Galdi G.P. An Introduction to the Mathematical Theory of the Navier-Stokes Equations: Steady-State Problems

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696 X Three-Dimensional Flow in Exterior Domains. Irrotational Case

Ω

(x − y)f

(y)|dy ≤ c (k|x − y|

−2

(x)

kf k

t,B

(x)

+ k|x − y|

−2

,Ω

(x)

kf k

r,Ω

(x)

)

(X.5.42)

and, by Theorem II.6.1,

Ω

(x − y)v

(y)D

(y)(y)|dy ≤ c (k|x − y|

−2

(x)

kv · ∇vk

t,B

(x)

+kv/|x − y|k

2,Ω

(x)

|v|

1,2,Ω

(x)

)

≤ c

(kv · ∇vk

t,B

(x)

+ |v|

1,2,Ω

(x)

(X.5.43)

By the hypothesis on f and Theorem X.1.1,

v ∈ W

2,t

(x)), t > 3

and so, the embedding Theorem II.3.4 yields

v, ∇v ∈ L

∞

(x)). (X.5.4 4)

From (X.5.42)–(X.5.44) we conclude that the integral

Ω

(x − y)F

(y)dy

is absolutely convergent a nd therefore

lim

R→∞

Ω

(x)

(x −y)F

(y)dy =

Ω

(x − y)F

(y)dy. (X.5.45)

Combining (X.5. 40), (X.5.41), and (X.5.4 5) furnishes for a.a. x ∈ Ω

ep(x) = p

−R

Ω

(x − y)F

(y)dy + σ(x) (X.5.46)

with

lim

R→∞

= p

and (X.5 .34) is proved. ut

Employing the same type of arguments used to show the asymptotic for-

mulas (V.3.19), (V.3.20) and (VII.6.18), (VII.6.19) (see also Exercise VII.6.3),

from Theorem X.5.2 we obtain the following result whose proof is left to the

reader as an exercise.

Theorem X.5.3 Let v be a generalized solution to the Navier–Stokes equa-

tions in a domain Ω of class C

, with

v ∈ W

2,r

(Ω

), for some R > δ(Ω

) and r ∈ (1, ∞),

X.5 Asymptotic Behavior: Preliminary Results and Representation Formulas 697

and let p be the asso ciated pressure ﬁeld. Then if f is of bounded suppo rt

and

f ∈ L

(Ω) for some s ∈ (1, ∞),

the following asymptotic representation formulas hold as |x| → ∞. If v

∞

= 0:

(x) = T

(x) + R

Ω

(x − y)v

(y)D

(y)dy + σ

(1)

(x)

(x) = T

(x) −R

Ω

(y)v

(y)D

(x − y)dy + σ

(2)

(x)

p(x) = p

−T

(x) − R

Ω

(x − y)v

(y)D

(y)dy + η(x)

(X.5.47)

where p

∈ R,

= −

∂Ω

(v, p)n

+ R

Ω

= −

∂Ω

(v, p) −Rv

+ R

Ω

(X.5.48)

and, for all |α| ≥ 0,

(k)

(x) = O(|x|

−2−|α|

), k = 1, 2

η(x) = O(|x|

−3−|α|

(X.5.49)

If v

∞

6= 0 (v

∞

= (1, 0, 0)), setting u = v − v

∞

(x) = M

(x) + R

Ω

(x − y)u

(y)D

(y)dy + s

(1)

(x)

(x) = m

(x) − R

Ω

(y)u

(y)D

(x − y)dy + s

(2)

(x)

p(x) = p

− M

∗

(x) − R

Ω

(x − y)u

(y)D

(y)dy + h(x)

(X.5.50)

where p

∈ R

= −

∂Ω

(u, p)n

+ Rδ

+ R

Ω

= −

∂Ω

(u, p) + R(δ

− u

+ R

Ω

∗

= −

∂Ω

(u, p)n

+ R[δ

−δ

]}n

+ R

Ω

(X.5.51)

and, for all |α| ≥ 0, all q ∈ (3/2, ∞] and j = 1, 2,

698 X Three-Dimensional Flow in Exterior Domains. Irrotational Case

(k)

(x) = O(|x|

−(3+|α|)/2

(k)

∈ L

(Ω)

h(x) = O(|x|

−3−|α|

(X.5.52)

Global Summ abi lity of Generalized Solutions when

∞

6= 0

A fundamental step in deriving the asymptotic structure of generalized solu-

tions is to establish “good” summability properties at large distances, that

is, in a domain Ω

, for suﬃciently large R. In this direction, in the three-

dimensional case, the only informati on that we have at the outset is that the

velocity ﬁeld v satisﬁes (v + v

∞

) ∈ D

1,2

(Ω) ∩ L

(Ω).

The objective of the present section is to show that if v

∞

6= 0 and if, for

some q

> 3, it is assumed that

f ∈ L

(Ω), for all q ∈ (1 , q

]

∗

∈ W

2−1/q

(∂Ω)

(X.6.1)

then

v + v

∞

∈ L

(Ω) for all r > 2 (X.6.2)

and, likewise,

∂v

∂x

∈ L

(Ω), for all s > 1

v ∈ D

1,t

(Ω), for all t > 4/3

p ∈ L

(Ω), for all σ > 3/2.

(X.6.3)

Conditions (X.6. 2) and (X.6.3) tell us, in particular, that under the assump-

tion (X. 6.1) on the data, any corresponding generalized solution and associ-

ated pressure ﬁeld have at large distances the same summability properties of

the Oseen fundamental solution E, e, or, what amounts to the same thing, of

the solution of the Oseen problem with the same data f and v

∗

. Moreover,

as in the linearized theory, if f ≡ v

∗

≡ 0 and v

∞

6= 0, we show that

v + v

∞

6∈ L

(Ω), for all r ∈ (1, 2]. (X.6.4)

Taking into account that for our model of liquid the density is a constant,

relation (X.6.4) with r = 2 shows that the kinetic energy in a steady motion

of a liquid in which a body moves with a constant velocity is, in general,

inﬁnite, a fact ﬁrst pointed out by Finn (1960).

p is possibly modiﬁed by addition of a constant.

X.6 Global Summability of Generalized Solutions when v

∞

6= 0 699

In order to show all the above, we introduce the following function space

(Ω) =

(u, φ) ∈L

loc

(Ω) : u ∈ D

2,q

(Ω) ∩ D

4q/(4−q)

(Ω) ∩ L

2q/(2−q)

(Ω) ;

∂u

∂x

∈ L

(Ω); φ ∈ D

1,q

(Ω) ∩L

3q/(3−q)

(Ω)

, q ∈ (1, 2) .

(X.6.5)

It is readily checked that X

becomes a Banach space when endowed with the

“natural” norm

k(u, φ)k

:= kuk

2q/(2−q)

+|u|

1,4q/(4−q)



∂u

∂x



+|u|

+kφk

3q/(3−q)

+|φ|

1,q

(X.6.6)

The following result holds

Lemma X.6.1 Let Ω be a C

-smooth exterior three-dimensional domain,

and assume, for some q ∈ (1, 2), that

f ∈ L

(Ω) ∩ L

3/2

(Ω) , v

∗

∈ W

2−1/q,q

(∂Ω) ∩ W

4/3,3/2

(∂Ω) . (X.6.7)

Then, every generalized solution v to the Navier–Stokes problem correspond-

ing to f , v

∗

and to v

∞

6= 0, a nd the associated pressure ﬁeld p

satisfy

(v + v

∞

, p) ∈ X

(Ω).

Proof. Without loss, we a ssume v

∞

= e

. Also, since the actual value of

R is irrelevant in the proof, we set R = 1, for simplicity. Recalling that

v ∈ D

1,2

(Ω), we m ay ﬁnd a sequence of second-order tensors {G

} with

components in C

∞

(Ω) such that G

→ ∇v in L

(Ω). Consider now the

problem

∆u +

∂u

∂x

= u · A

+ ∇φ + F

∇ · u = 0











in Ω

u = u

∗

:= v

∗

+ e

at ∂Ω ,

(X.6.8)

where A

:= ∇v −G

, F

:= v ·G

+ f , while the i nteger k will be speciﬁed

successively. Clearly, (v + e

, p) is a solution to (X.6.8), f or all k ∈ N. Our

plan is to show the existence of a solutio n to (X.6.8) in the class X

(Ω),

and then prove that such a solution coincides with (v + e

, p). To this end,

we begi n to observe tha t, in view of the assumption on f and the fact that

(v + e

) ∈ L

(Ω), it follows that F

∈ L

(Ω) ∩ L

3/2

(Ω), for all k ∈ N. We

now set X

q,3/2

(Ω) = X

(Ω) ∩ X

3/2

(Ω), endowed with the norm k ·k

q,3/2

k· k

+ k ·k

3/2

, and consider the map

M : (w, τ) ∈ X

q,3/2

(Ω) → (u, φ) := M(w, τ)

where (u, φ) satisﬁes:

Possibly modiﬁed by the addition of a constant.

700 X Three-Dimensional Flow in Exterior Domains. Irrotational Case

∆u +

∂u

∂x

= w · A

+ ∇φ + F

∇ · u = 0











in Ω

u = u

∗

at ∂Ω .

(X.6.9)

Observing that, by the H¨ol der inequality and the fact that v ∈ D

1,2

(Ω),

kw · A

≤ kwk

2s/(2−s)

< ∞, s ∈ (1, 2) , (X.6.10)

with the help of Theorem VII.7.1 we deduce, on the one hand, that the map

M is well-deﬁned and, on the other hand, that there exists a (unique) solution

(u, φ) ∈ X

q,3/2

(Ω) to (X.6.9) obeying the estimate (with q

= q, q

= 3/2)

k(u, φ)k

q,3/2

≤ c

i=1



kwk

/(2−q

)

+ kF

+ ku

∗

2−2/q

(∂Ω)



≤ c

k(w, τ)k

q,3/2

i=1



+ ku

∗

2−2/q

(∂Ω)



(X.6.11)

Thus, i f we choose k such that

≤ 1/(2c) (X.6.12)

and deﬁne

δ := 2c

i=1



+ ku

∗

2−2/q

(∂Ω)



from (X.6.11 ) it follows at once that M maps the closed ball { (u, φ) ∈

q,3/2

(Ω) : k(u, φ)k

q,3/2

≤ δ} into itself. Moreover, in view of the linear-

ity of the map M and of (X.6.12 ), from (X.6 .11) with kF

= 0, i = 1, 2,

we deduce that M is a contraction and, therefore, there exists one and only

one (u, φ) ∈ X

q,3/2

solution to (X.6.8). Next, set (w, τ) := (u −v −e

, φ −p).

We thus ﬁnd

∆w +

∂w

∂x

= w ·A

+ ∇τ

∇ · w = 0











in Ω

w = 0 at ∂Ω .

(X.6.13)

Recalling that, by the assumption on v and (X.6.10), it i s g := w · A

∈

3/2

(Ω), with the help of T heorem VII. 7.1 we inf er that the problem

∆

w +

∂

∂x

= g + ∇eτ

∇ ·

w = 0











in Ω

w = 0 at ∂Ω .

(X.6.14)

X.6 Global Summability of Generalized Solutions when v

∞

6= 0 701

has a (unique) solution (

w, eτ ) in the cla ss X

3/2

(Ω). We claim that (

w, eτ) =

(w, τ ). In fact, the ﬁelds z :=

w − w and χ := eτ − τ solve the homogeneous

Oseen problem (X.6.14) with g ≡ 0. Furthermore, z ∈ L

(Ω), because

w, u ∈

3/2

(Ω), while v + e

∈ L

(Ω) by assumption. Therefore, from Theorem

VII.6.2 and Exercise VII.6.2 we obtain z ≡ ∇χ ≡ 0. Consequently, w ∈

3/2

(Ω), and so, again by Theorem VII.7.1 applied to (X. 6.13), we obtain

k(w, φ)k

3/2

≤ c kw · A

3/2

≤ c kA

k(w, φ)k

3/2

Thus, by using (X.6.12) into this l atter inequality, we deduce w ≡ ∇τ ≡ 0,

that is (u, φ) = (v + e

, p + C), for some C ∈ R, and the proof of the lemma

is complete.

Combining the result of the previous lemma with those of Theorem X.5.1

we prove the following.

Theorem X.6.4 Let Ω be a C

-smooth, exterior three-dimensional domain

and assume that

f ∈ L

(Ω), v

∗

∈ W

2−1/q

,1/q

(∂Ω), v

∞

6= 0,

for some q

> 3, and all q ∈ (1, q

]. Then every corresponding g eneralized

solution v to the Navier–Stokes problem (X.0.8), (X.0.4) sati sﬁes the following

summability properties

(v + v

∞

) ∈ L

(Ω) ,

∂v

∂x

∈ L

(Ω) , v ∈ D

1,t

(Ω) , p ∈ L

(Ω),

for a ll r ∈ (2, ∞], s ∈ (1 , ∞], t ∈ (4/3, ∞] and σ ∈ (3/2, ∞] where p is (up to

a constant) the pressure ﬁeld associated to v by Lemma X.1.1. If i n addition,

Ω is of class C

, and f ∈ W

1,q

(Ω), v

∗

∈ W

3−1/q

,1/q

(∂Ω), then we have

also

v ∈ D

2,τ

(Ω) , p ∈ D

1,τ

(Ω) , (X.6.15)

for a ll τ ∈ (1, ∞].

Proof. The stated summability properties, in the domain Ω

, follow at once

from Lemma X.6.1 and Theorem X.5.1. On the other hand, in the domain

Ω

, they are a consequence of Theorem IV.5.1. ut

Remark X.6.2 It is worth emphasizing that T heorem X.6.4 does not require

the vanishing of the ﬂux of v

∗

through the boundary ∂Ω. 

Remark X.6.3 Summabi lity properties at large distances for higher order

derivatives can be likewise obtained by using Theorem VII.7.1, with f re-

placed by f + (v + v

∞

) · ∇v. To show this, let us assume, fo r simplicity, f

of bounded support i n Ω. Observing that, by Theorem X.6.4 and by (X.6.15)

for suﬃciently large R > δ(Ω

), we have

702 X Three-Dimensional Flow in Exterior Domains. Irrotational Case

N ≡ (v + v

∞

) · ∇v ∈ W

1,τ

(Ω

from Theorem VII.7.1, it follows that

v ∈ D

3,τ

(Ω

), p ∈ D

2,τ

(Ω

for a ll τ > 1. Then

N ∈ W

2,τ

(Ω)

for a ll τ > 1, and so on. Therefore, we can conclude, by iteration,

v ∈ D

m+2,τ

(Ω

), p ∈ D

m+1,τ

(Ω

for a ll m ≥ 0 and all τ > 1. 

The remaining part of this section is devoted to investigate the ﬁniteness

of the kinetic energy of the liquid. To this end, we begin to observe that, from

Theorem X.5.3, it follows that for f ∈ L

(Ω), s > 1, of bo unded support in

Ω, the ﬁeld u = v + v

∞

= e

) admits the following representation:

(x) = E

(x)m

−R

Ω

(y)u

(y)D

(x − y)dy + s

(x) (X.6.16)

where

= −

∂Ω

(u, p) + R(δ

− u

)] n

+ R

Ω

(X.6.17)

and

s ∈ L

(Ω

), for a ll τ > 3/2, (X.6.18)

see Exercise VII.6.3. Observing that, by (VII.3.21) and (VII.3.33),

∇E ∈ L

) f or all r ∈ (4/3, 3/2),

and that, by Theorem X.6.4,

∈ L

(Ω) for all s > 1,

from Young’s theorem on convolutions it follows that

Ω

(y)u

(y)D

(x −y)dy ∈ L

(Ω), for all t > 4/3. (X.6.19)

In view of (X.6.16)–(X.6.19), we then conclude that, for suﬃciently large R,

u ∈ L

(Ω

), q ∈ (3/2, 2] (X.6.20)

if and only if

∈ L

(Ω

), q ∈ (3/2, 2]. (X.6.21)

X.6 Global Summability of Generalized Solutions when v

∞

6= 0 703

Consider now the quadratic form

Q ≡ E

(x)E

(x)m

Starting from (VII.3 .20) a nd using the symmetry properties of the tensor ﬁeld

(x), it is not hard to show that, for any ρ > 0,

∂B

(x)E

(x) = 0, j 6= k,

and so, the i ntegrability of Q is reduced to that of

= m

+ m

+ (m

+ m

+(m

+ m

+ (m

+ m

(X.6.22)

However, as we know from (VII.3.30), for m

6≡ 0, no term in the sum (X.6.2 2)

is i ntegrable over Ω

and so (X. 6.21) with q = 2 can not hold unless m

≡ 0.

In fact, we can say more. Actually, since E(x) tends to zero as |x| → ∞,

(X.6.21) cannot hold for any of the speciﬁed values of q, unless m

≡ 0. We

thus conclude that property (X.6 .20) can hold if and only if some restrictions

are imposed on the motion itself and which are described by the conditi ons

≡ −

∂Ω

(u, p) + R(δ

−u

)] n

+ R

Ω

= 0, i = 1, 2, 3.

(X.6.23)

From the physical point of view, (X.6.23) means that there is no net external

force applied to the “body” Ω

. This circumstance occurs in the case of steady

ﬂow around a body which, for instance, propels itself either by ma intaining a

mom entum ﬂux across portion of its boundary or by moving tangentially por-

tions of its boundary (as by belts). However, the existence theory related to

problems of this kind may be completely diﬀerent than that developed so far

for the “classical” problem (X.0.3)–(X.0.4), in that the solution must obey the

extra conditions expressed by (X.6.23). As a consequence, one has to intro duce

another unknown into the problem which, as suggested by physics, can be ei-

ther the velocity at the boundary, or the (nonzero) velocity at inﬁnity. This

type of questions has been considered by several authors. Among others, we

refer to Sennitskii (1978, 1984) for ﬂow around symmetric self-propelled bod-

ies, and to Galdi (1999a, 2002) for a general exi stence and uniqueness theory.

The asymptotic behavior of velocity and pressure ﬁelds has been investigated

in full detail by Pukhnacev (1989).

Another worth of mentioning circumstance w here (X.6.23) occurs is the

case when Ω = R

and f has zero average on Ω. In such a situation we

thus obtain, in particular, that the kinetic energy of the liquid is ﬁnite. For

this type of problems we refer to the papers of Bjorland & Schonbek (2009),

Bjorland, Brandolese, Iftimie & Schonbek (2011), and Sil vestre (2009), this

latter also considering a more general choice of v

∞

704 X Three-Dimensional Flow in Exterior Domains. Irrotational Case

However, there is also a very signiﬁcant case where (X.6.23) can not hold

and, as a consequence, the total kinetic energy of the liquid is inﬁnite. Specif-

ically, consider the situation when Ω is exterior to just one compact body B

(say)

and that v

∗

≡ f ≡ 0. Physically, this means that B is steadily moving

into the liquid with velocity v

∞

. In such a case (X.6 .23) is equivalent to

∂Ω

T (u, p) ·n = 0. (X.6.24)

On the other hand, in Theorem X.7.1 of the following section it will be proved

that v and p obey the energy equation

Ω

∇v : ∇v = v

∞

∂Ω

T ( u, p) ·n , (X.6.25)

and , being v

∞

6= 0, from (X.6.24) and (X.6.25) it follows v ≡ 0 in Ω, which

gives an absurd conclusion. Therefore, (X.6.24) cannot ho ld and, consequently,

u (≡ v+v

∞

) cannot satisfy (X.6.20) for any o f the speciﬁed values of q, which

proves, in particular, that the kinetic energy of the liquid is inﬁnite.

The above considerations are collected in the following.

Theorem X.6.5 Let v be a generalized solution to the Navier–Stokes prob-

lem (X.0.8), (X.0.4) in a three-dimensional exterior domain of class C

with

∞

6= 0 (v

∞

= e

). Assume that, for some r, s > 1

v ∈ W

2,r

loc

(Ω) f ∈ L

(Ω),

with f of bounded support and let Ω

⊃ supp (f ). Then,

u ≡ v + v

∞

∈ L

(Ω

), for some q ∈ (3/2, 2] (X.6.26)

if and only if

≡ −

∂Ω

(u, p) + R(δ

− u

)] n

+ R

Ω

= 0, i = 1, 2, 3.

Moreover, if f ≡ v

∗

≡ 0, it follows that m

6≡ 0, and therefore (X.6.26) can

not hold. Thus, in particular, under these latter conditions on the data, the

total kinetic energy of the liquid is inﬁnite.

Exercise X.6.1 U nder the assumptions of Theorem X. 6.5, prove that if

v + v

∞

∈ L

(Ω), for some q ∈ (1, 2], (X. 6.27)

then m

≡ 0. Thus, if f ≡ v

∗

≡ 0, (X.6.27) can not hold.

Exercise X.6.2 Let the assumptions of Theorem X.6.5 be satisﬁed. Suppose, also,

that f ≡ 0, v

∗

≡ v

= const. Show that (X.6.27) holds if and only if v ≡ v

≡ e

This restriction is, of course, unnecessary. As assumed so far, Ω can be exterior

to s ≥ 1 compact bodies.

X.7 Energy Equation and Uniqueness when v

∞

6= 0 705

X.7 The Energy Equation and Uniqueness for

Generalized Soluti ons when v

∞

6= 0

Our objective here is two-fol d. On one hand, we show that, under suitable

regularity assumption on the data, weak solutions corresponding to v

∞

6= 0

satisfy the energy equation, and, on the other hand, that if the data are

“suﬃciently small”, they are uni que in their own class.

We begin to give a suﬃcient condition for the validity of the energy equal-

ity.

Theorem X.7.1 Let Ω be a C

-smooth exterior three-dimensional domai n,

and let v be a generalized solution to the Navier–Stokes problem (X.0.8),

(X.0.4) corresponding to the data

f ∈ L

4/3

(Ω) ∩ L

3/2

(Ω) , v

∗

∈ W

4/3,3/2

(∂Ω) , v

∞

6= 0 . (X.7.1 )

Then v veriﬁes the energy equation (X.2.29), where p is the pressure ﬁeld

associated to v by Lemma X.1.1.

Proof. Under the assumption (X.7.1), from Lemm a X.6.1 we have, in partic-

ular,

v + v

∞

∈ L

(Ω), v

∞

· ∇v ∈ L

4/3

(Ω),

so that (X.2.11 ) ho lds with q = 4. Thus, the result follows from Remark X.2.6

and Exercise X.2.2. ut

An im portant corollary to Theorem X.7.1 is the following result of Liouville-

type.

Theorem X.7.2 Let v be a generalized solution to the Navier–Stokes prob-

lem (X.0.8), (X.0.4) in R

corresponding to f ≡ 0 and v

∞

6= 0. Then

v(x) = −v

∞

for all x ∈ R

Remark X.7.1 Theorem X.7.2 necessitates the condition v

∞

6= 0. It is not

known if the result continues to be valid when v

∞

= 0. In this respect,

cf. Remark X. 9.4. 

We shal l now investigate the uniqueness problem. To this end, we begin

to show that if f (at large distances) and v

∗

have some property in addition

to those required i n Theorem X.7.1, and if the da ta are suitably “small,” the

corresponding generalized solution sati sﬁes conditions (X.3. 3) and (X.3.5 ).

Lemma X.7.1 Let the assumptions of Theorem X.7.1 be satisﬁed and as-

sume, in addition, that

f ∈ L

6/5

(Ω)

Rkfk

6/5

+ kv

∗

+ v

∞

7/6,6/5(∂Ω)

min

(

√

)

(X.7.2)