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

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

where, in the last inequality, we have used Lemma I I.9.2. The desired estimate

for the pressure then follows from (X.9.15), (X.9.16), which completes the

proof of the lemma. ut

Collecting the results of Lemma X.9.1 and L emma X.9.2, we prove the

following result that provides, at least for “small” data, the asymptotic be-

havior of a weak solution corresponding to v

∞

= 0 and verifying the energy

inequality.

Theorem X.9.1 Let Ω be as in Lemma X.9. 1 and let v be a generalized

solution to the Navier–Stokes problem (X.0.8), (X.0.4) corresponding to v

∗

≡

∞

≡ 0 and f = ∇ · F

with

(1 + |x|

)F ∈ L

∞

(Ω).

Assume, further, that v obeys the energy inequality

|v|

1,2

≤ R(F , ∇v).

Then:

(i) If

[|F |]

min{A, 1/2B} (X.9.17)

with A and B given in (X.9.1) and (X.9.2), respectively, we have

v ∈ D

1,q

(Ω), p ∈ L

(Ω) for each q > 3/2 ,

(1 + |x|)v ∈ L

∞

(Ω).

(ii) If, in addition to (X.9.17), f is of bounded support then

v(x) = O(|x|

−|α|−1

) ,

p(x) = O(|x|

−|α|−2

)

(X.9.18)

for a ll |α| ∈ N .

Proof. In view of (X.9. 17), by Lemma X. 9.1 we can construct a generalized

solution w (say) tha t, together wi th the associate pressure ﬁeld π, satisﬁes

w ∈ D

1,q

(Ω), π ∈ L

(Ω) for each q > 3/2,

[|w|]

< ∞,

and

[|w|]

+ |w|

1,q

+ kπk

≤ BR[|F |]

However, again by (X.9.17), with the help of the uniqueness Theorem X.3.2

and Lemma X.1.1, we conclude that v ≡ w, p ≡ π and the ﬁrst part of the

theorem follows. The second part is an obvious consequence of Lemma X.9.2.

In the weak sense.

X.9 On the Asymptotic Structure of Generalized Solutions when v

∞

= 0 727

Remark X.9.2 By the same a rgument employed in the proof of Theorem

X.9. 1, in view of Remark X.9.1 we can show that in dimension n ≥ 4 ev-

ery generalized solution satisfying the energy inequality and corresponding to

“small” f, behaves at large distances as 1/|x|. 

Let us p oint out two important consequences of Theorem X.9.1. Observing

that the generalized solution v of Theorem X.9.1 satisﬁes

v ∈ L

(Ω),

from Theorem X.2.1 we at once o bta in the following theorem.

Theorem X.9.2 Let the assumptions of Theorem X.9.1(i) be satisﬁed. Then

v obeys the energy equation

|v|

1,2

= R(F , ∇v).

Likewise, from Theorem X.3.2, it follows.

Theorem X.9.3 Let the assumptions of Theorem X.9.1(i) be veriﬁed. Then

v is the only generalized solution corresponding to f and satisfying the energy

inequality.

The results of Theorem X.9.1(i) show that every generalized soluti on obey-

ing the energy inequality and corresponding to “small” data of bounded sup-

port has the same asymptotic b ehavior as the Stokes fundamental solution

(U, e). It is then natural to wonder whether, in analogy with the case v

∞

6= 0

(see (X.8.17)), the velocity ﬁeld v admits an asymptotic expansion of the type

v(x) = α · U (x) + O(1/|x|

1+β

) as |x| → ∞, (X.9.19)

for som e α ∈ R

and β > 0. This problem has been considered by Deuring &

Galdi (2000), who showed that, in fact, an expansion like (X.9.19) is possible

only if α = 0. More precisely, we have the following result for whose proof we

refer to Theorem 3.1 of the paper by Deuring & Galdi.

Theorem X.9.4 Let v be a generalized solution to the Navier–Stokes prob-

lem (X.0.8), (X.0.4) corresponding to v

∞

≡ 0 and f ∈ L

loc

(Ω). Assume there

is R > δ(Ω

) such that for all x ∈ Ω

the following conditions hold:

(i) |f (x)| ≤ C

/|x|

for som e γ > 3, and C

> 0 ;

(ii) |v(x)| ≤ C

/|x| f or some C

> 0 .

Then, if there is a constant M > 0 such that

|x|

1+β

|v(x) − α ·U (x)| ≤ M ,

for som e α ∈ R

, some β > 0 a nd all x ∈ Ω

, necessarily α = 0.

See also Exercise X.9.2.

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

Remark X.9.3 Theorem X.9.4 leaves open the intriguing question of whether

a generalized solution satisfyi ng the assumption (ii) of Theorem X.9.4 can still

admi t an asymptotic expansion of the type (X.9.19), but with α·U(x) replaced

by some other vector ﬁeld that behaves like 1/|x| as |x| → ∞.

Such a question ﬁnds an answer in the work of Nazarov & Pileckas (2000,

Theorem 3.2 and Remark 3.3 ) who proved the following result.

Under the

assumptions that Ω, v

∗

, f are suﬃciently smooth with v

∗

and f suﬃciently

small in suitable norms, and f of bo unded support, there exists a (uniq ue)

corresponding (smooth) solution (v, p) to the Navier–Stokes problem (X.0.8),

(X.0.4) with v

∞

= 0, such that the following asymptotic representation holds:

v(x) =

|x|

+ O(1/|x|

1+β

p(x) =

|x|

+ O(1/|x|

2+β

) ,

(X.9.20)

where V and P are functions deﬁned on the uni t sphere S

, and β ∈ (0, 1).

This interesting result has been further clariﬁed by

Sver´ak (2006) and,

successively, its proof simpl iﬁed by Korolev &

Sver´ak (2007, 20 11). Specif-

ically, these authors show that the terms of order |x|

−1

in the expansion

(X.9.20) must coincide with a speciﬁc velocity and pressure ﬁeld of the family

of solutions to the Navier–Stokes equations obtai ned by Landau (1944) and,

independently, by Squire (1951); see (X.9.21). More precisely, let b ∈ R

−{0},

and let (r, θ, φ) be a system of polar coordinates, with polar axis oriented in

the direction b/|b| which, without loss, we may take coinciding with the pos-

itive x

−direction. The Landau solution (U

, P

) corresponding to b is then

deﬁned as follows



−1

(A − cos θ)

−1



= −

2 sin θ

r(A −cos θ)

= 0 ,

4(A cos θ − 1)

(A − cos θ)

(X.9.21)

where A ∈ (1, ∞) is a parameter chosen in such a way that

As a matter of fact, the t ime line of the events goes as follows. I learned the main

ideas of the proof of Nazarov & Pileckas result in April 1996, after a conversation

I had with Professor Serguei Nazarov, while he was visiting Professor Padula

and me at the University of Ferrara. I then got the feeling of the invalidity of

(X.9.19) with α 6= 0 and discussed the matter with Professor Paul Deuring, at

an Oberwolfach meeting in August 1997. However, b oth papers of Nazarov &

Pileckas and Deuring & Galdi were published only later on, in the year 2000.

X.9 On the Asymptotic Structure of Generalized Solutions when v

∞

= 0 729

16π



A +

log

A −1

A + 1

3(A

− 1)



= b (X.9.22)

By direct inspection, we ﬁnd that the function on the left-hand side is mono-

tonically decreasing in A ∈ (1, ∞) and that its range covers the entire positive

line (0, ∞). Therefore, for any given b (> 0) we ﬁnd one and only one A sat-

isfying (X.9.22), namely, one and only one Landau solution (U

, P

Suppose now (v, p) is a regular solution to (X.0.8)

1,2

with Ω suﬃciently

smooth, and set

b := −

∂Ω

T ( v, p) · n ,

Then in T heorem 1 of Korolev &

Sver´ak (2007, 2011) it is proved that for

each β ∈ (0, 1) there exists ε > 0 such that, if v satisﬁes the assumption of

Theorem X.9.4(ii) with a (suﬃciently small) C

= C

(ε), necessarily v and p

have the fol lowing asymptotic behavior as |x| → ∞

v(x) = U

(x) + O(1/|x|

1+β

p(x) = P

(x) + O(1/|x|

2+β

) .

Whether or not in these formulas (or in (X.9.20)) we may take β = 0 is open.



Remark X.9.4 The following L iouville problem is open: Are there noniden-

tically vanishing smooth solutions v, p to the Navier–Stokes problem

∆v = v · ∇v + ∇p

∇v = 0

)

in R

lim

|x|→∞

v(x) = 0

(X.9.23)

such that

∇v : ∇v < ∞.

(X.9.24)

Even though an answer to this question is not yet available, we can look for

conditions on v under which (X.9.23) admits the null solution only. In this

respect, we have the following theorem.

Theorem X.9.5 Suppose v is a smooth solution

to (X.9.23) such that

Temporarily, we set R = 1.

We recall that the analogous question when the limiting velocity at inﬁnity is

nonzero is solved in Theorem X.7.2.

We may equivalently require v ∈ L

loc

). Actually, by Theorem X.1.1 this would

imply v, p ∈ C

∞

) as required. Notice that condition (X.9.24) is not needed.

In view of (X.9.23)

and the regularity of v, we may equally assume

v ∈ L

) for some q ∈ [1, 9/2].

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

v ∈ L

9/2

) (X.9.25)

then v ≡ 0.

Proof. For R > 0 , let ψ

be a real nonincreasing smooth function deﬁned in

such that ψ

(x) = 0 for |x| ≥ 2R, ψ

(x) = 1 for |x| ≤ R and satisfying

|∇ψ

(x)| ≤ M/R,

for some (positive) constant M independent of x ∈ R

and R. Setting B(R) =

, from this latter inequality it follows that

Ψ ≡

B(R)

|∇ψ

≤ C

for some C indep endent of R. Multiplying (X.9.23)

by ψ

v, i ntegrating by

parts over R

and taking into account (X.9.23)

yields

∇v : ∇v =



−∇ψ

· ∇v · v +

v · ∇ψ

+ pv · ∇ψ



≡ I

+ I

(X.9.26)

We have

| ≤ Ψ

1/3

|v|

1,9/4,B(R)

kvk

9/2,B(R)

| ≤ Ψ

1/3

kvk

9/2,B(R)

| ≤ Ψ

1/3

kpk

9/4,B(R)

kvk

9/2,B(R)

(X.9.27)

Observing that

v · ∇v ∈ D

−1,9/4

from Theorem IV.2.2, Theorem V.3.2, and Theorem V.3.5, it follows that

∇v ∈ L

9/4

), p ∈ L

9/4

and so from inequali ty (X.9.27) we ﬁnd

lim

R→∞

= 0, i = 1, 2, 3. (X.9.28)

Relations (X.9. 26) and (X.9.28) imply, by the monotone convergence theorem,

∇v ≡ 0. Since v satisﬁes (X.9.25), this latter condition delivers v ≡ 0 and the

proof of the theorem is complete. ut

It is worth notici ng tha t, in view o f the Sobolev inequality (II.3.7), a

generalized solution v to (X.9.23) (in dimension 3) satisﬁes only

v ∈ L

)

X.10 Limit of Vanishing Reynolds Number and Stokes Problem 731

which, in the sense of the behavior at large distances, is weaker than (X.9.25).

Theorem X.9.5 can be extended, with formal changes in the proof, to the

n-dim ensional case, n ≥ 4, provided we replace (X.9.25) with the assumption

that

v ∈ L

3n/(n−1)

). (X.9.2 9)

Now if v satisﬁes (X.9. 24), from the Sobolev inequality (II.3.7 ) we deduce

v ∈ L

2n/(n−2)

), (X.9.3 0)

and since v is smooth and obeys the vanishing condition (X.9.23)

, it follows

that (X.9.30) implies (X.9.29) and we may conclude that every smooth so-

lution to (X.9. 23) in R

with n ≥ 4, satisfyi ng (X.9.24) is identically zero.

As we shall see in Chapter XII (cf. Theorem XII.3.1), this property continues

to hold also for n = 2 (pla ne ﬂows), so that the three-dimensional case is the

only one that remains open. 

Exercise X.9.2 (K ozono, Sohr & Yamazaki 1997) Let v be a generalized solution

to (X.0.8), (X.0.4) corresponding to v

∗

≡ v

∞

≡ 0 and f ∈ D

−1,2

(Ω). Show that

if, in addition, v ∈ L

9/2

(Ω), then v satisﬁes the energy equality (X.2.4). Hint: Use

Theorem V.5.1, along with the argument adopted in the proof of Theorem X.9.5.

X.10 Limit of Vanishing Reynolds Number: Transition

to the Stokes Problem

The aim o f this section is to show that every generali zed solution v to the

Navier–Stokes problem (X.0.8)–(X.0.4), corresponding to suﬃciently smooth

data, tends in an appropriate sense, as the Reynolds number R → 0, to the

solution v

of the linearized Stokes system corresponding to the same data.

If the lim iting velocity v

∞

is zero, such a problem has been already con-

sidered in Exercise X.9. 1. Actually, from the results of this exercise, it follows

that any generalized solution can be expressed (for small R) as a perturbation

series in R around v

, that is,

v(x) = v

(x) +

∞

k=1

(x). (X.10.1)

We shall, therefore, turn our attention to the more involved case where v

∞

0. As a matter of fact, in such a situation an expansion of the type (X.10.1) is

no longer expected, as we are going to show. Actually, assume that we try to

express v in the form (X.10.1), with v

solving the following Stokes problem

In dimension n = 4, this also follows from Remark X.2.4.

For the sake of simplicity, we assume that there are no body fo rces acting on the

liquid.

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

∆v

= ∇p

∇ · v

= 0

)

in Ω

= v

∗

at ∂Ω , lim

|x|→∞

(x) = e

(X.10.2)

Then, at ﬁrst order in R, we should have

∆v

= v

· ∇v

+ ∇p

∇ · v

= 0

)

in Ω

= 0 a t ∂Ω

(X.10.3)

and

lim

|x|→∞

(x) = 0. (X.10.4)

In view of the asym pto tic properties established in Theorem V.3.2 (cf. also

Exercise V.3.4), we have

·∇v

∈ L

(Ω), for q > 3/2,

whereas

·∇v

6∈ L

(Ω), for q ≤ 3/2.

Thus, from Theorem V.4.7 we infer the existence of a solution v

, p

to the

system (X.1 0.3) such that

∈ D

2,q

(Ω), for q > 3/2.

However, this property is not enough to control the behavior of v

at large dis-

tances and, consequently, we cannot prove the validity of (X.10.4 ). The situa-

tion just described is a form of the so-called Whitehead paradox; cf. Whitehead

(1888) and Oseen (1927, p. 163).

In view of all this, we expect that v

, p

should approximate v, p as R → 0

in a form weaker than that required by the expansion (X.10.1). We shall, in

fact, prove that this approximation holds in the sense of uniform convergence;

cf. Theorem X.10.1.

In order to reach our objective, we need several preliminary results. For the

sake of formal simpl icity, we assume that there are no body forces acting on the

liquid. For the same reason, the assumption we shall make on the regularity

of Ω and v

∗

will be somewhat stronger than that actually needed. Weakening

of these hypotheses will be left to the interested reader as an exercise.

Lemma X.10.1 Let Ω be an exterior domain of R

of class C

and let v be

a generalized solution to (X.0.8), (X.0.4) corresponding to v

∞

= e

, f ≡ 0

and R ∈ (0, B], for some B > 0. Then, if

∗

∈ W

11/2,2

(∂Ω),

X.10 Limit of Vanishing Reynolds Number and Stokes Problem 733

there exists C = C( Ω, v

∗

, B, R) > 0 such that

kpk

2,Ω

+ |v|

1,2

+ kvk

2,∞

≤ C (X.10.5)

for a ll R > δ(Ω

Proof. Throughout the proof, by the letter C we mean any positive constant

depending o n Ω, v

∗

, B, and R, but otherwise independent of R. The value

of C may, however, change within the same context. By Theorem X.4.1 and

Theorem X.7.3, we know that there is a B > 0 such that

kvk

1,2,Ω

+ kpk

2,Ω

+ |v|

1,2

≤ C (X.10.6)

where p has been modiﬁed by the addition of a suitable constant. From The-

orem IV.5.1 we easily derive, for all R > r > δ(Ω

kvk

m+2,q,Ω

≤ C(kv · ∇vk

m+2,q,Ω

+ kvk

1,q,Ω

+ kpk

q,Ω

+ 1)

with m = 0, 1, 2, q > 1 and so, by (X.10.6), it follows that

kvk

m+2,q,Ω

≤ C(kv · ∇vk

m+2,q,Ω

+ 1). (X.10.7)

Since

kv · ∇ vk

m+2,q,Ω

≤ k(v − v

∞

) · ∇vk

m+2,q,Ω

+ kv

∞

· ∇vk

m+2,q,Ω

with the aid of the H¨older inequality and the inequali ty

kv − v

∞

≤ γ

|v|

1,2

(X.10.8)

(cf. Theorem II.6.1), we obtain

kv · ∇vk

3/2,Ω

≤ kv − v

∞

|v|

1,2

≤ C,

where use has been made of (X.10.6 ). Inserting this info rmation into (X.10.7)

with m = 0, q = 3/2, and r = R

furnishes

kvk

2,3/2,Ω

≤ C.

With the help of the embedding Theorem II.5 .2, we then easily prove

kv · ∇vk

2,Ω

≤ C

and (X.1 0.7) implies

kvk

2,2,Ω

≤ C,

for R

< R

. Agai n using Theorem II.5.2, together with this latter relation,

we show

kv · ∇vk

1,2,Ω

≤ C

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

and (X.1 0.7) then yields

kvk

3,2,Ω

≤ C,

for R

< R

. Iterating this procedure one more tim e, we ﬁnally arrive at

kvk

4,2,Ω

≤ C,

for all R > δ(Ω

). This condition with the aid of Theorem II.3.4 then furnishes

kvk

2,∞,Ω

≤ C. (X.10.9)

Now, from Theorem X.5. 1 we have, for suﬃciently large R,

kvk

2,∞,Ω

R ≤ C, (X.10.10)

and the lemma is thus a consequence of (X.10.6), (X.10.9), and (X.10.10). ut

Lemma X.10.2 Let the assumptions of Lemma X.10.1 be satisﬁed. Then for

all R ∈ (0, B] and all q ∈ [3/2, 2] we have

k(v − v

∞

) · ∇vk

2,q

≤ C,

where C = C(q, Ω, v

∗

, B).

Proof. Throughout the proof, by the letter C we mean any positive constant

depending, at most, on q, Ω, v

∗

, and B. Set

u = v − v

∞

From Lemma V.4.3 it follows, in particular, with m = 0, 1, 2, that

|v|

m+2,2

≤ C(ku · ∇vk

m+2,2

+ kvk

2,Ω

+ kpk

2,Ω

+ 1).

Using Lemm a X.10.1 in this inequality we ﬁnd

|v|

m+2,2

≤ C(ku · ∇vk

m+2,2

+ 1). (X.10.11)

Again employing Lemma X.10.1 we ﬁnd

ku · ∇vk

≤ C

so that (X.10.11) furnishes

|v|

2,2

≤ C. (X.10.12)

Using this inequality, together with Lemma X.10.1, we i nfer that

ku · ∇vk

1,2

≤ C

and, therefore, (X.10.11) implies

|v|

3,2

≤ C. (X.10.13)

X.10 Limit of Vanishing Reynolds Number and Stokes Problem 735

With the help of this estimate and Lemma X. 10.1, we derive

ku · ∇vk

2,2

≤ C. (X.10.1 4)

From (X.10.8) and Lemma X.10.1, we ﬁnd

ku · ∇vk

3/2

≤ kuk

|v|

1,2

≤ C. (X.10.15)

Moreover, by Theorem I I.6.1 and (X.10.11), i t follows that

|v|

1,6

≤ γ

|v|

2,2

≤ C (X.10.1 6)

and so, by Lemma X.10.1 and interpo lation, we derive

|v|

1,3

≤ C.

Consequently,

|u · ∇v|

1,3/2

≤ |v|

1,3

+ kuk

|v|

2,2

≤ C. (X.10. 17)

In addition, from (X.10 .8), (X.10.12), (X.10.13), and (X.10.16) we obtain

|u · ∇v|

2,3/2

≤ 3|v|

1,6

|v|

2,2

+ kuk

|v|

3,2

≤ C. (X.10.18)

Collecting (X.10.15)–(X. 10.17) we ﬁnd

ku ·∇vk

2,3/2

≤ C,

and the l emma becomes a consequence of this inequality, (X.10.14), and the

elementary L

convexity i nequality (II.2 .7). ut

Lemma X.10.3 Let the assumptions of Lemma X.10.1 be satisﬁed. Denote

by u

, π

the solution to the f ollowing Oseen problem

∆u

−R

∂u

∂x

= ∇π

∇ · u

= 0











in Ω

lim

|x|→∞

(x) = 0

= v

∗

−e

≡ u

∗

at ∂Ω.

(X.10.19)

Then, there exists B > 0 such that for any R ∈ (0, B], all q ∈ [3/2, 2], all

t ∈ [12/5, 3), and all s ∈ (2, 3) we have

|v − u

1,3q/(3−q)

+ kD

(v − u

2,q

+ kp − π

3q/(3−q)

+ k∇(p − π

2,q

≤ C R

kv − u

−e

3t/(3−t )

≤ C R

2−3/t

1,3q/(3−q)

+ kD

2,q

+ k∇π

2,q

≤ C

≤ C R

1−3/s

(X.10.20)

where C = C(Ω, v

∗

, B, q, t, s).