx dt ∂t Ω Z TZ Z ∂ω ∂ω v + ∇ω · ∇v dx dt − ϕ dsx dt, ∂t Ω ∂Ω ∂n 0 Z
and Z Z T ∇ω · ∇v dx dt ≤ Ckvk−s+ 52 ,− s2 + 54 kωks− 12 , 2s − 14 , 0 Ω Z Z T ∂ω ∂ω v dx dt ≤ Ckvk − s2 + 54 k k s 5 H (0,T ;L2 (Ω)) ∂t H 2 − 4 (0,T ;L2 (Ω)) 0 Ω ∂t ≤ Ckvk−s+ 52 ,− s2 + 54 kωks− 12 , s2 − 14 . Hence k
∂ω ks−2, s2 −1,∂Ω×(0,T ) ≤ Ckωks− 12 , s2 − 14 , ∂n
which yields (35). It follows from (33)-(35) that Z 0
T
Z Ω
ωh dx dt ≥ Ckρks−2, s2 −1 kϕk−s,− 2s .
It is easy to see that Z
T
Z ωh dx dt,
a(ρ, ϕ) = 0
Ω
so (28) is verified. To verify the other conditions of Proposition 3.1 is straightforward.
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
315
Using the fundamental solution K, the solution ωk of the scheme (14)-(22) can be expressed as Z K(x − ξ, ti )ω0 (ξ) dξ ωk (x, ti − 0) = R2 i−1 Z X
−
(36)
j=0
+
Z K(x − ξ, ti − tj )
Ω
F (ξ, τ ) dξdτ tj
Z
i−1 Z X j=0
tj+1
K(x − ξ, ti − tj )
∂Ω
tj+1
ρk (ξ, τ ) dτ dsξ , tj
ωk and ρk is the solution to (18). where F = u˜k · ∇˜
4. Upper bound estimate for linear problems To analyse the above scheme for the Navier-Stokes equations, we study the same scheme for linear problems first. We replace (5) by ∂ω + F (x, t) = 4ω, ∂t
(37)
where F is a known function. This makes (37) (6)-(9) a linear problem. We use the same scheme to solve it, and estimate the approximate solution in this and the next section. Likewise, we assume that the exact solution is sufficiently smooth. To estimate the approximate solution, let us first define some operators. We define A = − 4 +I, where I is the identity. The semigroup of operators generated 1 from A in L2 (R2 ) is denoted by e−At . We define B : H s (Ω) → H s+ 2 (∂Ω) by B : ω → g, Z M (x, η)ω(x) dx, g(η) = Ω
and the restriction operator from H s (R2 ) to H s (Ω) is denoted by R. Let us estimate fk first. The first term in (20) is sufficiently regular, so we need to estimate the second term only. Let i0 = [ Tk ], and hi =
i−1 X
BRe
j=0
−(A−I)(ti −tj )
Z
tj+1
F (τ ) dτ, tj
where we understand F (τ ) to be the zero extension of F to the space R2 . Lemma 4.1. If 2 ≤ s < 3, then i0 X i=1
where
1 2
> β > s − 52 .
khi k2s,∂Ω k ≤ CkF k2L2 (0,T ;H β (Ω)) ,
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LUNG-AN YING
Proof. By the properties of analytic semigroups of operators (see [6]) we have
2
Z tj+1 i0 i0 X i−1 X X
−(A−I)(ti −tj )
khi k2s,∂Ω k ≤ C e F (τ ) dτ k
tj
1 i=1 i=1 j=0 s− 2
2
Z i0 X X
i−1 s − 1 −(A−I)(t −t ) tj+1 i j
2 4 A e F (τ ) dτ ≤C
k
tj
i=1 j=0 0 2 Z i i−1 0 X X tj+1 s−β β 1 (ti − tj )− 2 + 4 kA 2 F (τ )k0 dτ k ≤C i=1
tj
j=0
2
i0 i−1 Z X X ≤C i=1
≤C
j=0
·
tj+1
1 − s−β 2 +4
(ti − tj )
i−1 Z X
j=0
tj+1
iX 0 −1 Z tj+1 j=0
kF (τ )kβ,Ω dτ k
tj+1
dτ
1 − s−β 2 +4
(ti − tj )
kF (τ )k2β,Ω dτ
tj
(ti − tj )−
s−β 1 2 +4
k
kF (τ )k2β,Ω dτ k
tj
i=1 j=0
=C
(ti − tj )
tj
j=0
i0 X i−1 Z X
1 − s−β 2 +4
tj
i0 X i−1 Z X i=1
≤C
tj+1
kF (τ )k2β,Ω dτ
tj
i0 X
(ti − tj )−
s−β 1 2 +4
k
i=j+1
≤ CkF k2L2 (0,T ;H β (Ω)) .
To estimate kfk kH 2s (0,T ;L2 (∂Ω)) we prove one auxiliary lemma first. Lemma 4.2. If β ∈ (0, 12 ) and g : [0, ti0 ] → R is a piecewise constant function, g(t) = gi , t ∈ (ti , ti+1 ), then Z ti0 Z ti0 X (gi − gj )2 (g(t) − g(τ ))2 dt dτ ≤ C k2 . 1+2β 1+2β |t − τ | |t − t | i j 0 0 i6=j
Proof. The left hand side can be divided into two parts: iX 0 −1 Z ti Z ti+1 k 1−2β X (gi−1 − gi )2 dt dτ ≤ (gi+1 − gi )2 , 1+2β |t − τ | 2β(1 − 2β) t t i−1 i i=1 i X Z |i−j|≥2
ti+1
ti
Z
tj+1
tj
Then the conclusion follows.
X (gi − gj )2 (gi − gj )2 dt dτ ≤ C k2 . 1+2β |t − τ | |ti − tj |1+2β |i−j|≥2
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
By (19),
∂fk ∂τ
317
is a piecewise constant function, 1 ∂fk = (fk i+1 − fk i ), ∂τ k
τ ∈ (ti , ti+1 ).
We define (38)
gi =
Lemma 4.3. If
5 2
1 (hi+1 − hi ). k
< s < 3, then
X kgi − gj k2 0 2 k ≤ CkF k2 s −1 , 1 H 2 (0,T ;H − 2 +ε (Ω)) |ti − tj |s−1 i6=j
where ε > 0. Proof. The left hand side is equal to 2
X kgi − gj k2 0 2 k s−1 |t − t | i j j
=2
X j
1 |ti − tj |s−1
i
X
BRe−(A−I)(ti+1 −tl )
l=0
−
i−1 X
BRe
−(A−I)(ti −tl )
l=0 j X
−
BRe−(A−I)(tj+1 −tl ) −
l=0
j−1 X
BRe−(A−I)(tj −tl )
!Z
F (τ ) dτ
!Z
tl tl+1
tl
l=0
tl+1
2
F (τ ) dτ .
0
We extend F by zero for τ < 0; then the right hand side is equal to
2
X j
1 |ti − tj |s−1
X Z tl+1
j −(A−I)(tj+1 −tl )
BRe (F (τ + ti − tj ) − F (τ )) dτ
tl
l=j−i −
j−1 X
BRe−(A−I)(tj −tl )
X j
1 |ti − tj |s−1
X
j−1
BR(e−(A−I)(tj+1 −tl )
l=j−i
X j
1 |ti − tj |s−1
≡ a1 + a2 .
2
(F (τ + ti − tj ) − F (τ )) dτ
0
2
−(A−I)(tj −tl ) −e ) (F (τ + ti − tj ) − F (τ )) dτ
tl
0
Z tj+1
2
(F (τ + ti − tj ) − F (τ )) dτ
BRe−(A−I)k
tj Z
+4
tl+1
tl
l=j−i
≤4
Z
tl+1
0
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Due to the properties of analytic semigroups we have
BR(e−(A−I)(tj+1 −tl ) − e−(A−I)(tj −tl ) )(F (τ + ti − tj ) − F (τ )) 0,∂Ω
3
4 −(A−I)(tj −tl )
≤ Ck A e (F (τ + ti − tj ) − F (τ )) 0
≤ Ck(tj − tl )−1+ 2 kF (τ + ti − tj ) − F (τ )k− 12 +ε ; ε
then, following the same lines of the proof of Lemma 4.1, we get j−1 Z tl+1 X X k2 ε (tj − tl )−1+ 2 kF (τ + ti − tj ) − F (τ )k2− 1 +ε dτ. a1 ≤ C s−1 2 |t − t | i j tl jl
0 i0 −1−i X
i0 =1
l=−i0
≤ CkF k2
Z
tl+1
kF (τ + i0 k) − F (τ )k2− 1 +ε 2
(i0 k)s−1
tl
s
1
H 2 −1 (0,T ;H − 2 +ε (Ω))
dτ
.
For the second term,
!2 Z tj+1 1 kF (τ + ti − tj ) − F (τ )k− 12 +ε dτ a2 ≤ C |t − tj |s−1 tj j 0. Proof. We get from (11) and (36) that
ω(x, ti ) − ωk (x, ti − 0) i−1 Z tj+1 Z X (K(x − ξ, ti − τ ) − K(x − ξ, ti − tj ))F (ξ, τ ) dξdτ =−
+ =−
j=0
tj
j=0
tj
i−1 Z X
i−1 Z X
i−1 Z X j=0
+
j=0
tj+1
Z (K(x − ξ, ti − τ ) − K(x − ξ, ti − tj ))F (ξ, τ ) dξdτ Ω
tj+1
Z (K(x − ξ, ti − τ ) − K(x − ξ, ti − tj ))ρk (ξ, τ ) dsξ dτ
tj
i−1 Z X
(K(x − ξ, ti − τ )ρ(ξ, τ ) − K(x − ξ, ti − tj )ρk (ξ, τ )) dsξ dτ
∂Ω
tj
j=0
+
Ω
tj+1 Z
∂Ω
tj+1
Z
tj
K(x − ξ, ti − τ )(ρ(ξ, τ ) − ρk (ξ, τ )) dsξ dτ ∂Ω
≡ a 1 + a2 + a3 . The sum in a1 , a2 can be divided into two parts: j = 0, · · · , i − 2, and j = i − 1. For j = i − 1 we have
Z
ti
−(A−I)(ti −τ ) −(A−I)k −e e F (τ ) dτ
ti−1 σ−2 Z ti ≤C kF (τ )kβ dτ ≤ Ck, ti−1
and for j = 0, · · · , i − 2 we have
X Z
i−2 tj+1 −(A−I)(t −τ ) −(A−I)(ti −tj ) i
−e e F (τ ) dτ
j=0 tj σ−2
i−2 Z tj+1 Z ti −τ
X
≤C e−(A−I)ζ dζF (τ ) dτ
tj ti −tj j=0
≤ Ck
i−2 Z X j=0
σ
tj+1
tj
(ti − τ )−
σ−β 2
kF (τ )kβ dτ ≤ Ck.
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We get for the first part of a2 that
Z
ti 1
RA 2 e−(A−I)(ti −τ ) − e−(A−I)k χk (τ ) dτ ka2 1 kσ−2 =
ti−1
Z ti
R e−(A−I)(ti −τ ) − e−(A−I)k ∇ · χk (τ ) dτ +
ti−1 σ−2 Z ti
σ2 − 12 −(A−I)(ti −τ ) ≤C − e−(A−I)k χk (τ ) dτ e
A 0
ti−1 Z ti
σ
2 −1 −(A−I)(ti −τ ) − e−(A−I)k ∇ · χk (τ ) dτ e
A
+C Z
0
ti−1 ti
(ti − τ )
≤C
ti−1 Z ti
(ti − τ )
+C Z ≤C
s−σ 3 ε 2 −4−2
s−σ 1 2 −4
ti−1 ti
(ti − τ )
kχk (τ )ks− 32 dτ
s−σ 3 ε 2 −4−2
dτ sup kχk (τ )ks− 52 −ε τ
ti−1
Z
kχk (τ )ks− 52 −ε dτ
! 12
ti
s−σ− 12
(ti − τ )
+C
Z
dτ
ti−1
We take s = σ +
3 2
ti
ti−1
! 12 kχk (τ )k2s− 3 2
dτ
.
and notice (41), then get ε
ka2 1 kσ−2 ≤ Ck 1− 2 . For the second part of a2 we can repeat the proof of Theorem 4.1 and get an upper bound of Ck. The details are omitted here. Finally, by potential theory and (47) we have 7
ka3 k1+ε, 12 + ε2 ≤ Ck 16 +
3β ε 8 −2
.
Then by the embedding theorem we get 7
ka3 k0,Ω ≤ Ck 16 +
3β ε 8 −2
.
Combining the above results, we obtain (48). 6. Estimates for the Navier-Stokes equation For the Navier-Stokes equation we denote F0 = u · ∇ω, where u and ω are exact solutions; then we regard F0 as a known function. We denote by f ∗ , ρ∗ and ω ∗ the approximate solutions to ∂ω + F0 = 4ω ∂t and (6)-(9). Then we define Z (49) ω ˜ ∗ (τ ) = −
τ
ti
F0 dτ + ω ∗ (ti − 0),
τ ∈ [ti , ti+1 ).
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
325
The estimates in Section 4 and Section 5 are valid for ω ∗ and ω ˜ ∗ , namely, we have (50)
ω ∗ (t)kσ ≤ C1 , kω ∗ (ti − 0)kσ , k˜
(51)
kω(ti ) − ω ∗ (ti − 0)kσ−2,Ω ≤ C2 k 16 + 7
3β ε 8 −2
,
for 0 ≤ i ≤ i0 and 0 ≤ t ≤ T . Certainly the constants C1 , C2 depend on the exact solution. Using this auxiliary solution we estimate ω ∗ − ωk ; then we obtain the estimates for ω − ωk . We need one fact about Sobolev spaces. For the reader’s convenience we recall it here: Lemma 6.1. If h, g ∈ H r (Ω) with 1 < r < 2, then khgkr ≤ Ckhkr kgkr ,
(52)
and if h ∈ H 1+ε (Ω), g ∈ H r (Ω) with ε > 0, 0 ≤ r ≤ 1, then khgkr ≤ Ckhk1+ε kgkr .
(53)
To estimate the upper bound and error for non-linear problems, let us prove the following lemmas first. Lemma 6.2. Let gi be defined by (38). Then i0 X
(54)
kgi k20,∂Ω k ≤ CkF k2 2
1
L (0,T ;H − 2 +ε (Ω))
i=1
,
where ε > 0. Proof. We notice that i−1 Z tj+1 1X BR e−(A−I)(ti+1 −tj ) − e−(A−I)(ti −tj ) F (τ ) dτ k j=0 tj Z ti+1 1 −(A−I)k F (τ ) dτ + BRe k ti Z tj+1 Z i−1 X 1 ti+1 −tj −(A−I)ζ BR (A − I)e dζ F (τ ) dτ =− k ti −tj tj j=0 Z 1 ti+1 F (τ ) dτ. + BRe−(A−I)k k ti
gi =
Following the same lines as the proof of Lemma 4.1, we get (54). For the Navier-Stokes equation, F = u ˜k ·∇˜ ωk . We apply Lemma 4.1 and Lemma 6.2 to obtain (55)
kfk k2,1 ≤ CkF k
1
+ C(ω0 );
1
+ C(ω0 ).
L2 (0,T ;H − 2 +ε (Ω))
then Lemma 3.1 gives (56)
kρk k0,0 ≤ CkF k
L2 (0,T ;H − 2 +ε (Ω))
Lemma 6.3. ωk k− 12 +ε ≤ Ck˜ ωk k21 +ε . k˜ uk · ∇˜ 2
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Proof. We have ωk = −∇ ∧ ((˜ uk · ∇)˜ uk ) = −∇ ∧ (∇(˜ uk ⊗ u ˜k )). u ˜k · ∇˜ Lemma 6.1 implies ωk k− 12 +ε = k˜ uk ⊗ u ˜k k 32 +ε ≤ Ck˜ uk k23 +ε ≤ Ck˜ ωk k21 +ε . k˜ uk · ∇˜ 2
Lemma 6.4. Assume that 1 < σ < 32 , 0 < ε ≤
and
ωk kσ−1−ε ≤ M0 . k˜ ωk kL2 (0,T ;H σ−ε (Ω)) + k˜
(57) Then (58)
σ−1 2 ,
2
i X
! 12 kωk (tl −
0)k2σ k
+ kωk (ti − 0)kσ−1 + kkωk (ti − 0)kσ+1 ≤ M1 ,
l=0
where M1 depends on M0 . Proof. We apply (36) and (42) to estimate ωk (ti − 0). We take β ∈ (σ − 1, 12 ), then get
X Z i−1 Z tj+1 X
i−1 tj+1 1 −(A−I)(t −t ) 1+σ−β i j
2 A e χk (τ ) dτ ≤ C (ti − tj )− 2 kχk (τ )kβ dτ.
j=0 tj j=0 tj σ
Then
2
i−1 Z tj+1 X
X 1
2 e−(A−I)(ti −tj ) χ (τ ) dτ k A k
i j=0 tj σ Z T kχk (τ )k2β dτ = Ckχk k2L2 (0,T ;H β (Ω)) . ≤C 0
The other terms in (36) and (42) can be estimated in the same way. By Lemma 4.4, kχk kL2 (0,T ;H β (Ω)) ≤ Ckρk kβ− 1 , β − 1 .
(59)
2
Now (56) can be applied. We obtain X kωk (tl − 0)k2σ k ≤ CkF k2 2
2
4
1
+ C(ω0 ).
1
+ C(ω0 ).
L (0,T ;H − 2 +ε (Ω))
l
Due to Lemma 6.3, X kωk (tl − 0)k2σ k ≤ Ck˜ ωk k 4 4
L (0,T ;H 2 +ε (Ω))
l
We apply the interpolation inequality to obtain Z T Z T 4 k˜ ωk k 1 +ε dt ≤ k˜ ωk k6−4σ+8ε k˜ ωk k4σ−2−8ε σ−ε σ−1−ε dt. 0
2
0
We notice that 6− 4σ + 8ε ≤ 2, so the assumption (57) implies the right hand side is bounded by a constant depending on M0 . Following the proof of Theorem 4.1, the ωk (ti − 0)kσ+1 can also be obtained. upper bounds for k˜ ωk (ti − 0)kσ−1 and kk˜
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
327
Lemma 6.5. If (58) holds for i ≤ i1 , then there exists a constant k0 > 0, depending on M1 , such that if k ≤ k0 , then k˜ ωk (τ )kr ≤ C3 kωk (ti − 0)kr ,
(60)
ti ≤ τ < ti+1 ,
for 0 ≤ r ≤ 2, where C3 depends on M1 . Proof. It can be proved (see [10]) that for the Euler equation ∂u + (u · ∇)u + ∇p = 0, ∂t ∇ · u = 0, there are constants C4 , C5 , such that if t ≤
1 C4 ku(0)k3 ,
then ku(t)kr ≤ C5 ku(0)kr
for 0 ≤ r ≤ 3. By (58) and interpolation we have kωk (ti − 0)k2 ≤ M1 k σ−3 kuk (ti − 0)k3 ≤ C(M1 )k 2 . Thus we take 2 σ−1 1 . k0 = C4 C(M1 )
σ−3 2
; hence
If k ≤ k0 , then (60) holds. Lemma 6.6. Under the assumptions of Lemma 6.5, ! 12 iX 1 +1 5 ∗ 2 (61) k(ω − ωk )(ti − 0)kσ k ≤ C6 k 8 (σ−1)−2ε , i=0
where ε > 0, k ≤ k0 , and C6 depends on M1 . ωk . We obtain by Lemma 6.1 and interpolation that Proof. Let F˜ = u · ∇ω − u ˜k · ∇˜ ˜k ) ⊗ u + u ⊗ (u − u˜k ) − (u − u ˜k ) ⊗ (u − u ˜k )))k− 12 +ε kF˜ k− 12 +ε = k∇ ∧ (∇((u − u ≤ Cku − u ˜k k 23 +ε + Cku − u˜k k23 +ε 2
≤ Ckω − ω ˜ k k 12 +ε + Ckω − ω ˜ k k21 +ε 2
≤ Ckω − ω ˜ k k 12 +ε + Ckω − ω ˜ k kσ−1 kω − ω ˜ k k2−σ+2ε . Then we notice (58) and (60), to get for τ < ti1 +1 that ˜ k k2−σ+2ε kF˜ k− 12 +ε ≤ Ckω − ω
≤ Ck˜ ω∗ − ω ˜ k k2−σ+2ε + Ckω − ω ˜ ∗ k2−σ+2ε .
Since ω is sufficiently smooth, (49) and (51) imply that kω(τ ) − ω ˜ ∗ (τ )kσ−2 ≤ Ck 16 + 7
3β ε 8 −2
,
τ < ti1 +1 .
Because β ∈ (0, 12 ) and ε > 0 are arbitrary, we take β close to the interpolation between it and (50) gives
1 2
kω(τ ) − ω ˜ ∗ (τ )k2−σ+2ε ≤ Ck 8 (σ−1)−2ε , 5
and consequently (62)
ω∗ − ω ˜ k k2−σ+2ε + Ck 8 (σ−1)−2ε . kF˜ k− 12 +ε ≤ Ck˜ 5
and ε small; then
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It follows from (36) that for i ≤ i1 + 1 Z
i−1 Z X
∗
(ω − ωk )(x, ti − 0) = −
Ω
j=0
Z K(x − ξ, ti − tj )
∂Ω
j=0
F˜ (ξ, τ ) dξdτ
tj
i−1 Z X
+
tj+1
K(x − ξ, ti − tj )
tj+1
(ρ∗ − ρk )(ξ, τ ) dsξ dτ.
tj
Then we get i−1 Z X
k(ω ∗ − ωk )(ti − 0)kσ ≤ C
tj
j=0
(63)
+C
tj+1
i−1 Z X
σ 1 ε (ti − tj )− 2 − 4 + 2 kF˜ k− 12 +ε dτ
tj+1
(ti − tj )−
β 1+σ 2 +2
kχ∗ − χk kβ dτ,
tj
j=0
where σ − 1 < β < 12 . Analogously to (55), (56) and (59), we have kχ∗ − χk kβ, β ,Ω×(0,ti ) ≤ CkF˜ k
(64)
2
1
L2 (0,ti ;H − 2 +ε (Ω))
,
where we notice that the term corresponding to ω0 is cancelled out. Let i2 be an index to be determined. We take the sum i2 X
k(ω ∗ − ωk )(ti − 0)k2σ k
i=0
≤C
i2 X i=0
+C
2
i−1 Z X
i2 X
j=0
i=0
tj+1
i−1 Z X
·
i−1 Z X j=0
+C ·
i2 X
2 tj+1
β − 1+σ 2 +2
(ti − tj )
tj+1
(ti − tj )
1 ε −σ 2 −4+2
(ti − tj )
i−1 Z X
tj+1
dτ
tj
j=0
i−1 Z X
kχ∗ − χk kβ dτ k
1 ε −σ 2 −4+2
kF˜ k2− 1 +ε dτ k 2
tj
i=0
j=0
≤ Ct
tj+1
kF˜ k− 12 +ε dτ k
tj
j=0
j=0
(ti − tj )
tj
i2 i−1 Z X X ≤C i=0
1 ε −σ 2 −4+2
tj+1
β − 1+σ 2 +2
(ti − tj )
tj
dτ
β − 1+σ 2 +2
(ti − tj )
kχ∗ − χk k2β dτ k
tj
3 2 −σ+ε i2
kF˜ k2 2
1
L (0,ti2 ;H − 2 +ε (Ω))
+ Ct1−σ+β kχ∗ − χk k2L2 (0,ti i2
2 ;H
β (Ω))
.
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
329
Then by (64) (65)
i2 X
˜ 2 k(ω ∗ − ωk )(ti − 0)k2σ k ≤ Ctα i2 kF k 2
1
L (0,ti2 ;H − 2 +ε (Ω))
i=0
,
where α = min( 32 − σ + ε, 1 − σ + β) > 0. Let us estimate the right hand side of (65) through (62). For ti ≤ τ < ti+1 we have Z τ ˜ k )(τ )k2−σ+2ε ≤ kF˜ k2−σ+2ε dτ + k(ω ∗ − ωk )(ti − 0)k2−σ+2ε . k(˜ ω∗ − ω ti
Then applying Lemma 6.1 we obtain ˜ k )(τ )k2−σ+2ε k(˜ ω∗ − ω Z τ ≤ (C + Ck˜ uk k1+ε k˜ ωk k3−σ+2ε ) dτ + k(ω ∗ − ωk )(ti − 0)k2−σ+2ε . ti
The inequalities (60) and (58) imply (66)
˜ k )(τ )k2−σ+2ε ≤ Ck σ−1−2ε + k(ω ∗ − ωk )(ti − 0)k2−σ+2ε . k(˜ ω∗ − ω
We substitute (66) and (62) into (65), and obtain i2 X
k(ω ∗ − ωk )(ti − 0)k2σ k
i=0
≤
Ctα i2
(i −1 2 X
) ∗
0)k22−σ+2ε k
∗
0)k2σ k
k(ω − ωk )(ti −
+k
5 4 (σ−1)−4ε
i=0
≤
Ctα i2
(i −1 2 X
) k(ω − ωk )(ti −
+k
5 4 (σ−1)−4ε
.
i=0 1 We take ti2 small enough so that Ctα i2 ≤ 2 , then get i2 X
k(ω ∗ − ωk )(ti − 0)k2σ k ≤ Ck 4 (σ−1)−4ε . 5
i=0
We notice that ti2 is independent of k; thus we can carry out an induction in a finite number, T /ti2 , of steps. Suppose (61) is true for i ≤ N i2 ; then X
(N +1)i2
k(ω ∗ − ωk )(ti − 0)k2σ k
i=N i2 +1
X
(N +1)i2
≤C
i=N i2 +1
X
(N +1)i2
+C
i=N i2 +1
i−1 Z X j=0
2 tj+1
1 ε −σ 2 −4+2
(ti − tj )
tj
i−1 Z X j=0
kF˜ k− 12 +ε dτ k 2
tj+1
tj
β − 1+σ 2 +2
(ti − tj )
kχ∗ − χk kβ dτ k.
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By (62),(64) and the estimate for i ≤ N i2 we get X
(N +1)i2
k(ω ∗ − ωk )(ti − 0)k2σ k
i=N i2 +1
X
(N +1)i2
≤C
i=N i2 +1
j=N i2 +1
X
i−1 X
(N +1)i2
+C
i=N i2 +1
+ Ck
2
Z
i−1 X
tj+1
(ti − tj )
tj
kF˜ k− 12 +ε dτ k 2
Z
j=N i2 +1
5 4 (σ−1)−4ε
1 ε −σ 2 −4+2
tj+1
(ti − tj )−
β 1+σ 2 +2
kχ∗ − χk kβ dτ k
tj
.
Following the same lines, we get the estimate for i ≤ (N + 1)i2 . After a finite number of steps we get the estimate up to i ≤ i1 + 1. Lemma 6.7. Under the assumptions of Lemma 6.5, k(ω ∗ − ωk )(ti − 0)kσ−1,Ω ≤ C7 k 8 (σ−1)−2ε , 5
(67) for i ≤ i1 + 1.
Proof. Analogously to (63), we have ∗
k(ω − ωk )(ti − 0)kσ−1 ≤ C
i−1 Z X j=0
+C
tj+1
tj
i−1 Z X j=0
1 σ ε (ti − tj ) 4 − 2 + 2 kF˜ k− 12 +ε dτ
tj+1
(ti − tj )− 2 + 2 kχ∗ − χk kβ dτ, β
σ
tj
and then k(ω ∗ − ωk )(ti − 0)kσ−1 12 i−1 Z tj+1 i−1 Z X X 1 ≤C (ti − tj ) 2 −σ+ε dτ j=0
tj
j=0
i−1 Z X +C j=0
tj+1
12 (ti − tj )−σ+β dτ
tj
tj+1
j=0
kF˜ k2− 1 +ε dτ 2
tj
i−1 Z X
12
tj+1
12 kχ∗ − χk k2β dτ .
tj
Then we use (62),(64) to get the desired estimate. Theorem 6.1. If 1 < σ < 32 , then (68)
i X
! 12 k(ω ∗ − ωk )(tl − 0)k2σ k
+ k(ω ∗ − ωk )(ti − 0)kσ−1 ≤ Ck 4 1
l=0
for 0 ≤ i ≤ [ Tk ], provided k ≤ k0 , where k0 > 0 is a fixed constant and C is independent of k.
CONVERGENCE STUDY OF THE CHORIN-MARSDEN FORMULA
331
Proof. We prove (58),(67) and (69)
i X
! 12 k(ω ∗ − ωk )(tl − 0)k2σ k
5
≤ C8 k 8 (σ−1)−2ε
l=0
by induction. They are obviously satisfied for i = 0. Let us fix the constant M0 in (57) first. By (50) it is easy to see that ω ∗ kσ−1−ε ≤ C9 . k˜ ω ∗ kL2 (0,T ;H σ−ε (Ω)) + k˜ We take M0 = 2C9 . We assume that (58), (67), and (69) are satisfied for i ≤ i1 . Analogously to (66), we have ˜ k )(τ )kσ−1 ≤ Ck 2 + k(ω ∗ − ωk )(ti − 0)kσ−1 k(˜ ω∗ − ω 1
for τ ∈ [ti , ti+1 ), i ≤ i1 . We substitute (67) in it and obtain ˜ k )(τ )kσ−1 ≤ Ck 8 (σ−1)−2ε . k(˜ ω∗ − ω 5
(50),(60) and (58) imply ˜ k )(τ )kσ ≤ C. k(˜ ω∗ − ω Then by interpolation we obtain ˜ k )(τ )kσ−ε ≤ Ck ε( 8 (σ−1)−2ε) . k(˜ ω∗ − ω 5
We take k0 small enough so that (57) holds for k ≤ k0 and τ < ti1 +1 . Then from Lemma 6.4, Lemma 6.6, and Lemma 6.7 we conclude that (58),(67),(69) are satisfied for i ≤ i1 + 1. Thus the induction is complete. We take σ close to 32 and ε small enough. Then the exponent is greater than 1 4. Finally let us prove our main result, Theorem 1.1. Proof of Theorem 1.1. We take β close to
1 2
and ε small enough in (48). Then
kω(ti ) − ω ∗ (ti − 0)kσ−2,Ω ≤ Ck 2 . 1
Theorem 4.1 implies kω ∗ (ti − 0)kσ,Ω ≤ C. Then by interpolation we get kω(ti ) − ω ∗ (ti − 0)kσ−1,Ω ≤ Ck 4 , 1
which in conjunction with (68) yields (23). 7. A remark on the vortex sheets
Rt We have shown in (36) that the density of vortex sheets is tjj+1 ρk (ξ, τ ) dτ . Let us make a heuristic analysis here to compare our result with the others in the literature. The vorticity in (36) is divided into three parts: by initial data, by convection, and the vortex sheet, ω = ω1 +ω2 +ω3 . Suppose the velocity generated by ω1 +ω2 is u = un n+uν ν on the boundary ∂Ω. Then un = 0, due to the boundary condition of the Euler equation. The tangential component of velocity generated by the vortex sheets should be −uν , to keep the no-slip boundary condition.
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LUNG-AN YING
On the other hand, potential theory tells us that for x0 ∈ ∂Ω, Z tZ ∂ K(x − ξ, t − τ )ρk (ξ, τ ) dsξ dτ lim x→x0 ∂n x ∂Ω x∈Ω 0 (70) Z tZ ∂ 1 K(x0 − ξ, t − τ )ρk (ξ, τ ) dsξ dτ. = ρk (x0 , t) + 2 ∂Ω ∂nx 0 The left hand side is
¯ν ∂ ∂u ¯n ∂ u ∂ω3 + = − , ∂n ∂Ω ∂n ∂ν ∂n
u ¯n u ¯ν ¯ν ) is generated by ω3 . Since ∂∂n + ∂∂ν = 0, we have where u ¯ = (¯ un , u ∂ 2 u¯ν ∂2u ¯ν ∂ω3 = + = 4¯ uν . 2 ∂n ∂Ω ∂ν ∂n2
¯ satisfies the Stokes equation, but since ω3 satisfies the heat equation on R2 . u ∇·u ¯ = 0 at t = 0, u¯ also satisfies the heat equation and the pressure equals to zero. We get ∂u ¯ν ∂uν ∂ω3 = = − . ∂n ∂Ω ∂t ∂Ω ∂t ∂Ω Let us omit the second term in (70), to get
∂ω3 ∂n |∂Ω
∂uν . ρk = −2 ∂t ∂Ω
= 12 ρk . Then
When we set a vortex sheet at t = tj+1 , we assume that the no-slip condition is satisfied at t = tj . Therefore Z tj+1 Z tj+1 ∂uν dt = −2uν (ξ, tj+1 ). ρk dτ = −2 ∂t tj tj We notice that uν (ξ, tj+1 ) is the tangential component of velocity after the convection step. This is exactly the strength of vortex sheets given in some works. By the above analysis the jump of velocity across the boundary is Z tj+1 ∂ω3 ∂ω3 − lim dτ lim c [uν ] = x→x0 ,x∈Ω ∂n x→x0 ,x∈Ω ∂n tj Z tj+1 ρk (ξ, t) dt = −2uν , =− tj − where Ωc is the exterior of Ω. Therefore it is assumed that u+ ν |∂Ω = −uν |∂Ω , where “ + ” and “ − ” indicate the exterior and the interior. If Ω is a half plane, the second term of (70) is exactly zero. This observation coincides with that in [3]. However, in general, this term does not vanish.
Acknowledgment The author thanks the referees for their valuable comments.
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