Tohoku Math. J.
58 (2005), 129–147
AN L
q
-ANALYSIS OF VISCOUS FLUID FLOW
PAST A ROTATING OBSTACLE
Dedicated to Professor Hermann Sohr on his sixty-fifth birthday
REINHARD FARWIG
(Received March 15, 2004)
Abstract. Consider the problem of time-periodic strong solutions of the Stokes and
Navier-Stokes system modelling viscous incompressible fluid flow past or around arotating
obstacle in Euclidean three-space. Introducing a rotating coordinate system attached to the
body, a linearization yields a system of partial differential equations of second order involving
an angular derivative not subordinate to the Laplacian. In this paper we find an explicit solution
for the linear whole space problem when the axis of rotation is parallel to the velocity of the
fluid at infinity. For the analysis of this solution in L
q
-spaces, 1 <q<∞, we will use tools
from harmonic analysis and a special maximal operator reflecting paths of fluid particles past
or around the obstacle.
1. Introduction. In recent years the analysis of the Navier-Stokes equations and of
models of non-Newtonian fluids describing the flow around or past a rotating body has at-
tracted much attention. Here we consider the Navier-Stokes equations modelling viscous
flow either past a rotating body K in Euclidean 3-space R
3
with axis of rotation ω =˜ωe
3
=
˜ω(0, 0, 1)
T
, ˜ω = 0, and with velocity u
∞
= ke
3
= 0 at infinity or around a rotating body K
which is moving in the direction of its axis of rotation. In each case a coordinate transform
and a linearization yield the system of partial differential equations
u
t
− νu + k∂
3
u −(ω ∧ x) ·∇u + ω ∧ u +∇p =f,
div u =0
(1.1)
in a time-independent exterior domain Ω = R
3
\K together with the initial-boundary condi-
tion
u(x, t) = ω ∧ x −u
∞
,u(x,0) = u
0
,u→ 0as |x|→∞.
Here u = (u
1
,u
2
,u
3
)
T
and p denote the velocity and pressure of the fluid, resp., f is a given
external force, and ν>0 is the constant coefficient of viscosity. In the stationary case to
be analyzed in this paper, we are led to an elliptic equation in the sense of Agmon-Douglis-
Nirenberg in which the term (ω ∧ x) ·∇u is not subordinate to −νu in the exterior domain
Ω. Note that a stationary solution (u, p) of (1.1) will lead to a time-periodic solution of the
original linearized problem.
2000 Mathematics Subject Classification. Primary 76D05; Secondary 35C15, 35Q35, 76D99, 76U05.
Key words and phrases. Littlewood-Paley theory, maximal operators, Oseen flow, rotating obstacles, singular
integral operator, Stokes flow.
130 R. FARWIG
To be more precise, consider the Navier-Stokes equations
v
t
− νv + v ·∇v +∇q =
˜
f in Ω(t), t > 0 ,
div v =0inΩ(t), t > 0 ,
v(y, t) =ω ∧ y on ∂Ω(t), t > 0 ,
v(y, t) → u
∞
=0as|y|→∞
(1.2)
with an initial value v(y, 0) = v
0
(y) and ω =˜ωe
3
= 0 in the time-dependent exterior domain
Ω(t) = O
ω
(t)Ω ,
where O
ω
(t) denotes the orthogonal matrix
O
ω
(t) =
cos ˜ωt −sin ˜ωt 0
sin ˜ωt cos ˜ωt 0
001
.
Then, introducing
x = O
T
ω
(t)y , u(x, t) = O
T
ω
(t)(v(y, t) − u
∞
), p(x,t) = q(y,t),(1.3)
(u, p) will satisfy the modified Navier-Stokes system
u
t
− νu + u ·∇u +(O
T
ω
(t)u
∞
) ·∇u
− (ω ∧ x) ·∇u +ω ∧ u +∇p = f in Ω × (0, ∞),
div u = 0inΩ × (0, ∞),
u(x, t) = ω ∧ x − O
T
ω
(t)u
∞
on ∂Ω × (0, ∞),
u(x, t) → 0as|x|→∞.
(1.4)
For details of the elementary calculation, see [9] when u
∞
= 0; for u
∞
= 0 the additional
term u
∞
·∇
y
u(O
T
ω
(t)y, t) = (O
T
ω
(t)u
∞
)·∇
x
u will appear. In the case, u
∞
ω,sayu
∞
= ke
3
,
to be considered here, O
T
ω
(t)u
∞
= ke
3
for all t>0. Thus (1.4) will lead to the system
u
t
− νu + u ·∇u + k∂
3
u
− (ω ∧ x) ·∇u + ω ∧ u +∇p = f in Ω,
div u = 0inΩ,
u = ω ∧ x −ke
3
on ∂Ω ,
u →0as|x|→∞,
(1.5)
a stationary solution of which corresponds to a time-periodic solution of the original system
(1.2). However, if u
∞
is not parallel to the axis of rotation e
3
,thetermO
T
ω
(t)u
∞
depends on
t. Therefore, in this paper, we will study the linearized and stationary version of (1.4) only
when u
∞
is parallel to ω.
Finally, we may consider the problem of a rotating body with axis of rotation ω and
with an additional translational velocity −u
∞
. In this case, Ω(t) = O
ω
(t)Ω − u
∞
t and
v(y, t) → 0as|y|→∞in (1.1). Then the transformation
x = O
T
ω
(t)(y +u
∞
t), u(x,t) = O
T
ω
(t)v(y, t), p(x, t) = q(y,t)
L
q
-ANALYSIS OF VISCOUS FLUID FLOW 131
again will lead to (1.4) (observer invariance) with the same fundamental difference between
the cases u
∞
ω and u
∞
∧ ω = 0.
The mathematical analysis of viscous flow past or around rotating obstacles started with
[1], where weak instationary solutions have been constructed in an even more general setting
allowing for time-dependent functions ω(t) and u
∞
(t). Decay results for this problem in the
whole space are discussed in [2]. Using semigroup theory, local mild and unique solutions
are constructed in [9], [10] when u
∞
= 0; since the corresponding semigroup is strongly
continuous, but not analytic, it is not clear whether the mild solution is a strong one. A
different approach in homogeneous Besov spaces is used in [13], where the term (ω ∧x) ·∇u
has been replaced by the more general term (Mx) ·∇u with an arbitrary traceless 3×3-matrix
M; here a local classical and unique solution is found for nondecaying initial data. Several
linear and stationary auxiliary problems in the whole space and in exterior domains have
been analyzed in [11]; to some extent the results are generalized to the nonlinear case and
the problem including the term (ω ∧ x) ·∇u in [12]. Several advanced aprioriestimates
of stationary and instationary solutions can be found in [7], including even non-Newtonian
fluids; in particular L
2
-estimates for (1.1) are established. Pointwise estimates yielding decay
rates such as |v(x)|≤c(1 +|x|)
−1
are obtained in [8] for the stationary nonlinear problem
when u
∞
= 0. With regard to further developments, e.g., to the discussion of stability, L
q
-
estimates, 1 <q<∞, are presented in [3] for the linearized whole space problem (1.1)
when u
∞
= ke
3
= 0. For the physical background and for applications to the free fall
of particles in fluids, see [7] and references therein. In [17] the time-dependent fundamental
solution (Green’s function) Γ(z,y;t) is calculated for the case u
∞
= 0, and several pointwise
estimates are given for t → 0,t →∞and for small and large spatial data z, y.
The main results of this paper are the following.
T
HEOREM 1.1. (1) Let 1 <q<∞, f ∈ L
q
(R
3
)
3
and g ∈ W
1,q
(R
3
) such that even
|(x
1
,x
2
)|g ∈ L
q
(R
3
).Furthermore, let ν>0, k ∈ R and ω = (0, 0, ˜ω)
T
∈ R
3
\{0}. Then the
linear problem in R
3
,
−νu + k∂
3
u −(ω ∧ x) ·∇u + ω ∧ u +∇p = f, div u = g ,(1.6)
has a solution (u, p) ∈
ˆ
W
2,q
(R
3
)
3
×
ˆ
W
1,q
(R
3
) satisfying the a priori estimates
ν∇
2
u
q
+∇p
q
≤ c(f
q
+ν∇g + (ω ∧ x)g − kge
3
q
),(1.7)
k∂
3
u
q
+(ω ∧ x) ·∇u − ω ∧ u
q
(1.8)
≤ c
1 +
k
4
ν
2
|ω|
2
(f
q
+ν∇g + (ω ∧ x)g − kge
3
q
)
with a constant c>0 independent of ν,k and ω.
(2) In addition to the assumptions in (1) and given a solution (u, p) ∈
ˆ
W
2,q
(R
3
)
3
×
ˆ
W
1,q
(R
3
) of (1.6), suppose that f ∈ L
r
(R
3
)
3
, g ∈ W
1,r
(R
3
), |(x
1
,x
2
)|g ∈ L
r
(R
3
), and let
(u
1
,p
1
) ∈
ˆ
W
2,r
(R
3
)
3
×
ˆ
W
1,r
(R
3
) be another solution of (1.6).Thenp − p
1
is constant and
u − u
1
equals αe
3
+ βω ∧ x, α, β ∈ R.
132 R. FARWIG
COROLLARY 1.2. (1) Let 1 <q<4, f ∈ L
q
(R
3
)
3
and g ∈ W
1,q
(R
3
) such that
|(x
1
,x
2
)|g ∈ L
q
(R
3
), and let (u, p) ∈
ˆ
W
2,q
(R
3
)
3
×
ˆ
W
1,q
(R
3
) be the solution of (1.6).Then
there exists β ∈ R such that
∇
(u − βω ∧ x) ∈ L
r
(R
3
)
6
for all r>1,
1
r
∈
1
q
−
1
4
,
1
3
.
Moreover,
∇
(u − βω ∧ x)
r
≤ C(f
q
+ν∇g + (ω ∧ x)g − kge
3
q
)
with a constant C = C(ν,k,ω;r) > 0.
(2) In (1) assume that even 1 <q<2. Then there exist α, β ∈ R such that
u − βω ∧ x − αe
3
∈ L
s
(R
3
)
3
for all s>1,
1
s
∈
1
q
−
1
2
,
2
3
.
Moreover,
u −βω ∧ x − αe
3
s
≤ C(f
q
+ν∇g + (ω ∧ x)g − kge
3
q
)
with a constant C = C(ν,k,ω;s) > 0.
R
EMARK 1.3. (1) In Theorem 1.1 fix f and g and let (u
ν,k,ω
,p
ν,k,ω
) denote a solu-
tion of (1.6) for ν>0, k = 0andω =˜ωe
3
, ˜ω = 0. Furthermore, let ν
0
> 0,k
0
∈ R and
˜ω
0
=˜ω
0
e
3
, ˜ω
0
∈ R.Then
u
ν,k,ω
u
ν
0
,k
0
,ω
0
in
ˆ
W
2,q
(R
3
)
3
,p
ν,k,ω
p
ν
0
,k
0
,ω
0
in
ˆ
W
1,q
(R
3
)
weakly as (ν,k,ω) → (ν
0
,k
0
,ω
0
),where(u
ν
0
,k
0
,ω
0
,p
ν
0
,k
0
,ω
0
) solves (1.6) with ν
0
replacing
ν, k
0
replacing k and ω
0
replacing ω. This result extends to the case of f, g depending on
ν, k, ω such that f
ν,k,ω
f
ν
0
,k
0
,ω
0
and g
ν,k,ω
g
ν
0
,k
0
,ω
0
in suitable weak topologies.
(2) Compared to the case k = 0 considered in [3], the results in Theorem 1.1 are
stronger. The uniqueness assertion does not allow for a term γ(x
1
,x
2
, −2x
3
)
T
,γ ∈ R, as in
[3] due to the term k∂
3
u.
(3) In (1.6) it is not possible to estimate the terms (ω ∧x) ·∇u and ω ∧u separately in
L
q
unless f and g satisfy an infinite set of compatibility conditions. The argument is based
on the simple identity
(ω ∧ x) ·∇u − ω ∧ u =˜ωO
e
3
(θ)∂
θ
(O
T
e
3
(θ)u) ;
for more details see Remark 2.3, Proposition 2.4 in [3] when k = 0.
(4) The fundamental solution of (1.5) which will be computed “explicitly" in Section
2 below will not lead to a classical Calderón-Zygmund integral operator, when considering
u in terms of f (and g ). See Section 2 in [3] for more details when k = 0; in this case the
fundamental solution has a slightly simpler form.
(5) It is not evident that for the solution u of (1.6) both lower order terms k∂
3
u and
(ω ∧ x) ·∇u − ω ∧ u can be estimated in L
q
-norms from the right-hand side. On the other
hand, it is remarkable that the aprioriestimate (1.8) depends on (k/
√
ν|ω|)
4
. The proof
in Section 2 using Marcinkiewicz’ multiplier theorem implies that for q = 2theterm
L
q
-ANALYSIS OF VISCOUS FLUID FLOW 133
C(1 + (k/
√
ν|ω|)
2
) will suffice. Thus complex interpolation will slightly improve (1.8).
Finally, an explicit example will indicate that the term C(1 + (k/
√
ν|ω|)
2
) is optimal, see
the end of Section 2.
In this paper we use standard notation for Lebesgue spaces and Sobolev spaces, namely
L
q
(Ω) and W
k,q
(Ω), 1 ≤ q ≤∞, for bounded and unbounded domains Ω ⊂ R
3
.To
control problems also in unbounded domains we need the space L
q
loc
(
¯
Ω) of functions which
are L
q
-integrable on every compact subset of
¯
Ω and homogeneous Sobolev spaces
ˆ
W
k,q
(Ω) ={u ∈ L
1
loc
(
¯
Ω)/Π
k−1
; ∂
α
u ∈ L
q
(Ω) for all α ∈ N
n
0
, |α|=k},
where ∂
α
= ∂
α
1
1
·····∂
α
n
n
for a multi-index α = (α
1
,...,α
n
) ∈ N
n
0
and Π
k−1
denotes the
set of all polynomials on R
n
of degree ≤ k − 1. The space
ˆ
W
k,q
(Ω) consists of equivalence
classes of L
1
loc
-functions being unique only up to elements from Π
k−1
and is equipped with
the norm
|α|=k
∂
α
u
q
.Since
ˆ
W
k,q
(Ω) can be considered as a closed subspace of L
q
(Ω)
N
for some N = N(k,n) ∈ N, it is reflexive and separable for every q ∈ (1, ∞).Formore
details on these spaces see Chapter II in [6], Chapter III in [14] and also [4], [5]. However,
sometimes being less careful, we will consider v ∈
ˆ
W
k,q
(Ω) as a function (representative)
rather than an equivalence class of functions, i.e., v ∈ L
1
loc
(Ω) such that ∂
α
v ∈ L
q
(Ω) for
every multi-index α with |α|=k.
The Fourier transform on R
3
of a function or distribution u is denoted by F u =ˆu, i.e.,
formally
ˆu(ξ) = (2π)
−3/2
R
3
e
−ix·ξ
u(x)dx , ξ ∈ R
3
.
The Fourier transform and its inverse F
−1
will be needed in particular on Schwartz’s space
S(R
3
) of rapidly decreasing functions and on its dual space S
(R
3
) of tempered distributions.
Furthermore, D(Ω) = C
∞
0
(Ω), and D
(Ω) denotes the set of all distributions on Ω. The
application of a distribution T on a test function u (or of a functional T ∈ X
in a dual space
X
of a given Banach space X on an element u ∈ X) is denoted by T,u.Givenq ∈ (1, ∞),
let
ˆ
W
−1,q
(Ω) be the dual space of
ˆ
W
1,q
(Ω), q
= q/(q −1).
Finally B
r
(y) ={x ∈ R
3
;|x − y| <r},r>0, denotes a ball in R
3
with respect to the
Euclidean norm |·|;moreover, B
r
= B
r
(0) and B
c
r
= R
3
\B
r
. The vector y = (y
1
,y
2
) ∈ R
2
rotated through +π/2 is denoted by y
⊥
= (−y
2
,y
1
). If y = (y
1
,y
2
,y
3
) ∈ R
3
, then y
=
(y
1
,y
2
), and ∇
= (∂
1
,∂
2
) is the corresponding partial gradient on R
3
. As usual, c denotes a
generic positive constant which may change its value from line to line.
2. The whole space problem. To solve the whole space problem (1.6), i.e.,
−νu + k∂
3
u − (ω ∧ x) ·∇u + ω ∧ u +∇p = f, divu = g in R
3
,(2.1)
we eliminate the pressure term. Applying div to (2.1)
1
, p is seen to be a weak solution of the
equation
p = divf + νg + (ω ∧ x) ·∇g − k∂
3
g ,