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

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196 III The Function Spaces of Hydrodynamics

III.4.1 Bounded Domains

The simplest situation occurs when Ω is star-shaped (with respect to the

origin, say). For ρ ∈ (0, 1) we know from Exercise II.1.3 that Ω

(ρ)

⊂ Ω. Thus,

given a vector v ∈

(Ω), by extending it by zero outside Ω, we deduce that

the family of vectors v

(x) ≡ v(x/ρ), x ∈ R

, belongs to

(Ω) and is of

compact support in Ω. Set

= min

x∈∂Ω

|x|, ε(ρ) = h

(1 −ρ)/2

and consider the molliﬁcation (v

)

ε(ρ)

≡ V

. By the results ofSection II.2

on regularizations we obtain that, for each ρ ∈ (0, 1), V

∈ D(Ω) and that

→ v in W

1,q

(Ω) as ρ → 1, for all q ∈ [ 1, ∞). Therefore, for these values

of q,

(Ω) ⊂ H

(Ω) and the two spaces coincide.

Using an idea of Bogovski

i (1980, Lemma 4), this result can be generalized

to a fairly reach class of bounded domains. Speciﬁcally, we have the following.

Theorem III.4.1 Assume that the bounded domain Ω ⊂ R

, n ≥ 2, is

the union of a ﬁnite number of domains Ω

each of which is star-shaped with

respect to an open ball B

with B

⊂ Ω

(e.g., Ω satisﬁes the cone condition).

Then

(Ω) = H

(Ω).

Proof. Let us ﬁrst consider the case q ∈ (1, ∞). Given v ∈

(Ω), let

} ⊂ C

∞

(Ω) be a sequence approximating v in W

1,q

(Ω) and denote by

a solution to (III.3.2) with f = −∇ · v

. Since f satisﬁes (III.3.1 ), by

Theorem III.3.1 w

exists, belongs to W

1,q

(Ω), and can be taken to b e of

compact support i n Ω. The functions u

= v

+ w

, k ∈ N, are thus diver-

gence free and of compact support in Ω. Furthermore, using (III.3.3) for w

we have

−vk

1,q

≤ kv

−vk

1,q

+ kw

1,q

≤ kv

−vk

+ ck∇· v

with c independent of k. Since ∇ · v = 0, it fo llows that

k∇· v

→ 0 as k → ∞,

and so the previous inequality furnishes

→ v in W

1,q

(Ω), as k → ∞.

Let (u

)

be the regularization of u

. For suﬃciently small ε, (u

)

belongs

to D(Ω) and approaches u

in W

1,q

(Ω) (see Section II.2 and Exercise II.3.2),

which proves v ∈ H

(Ω). Let us now prove H

(Ω) =

(Ω). To this end,

See Remark III.3.4 and Remark III.3.5.

III.4 The Spaces H

197

it is suﬃcient to show that every linear functional F deﬁned in

(Ω) and

vanishing in H

(Ω) is identically zero. Since H

(Ω) ⊂ H

(Ω) and

(Ω) =

(Ω) for q > 1, we have F ∈ S where

S =



` ∈



(Ω)



: `(v) = 0 for all v ∈ H

(Ω)



On the other hand, by what we already proved, S ≡ 0 and hence F ≡ 0,

which completes the proof. ut

III.4.2 Exterior Do mains

Following the argument of Ladyzhenskaya & Solonnikov (1976), we can prove

the following.

Theorem III.4.2 Let Ω ⊆ R

, n ≥ 2, be an exterior domain such that, for

some ρ > δ(Ω

), Ω

satisﬁes the assumption of Theorem II I.4.1. Then,

(Ω) = H

(Ω), 1 < q < ∞.

Proof. We begin with the obvious observation that if Ω

satisﬁes the assump-

tion of Theorem III.4 .1, then also Ω

does, for all R > ρ. Now, let ψ ∈ C

(R)

with ψ(ξ) = 1 if |ξ| ≤ 1 and ψ(ξ) = 0 if |ξ| ≥ 2 and set ψ

(x) = ψ(|x|/R),

R > ρ. For v ∈

(Ω), denote by w

(R)

a solution to the problem

∇ · w

(R)

= −v · ∇ψ

(R)

∈ W

1,q

(Ω

R,2R

)

(R)

1,q,Ω

R,2R

≤ ckv · ∇ψ

q,Ω

(III.4.2)

Since the compa tibility condition

Ω

R,2R

v · ∇ψ

Ω

R,2R

∇· (vψ

) = 0

is satisﬁed, such a ﬁeld exists, by Theorem III.3.1. Moreover, by Lemma III.3.3

the constant c does not depend on R. Also, since ∇ψ

= O(1/R) uniformly

in x, using inequality (II.5 .5) we deduce for some c

, c

independent of R

(R)

q,Ω

R,2R

≤ c

R|w

(R)

q,Ω

R,2R

≤ c

kvk

q,Ω

R,2R

. (II I.4.3)

Setting w

(R)

≡ 0 in the complement of Ω

R,2R

, we deﬁne

(R)

= ψ

v + w

(R)

198 III The Function Spaces of Hydrodynamics

Due to (III.4.2), v

(R)

∈

(Ω

). Since Ω

satisﬁes the assumption of The-

orem III.4.1, we have

(Ω

) = H

(Ω

), for all R > δ(Ω

and therefore, given ε > 0, we can ﬁnd v

ε,R

∈ D(Ω

) ⊂ D (Ω) such that

(R)

−v

ε,R

1,q,Ω

< ε.

So,

kv − v

ε,R

q,Ω

≤ kv

ε,R

−v

(R)

q,Ω

+ kv − v

(R)

q,Ω

< ε + k(1 − ψ

)vk

q,Ω

+ kw

(R)

q,Ω

R,2R

Because of (III.4.3) and the properties of ψ

, by taking R suﬃciently large

and ε suﬃciently small, we can make the rig ht-hand side of this inequality a s

small as we please, thus proving

ε,R

−vk

q,Ω

→ 0 as ε → 0, R → ∞.

By the some token one shows

ε,R

−v|

1,q,Ω

→ 0 as ε → 0, R → ∞,

which completes the proof of the coincidence. ut

III.4.3 Domains with Noncompact Boundary

It is easy to convince oneself that the method of proof just employed for the

exterior case applies with no signiﬁcant changes to show

(Ω) = H

(Ω),

1 < q < ∞, for domains Ω with noncompact boundary, provided the following

conditions are satisﬁed for all R greater than a ﬁxed number R

(i) Ω

= {x ∈ Ω : |x| < 2R} and Ω

R,2R

= {x ∈ Ω : R < |x| < 2R} are do-

mains;

(ii) Problem (III.4.2) is solvable with a constant c independent o f R;

(iii) Inequality (III.4.3) holds wi th a constant c

independent of R;

(iv)

(Ω

) = H

(Ω

);

(see Ladyzhenskaya & Solonni kov 1 976, Theorem 4.1). In particular, condi-

tions (i)-(iv) are certainly fulﬁlled if Ω = R

. Thus, we have.

Theorem III.4.3

) = H

), 1 < q < ∞.

Our next task, throughout the rest of this section, will be to i nvestigate the

question of coincidence when Ω has “exits” to inﬁnity, namely, when outside

a connected, com pact subset Ω

(say) Ω splits into m disjoint comp onents

Ω

, which in possibly diﬀerent coordinate systems (depending on i), can be

represented as

III.4 The Spaces H

199

Ω

= {x ∈ R

: x

> 0, x

∈ Σ

)},

where Σ

) is a do main in R

n−1

smoothly varying with x

and x

, . . . , x

n−1

). Familiar doma ins Ω having these properties are, for instance,

inﬁnitely long pipes and tubes of possibl y varying cross section.

As we have noted at the beginning of this section, the coincidence of

(Ω)

and H

(Ω) can be tightly related to the way in which Σ

) “widens” a s

→ ∞. It would thus be useful to establish a characterization of the class of

cross sections for which the coincidence holds but, to the best of our knowl-

edge, such a result is not available yet; nonetheless, one can certainly select

certain classes of Σ

) and establish whether or not

(Ω) = H

(Ω). For

example, if all Σ

, i = 1, . . ., m, are independent of x

, i.e., each Ω

reduces

to a straight semi-inﬁnite cylinder, then, by the same technique used for the

case of an exterior domai n, it is not ha rd to show that

(Ω) = H

(Ω),

1 < q < ∞, provided Ω has a mild degree of regularity; see Exercise III.4.4.

On the other hand, if some o f the domains Σ

) become unbounded as

→ ∞ with a suitable growth rate, then, for the corresponding Ω we may

have

(Ω) 6= H

(Ω).

An example of such domains was gi ven for the ﬁrst time by Heywood

(1976), who proved noncoincidence o f

(Ω) and H

(Ω) when Ω is a n “aper-

ture domain,” namely, a domain of R

, n > 2, of the type

Ω = Ω

∪ Ω

with Ω

a bounded subset of the plane x

= 0 containing a uni t disk C and,

in the same coordinate system,

Ω

= R

−

, Ω

= R

so that the two cross sections Σ

, Σ

reduce to the entire space R

n−1

. Stated

equivalently,

Ω = {x ∈ R

: x

6= 0 or x

∈ Ω

}, (III.4.4)

with x

= (x

, . . . , x

n−1

). By using the ideas of Heywood, we now show the

following.

Theorem III.4.4 Let Ω be the “aperture” domain (III.4.4). Then

(Ω) 6= H

(Ω) .

Moreover,

d ≡ dim



(Ω) / H

(Ω)



= 1.

Proof. Take the origin of coordinates at the center of C, denote by Γ

, i = 1, 2,

the cones {x ∈ Ω

: x

< |x

|} and set

= Γ

∩ {|x| = 1}, i = 1, 2.

200 III The Function Spaces of Hydrodynamics

We then let

∗

(x) =

ω(θ)x

|x|

with

ω(θ) =











(cos 2θ)

for θ ∈ [0, π/4]

0 for θ ∈ [π/4, 3π/4]

−(cos 2θ)

for θ ∈ [3π/4, π],

where θ is the angle between the positive x

axis and the ray joining the point

x with the origin. Evidently,

∗

∈ L

loc

∇ · b

∗

= 0, for a ll x ∈ Ω − {0}

supp (b

∗

) ⊆ Γ

∪ Γ

Furthermore, by indicating with S

i,R

the surface Ω

∩ {|x| = R}, it follows

that

1,R

∗

(x) · n =

ω(θ) = −

ω(θ)

= −

2,R

∗

(x) · n ≡ φ = const.> 0.

(III.4.5)

Setting

b = (b

∗

)

, 0 < ε < 1/2,

the regularization of b

∗

, by the properties of regularizations (see Section II.2)

we readily deduce b ∈ C

∞

) and that the following conditions hold for all

x ∈ Ω

∇ · b(x) = 0

|b(x)| ≤ c |x|

−n+1

|∇b(x)| ≤ c |x|

−n

|∆b(x)| ≤ c |x|

−n−1

(III.4.6)

Furthermore,

supp (b) ⊂

| < |x

| + 1/

√

. (III.4.7)

Using (III.4.6)

2,3

, (III.4.7) along with a now standard “cut-oﬀ” argument, it

follows that

b ∈ W

1,2

(Ω)

and so, in virtue of (III.4.6)

, we may infer

b ∈

(Ω).

III.4 The Spaces H

201

However, by the solenoidality of b and (III.4.5), using the properties of regu-

larizati ons we can select ε so small that

Ω

b · n ≡ const.> φ/2 > 0. (I II.4.8)

Condition (III.4.8) then tells us that b 6∈ H

(Ω). In fact, since every element u

from H

(Ω) is approximated by functions from D( Ω) it is immediately shown

that

Ω

u · n = 0. (III.4.9)

We may then conclude

(Ω) 6= H

(Ω). We shall now prove the second

part of the theorem. For si mplicity, we shall take the bounded dom ain Ω

just coincident with the unit circle C and begin to show that if u ∈

(Ω)

satisﬁes (III.4.9) then u ∈ H

(Ω). Actually, since Ω = Ω

∪ Ω

with

Ω

≡ R

−

and Ω

≡ R

, we set

= Ω

∩B

= Ω

∩B

and denote by v

a solenoidal vector ﬁeld in D

that equals u at Ω

, vanishes

at Ω

∩∂B

and belongs to W

1,2

). Since u satisﬁes (III.4.9), such a ﬁeld

exists in virtue of Exercise III.3.5. Furthermore, it is a simple task to show

that, by extending v

to zero outside D

, the vector ﬁeld v

+ v

is solenoidal

in B

and belongs to W

1,2

). Denoting by u

the restriction of u to Ω

, we

may thus write

u = (u

−v

) + (u

− v

) + (v

+ v

) ≡ w

+ w

Employing the results on the coincidence of the two spaces for the half -space

and for a bounded domain, we deduce

∈ H

−

), w

∈ H

), w

∈ H

and so, since each of these latter spaces is embedded in H

(Ω) we conclude

u ∈ H

(Ω). Take now the vector b constructed before and multiply it by a

constant in such a way that, denoting this new vector by b,

Ω

b · n = 1. (III.4.10)

(This is allowed, in virtue of (III.4.8).) Take any v ∈

(Ω) and let Φ indicate

its ﬂux through Ω

. Then, by (III.4.10),

v − Φb ≡ u

satisﬁes (III.4.9) a nd so u ∈ H

(Ω), which proves d = 1.

202 III The Function Spaces of Hydrodynamics

Exercise III.4.2 By means of the arguments described in the proof of the preced-

ing theorem, show that f or Ω given in (III.4.4):

(Ω) 6= H

(Ω), for all q > n/(n − 1), n ≥ 2.

Moreover, for the above values of q, show dim(

(Ω) / H

(Ω)) = 1 .

It is not diﬃcult to convince oneself that Heywood’s procedure described

in the proof of Theorem III.4.4 should also work for a general class of domains

with a ﬁnite number of exits to inﬁnity, provided each exit contains a semi-

inﬁnite cone. This problem is taken up by Ladyzhenskaya & Solonnikov (1976,

Theorem 4.2) who prove, in fact, that

(Ω) 6= H

(Ω), whenever Ω ⊂ R

n ≥ 3, has m ≥ 2 (disjoint) exits Ω

to inﬁnity and provided each Ω

contains

a semi-inﬁnite cone. Furthermore,

dim(

(Ω) / H

(Ω)) = m −1 .

However, if n = 2, then

(Ω) = H

(Ω); see also Kapitanski

i (1 981).

Remark III.4.1 In the l ight of the proo f of Theorem III.4.4, it is easy to

show that for the domain (III.4.4), H

(Ω) 6= H

(Ω), for all q > n/(n − 1),

n ≥ 2, where H

(Ω) and H

(Ω) are deﬁned in Section III.2 and Theorem

III.2.3. Actually, in view of Lemma III.2.1 , it is enough to prove the existence

of v ∈ H

(Ω) such that

Ω

v ·∇φ = κ 6= 0, for some φ ∈ D

1,q

(Ω). (III.4.11)

For simplicity, we shall take Ω

= C, with C unit disk of R

n−1

. Putting the

origin of coordinates at the center of C and setting r = |x|, we choose

φ(x) =











exp(−r) if r ≥ 1 and x

> 0

exp(−1) if r ≤ 1

1 − (1 − exp(−1))exp(−r + 1) if r ≥ 1 and x

< 0.

Clearly, φ ∈ D

1,q

(Ω) for any q

≥ 1, and

lim

r→∞

φ(x) =

(

0 if x

> 0

1 if x

< 0.

(III.4.12)

Taking into account the properties of the ﬁeld b introduced in the proof of

Theorem III.4.4, it immedia tely fo llows that

b ∈ H

(Ω) for all q > n/(n − 1), n ≥ 2.

Furthermore, by (III.4.6)

and (III.4.7), we ﬁnd for all R > 0

III.4 The Spaces H

203

Ω∩B

b·∇φ =

φb·n+

−

φb·n =

φb·n+

−

(φ−1)b·n+

Ω

b·n,

(III.4.13)

where

= ∂B

∩ R

, Σ

−

= ∂B

∩ R

−

Thus, letting R → ∞ into (III.4.13) and recalling (III.4.12), (I II.4.6)

and

(III.4.8) we recover (III.4.11) with v = b and κ =

Ω

b · n. 

The aim of the remaining part of this subsection is to analyze all previous

problems when the exits Ω

are b odies of rotation, i.e., in po ssibly diﬀerent

coordinate systems,

Ω

= {x ∈ R

: x

> 0, |x

| < f

)}, (III.4.14)

where f

are suitable strictly positive functions. This class of domains is

interesting in that, provided f

satisﬁes a global Lipschitz condition and

Ω

≡ Ω ∩ B

has a mild degree of regularity for all R, we can completely

characterize the class of f

for whi ch the coincidence o f

and H

holds,

1 < q < ∞, and, in the case where it doesn’t, we can establish the dimension

of the quoti ent space. The above study is essentially due to Solonnikov &

Pileckas (1977), Pileckas (1983), and independently to Bogovski

i & Maslen-

nikova (1978) and Maslennikova & Bogovski

i (1978, 1981a, 1981b, 1983). We

begin to show the following preliminary result.

Lemma III.4.1 Let

Ω =

[

i=1

Ω

be a domain in R

, n ≥ 2, where Ω

is a compact set and Ω

, i = 1, . . ., m, are

disjoi nt domains that (in possibly diﬀerent coordinate systems) are of type

(III.4.14) with f

satisfying the following conditions:

(i) f

(t) ≥ f

> 0;

(ii) |f

) − f

)| ≤ M| t

− t

|, for some constants f

, M and for all t, t

> 0.

Suppose, further, that

Ω

≡ Ω ∩ B

satisﬁes the cone condition for all R > δ(Ω

) (the orig in of coordinates is

taken in Ω

). Then

(Ω) = H

(Ω), 1 < q < ∞, if and only if all vectors

v ∈

(Ω) have zero ﬂux through the planar cross sections Σ

= Σ

) of

Ω

perpendicular to the axis x

= 0 and passing through the point (0, x

that is,

v · n = 0, i = 1, . . ., m. (III.4.15)

Of course, the ﬂux of v through Σ

) is independent of x

204 III The Function Spaces of Hydrodynamics

Proof. For ﬁx ed i and a ll R > 0, put

= R + f

(R)/2M

and

Ω

= Ω

∩

R < x

in the system o f coordinates to which the exit Ω

is referred. Let ζ ∈ C

(R)

be such that ζ(ξ) = 1 for ξ ≤ 1 and ζ(ξ) = 0 for ξ ≥ 2 and set

(x) = ζ (2M [x

+ (f

(R)/2M ) −R] /f

(R)) .

Obviously, ζ

is equal to one fo r x

≤ R and is zero for x

≥

and,

moreover,

|∇ζ

| ≤ C/f

(R), (III.4.16)

for som e C independent of R. Denote by w

a solution to the problem

∇ · w

= −∇ζ

·v in Ω

∈ W

1,q

(Ω

)

1,q,Ω

≤ Ck∇ζ

·vk

q,Ω

(III.4.17)

In view of Theorem III.3.1, (III.4.17) is solvable since, as a consequence of

(III.4.15), the compatibility condition

Ω

∇ζ

·v = 0

is satisﬁed. Moreover, the constant C in (III.4.17)

can be taken independent

of R, see Exercise III.4.3. Extend w

to zero outside Ω

and set

≡

i=1

, w

≡

i=1

, v

≡ ζ

v + w

We have that v

is a solenoidal vector ﬁeld belonging to W

1,q

(Ω

) for all

suﬃciently large R

. So, v

∈

(Ω

) and, by the result of this subsection

(a) and the a ssumption made on Ω

, v

can be approximated by elements of

D(Ω

) ⊂ D (Ω). Therefore, given ε > 0, we can ﬁnd v

ε,R

∈ D(Ω) such that

− v

ε,R

1,q,Ω

< ε.

We then have

kv − v

ε,R

q,Ω

≤ kv

− v

ε,R

q,Ω

+ kv

−vk

q,Ω

< ε + k(1 − ζ

)vk

q,Ω

i=1

q,Ω

(III.4.18)

III.4 The Spaces H

205

From (I II.4.16), (III.4.17)

and (II.5.5) we deduce for some constants c

, c

independent of R, and f or i = 1, . . . , m,

q,Ω

≤ c

(R)|w

1,q,Ω

≤ c

kvk

q,Ω

which, once replaced into (III.4.18), shows that v

ε,R

approaches v in L

(Ω)

when ε → 0 and R

→ ∞. Likewise, one shows ∇v

ε,R

→ ∇v in L

(Ω), thus

proving the lemma. ut

Remark III.4.2 If the domain Ω is itself a body of rotation, i.e.,

Ω = {x ∈ R

: x

∈ R, |x

| < f(x

)}

with f satisfying assumptions (i) and (ii) of Lemma III.4.1, this lemma con-

tinues to hold for q = 1; see Maslennikova & Bogovski

i (1 981b, Section 2).



Exercise III.4.3 Show that the constant C in (III.4.17)

can be taken independent

of R. Hint: Make the change of variables:

= 2Mx

(R), y

= 2M(x

− R)/f

(R).

By hypothesis (ii) on f

, the domain Ω

goes into

Ω

∗

(

y ∈ R

: |y

| ≤ g

) ≡ 2M

R + (2M)

−1

(R)y

(R)

, 0 < y

< 1

)

with

M ≤ g

(t) ≤ 3M, for all t > 0.

Ω

∗

is thus contained in a ball of ﬁxed radius for every R. One t hen solves (III.4.17)

in Ω

∗

with a constant independent of R and retransform the solution to Ω

obtaining the desired result.

Let us now consider some consequences of Lemma III. 4.1. First of all, if

m = 1, i.e., Ω has only one exit to inﬁnity, then

(Ω) = H

(Ω) for all

q ∈ (1, ∞). Actually, in this case we may take Ω

= Ω ∩R

−

and so, denoting

by {v

} ⊂ C

∞

(Ω) a sequence of functions approximating v in W

1,q

(Ω) and

by the intersection of Ω with the plane x

= 0, it follows that

·n =

Ω

∇ · v

Since ∇ · v = 0, the right-hand side of this identity tends to zero as k → ∞

and, by Theorem II.4.1, we deduce (III.4.1 5), which proves the coincidence.

In the following, we shall assume m ≥ 2. On every cross section Σ

) we have

|Φ

≡



≤ Cf

(n−1) (q−1)

)

|v|