Journal of Mathematical Sciences, Vol. 106, No. 3, 2001
L p-ESTIMATES FOR A SOLUTION TO THE NONSTATIONARY STOKES EQUATIONS UDC 517.9
H. Koch and V. A. Solonnikov L p-estimates for the nonstationary Stokes problem are established. Bibliography: 14 titles.
§ 1. Introduction
We consider the initial-boundary-value problem ∂~v − ∆~v + ∇p = ∇ · F(x, t), ∇ · v = 0, x ∈ Ω ⊂ Rn, t ∈ (0, T ), ∂t ~v(x, 0) = 0, ~v(x, t) x∈S = ~a(x, t)
(1.1)
n 2 in a bounded domain Ω ⊂ R , n > 2,with boundary S ∈ C , where F(x, t) is a matrix with entries Fik (x, t), n ∂F i, k = 1, . . . , n, and ∇ · F = ∑ ik . We assume that F and ~a satisfy the following conditions: i=1 ∂xi k=1,...,n
QT = Ω × (0, T ),
F ∈ L p (QT ),
Z
~a ·~n dS = 0,
p > 1,
1−1/p,1/2−1/(2p)
~a ∈ Wp,0
~ ~a ·~n = divS A,
ΣT = S × (0, T ),
(ΣT ),
(1.2)
~ ∈ W 1/2(0, T ;Wp1−1/p(S)). A p,0
S l,l/2
By Wpl (Ω), Wp
(QT ) and so on we mean the standard Sobolev–Slobodetskii isotropic and antisotropic l,l/2
l,l/2
spaces defined, for example, in [1]. We denote by Wp,0 (QT ) and Wp,0 (ΣT ) the sets of elements in l,l/2
l,l/2
Wp (QT ) and Wp (ΣT ) that admit the zero extension to Ω × (−∞, T) and Σ × (−∞, T) such that the extended functions are of the same class and the following estimates hold: kukW l,l/2 (Q ) = ku0 kW l,l/2 (Ω×(−∞,T)), p,0
p
T
kukW l,l/2 (Σ ) = ku0kW l,l/2 (S×(−∞,T )), p,0
p
T
l,l/2
where u0(x, t) = u(x, t) for t > 0 and u0 (x, t) = 0 for t < 0. If u ∈ Wp l/2−r/2
u ∈ Wp
l/2−r/2
(0, T ;Wpr (Ω)) or u ∈ Wp,0
l,l/2
(QT ) or u ∈ Wp,0 (QT ), then
(0, T ;Wpr (Ω)), r < l , respectively.
~. We refer the reader to [2] for representations of functions on S in the form divS A The main result of the paper is contained in the following assertion. Theorem 1.1. Let Ω be a bounded convex domain with boundary S ⊂ C2 . Then the solution to the problem (1.1) satisfies the inequality ~ 1/2 k~vkW 1,1/2(Q ) 6 c (T )( kFkL p (QT ) + k~akW 1−1/p,1/2−1/(2p)(Σ ) + kAk ) ≡ cN. 1−1/p W (0,T ;W (S)) p,0
T
p,0
T
p,0
p
Translated from Problemy Matematicheskogo Analiza, No. 22, 2001, pp. 197–239.
3042
1072-3374/01/1063-3042$25.00
c 2001 Plenum Publishing Corporation
(1.3)
The pressure p(x, t) can be represented in the form p(x, t) = p1 (x, t) +
∂P(x, t) , ∂t
(1.4)
where P(x, t) is a harmonic function and kp1 kL p(QT ) + k∇PkW 1,1/2(Q ) 6 cN.
(1.5)
T
p,0
~ ∈ W 1−1/(2p)−r/2(0, T ;Wpr (S)), r ∈ (0, 1 − 1/p] ( in particular, if ~a ·~n = 0 and A ~ = 0), then Moreover, if A p,0 ~ 1−1/(2p)−r/2 kPkW 1−1/(2p)−r/2(0,T ;W 1/p+r(Ω)) 6 c (T )(N + kAk ). W (0,T ;W r (S)) p
p,0
p,0
(1.6)
p
The constants in (1.3)–(1.5) are nondecreasing functions of T .
The assumption that Ω is convex is explained by the fact that for the representation of a solution to the problem (1.1) with F = 0 the hydrodynamical potentials are used [3–5]. In the case of a half-space, the explicit formulas for a solution to this problem allow us to obtain a stronger result, namely, the boundedness of the norm kPkW 1−1/(2p)(0,T ;L (S)) (in this case, S = Rn−1). The problem (1.1) with ~a = 0 was p
p,0
considered in [6, 7]. In [6], the estimate (1.3) was obtained under the assumption that all the vector fields ~Fn = (Fik , . . . , Fnk ), k = 1, . . . , n, are solenoidal and ~Fk ·~n = 0 (in the weak sense). As was shown in [7], S in the case p = 2, the pressure can be defined as the sum p(x, t) = p1 (x, t) + p2 (x, t), where p1 ∈ L p (QT ) −1/2 and p2 ∈ W2 (0, T ;W21 (Ω)). In Secs. 2–6, we analyze the problem (1.1) with F = 0. In Sec. 7, we prove Theorem 1.1 and discuss possible generalizations of this theorem, an elementary proof of the representation (1.4) for pressure (perhaps, with some other p1 and P), and the estimate (1.5) (but not (1.6)) for p = 2. § 2. Problem (1.1) with F = 0
We consider the problem ~vt − ∆~v + ∇p = 0, ∇ ·~v = 0, ~v = 0, ~v = ~a(x, t), t=0
x ∈ Ω, t ∈ (0, T ),
(2.1)
x∈S
where ~a(x, t) is a smooth function such that Z
~a ·~ndS = 0,
~a(x, 0) = 0.
S
As is known [3–5], in the case of a convex domain Ω, the solution to the problem (2.1) can be written in the form ~ t) + ∇V (x, t), ~v(x, t) = U(x,
(2.2) 3043
where V (x, t) is the electrostatic potential of a single layer, i.e., Z
E(x − y)ψ(y, t) dSy ,
V (x, t) = 2 S
− E(x) =
1 , |S1|(n − 2)|x|n−2
1 log |x|, 2π
n > 2,
(2.3)
n = 2,
is the fundamental solution of the Laplace equation (|S1| is the area of the unit sphere in Rn ) and Zt Z
G(x, y, t − τ)~Φ(y, τ) dSy dτ
~ (x, t) = U
(2.4)
0 S
is the hydrodynamical potential. The matrix kernel G is expressed in terms of the “second fundamental solution” to the nonstationary Stokes equations defining a solution to the problem (2.1) in a half-space, i.e., the problem ~ut − ∆~u + ∇q = 0, div~v = 0, x ∈ Rn+ = (xn > 0), ~u t=0= 0, (2.5) 0 0 0 0 = ~a(x , t) = (a1 (x , t), . . . , an−1(x , t), 0), x = (x1 , . . . , xn−1), ~u xn =0
and the density ~Φ satisfies the condition ~Φ ·~n = 0. Regarding the solution of the problem (2.5), there is an explicit formula obtained in [8] in the three-dimensional case: n−1 Z
ui (x, t) =
t
∑
j=1
Gi j (x, t) = −2 (0) Qi (x, t) =
2 /(4t)
∂Γ ∂ (0) δi j + 4 Q (x, t), ∂xn ∂x j i
Zxn
Z
dyn 0
where Γ(x, t) = (4πt)−n/2 e−x determined by the formula
Gi j (x0 − y0 , xn, t − τ)a j (y, τ) dy0 dτ, (0)
Rn−1
0
(0)
Z
Rn−1
∂Γ(y, t) ∂E(x − y) 0 dy , ∂yn ∂xi
(2.6) (2.7)
is the fundamental solution of the heat equation. The pressure q(x, t) is
n Zt Z ∂ (0) q(x, t) = −4 − ∆ ∑ dτ Q j (x0 − y0 , xn, t − τ)a j (y0 , τ) dy0. ∂t j=1 0
Rn−1
(2.8)
(0)
Setting Gi j = 0 for j = n, we can write (2.6) in the form Zt Z
u(x, t) = 0
Rn−1
G(0)(x0 − y0 , xn, t − τ)~a(y0 , τ) dy0 dτ,
T ∂Γ ∂ (0) (0) ~ G (x, t) = −2 (I − en ⊗ en ) + 4 ∇ −~en ⊗ Q (x, t) . ∂xn ∂xn 3044
(2.9)
The tensor G in (2.4) coincides with the tensor (2.9) for the tangent plane T (y) at the point y ∈ S, i.e., T ∂ ∂Γ(x − y, t) ~ y, t) , G(x, y, t) = −2 (I −~n(y) ⊗~n(y)) + 4 ∇ −~n(y) ⊗ Q(x, ∂n ∂n Z (2.10) ∂Γ(z − y, t) ~ y, t) = Q(x, ∇E(x − z) dz, ∂n Π(x,y)
∂ ∂Γ =~n(y) · ∇x , =~n(y) · ∇z Γ(z − y, t), and Π(x, y) is the layer ∂n ∂n ~ in the form between the plane T (y) and the parallel plane passing through the point x. We also can write Q
where ~n(y) is the unit inward normal to S,
~n(y)·(x−y) Z
~ y, t) = Q(x,
Z
dηn
Rn−1
0
∂Γ(η, t) ∇E(C(y)(x − y) − η) dη, ∂ηn
(2.11)
where C(y) is an arbitrary orthogonal matrix such that Cn j (ξ) = n j (ξ),
j = 1, . . . , n.
(2.12)
However, one should keep in mind that the continuous matrix C(y) defined on the entire surface S and possessing such properties does not always exist. The pressure p(x, t) corresponding to the velocity field (2.2) is expressed by the formula Zt Z Z ∂ ∂ ~ ~ p(x, t) = −4 −∆ Q(x, y, t − τ)Φ(y, τ) dSy dτ − 2 E(x − y)ψ(y, t) dSy . ∂t ∂t 0 S
(2.13)
S
The functions ~Φ(y, t) and ψ(y, t) in (2.3) and (2.4) satisfy the conditions ~Φ(y, t) ·~n(y) = 0, Z
(2.14)
ψ(y, t) dSy = 0.
(2.15)
S
The boundary conditions get the following Volterra–Fredholm-type equations ~Φ(x, t) + Gtan [~Φ] + ∇SV [ψ] = ~atan (x, t), ∂V [ψ] ψ(x, t) + +~n · G[~Φ] = ~a(x, t) ·~n(x), x ∈ S. ∂nx
(2.16)
where G[Φ] is the direct value of the potential (2.4) on S, Gtan [~Φ] = G[~Φ] −~n(~n · G[~Φ]) is its tangent ∂V component, ∇SV = ∇V −~n is the tangent component of the gradient of V on S, and ~atan = ~a −~n(~a ·~n). ∂n r,r/2
§ 3. Solvability of (2.16) in L p(ΣT ) and Wp,0 (ΣT )
The solvability of the system (2.16) in the space of continuous functions and functions satisfying the r,r/2 H¨older condition was established in [3–5], where this system was considered in L p(ΣT ) and Wp (ΣT ), 3045
r ∈ (0, 1). We need point-wise estimates for the elements Gi j of the tensor G. They follow from the (0) estimates for Gi j (x, t) obtained for n = 3 in [8] (the same arguments are valid in the case n > 2): cxλn ∀λ ∈ (0, 1), t m+(1+λ)/2(|x|2 + t)(n+l)/2 c (0) |Dlx0 Dkxn Dtm Gi j (x, t)| 6 m+1/2 2 . t (|x| + t)(n+l)/2(x2n + t)k/2 (0)
|Dlx0 Dtm Gi j (x, t)| 6
(3.1) (3.2)
Hence the kernels Gi j (x, y, t) are weakly singular in ΣT = S × (0, T ). Indeed, we assume that a point x in (3.1) lies in the surface given by the equations xn = F(x0 ),
x0 = (x1 , . . . , xn−1) ∈ Kd = {|x0 | < d},
(3.3)
where F ∈ C2 (Kd ), F(x0 ) > 0, F(0) = 0, ∇F(0) = 0. Hence |∇F(x0 )| 6 c |x0 |, |F(x0 )| 6 c |x|2 . Then (0)
|Gi j (x, t)| 6
cF λ (x0 ) c 6 . t (1+λ)/2(|x|2 + t)n/2 t (1+λ)/2(|x|2 + t)(n−2λ)/2
This equation and (2.10) imply the estimate |Gi j (x, y, t)| 6
c t (1+λ)/2(|x − y|2 + t)(n−2λ)/2
∀x, y ∈ S.
(3.4)
By (3.1) and (3.2), for any x and z lying on the same surface (3.3) and satisfying the conditions |x − z| 6 |x − y|/2 we have (0)
(0)
(0)
(0)
(0)
(0)
|Gi j (x, t) − Gi j (z, t)| 6 |Gi j (x, t) − Gi j (ξ, t)| + |Gi j (ξ, t) − Gi j (z, t)| |x0 − z0 |λF λ(x0 ) |F(x0 ) − F(z0 )|λ c |x0 − z0 |λ 6 c (1+λ)/2 2 + 6 , t (|x| + t)(n+λ)/2 t (1+λ)/2(|x|2 + t)n/2 t (1+λ)/2(|x|2 + t)(n−λ)/2
where ξ = (z0 , F(xn )). Consequently, for any x, y, z ∈ S such that |x − z| 6 |x − y|/2 we have |Gi j (x, y, t) − Gi j (z, y, t)| 6
c |x − z|λ . t (1+λ)/2(|x − y|2 + t)(n−λ)/2
(3.5)
We prove the following assertion. Proposition 3.1. For any ~a ∈ L p(ΣT ) satisfying the condition Z
~a ·~ndSx = 0
S
the system (2.16) has a unique solution ~Φ, ψ ∈ L p(ΣT ) satisfying the conditions (2.14) and (2.15) and the following inequality: k~ΦkL p (Σt ) + kψkL p(Σt ) 6 c (T )k~akL p(Σt ) 3046
∀t ∈ (0, T ).
(3.6)
Proof. Any potential Zt Z
v(x, t) =
K(x, y, t − τ) f (y, τ) dSy dτ
0 S
with weakly polar kernel satisfies the inequality kvkL p(Σt ) 6
Zt
k(t − τ)k f kL p (Στ) dτ,
(3.7)
0
where 1/p0
1/p
k(t) = k1 (t), k2 Z
1/p0 = 1 − 1/p,
(t),
k1 (t) = sup
|K(x, y, t)| dSx, k2 (t) = sup
y∈S
x∈S
S
Z
|K(x, y, t)| dSy.
S
It can be easily proved with the help of the H¨older inequality and the Minkowski inequality. Indeed, 1/p Z 1/p0 Zs Z p |v(x, s)| 6 dτ |K(x, y, s − τ)| | f (y, τ)| dSy |K(x, y, s − τ)| dSy 0
6
Zs
S 1/p k2 (s − τ) dτ
0
kv(·, s)kL p (S) 6
Zs
Z
S
|K(x, y, s − τ)| | f (y, τ)| dSy p
S
k(s − τ)k f (·, τ)kL p (S) dτ =
0
Z∞
1/p ,
k(τ)k f0 (·, s − τ)kL p(S) dτ,
0
where f0 (x, t) = f (x, t) for t > 0 and f0 (x, t) = 0 for t < 0. From this estimate and the Minkowski inequality we easily obtain (3.7). In particular, by (3.4), we have kG[~Φ]kL p(Σt ) 6 c
Zt
k~ΦkL p (Στ)
0
dτ , (t − τ)(1−λ)/2
λ ∈ (0, 1/2).
(3.8)
The same inequalities are valid for Gtan [~Φ] and ~n · G[~Φ]. We consider the Fredholm equation ψ(x) +
∂V [ψ] = f (x) ∂n
(3.9)
and the Volterra-type equation ~ϕ(x, t) + Gtan[~ϕ] = ~b(x, t)
(3.10)
on S. As is well known from the potential theory, if f ∈ L p(S) satisfies (2.15), then Eq. (3.9) has a unique solution satisfying the same condition and the inequality kψkL p(S) 6 c k f kL p (S).
(3.11) 3047
The system (3.10) is uniquely solvable for any ~b ∈ L p(ΣT ); moreover, the solution can be constructed by the method of successive approximations ~ϕm+1 (x, t) = ~b(x, t) − Gtan [~ϕm ],
~ϕ1 = ~b.
m = 1, . . . ,
By (3.8), ~ωm+1 = ~ϕm+1 −~ϕm satisfy the inequality k~ωm+1 kL p(Σt ) 6 c
Zt
(t − τ)−1+λ/2k~ωm kL p(Στ ) dτ,
0
λ ∈ (0, 1/2), which guarantees the convergence of the sequence {~ϕm }. We have k~ωm+1 kL p (Σt ) 6 c
m
Zt 0
dt1 ··· (t − t1)1−λ/2
t 6 cm k~bkL p(Σt )
Zt
kω1 kL p (Σtm )
0
dtm (tm−1 − tm)1−λ/2
Γ(λ/2)m−1 , Γ(1 + mλ/2)
mλ/2
where ~ω1 = ~ϕ1 = ~b and, consequently, k~ϕm+1 kL p(Σt ) 6
m
∑ k~ωk+1kLp (Σt ) 6 c (T )k~bkLp (Σt ).
k=0
The solution ~ϕ = lim ~ϕm satisfies the inequality m→∞
k~ϕkL p(Σt ) 6 c (T )k~bkL p(Σt ) ,
t ∈ (0, T ).
(3.12)
The uniqueness of a solution follows from the estimate k~ϕ0kL p (Σt ) 6 kGtan [ϕ0 ]kL p(Σt ) 6 c
Zt
k~ϕ0 kL p (Στ)
0
dτ (t − τ)1−λ/2
for the solution to the homogeneous equation. Consequently, ~ϕ0 = 0. It is clear that ~ϕ ·~n = 0 if ~b ·~n = 0. The system (2.16) can be solved by the method of successive approximations as follows. If a vector ~ field Φm satisfies the condition (2.14), then ψm+1 (x, t) can be found as a solution to Eq. (3.9) with f ≡ fm = −~n · G[~Φm ] +~a ·~n.
Since fm satisfies the necessary condition (2.15), we conclude that ψm+1 (x, t) satisfying the same condition is uniquely defined and, by (3.11) and (3.8), we have Zt kψm kL p (Σt ) 6 c k fm kL p (Σt ) 6 c k~Φm kL p (Στ) 0
3048
dτ + k~a ·~nkL p(Σt ) . (t − τ)1−λ/2
Further, ~Φm+1 is determined from Eq. (3.10) with ~b = ~bm = ~atan − ∇SV [ψm+1]. It is clear that ~bm ·~n = 0. Hence ~Φm+1 ·~n = 0 and we have k~Φm+1 kL p (Σt ) 6 c k~bm kL p(Σt ) .
For the first approximation we take ~Φ1 =~atan . The above arguments lead to the conclusion that ψm+1 − ψm and ~Φm+1 − ~Φm satisfy the inequalities kψm+1 − ψm kL p(Σt ) 6 c
Zt
(t − τ)−1+λ/2k~Φm − ~Φm−1kL p (Στ) dτ,
0
k~Φm+1 − ~Φm kL p (Σt ) 6 k∇SV [ψm+1 − ψm ]kL p (Σt ).
Since ∇SV [ψ] is a singular integral operator on S, by the Calderon–Zygmund theorem, we have k∇SV [ψ]kL p (S) 6 ckψkL p (S)
(3.13)
and, consequently, k~Φm+1 − ~Φm kL p(Σt ) 6 c
Zt
(t − τ)−1+λ/2k~Φm − ~Φm−1kL p (Στ) dτ,
0
which guarantees the convergence of the sequences {~Φm } and {ψm } to a unique solution to the system (2.16) and the estimate (3.6) for this solution. The proposition is proved. We pass to the study of differential properties of solutions to the system (2.16). To this end, we need ∂V [ψ] , ∇SV , and G[~Φ]. to study properties of the operators ∂n ∂V [ψ] Proposition 3.2. If S ⊂ C2, then is a continuous operator from L p(S) into Wpr (S) and ∇SV is ∂n continuous in Wpr (S) for all r ∈ (0, 1). The following inequalities hold:
∂V [ψ]
(3.14)
∂n r 6 c kψkL p(S), Wp (S) k∇SV kWpr (S) 6 c kψkWpr (S).
(3.15)
Proof. The inequality (3.14) follows from the estimates |L(x, y)| 6 c |x − y|−n+2,
|L(x, y) − L(z, y)| 6
c |x − z|λ , |x − y|n−2+λ
1 2 x−y ∂V [ψ] (λ ∈ (0, 1], |x − z| 6 |x − y|) for the kernel L(x, y) = − ~n(x) · of the operator . We prove 2 |S1 | |x − y|n ∂n only the inequality (3.15) which follows from the estimate p Z 2 µp ∂ p (3.16) ∂xi ∂x j V [ψ] ρ (x) dx 6 c kψkWpr (S), Ω
3049
(µ = 1 − 1/p − r ∈ (−1/p, 1 − 1/p), ρ(x) = dist(x, S)) and the theorem about traces on S of functions in weight Sobolev spaces. The domain Ω is the union of domains Ωh , ω1 , . . . , ωM : Ω = Ωh ∪
M ∪ ωk ,
(3.17)
k=1
where Ωh = {x ∈ Ω : ρ(x) > h} and ωk = {Fk (ξ0 ) < ξn < Fk (ξ0 )+ 2h, |ξ0| 6 2h}, ξ = (ξ0 , ξn) are the Cartesian coordinates with origin x(k) ∈ S such that the xn -axis is parallel to ~n(x(k)), and Fk is a function of class C2(Kd ) that describes S in a neighborhood of x(k) by the equation ξn = Fk (ξ0 ). The parameter h < d/3 will be chosen later. It is obvious that
2
∂
µ
V [ψ] · ρ 6 c (h) kψkL p(S).
∂xi ∂x j
L p (Ωh)
2
∂
µ
To estimate
∂xi ∂x j V [ψ]ρ
, we consider the local coordinates {ξ j } j=1,... ,m . We note that the kernel
L p (ωk )
∂2E(ξ − η) satisfies the inequality ∂ξi ∂ξ j 2 ∂ E c ∂ξi ∂ξ j 6 |ξ − η|n ;
moreover, the integral Z
Z
n ∂2 E ∂2E(ξ − η) 2 Ii j = dSη = ∑ nk (η) dSη ∂ξi∂ξ j ∂ηi ∂η j k=1 S S Z n Z n ∂ ∂ ∂E ∂ ∂E(ξ − η) = ∑ nk nk −nj dSη − ∑ nj dSη ∂η j ∂ηk ∂ηi ∂nη k=1 k=1 ∂ξi S n
Z
∂ =−∑ k=1 ∂ξi
S
S
∂n (η) ∂ E(ξ − η)n j (η) k dSη − ∑ ∂ηk k=1 ∂ξi n
Z
nj S
∂E dS ∂nη
does not exceed c (1 + | logρ(ξ)|). Let ξ ∈ ωk . We use the relation ∂2V = ∂ξi ∂ξ j
Z S
∂2 E(ξ − η) (ψ(η) − ψ(e ξ)) dSη + ψ(e ξ)Ii j ≡ W1 + W2, ∂ξi ∂ξ j
where e ξ = (ξ0 , Fk (ξ0 )) ∈ S. For any ξ ∈ ωk , η ∈ σk ≡ {ξ ∈ S : |ξ0| 6 2h} we have |ξ − η|2 = |ξ0 − η0 |2 + |ξn − Fk (η0 )|2 .
If h is sufficiently small so that |Fk (ξ0 ) − Fk (η0 )| 6 sup |∇Fk | |ξ0 − η0 | 6 3050
1 0 |ξ − η0 |, 4
then |ξ − η|2 = |ξ0 − η0 |2 + |ξn − F(ξ0 )|2 + 2(ξn − Fk (ξ0 ))(Fk (ξ0 ) − Fk (η0 )) + |Fk (ξ0 ) − Fk (η0 )|2 1 > (|ξ0 − η0 |2 + (ξn − F(ξ0 ))2 ). 2
(3.18)
In particular, if η = ξ is the nearest point of S to ξ, then ρ(ξ) > c (ξn − F(ξ0 )).
(3.19)
Therefore, Z
Z
|W2| ρ (ξ) dξ = p pµ
ωk
dξ
0
6
|ξ0 | 2h
6c
Z
6
|ψ(e ξ)| p dξ0
|ξ0 | 2h
FZ k +2h
|W2| p ρ pµ|Ii j | p dξn
Fk
FZ k +2h
(1 + | log(ξn − F)|) (ξn − Fk ) dξn 6 c p
Z
pµ
|ψ(ξ)| p dS;
(3.20)
σk
Fk
moreover, Z Z 1/p0 |ψ(η) − ψ(e ξ)| p dSη 1/p dSη |ψ(η) − ψ(e ξ)| p |W1(ξ)| 6 c dSη 6 |η − ξ|n |η − ξ|n−1+p−εp |η − ξ|n−1+pε S S S Z |ψ(η) − ψ(e ξ)| p dSη 1/p c 6 ε , ρ (ξ) |η − ξ|n−1+p−εp Z
S
Z
|W1(ξ)| p ρµp(ξ) dξ 6 c
ωk
Z
ρµp−εp (ξ) dξ
ωk
Z Z
+c
|ψ(η) − ψ(e ξ)| p dSη dSξ
σk σk
6c
Z Z
S\σk FZ k +2h Fk
|ψ(η) − ψ(ξ)| p
S σk
Z
|ψ(η) − ψ(e ξ)| p dSη |η − ξ|n−1+p−εp
(ξn − Fk )(µ−ε)p dξn (|ξ0 − η0 |2 + (ξn − Fk )2 )(n−1+p−εp)/2
dSη dSξ |ξ − η|n−1+pr
(3.21)
since |η − ξ| > c (h) for η ∈ S \ σk , ξ ∈ ωk . Combining the inequalities (3.20) and (3.21), we arrive at (3.16). The proof of the inequality Z
p |∇V [ψ]| pρµp(x) dx 6 c kψkW r (S) p
Ω
(3.22)
is much simpler. By the trace theorem for weight Sobolev spaces and (3.16), (3.22), we find k(∇V )i kWpr (S) 6 c kψkWpr (S), 3051
where (∇Vi ) are the limit values of the gradient of V on S, in particular, k∇S1V kWpr (S) 6 c kψkWpr (S).
The proposition is proved.
We will use the estimate (3.16) with µ = 0, r = 1 − 1/p. We consider the direct value of the hydrodynamic potential (2.4) Zt Z
~u(x, t) =
G(x, y, t − τ)~Φ(y, τ) dSy dτ, x ∈ S.
(3.23)
0 S
Proposition 3.3. If ~Φ ∈ L p (ΣT ), then k~ukL p (0,T ;Wpr (S)) 6 c k~ΦkL p (ΣT ) , k~uk
r /2
1 (0,T ;L (S)) Wp,0 p
(3.24)
6 c k~ΦkL p (ΣT )
(3.25)
for any r ∈ (0, 1) and r1 ∈ (0, 1/2). Proof. By (3.4) and (3.5) we have Z∞
Z
|Gi j (x, y, τ) − Gi j (z, y, τ)| dSy
dτ 0
6
Z∞
S
Z
(|Gi j (x, y, τ)| + |Gi j (z, y, τ)|) dSy +
dτ 0
Sx (2ρ)
Z 6c
+ S\Sx(2ρ)
Z
dτ 0
Z∞
dSy
Sx (2ρ)
Z
Z∞
0 λ
|Gi j (x, y, τ) − Gi j (z, y, τ)| dSy
S\Sx(2ρ)
1 1 + dτ τ(1+λ)/2(|x − y|2 + τ)(n−2λ)/2 τ(1+λ)/2(|z − y|2 + τ)(n−2λ)/2
|x − z| dSy
Z∞ 0
dτ τ(1+λ)/2(|x − y|2 + τ)(n−λ)/2
6 c |x − z|θ,
where ρ = |x − z|, Sx (r) = {y ∈ S : |x − y| < r}, θ < λ < 1 (thus, θ is an arbitrary number in the interval (0, 1)). Estimating ~u(x, t) −~u(z, t) with the help of the H¨older inequality, we find |~u(x, t) −~u(z, t)| 6 c |x − z|
θ/p0
Zt Z
|G(x, y, t − τ) − G(z, y, t − τ)| |~Φ(y, τ)| p dSy dτ)1/p,
0 S
ZT
Z Z
dt 0
3052
S S
dSx dSz |~u(x, t) −~u(z, t)| p 6c |x − z|n−1+pr
ZT Z 0 S
(3.26) |~Φ(y, τ)| p J(y, τ) dSy dτ,
where J(y, τ) =
ZT
Z Z
|G(x, y, t − τ) − G(z, y, t − τ)| dSx dSz |x − z|n−1+pr−θ(p−1)
dt τ
S S
ZZ
6c
|x − z|
−n+1−pr+θ(p−1)
ZT
dSz dSz
> 21 |x−y|
|x−z|
+
1
dt
(t − τ)(t+θ)/2(|z − y|2 + t − τ))(n−2θ)/2 ZZ
|x − z|−n+1−pr+θp dSx dSy
+c
6 21 |x−y|
|x−z|
6c
> 12 |x−y|
|x−z|
ZZ
+
> 31 |z−y|
|x−z|
ZZ
+
6 21 |x−y|
|x−z|
6c
Z Z S S
ZT
dt (t − τ)(1+θ)/2(|x − y|2 + t − τ)(n−θ)/2
τ
ZZ
τ
1 (t − τ)(1+θ)/2(|x − y|2 + t − τ)(n−2θ)/2
dSx dSz n−1+p(r−θ)+θ |x − z| |x − y|n−1−θ
dSx dSz n−1+p(r−θ)+θ |x − z| |z − y|n−1−θ dSx dSz n−1+p(r−θ) |x − z| |x − y|n−1
1 1 dSx dSz + 6 c, n−1−ε n−1−ε |x − y| |z − y| |x − y|n−1+p(r−θ)+ε
if we choose θ ∈ (r, 1) and ε ∈ (0, min(1, p(θ − r)). This estimate and (3.26) imply (3.24). To prove (3.25), we use the inequalities Z Z∞
|Gi j (x, y, τ − h) − Gi j (x, y, τ)| dτ dSy 6
S 0 Z∞ Z
+ 2h S Z2hZ
+
(|Gi j (x, y, τ − h)| + |Gi j (x, y, τ)|) dSy dτ
S 0
Zh Z |Gi j (x, y, τ − h) − Gi j (z, y, τ)| dSy dτ 6 c 0 S Z∞ Z
dτ dSy +c (1+µ)/2 (τ − h) (|x − y|2 + τ − h)(n−2µ)/2
h S µ/2
6 ch
Z Z2h
2h S
dτ dSy (1+µ)/2 τ (|x − y|2 + τ)(n−2µ)/2 h dτ dS (3+µ)/2 (τ − h) (|x − y|2 + τ − h)(n−2µ)/2
,
(3.27)
where µ ∈ (0, 1/2), Gi j (x, y, τ − h) = 0 for τ < h. The last inequality follows from (3.2) and (2.10). Similarly, we can prove that Z∞
Z
dτ 0
|Gi j (x, y, τ − h) − Gi j (x, y, τ)| dSx 6 chµ/2 .
(3.28)
S
3053
Therefore, the difference ~u(x, t − h) −~u(x, t) =
Zt Z
(G(x, y, τ − h) − G(x, y, τ))~Φ(y, t − τ) dSy dτ
0 S
can be estimated as follows: Zt Z
|~u(x, t − h) −~u(x, t)| 6 ch
µ/2p
|G(x, y, τ − h) − G(x, y, τ)| |~Φ(y, t − τ)| dSy dτ p
1/p ,
0 S
Z1 0
ZT Z
dh h1+pr1/2 6c
0 S
Z1 0
6c
|~u(x, t − h) −~u(x, t)| p dSx dτ
h1+p(r1/2−µ/2)+µ/2
Z1 0
ZT Z
dh ZT Z
dh h1+p(r1−µ)/2
|~Φ(y, τ)| p dSy dτ
ZT Z
|G(x, y, t − τ − h) − G(x, y, t − τ)| dSy dt
τ S
0 S
|~Φ(y, τ)| p dSy dτ.
0 S
Therefore, taking µ ∈ (r1 , 1/2), from (3.6) we obtain (3.25). The proposition is proved.
Proposition 3.4. The potential (3.23) satisfies the inequality k~ukW r/2(0,T ;L p
p (S))
6 c k~ΦkW (r−µ)/2(0,T ;L p
p (S))
,
(3.29)
where µ ∈ (0, 1/2). Proof. We use the relation ∆t2 (−h)~u(x, t)
Z∞ Z
=
∆t (−h)G(x, y, t − τ)∆τ (−h)~Φ(y, τ) dτ dSy,
−∞ S
assuming that G(x, y, t) and ~a(y, t) vanish for t < 0, where ∆t (−h) f ≡ f (t − h) − f (t) and ∆t2 (−h) f (t) = f (t − 2h) − 2 f (t − h) + f (t). Estimating ∆t2 (−h)~u in the same way as in Proposition 3.3 and using (3.27), (3.28), we find Z∞ 0
dh
k∆2 (−h)~ukLpp (ΣT ) h1+pr/2 t
6c
Z∞ 0
dh h1+p(r−µ)/2
k∆t2 (−h)~ΦkLpp (ΣT ) .
From this inequality and (3.25) and (3.6) we obtain (3.29). The proposition is proved.
By Propositions 3.1–3.4, the following assertion holds. r,r/2
Proposition 3.5. If ~a ∈ Wp,0 (ΣT ), r ∈ (0, 1), then the solution to the system (2.16) also belongs to r,r/2
Wp,0 (ΣT ) and satisfies the estimate k~ΦkW r,r/2 (Σ ) + k~ψkW r,r/2 (Σ ) 6 c k~akW r,r/2 (Σ ) . p,0
3054
T
p,0
T
p,0
T
(3.30)
Indeed, by (3.14), (3.15), and (3.29), we have kψkL p (0,T ;Wpr (S)) 6 c ( kψkL p(ΣT ) + k~ΦkL p (ΣT ) + k~a ·~nkL p(0,T ;Wpr (S))) 6 c k~akL p(0,T ;Wpr (S)), k~ΦkL p (0,T ;Wpr (S)) 6 c ( k~ΦkL p(ΣT ) + kψkL p(0,T ;Wpr (S)) + k~atankL p (0,T ;Wpr (S))) 6 c k~akL p (0,T ;Wpr (S)).
The norms kψkW r/2(0,T ;L p,0
p (S))
and k~ΦkW r/2(0,T ;L p,0
p (S))
are estimated in the same way in two steps: for
r = r1 ∈ (0, 1/2) and, at the second step, for r ∈ (0, 1). § 4. Estimate for Rate
We prove the estimate (1.3) in the case F = 0. Proposition 4.1. The potential (2.4) satisfies the inequality ~ L (Q ) 6 c k~Φk k∇Uk p T
1−1/p,1/2−1/(2p)
Wp,0
(ΣT )
.
(4.1)
Proof. We represent the domain Ω as a union of the domains (3.17). The inequality ~ L (Ωh×(0,T )) 6 c (h)k~ΦkL (Σ ) k∇Uk p T p
(4.2)
immediately follows from the estimate c |∇xGi j (x, y, t)| 6 √ 2 , t(ρ (x) + t)1/2 (|x − y|2 + t)n/2
x ∈ Ω, y ∈ S
(4.3)
~ kL (ω ×(0,T )). We pass to local coordinates at the which is a consequence of (3.2). Let us estimate k∇U p k ~ ∂ U point x(k) and write in the form ∂xm ~ ∂U = ∂xm
Zt Z 0 S Zt Z
+ −∞ S
∂G(x, y, t − τ) ~ (Φ(y, τ) − ~Φ(e x, τ)) dSy dτ ∂xm ∂G(x, y, t − τ) ~ ~ m(1) + U ~ m(2) + U ~ m(3), (Φ(e x, τ) − ~Φ(e x, t)) dSy dτ + Jm~Φ(e x, t) = U ∂xm
where xe ∈ S is a point defined for any x ∈ ωk as was mentioned in Proposition 3.2 and Jm =
Zt
Z
dτ −∞
S
∂G(x, y, t − τ) dSy. ∂xm 3055
By (4.3), we have ~ m(1)(x, t)| 6 c |U
Zt 0
dτ √ t − τ(ρ2 (x) + t − τ)1/2
Z S
|~Φ(y, τ) − ~Φ(e x, τ)| dSy (|x − y|2 + t − τ)n/2
1/p0 Zt Z dτ dSy 6c √ t − τ(ρ2 (x) + t − τ)1/2(|x − y|2 + t − τ)(n−1+εp0)/2 0 S
×
Zt 0
dτ √ t − τ(ρ2 (x) + t − τ)1/2
Z S
1/p |~Φ(y, τ) − ~Φ(e x, τ)| p dSy (|x − y|2 + t − τ)(n−1+p−εp)/2
Zt 1/p Z |~Φ(y, τ) − ~Φ(e x, τ)| p dSy dτ √ 6c , t − τ(ρ2 (x) + t − τ)1/2 (|x−y|2 +t −τ)(n−1+p−εp)/2ρεp (x) 0
S
where ε 1, and, consequently, ZT
Z
dt
6c
ωk
0
F(x0)+2h
×
~ m(1)(x, t)| p dt |U
Z
Z
6
|~Φ(y, τ) − ~Φ(e x, τ)| p dτ dSy dx0
0 S |x0 | 2h
ZT
dxn
F(x0)
ZT Z
τ
dt . (t − τ)1/2(ρ2 (x) + t − τ)1/2ρεp (x)(|x−y|2 +t −τ)(n−1+p−εp)/2
Since |x − y| > ρ(x), the integral with respect to t does not exceed the quantity c ρ2εp (x)|x − y|n−1+p−2εp
.
Using the inequalities (3.18) and (3.19), it is easy to show that for all y ∈ S F(xZ0)+2h F(x0)
dxn 2εp ρ (x)|x − y|n−1+p−2εp
6
c . |e x − y|n−2+p
Hence ZT Z 0 ωk
~ m(1)(x, t)| p dx dt |U
6c
ZT Z Z ~ |Φ(y, τ) − ~Φ(z, τ)| p 0 σk S
|x − z|n−2+p
dSy dSz dτ.
By (4.3), we have ~ m(2)(x, t)| 6 c |U
Zt
−∞
3056
Zt 1/p |~Φ(e x, τ) − ~Φ(e x, t)| |~Φ(e x, τ) − ~Φ(e x, t)| p dτ 1 √ dτ 6 c √ , ρε (x) t − τ(ρ2 (x) + t − τ) t − τ(ρ2 (x) + t − τ)(p+1)/2−εp −∞
(4.4)
ZT Z
(2)
~ m (x, t)| p dx dt |U
0 ωk
Z
6c
0
ZT
dx
6
−∞
|x0 | 2h
Z
6c
dτ
0
dx
6
√ t −τ
τ
ZT ZT ~ |Φ(e x, τ) − ~Φ(e x, t)| p
−∞ τ
|x0 | 2h
ZT ~ Z∞ |Φ(e x, τ) − ~Φ(e x, t)| p dt
(t − τ)(1+p)/2
F(x0)
dτ dt 6 c
dxn 0 εp (xn − F(x )) ((xn − F(x0 ))2 + t − τ)(p+1)/2−εp Z
ZT ZT
dSk
σk
|~Φ(z, τ) − ~Φ(z, t)| p
−∞ τ
dτ dt . (t − τ)(p+1)/2
If we show that |Jm | 6
c
(4.5)
ρε1 (x)
for sufficiently small ε1 , then ZT Z
~ m(3)(x, t)| p dx dt |U
0 ωk
6c
ZT Z
|~Φ(x, t)| p dS dt.
0 σk
Combining this estimate with (4.2) and (4.4), we arrive at (4.1). Since −2
Zt
−∞
∂Γ(x − y, t − τ) ∂E(x − y) dτ = 2 , ∂n ∂n
we have −2
Zt Z
−∞ S
The kernel
∂ ∂Γ(x − y, t − τ) (δi j − ni (y)n j (y)) dSy dτ = 2 ∂xm ∂n
Z S
∂ ∂E(x − y) ni (y)n j (y) dSy . ∂xm ∂n
(4.6)
∂ ∂E(x − y) can be written in the form ∂xm ∂n n ∂ n ∂E(x − y) ∂ ∂ ∂E(x − y) . ∑ nk (y) ∂xk = ∑ nk (y) ∂ym − nm ∂yk ∂xm k=1 ∂yk k=1
Therefore, the integral (4.6) is equal to the quantity n Z ∂E(x − y) ∂ ∂ −2 ∑ nk (y) − nm ni (y)n j (y) dSy ∂xk ∂ym ∂yk k=1 S
and does not exceed c (1 + | logρ(x)|). We consider Z∞ Z 0 S
∂ 0 G (x, y, t) dSy = I, ∂xm i j 3057
where G0i j are elements of the tensor T ∂ ~ y, t) . G (x, y, t) = 4 ∇ −~n(y) ⊗ Q(x, ∂n 0
For points y ∈ S close to x (for example, such that |y0 | 6 d ) we use formula (2.11) under the assumption that Cik (y0 ) are functions of class C1(Kd ) for |y0 | 6 d . For any t > 0, x ∈ ωk , and y ∈ σk we have G0i j (x, y, t) =
n ∂ ∂ − n j (y) ∑ nk (y) ∂x j ∂xk k=1 ~n(y)·(x−y) Z
n
=
∑
Z
Rn−1
0
0
Rn−1
Γqn (C(y)(x − y) − η, t)Ei (η) dη Cq j (y) − n j 0
dηn
q=1
~n(y)·(x−y) Z Z dηn Γn (C(y)(x − y) − η, t)Ei (η) dη0 ∑ nkCk j(y) , n
k=1
where the subscripts Γ and E denote the derivatives with respect to the corresponding variables, for ∂E(η) . Moreover, example, Ei = ∂ηi
n ∂ ∂ − n j (y) ∑ nk (y) ∂y j ∂yk k=1 n
=−∑
~n(y)(x−y) Z
dηn
q=1
n
+∑
q=1
Z
~n(y)(x−y) Z
Z
dηn 0 n
Rn−1
Rn−1
0
Γqn(C(y)(x − y) − η, t)Ei (η) dη Cq j (y) − n j 0
Rn−1
0
~n(y)·(x−y) Z Z dηn Γn(C(y)(x − y) − η, t)Ei (η) dη0 ∑ nkCk j(y) n
k=1 0
Γqn (C(y)(x − y) − η, t)Ei (η) dη
n
∑
l=1
∂Cql (y) − η j (y) ∑ nk (y) (xl − yl ) . ∂yk k=1
∂Cql (y) (xl − yl ) ∂y j
Thus, n n ∂Qq j (x, y, t) ∂ 0 ∂ ∂ ∂Qi Gi j (x, y, t) = − ∑ nk (y) nk (y) − n j (y) +∑ Hqi (x, y), ∂xm ∂y j ∂yk ∂xm q=1 ∂xm k, j=1
where n
Hqi (x, y) =
∑
k,l=1
∂ ∂ nk −nj Cql (y) · (xl − yl ), ∂y j ∂yk
~n(y)(x−y) Z
Qqi (x, y, t) =
dηn 0
3058
Z
Rn−1
Γqn (C(y)(x − y) − η, t)Ei (η) dη0 .
(4.7)
Let ζ(y) be a smooth function that is equal to 1 for |y0 | < 3h and 0 for |y0 | > d . By (4.7), we have Zt Z
ζ(y)
−∞ S n
=
∑
∂G0i j (x, y, t − τ) dSy dτ ∂xm
Zt Z
k, j=1−∞ S t n Z Z
+∑
∂Qi (x, y, t − τ) ∂ ∂ nk (y) − n j (y) (nk (y)ζ(y)) dSy dτ ∂xm ∂y j ∂yk
ζ(y)Hq j (x, y)
q=1−∞ S
Since
∂Qqi (x, y, t) dSy dτ. ∂xm
(4.8)
∂Qi c ∂xm 6 √t − τ(ρ2 (x) + t − τ)1/2(|x − y|2 + t − τ)(n−1)/2
(this follows from [8, Theorem 3]), the first term on the right-hand side does not exceed the quantity Z∞ Z
c 0 σ
dS dτ 6c √ 2 τ(ρ (x) + τ)1/2 (|x − y|2 + t)(n−1)/2
Z∞ Z 0 σ
dS dτ c 6 √ 2 ε τ(ρ (x) + τ)(1+ε)/2 |x − y|n−1+ε ρ (x)
for so small ε > 0 as desired (σ = {x ∈ S : |x0 | < d}). To estimate the second term, we use the relation ∂ Qni (x, y, t) = ∂t −
~n(y)(x−y) Z
Z
dηn
Rn−1
0
~n(y)(x−y) Z
Z
dηn 0
Rn−1
Γ(C(y)(x − y) − η, t)Ei (η) dη0
∆0η Γ(C(y)(x − y) − η, t)Ei (η) dη0 ,
(4.9)
where ∆0η =
n−1
∂2
∑ ∂η2 .
j=1
j
∂ If we integrate with respect to t , then the contribution of the term containing the derivative is zero and ∂t for q < n we have ∂Qqi c ∂xm 6 √t − τ(ρ2 (x) + t − τ)1/2(|x − y|2 + t − τ)n/2
(in view of [8, Theorem 3]). The same estimate is valid for the derivative of the last term in (4.9) (cf. [9, (2.27)]). Since |Hq j | 6 c |x − y|, the above arguments mean that n Zt Z ∂Q (x, y, t) c qi ∑ ζ(y)Hq j (x, y) dSy dτ 6 ε . ∂xm ρ (x) q=1 −∞ σ
3059
It is clear that for x ∈ ωk at least the same estimate holds for Zt Z ∂G0 ij −∞ S
∂xm
(1 − ζ) dS dτ.
Thus, the inequality (4.5) and Proposition 4.1 are proved. Proposition 4.2. The potential (2.4) satisfies the inequality ~ k 1/2 kU W (0,T ;L p,0
p (Ω))
6 c k~ΦkW 1/2−1/(2p)(0,T ;L p,0
(4.10)
p (S))
Proof. We estimate the second-order difference ~ (x, t) = U ~ (x, t − 2h) − 2U ~ (x, t − h) + U ~ (x, t) ∆2−h (t)U
with the help of the obvious formula ~ (x, t) = ∆2−h (t)U
Zt
Z
∆−h (t)G(x, y, t − τ)∆−h (τ)~Φ(y, τ) dτ dSy.
dτ −∞
S
From the estimate (3.2) and (3.4) it follows that for δ ∈ (0, 1) we have Zt
Z
|ρδ (x)| |∆−h (t)Gi j (x, y, t − τ)| dSy 6 chδ/2
dτ −∞
S
Zt Z dτ −∞
Ω
∀x ∈ Ω,
∆−h (t)Gi j (x, y, t − τ) ρ−δ(x) dx 6 ch−δ/2+1/2,
y ∈ S.
Consequently, for sufficiently small ε > 0 we have ~ (x, t)| 6 |∆2−h (t)U
Zt
Z
dτ −∞
6
Z
dτ −∞
6 ch
ρ
(x)|∆−h (t)G(x, y, t − τ)| dSy
1/p0
S
Zt
ε
εp0
Zt
−εp
|∆−h (t)G(x, y, t − τ)|ρ
S
Z
dτ −∞
(x)|∆−h (τ)~Φ(y, τ)| dSy
−εp
|∆−h (t)G|ρ
|∆−h (τ)~Φ| dSy
1/p
p
S
and ZT
ZT 2 p 1/2 dτ |∆−h (t)~Φ(x, t)| dx 6 ch −∞ −∞ Ω
3060
Z
Z
dτ S
|∆−h (τ)~Φ(y, τ)| p dSy.
1/p
This implies the inequality Z∞ 0
dh
~ p k∆2 Uk h1+p/2 −h L p(Ω×(−∞,T))
1/p Z∞ 6c 0
dh
k∆−h~ΦkLpp(S×(−∞,T )) (1+p)/2
1/p
h
which is equivalent to (4.9). We pass to the estimate for ∇V[ψ]. In view of the inequality (3.16) (for µ = 0), we have k∇V [ψ]kL p(0,T ;Wp1 (Ω)) 6 c kψkL
1−1/p (S)) p (0,T ;Wp
.
(4.11)
Now, we use the condition (1.2) for ~a·~n and the fact that ~n· G[~Φ], is represented in the form ~n· G[~Φ] = divS ~D because it is the normal component of the solenoidal vector field G[~Φ]; moreover, k~DkW 1−1/p(S) 6 c kG[~Φ]kL p(Ω) p
(cf. [2, Proposition 3.4]). The same estimate holds for ∆−h (t)~D and ∆−h (t)G[Φ]. Hence k~DkW 1/2(0,T ;W 1−1/p(S)) 6 c kG[~Φ]kW 1/2(0,T ;L p
p,0
p,0
p (Ω))
6 c k~ΦkW 1−1/p,1/2−1/(2p)(Σ ) . T
p,0
Thus, ~ − divS ~D, ψ0 ≡ ~a ·~n −~n · G[~Φ] = divS A V [ψ0 ] = 2
Z
∇E(x − y) · (~A(y) − ~D(y)) dSy ;
S
Using (3.16) again, we find k∇V [ψ0 ]kW 1/2(0,T ;L p,0
p (Ω))
~ 1/2 6 c( kAk + k~ΦkW 1−1/p,1/2−1/(2p)(Σ )). 1−1/p W (0,T ;W (S)) p
p,0
T
p,0
(4.12)
∂V ∂V Because of equation (2.16)2 , we have ψ = − [ψ] + ψ0; moreover, L[ψ] ≡ − [ψ] satisfies the same ∂n ∂n integral equation L[ψ] = L2[ψ] + L[ψ0 ]
and the condition Z
L[ψ] dSη = 0. S
Therefore, k∇V [ψ]kW 1/2(0,T ;L p,0
p (Ω))
6 k∇V [ψ0 ]kW 1/2(0,T ;L p,0
p (Ω))
+ k∇V [L[ψ]]kW 1/2(0,T ;L p,0
~ 1/2 6 c ( kAk + k~ΦkW 1−1/p,1/2−1/(2p)(Σ ) + kL[ψ]kW 1/2(0,T ;L 1−1/p W (0,T ;W (S)) p,0
p
p,0
T
p,0
p (Ω))
p (S))
,
(4.13) 3061
where the last norm does not exceed the quantity c kL[ψ0 ]kW 1/2(0,T ;L p,0
0
−L[ψ ] = 2
Z S
= −2
∂E(x − y) ~ − ~D) dS = −2 divS (A ∂nx Z
∇S
S
Z
∇S
S
p (S))
. We use the relations
∂E(x − y) ~ (A(y, t) − ~D(y, t)) dSy ∂nx
∂E(x − y) ~ ~ t) + ~D(x, t)) dSy (A(y, t) − ~D(y, t) − A(x, ∂nx
~ t) − ~D(x, t)) · = −2(A(x,
Z
∇S
S
∂E(x − y) dSy. ∂nx
(4.14)
∂E(x − y) are the functions ∂nx n ∂ ∂ 1 nl (x)(yl − xl ) H j (x, y) ≡ ∑ nk (y) nk (y) − n j (y) ∂y j ∂yk |S1 | |x − y|n k,l=1
The components of the kernel −∇S
=−
n |S1|
n
∑
nk (y)(nk (y)(y j − x j ) − n j (y)(yk − xk ))
k,l=1
n
+ ∑ nk (y)(nk (y)n j (x) − n j (y)nk (x)) k=1
nl (x)(yl − xl ) |x − y|n+2
1 |S1| |x − y|n
(4.15)
satisfying the inequalities |H j (x, y)| 6 c |x − y|−n+1.
The integrals Z
n
H j (x, y) dSy =
∑
k=1
S
Z
˙ − x) ∂ ∂ 1 ~n(x)(y nk (y) − n j (y) nk (y) dS ∂y j ∂yk |S1| |x − y|n
S
are bounded. Therefore, from (4.14) it is easy to deduce the estimate kL[ψ0 ]kW 1/2(0,T ;L p,0
p (S))
~ − ~Dk 1/2 6 c kA . 1−1/p W (0,T ;W (S)) p
p,0
Together with (4.12) and (4.13), it yields k∇V [ψ]kW 1,1/2(Q ) 6 c ( kΦkL p,0
T
1−1/p (S)) p (0,T ;Wp
~ 1/2 + k~ΦkW 1−1/p,1/2−1/(2p)(Σ ) + kAk ). 1−1/p W (0,T ;W (S)) p,0
T
p,0
p
(4.16)
Combining this estimate with (4.9) and (3.30), we arrive at (1.3). § 5. Estimate for Pressure
We first consider the solution (2.6), (2.8) to the problem (2.5) in the half-space under the assumption that a j (x0 , t) are smooth functions and a j (x, 0) = 0. Integrating by parts, it is easy to show that Z ∂ (0) (0) 0 0 0 0 Q j (x, t) = Q (x, t) + Γ(x − η , xn , t)E(η , 0) dη − A(x, t) , ∂x j
Rn−1
3062
where Zxn (0)
Q (x, t) =
Z
dηn
Rn−1
0
Z
∂E(η) 0 Γ(x − η, t) dη , ∂ηn
A(x, t) =
Rn−1
Γ(η0 , 0, t)E(x0 − η0 , xn) dη0 ,
and, consequently, n−1
q(x, t) = −4 ∑
j=1
n−1
+4 ∑
j=1
Since
∂ ∂x j
∂ ∂ ∂x j ∂t
Zt Z ∂ −∆ Q(0)(x0 − y0 , xn, t − τ)a j (y0 , τ) dy0 dτ ∂t 0
Zt
Z
dτ
Rn−1
0
Rn−1
A(x0 − y0 , xn, t − τ)a j (y0 , τ) dy0.
Zt Z Z ∂ Q(0)(x0 − y0 , xn, t − τ)a j (y0 , τ) dy0 dτ = lim Q(0)(x0 − y0 , xn , ε)a j (y0 , t − ε) dy0 −∆ ε→0 ∂t 0
Zt−ε
+ lim
Z
dτ
ε→0 0
1 = 2
Rn−1
Z
Rn−1
Rn−1
Rn−1
∂ − ∆ Q(0)(x0 − y0 , xn, t − τ)a j (y0 , τ) dy0 ∂t
∂E(x0 − y, xn) a j (y0 , t) dy0 − ∂xn
Zt
Z
dτ
Rn−1
0
∂2A(x0 − y0 , xn, t − τ) a j (y0 , τ) dy0, ∂x2n
we have q(x, t) = q1(x, t) +
∂q0(x, t) , ∂t
where n−1
q1 (x, t) = −2 ∑
Z
Rn−1
j=1
∂ +4 ∑ j=1 ∂x j n−1
Zt
∂2 E(x0 − y0 , xn) a j (y0 , t) dy0 ∂x j ∂xn Z
dτ 0
n−1
q0 (x, t) = 4 ∑
j=1
Zt
Rn−1 Z
dτ 0
Rn−1
∂2A(x0 − y0 , xn, t − τ) a j (y0 , τ) dy0, 2 ∂xn ∂A(x0 − y0 , xn , t − τ) a j (y0 , τ) dy0. ∂x j
Arguing in the same way, we can write the pressure p(x, t) (2.12) in the form (1.4), where p1(x, t) = −2
Z S
Zt
P(x, t) = 4
~Φ(y, t) · ∇x ∂E(x − y) dSy + +4 ∂n Z
dτ 0
S
Zt
Z
dτ 0
~Φ(y, τ) · ∇x A(x, y, t − τ) dSy + +2
S
Z
2 ~Φ(y, τ) · ∇x ∂ A(x, y, t − τ) dSy, ∂n2
E(x − y)ψ(y, t) dSy = p0 + p00 ,
(5.1)
(5.2)
S
3063
Z
∂ = ~n(y) · ∇x , ∂n
A(x, y, t) =
E(x − η0 )Γ(η0 − y, t) dη0,
T (y)
and T (y) is the tangent plane to S at the point y. From the inequality (62) in [8] we obtain the estimate |Dxj Dtm A(x, y, t)| 6
c t m+1/2(|x − y|2 + t)(n−2+| j|)/2
.
(5.3)
Moreover, it should be taken into account that ~Φ ·~n = 0. Hence the operator ∇x in (5.1) can be replaced with n ∂ ∂ ∇S,x = − n j (y) ∑ nk (y) . ∂x j ∂x k j=1,... ,n k=1 Therefore, using the same arguments as in Sec. 4, we find kp1 kL p(QT ) 6 c k~ΦkW 1−1/p,1/2−1/(2p)(Σ ), T
p,0
k∇PkW 1,1/2(Q ) 6 c ( k~ΦkW 1−1/p,1/2−1/(2p)(Σ ) + kψkL p,0
T
T
p,0
1−1/p (S)) p (0,T ;Wp
~ 1/2 + kAk ). 1−1/p W (0,T ;W (S)) p,0
p
We pass to the proof of the inequality (1.6). We write p0 (x, t) in the form Zt
0
p (x, t) = 4
Z
dτ
0 Zt
+4
S
Z
dτ 0
~Φ(y, τ) · ∇A0 (x, y, t − τ) dSy ~Φ(y, τ) · ∇(A − A0 ) dSy ≡ p01(x, t) + p02(x, t),
S
where A0 (x, y, t) =
Z
E(x − η)Γ(η − y, t) dSη.
S
Hence p01(x, t) =
Z
∇E(x − η) · ~X (η, t) dSη,
Zt Z
~X (η, t) =
S
Γ(η − y, t − τ)~Φ(y, τ) dSy dτ.
0 S
By the well-known properties of the heat potential and the electrostatic potential of a single layer, we have kp01 kW 1−1/(2p)−r/2(0,T ;W 1/p+r (Ω)) 6 ck~X kW 1−1/(2p)−r/2(0,T ;W r (S)) 6 c k~ΦkW 1−1/p,1/2−1/(2p)(Σ ). p,0
p
p,0
p
p,0
(5.4)
T
One can show (we do not consider this in detail) that the same estimate is valid for p02 since the kernel A − A0 is less singular than A and A0 . We estimate p00 using the arguments from the proof of (4.16). 3064
We consider the kernel of the potential ~n · G[~Φ]. By (2.10), we have ∂Γ(x − y, t) n ∑ ni(y)(ni (y)nk (x) − ni(x)nk (y)) ∂n i=1 i,k=1 n ∂ ∂ − n j (y) + 4 ∑ (ni (x) − ni (y)) Qi (x, y, t) ∂x j ∂n i=1 n
∑ ni (x)Gi j (x, y, t) = −2
∂ n ∑ Qi (x, y, t)ni(y) ∂xk i=1 k=1 n n ∂ ∂ + 4 ∑ nk (x) nk (x) − n j (x) ∑ Qi (x, y, t)ni(y), ∂x j ∂xk i=1 k=1 n
+ 4 ∑ (n j (x)nk (x) − n j (y)nk (y))
i.e.,
(5.5)
∂ ∂ ∑ ni(y)Gi j (x, y, t) = 4 ∑ nk (x) nk ∂x j − n j (x) ∂xn Q0 (x, y, t) + G0j(x, y, t), i=1 k=1 n
n
(5.6)
n
where Q0 (x, y, t) = ∑ Qi (x, y, t)ni (y) and G0j are the first three terms in (5.5). Therefore, i=1
∂ ∂ ~ + 4 ∑ nk (x) nk (x) ψ ≡ ~a ·~n −~n · G[~Φ] = divS A − n j (x) ϕ j + χ, ∂x ∂x j k k=1 n
0
ϕ j (x, t) = χ(x, t) =
ZT Z
Q0 (x, y, t − τ)Φ j (y, τ) dSy dτ,
0 S Zt Z
n
∑ G0j (x, y, tτ)Φ j (y, τ) dSy dτ, j=1
0 S 0
R (x, t) ≡ 2
Z
0
E(x − y)ψ (y, t) dSy = −2
S
n
+8
∑
i, j=1
Z S
(5.7)
Z
~ t) dSy ∇yE(x − y)A(y,
S
∂ ∂ − nk (y) n j (y) (nk (y)E(x − y))ϕ0j (y, t) dSy + 2 ∂yk ∂y j
Z
E(x − y)χ(y, t) dSy ,
S
and ~ 1−1/(2p)−r/2 kR0 kW 1−1/(2p)−r/2(0,T ;W 1/p+r(Ω)) 6 c ( kAk + k~ϕkW 1−1/(2p)−r/2(0,T ;W r (S)) W (0,T ;W r (S)) p
p,0
p
p,0
+ kχkW 1−1/(2p)−r/2(0,T ;L p,0
p (S))
p
p,0
).
From the analysis of the kernels Qi given in [8] it follows that the functions G0i (x, y, t), x, y ∈ S, satisfy ~ ·~n(y), x, y ∈ S, these inequalities hold with (3.27) and (3.28) for any µ ∈ (0, 1/2), and for the functions Q µ = 1/2. Therefore, (0)
kχkW 1−1/(2p)−r/2(0,T ;L p,0
p (S))
6 c k~ΦkW 1/2−1/(2p)(0,T ;L p,0
k~ϕkW 1−1/(2p)−r/2(0,T ;W r (S)) 6 c ( k∇~ϕkW 1/2(0,T ;L p
p,0
6 c k~ΦkW 1/2−1/(2p)(0,T ;L p,0
p,0
p (S))
p (Ω))
,
+ k~ϕkW 1−1/(2p)(0,T ;L p,0
p (S))
)
p (S))
3065
and, consequently, ~ 1−1/(2p)−r/2 kR0 kW 1−1/(2p)−r/2(0,T ;W 1/p+r (Ω)) 6 c ( k~ΦkW 1−1/p,1/2−1/(2p)(Σ ) + kAk ). W (0,T ;W r (S)) p
p,0
T
p,0
The function
Z
00
R (x, t) = 2
(5.8)
p
p,0
E(x − y)L[ψ](y, t) dSy
S
satisfies the inequality kR00 kW 1−1/(2p)−r/2(0,T ;W r+1/p(Ω)) 6 c kL[ψ]kW 1−1/(2p)−r/2(0,T ;L p
p,0
p,0
p (S))
6 c kL[ψ]0kW 1−1/(2p)−r/2(0,T ;L p,0
p (S))
(5.9)
and, in view of (5.7), L[ψ0 ] can be written in the form close to (4.14), i.e., Z
∂E(x − y) ~ ~ t)) dSy +~I(x) · A(x, ~ t) (A(y, t) − A(x, ∂nx S n Z ∂ ∂ ∂E +8 ∑ nk −nj nk (ϕ j (y, t) − ϕ j (x, t)) dSy + ~J (x) ·~ϕ(x, t) ∂y j ∂yk ∂nx j,k=1
−L[ψ0] = −2
Z
+2 S
∇S
S
∂E(x − y) χ(y, t) dSy, ∂nx
where ~I and J~ are bounded functions. As in Sec. 4, we have kL[ψ0 ]kW 1−1/(2p)−r/2(0,T ;L p,0
+ kχkW 1−1/(2p)−r/2(0,T ;L p,0
p (S))
p (S))
~ 1−1/(2p)−r/2 6 c ( kAk + k~ϕkW 1/2−1/(2p)−r/2(0,T ;W r (S)) W (0,T ;W r (S)) p
p,0
p
p,0
) 6 c ( k~AkW 1−1/(2p)−r/2(0,T ;W r (S)) + k~ΦkW 1−1/p,1/2−1/(2p)(Σ ) ). p
p,0
p,0
(5.10)
T
From (5.4) and (5.8)–(5.10) we obtain the estimate (1.6). § 6. Proof of Theorem 1.1
We extend Fik (x, t) from Ω to Rn by setting, for example, Fik (x, t) = 0 for x ∈ Rn \ Ω. We set ~v(1)(x, t) =
n
∑
k=1
∂ ~wk (x, t), ∂xk
p(1)(x, t) =
n
∑
k=1
∂ rk (x, t), ∂xk
(6.1)
where (~wk , rk ) is defined as a solution to the Cauchy problem ∗ ∗ ~wkt − ∆~wk + ∇rk = ~Fk∗ = (Fk1 , . . . , Fkn ),
∇ · ~wn = 0, ~wk = 0,
x ∈ Rn, t ∈ (0, T ),
(6.2)
t=0
Fik∗ is the extension of Fik (cf. above). It is clear that k~wk kW 2,1(Rn×(0,T )) + k∇rkkL p (Rn×(0,T )) 6 c k~Fk kL p (QT ), p
k~v kW 1,1/2(Q ) + kp(1)kL p(QT ) 6 c kFkL p (QT ) . (1)
p,0
3066
T
(6.3)
Moreover the problem (1.1) takes the form (2)
~vt − ∆~v(2) + ∇p(2) = 0, ∇ ·~v(2) = 0, ~v(2) = 0, ~v(2) = ~a −~v(1) ≡ ~b t=0
S
(6.4)
S
for ~v(2) =~v −~v(1), p(2) = p − p(1). The vector field ~b satisfies the condition (1.2) since k~bkW 1−1/p,1/2−1/(2p)(Σ ) 6 k~akW 1−1/p,1/2−1/(2p)(Σ ) + c kFkL p (QT ) . T
p,0
T
p,0
(6.5)
Moreover, ~n ·~v
(1)
n ∂wkq ∂ ∂ ~, = ∑ nq = ∑ nq − nk wkq = divS W ∂xk ∂xk ∂xq k=1 k=1 n
n
Wk =
∑ nq (wkq − wqk ).
(6.6)
q=1
~ ∈ W 1−1/(2p)−ρ/2(0, T ;Wpρ (S)) for all ρ ∈ [0, 1] and (ΣT ), we have W p,0
2−1/p,1−1/(2p)
Since ~wk ∈ Wp,0
~ k 1−1/(2p)−ρ/2 kW 6 c kFkL p (QT ) . ρ W (0,T ;W (S)) p
p,0
(6.7)
The estimates (1.3)–(1.6) follow from (6.5), (6.7) and the above estimates for the solution to the problem (6.4). We discuss possible generalizations of Theorem 1.1. First of all, we can consider the problem with nonzero initial conditions ~vt − ∆~v + ∇p = ∇ · ~F, ∇ ·~v = 0, ~v =~v0 (x) ~v = ~a(x, t), t=0
1−2/p
where ~v0 is a given solenoidal vector field in Wp n
~v0(x) =
∑
k=1
(6.8)
S
∂~w0k , ∂xk
(Ω) (for p > 2). It can be represented as 2−2/p
~w0k ∈ Wp
(Ω), ∇ · ~w0k = 0
(6.9)
if for ~w0k we take ∂ ~w0k (x) = ∂xk
Z
Rn
E(x − y)~v∗0 (y) dy,
where ~v∗0 (y) is a solenoidal extension of ~v0 (y) to Rn that has compact support and satisfies the estimate k~v∗0 kW 1−2/p(Rn) 6 c k~v0kW 1−2/p(Ω) p
p
(cf. § 2.) It is clear that k~w0k kW 2−2/p(Ω) 6 c k~v0kW 1−2/p(Ω). p
p
3067
For p < 2 we can postulate the representation (6.9) of ~v0 and define ~v(1) by the relation (6.1), where ~wk is a solution to the problem ~wkt − ∆~wk + ∇rk = ~Fk , ∇ · ~wk = 0, x ∈ Rn, ~wk = ~w∗ (x) 0k
t=0
and ~w∗0k is the solenoidal extension of ~w0k from Ω to Rn . Instead of (6.3), we have k~wk kW 2,1(Q ) 6 c ( k~Fk kL p(QT ) + k~w0k kW 2−2/p(Ω)), p
T
p
k~v kW 1,1/p(Q ) 6 c ( kFkL p (QT ) + V ), (1)
p
T
where V=
k~v0 k 1−2/p , W )(Ω)
p > 2,
∑n k~w0k k 2−2/p , k=1 W (Ω)
p < 2.
p
p
For p < 2 the initial condition is satisfied in the weak sense. For any smooth function ϕ(x, t) with compact support that vanishes at t = T we have ZT Z
ϕt~v(x, t) dx dt =
0 Ω
ZT Z 0 Ω
Taking ϕ(x, t) = ϕ(x) f (t), f (t) = 1 − 1 δ
Zδ Z
n
∑ ϕxk (x, t)~wkt (x, t) dx dt +
k=1
Z Ω
n
∑ ϕxk (x, 0)~w0k (x) dx.
k=1
t for t ∈ (0, δ), f (t) = 0 for t > δ, we find δ
ϕ(x)~v(x, t) dx dt → −
0 Ω
Z Ω
n
∑ ϕxk (x)~w0k (x) dx,
δ → 0,
k=1
which can be interpreted as the initial condition n
~v(x, 0) =
∑
k=1
∂~w0k ∂xk
in the generalized sense. For ~b in (6.4) satisfy the condition (1.2), we need to require that ~v0 S = ~a t=0 if q > 3; moreover, for q = 3 these condition means the boundedness of the integral ZT Z Z
I= 0 Ω S
|~a(x, t) −~v0 (y)|3 dt dy dSx. (|x − y|2 + t)(n+2)/2
~ − divS W ~ . Analyzing the estimates (1.5) For ~v(1) ·~n we have the representation (6.6). Hence ~b ·~n = divS A ~ −W ~ does not necessarily vanish at t = 0 if we restrict ourselves to estimates for ~v and (1.6), we see that A and p on the interval (0, T ). We arrive at the inequalities k~vkW 1,1/2(Q ) + kp1kL p(QT ) + k∇PkW 1,1/2(Q ) 6 cN1, p
p
T
kPkW 1−1/(2p)−r/2(0,T ;W r+1/p(Ω)) 6 cN1, p
3068
p
T
where ~ 1/2 N1 = kFkL p (QT ) + V + k~akW 1−1/p,1/2−1/(2p)(Σ ) + kAk + J, 1−1/p W (0,T ;W (S)) p
p
T
p
J = 0 if p > 3, J = I 1/p if p = 3, and J = T −p/2+1/2k~akL p (ΣT ) if p < 3. We consider the problem with nonzero divergence ~vt − ∆~v + ∇p = ∇ · F, ∇ ·~v = g, ~v t=0 = ~v0 (x), ~v S = ~a. 1−2/p
We assume that p > 2 and ~v0 ∈ Wp
(Ω). We also assume that Z
1/2 g = ∇ ·~h(x, t), ~h ∈ Wp (0, T ; L p(Ω)),
g ∈ L p(QT ),
~ t), (~a −~h) ·~n S = divS A(x,
Z
g dx =
Ω 1/2 1−1/p ~ ∈ Wp (0, T ;Wp A (S)),
~a ·~ndS,
S
∇ ·~v0 = g(x, 0).
In addition, we suppose that ~a and ~v0 agree at t = 0, x ∈ S. We introduce the auxiliary vector field ~u0 = ∇Φ(x, t), where Φ(x, t) is a solution to the Neumann problem ∂Φ = ~a ·~n, ∇2Φ = g, ∂n S i.e., Z
Φ(x, t) =
N(x, y)g(y, t) dy −
Ω
=−
Z
Z
N(x, y)~a ·~ndS
S
∇N(x, y)~h(y, t) dy +
Ω
Z
~ t) dS, ∇SN(x, y)A(y,
S
where N is the Green function for the Neumann problem. Together with the pressure q0 = −~Φt , the vector field ~u0 satisfies the Stokes system ~ut0 − ∆~u0 + ∇g = −∇q,
∇ ·~u = g
and the following inequality: k∇~u0 kL p (QT ) 6 c ( kgkL p(QT ) + k~a ·~nkL k~u0 kW 1/2(0,T ;L p
p (Ω))
1−1/p (S)) p (0,T ;Wp
),
+ k∇ΦkW 1,1/p(Q ) 6 c ( k~hkW 1/2(0,T ;L p
p
T
p (Ω))
~ 1/2 + kAk ). 1−1/p W (0,T ;W (S)) p
p
One can estimate the norm kΦkW 1−1/(2p)−r/2(0,T ;W r (Ω)) under some additional assumptions on ~h. We do not p
p
discuss this in detail. For ~u = ~v −~u0 , q = p − p0 we have the above problem ~ut − ∆~u + ∇g = ∇F + ∇q, ∇ ·~u = 0, ~u =~v0 −~u0 (x, 0), ~u = ~b = ~a −~u0 t=0
S
x∈S
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with ~b ·~n = 0. To conclude the consideration, we briefly discuss the case /’;[p-0 p = 2, ~a = 0. In this case, there is 1,1/2 (QT ) to the problem (1.1) satisfying the integral identity a generalized solution ~v ∈ W2 ZT Z
(−~ηt ·~v(x, t) + ∇~v:∇~η) dx dt = −
0 Ω
ZT Z
F:∇~ηdx dt +
0 Ω
Z
~v0 (x) ·~η(x, 0) dx,
(6.10)
Ω
where ~η is an arbitrary smooth solenoidal vector field vanishing at x ∈ S and t = T , and the inequality k~vkW 1,1/2(Q ) 6 c ( kFkL2(QT ) + k~v0kL2 (Ω)) 2
T
(cf. [10, 1]). To introduce the pressure, it suffices to solve three stationary Stokes problems: ∇F + ∆~v = ∇p1 − ∆~Φ1 , ∇ · ~Φ1 = 0, x ∈ Ω, ~Φ1 S = 0, ~v = ∇P − ∆~Φ2 , ∇ · ~Φ2 = 0, x ∈ Ω, ~Φ2 S = 0, ~v0 = ∇P0 − ∆~Φ0 , ∇ · ~Φ0 = 0, x ∈ Ω, ~Φ0 S = 0. The function p1 satisfies the integral identity Z
(F:∇~ψ + ∇~v:∇~ψ) dx =
Ω
Z
p1 ∇ · ~ψ dx
(6.11)
Ω
◦
◦
for any ~ψ ∈ W 12 (Ω) H(Ω) ≡ H⊥(Ω), where H(Ω) is the space of solenoidal vector fields in W 12 (Ω) and R for the inner product we take ∇~u:∇~vdx. In H⊥ (Ω), we introduce the norm k div~ψkL2(Ω) equivalent to Ω
the norm k~ψkW 1(Ω). Therefore, the functional on the left-hand side of (6.11) can be represented as the R2 ~ inner product div ξ div~ψ dx. We set p1 = div~ξ (cf., for example, [11–14]). It is clear that kp1kL2 (QT ) 6 c ( kFkL2(QT ) + k∇~vkL2(QT )).
From (6.11) it follows that ZT Z
(F:∇~ψ + ∇~v:∇~ψ) dx dt =
0 Ω
ZT Z
p1∇ · ~ψ dx dt
(6.12)
0 Ω
for any smooth function ~ψ(x, t) such that ~ψ S = 0. In the same way, we find ZT Z
~v · ~ψ1 (x, t) dx dt =
Z0 Ω
Z
Ω
Ω
~v0 · ~ψ0 dx =
ZT Z
P(x, t)∇ · ~ψ1 dx dt,
0 Ω
(6.13)
P0 (x)∇ · ~ψ0 dx,
where ~ψ0 ∈ H⊥ (Ω), ~ψ1 ∈ H⊥(Ω) for all t ∈ (0, T ), ~ψ0 S = 0, ~ψ1 x∈S= 0, k∇PkW 1,1/2(Q ) 6 c k~vkW 1,1/2(Q ) , 2
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T
2
T
kP0 kW 1(Ω) 6 c k~v0kL2(Ω). 2
Since ∇ ·~v = 0, ∇ ·~v0 = 0, the functions P(x, t) and P0 (x) harmonic. Let ~ϕ(x, t) be an arbitrary smooth vector-valued function vanishing at x ∈ S and at t = T . We represent it in the form ~ϕ(x, t) = ~η(x, t)+~ψ(x, t), Rt
~η ∈ H(Ω), ~ψ ∈ H⊥ (Ω), ∀t ∈ (0, T ). Let ~ψ(x, t) = ~ψ(x, 0) + ~ψτ (x, τ) dτ. From (6.10)–(6.13) for ~ψ1 = ~ψt 0
and ~ψ0 = ~ψ(x, 0) it follows that ZT Z
(−~ϕt (x, t) ·~v(x, t) + ∇~v:∇~ϕ) dx dt −
0 Ω
Z
+ Ω
ZT Z
p1 (x, t)∇ ·~ϕdx dt +
0 Ω
P0 (x)∇ ·~ϕ(x, 0) dx = −
ZT Z 0 Ω
∇F:∇~ϕdx dt +
ZT Z
P(x, t)∇ ·~ϕt dx dt
0 Ω
Z
~v0 (x) ·~ϕ(x, 0) dx,
Ω
∂P i.e., the function p1 + can be regarded as the pressure in the problem (6.8) and P(x, 0) = P0 (x). It is ∂t clear that the convexity of the domain Ω was not required.
References 1. O. A. Ladyzhenskaya, V. A. Solonnikov, and N. N. Ural0 tseva, Linear and Quasi-linear Equations of Parabolic Type, Nauka, Moscow (1967). [English translation: American Mathematical Society, Providence (1968).] 2. V. A. Solonnikov, “L p -estimates for solutions to the initial-boundary-value problem for the generalized Stokes system in a bounded domain,” Probl. Mat. Anal., 21, 2448–2484 (2000). [English translation: J. Math. Sci., 105, No. 5, (2001).] 3. J. Leray, “Essay sur les movements plans d’un liquide visqueux que limitent des porois,” J. Math. Pures Appl., 14, No. 4, 331–418 (1934). 4. K. K. Golovkin, “The potential theory for nonstationary linear Navier–Stokes equations in the case of three spatial variables,” Tr. Mat. Inst. Steklova, 59, 87–99 (1960). 5. V. A. Solonnikov, “On the theory of nonstationary hydrodynamic potentials,” Proc. Conferen. on the Navier– Stokes equations, 2000. Varenna. 6. M. Giga, Y. Giga, H. Sohr, “L p estimates for the Stokes system,” Functional Analysis and Related Topics,” 1991, Kyoto; Lecture Notes in Math., 1540, 55–67 (1993). 7. M. Wolff, Einf¨uhrung des Drucks f¨ur die instation¨aren Stokes–Gleichungen mittels der Methode von Kaplan [Manuscript]. 8. V. A. Solonnikov, “Estimates for solutions to nonstationary linearized Navier–Stokes equations,” Tr. Mat. Inst. Steklova, 70, 213–317 (1964). 9. V. A. Solonnikov, “Estimates for a solution to an initial-boundary-value problem for the linear nonstationary Navier–Stokes equations,” Zap. Nauchn. Sem. LOMI, 59, 178–254 (1976). 10. O. A. Ladyzhenskaia, The Mathematical Theory of Viscous Incompressible Flow, Nauka, Moscow (1970). [English translation: Gordon and Breach, New York (1969).] 11. V. A. Solonnikov and V. E. Shchadilov, “On a boundary-value problem for a stationary system of Navier– Stokes equations,” Proc. Steklov Inst. Math., 125, 196–210 (1973).
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12. O. A. Ladyzhenskaya and V. A. Solonnikov, “Some problems of vector analysis, and general formulations of boundary-value problems for the Navier–Stokes equations,” Zap. Nauchn. Sem. LOMI, 59, 178–254 (1976). 13. V. A. Solonnikov, “On the Stokes equations in domains with non-smooth boundaries and on viscous incompressible fluid with a free surface,” Nonlinear partial differential equations and their applications. Coll´ege de France Seminars. Vol. III. H. Brezis and J. L. Lions (Eds). Research Notes in Mathematics, 70, 340–423 (1982). 14. V. A. Solonnikov, “Stokes and Navier–Stokes equations in domains with non-compact boundaries,” Nonlinear partial differential equations and their applications. Coll´ege de France Seminars. Vol. IV. H. Brezis and J. L. Lions (Eds.) Research Notes in Mathematics, 84, 240–349 (1983).
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