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

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216 III The Function Spaces of Hydrodynamics

v = u +

`−1

i=1

where u ∈ D

1,q

(Ω). Here,

j=1

and Φ

, j = 1, . . . , `, with

j=1

= 0, are given by

v · n,

with n normal to Σ

in the direction of increase of the coordinate x

in Ω

Moreover, {d

} is a basis in F

1,q

(Ω) constituted by C

∞

(Ω) vector ﬁelds that

vanish in a neighborhood of ∂Ω, in ω

(β), i = 1, . . ., `, for all suﬃciently

small β and in Ω

, i = ` + 1, . . ., m, for suﬃciently large |x|. Finally, for every

j = 1, . . . , ` and i = 1, . . . , ` −1, the vectors d

satisfy an estima te of the type

(III.4.35)

along with the condi tions

· n = δ

−δ

i+1j

Remark III.5.1 Theorem III.5.3 holds with the restriction from below q > 1

instead of q > n/(n − 1) as required in Theorem III.4.6. This is because,

unlike integrals (III.4.37), integrals (I II.5.2) certainly diverge only if q = 1;

see Theorem III.4.5. 

Remark III.5.2 The results of Lemma III.5.1 and Theorem III.5.2 continue

to hol d in the more general case when the domains Ω

are not necessarily

bodies of rotation but only satisfy the requirements listed in Remark III.4.3;

see Pileckas (1983, §3). 

Exercise III.5.1 Let Ω be as in Exercise III.4.4. Assume

1,q

(Ω

) = D

1,q

(Ω

)

for all suﬃciently large R. Show

1,q

(Ω) = D

1,q

(Ω).

Exercise III.5.2 Let Ω be the “aperture domain” (III.4.4). Show

1,2

(Ω) 6=

1,2

(Ω), for all n ≥ 2. (Notice that, unlike spaces

(Ω) and H

(Ω), the case

n = 2 is included.)

Exercise III.5.3 (Heywood 1976) Show that for any domain Ω ⊆ R

, n ≥ 2, we

have

1,2

(Ω) = D

1,2

(Ω) if and only if the only vector v ∈

(Ω) such that

Ω

∇v : ∇ϕ = 0, for all ϕ ∈ D(Ω)

is the null vector.

III.5 The Spaces D

1,q

217

Our last objective in this section is to give a representation of functionals

on D

1,q

(Ω) that identically vanish on

1,q

(Ω). As we shall see in the next

chapter, this question is tightly linked with the existence of the pressure ﬁeld

associated to the motion of a viscous, incompressible ﬂuid. We have (Pileckas

1983)

Theorem III.5.3 Assume Ω be such that problem (III.3.65)

is solvable for

any f ∈ L

(Ω) [respectively, f ∈

(Ω) ≡ L

(Ω) / R, if Ω is b ounded]. Then,

given any (bounded) linear functional F on D

1,q

(Ω), 1 < q < ∞, identically

vanishing on

1,q

(Ω), there exists a uniquely determined

p ∈ L

(Ω)

respectively, p ∈

(Ω)

such that F admits the following representation:

F(v) =

Ω

p∇ ·v, for all v ∈ D

1,q

(Ω). (III.5.3)

Proof. Consider the operator

A : v ∈ D

1,q

(Ω) → ∇ ·v ∈ L

(Ω).

Evidently, A is linear and bounded and, by assumption, its range R(A) co-

incides with the whole o f L

(Ω) (

(Ω), for Ω bounded). By a well-known

theorem on adjoint equatio ns (Banach closed range theorem), see, e.g., Brezis

(1983, Th´eor`eme I I .18) we then have

[ker(A)]

⊥

= R(A

∗

), (III.5.4 )

where ker(A) is the kernel of A, A

∗

is its adjo int and ⊥ means annihilator,

cf. Exercise III.1.1. However, it is obvious that

ker(A) =

1,q

(Ω)

so that (III.5 .4) delivers

1,q

(Ω)

⊥

= R(A

∗

)

and, consequently, all functionals F vanishing on

1,q

(Ω) must be in the

range of A

∗

. On the other hand, by deﬁnition, every element in R(A

∗

) is of

the form L(Av), where L is a functional on L

(Ω) [respectively,

(Ω), if Ω is

bounded]. We may then employ the Riesz representation theorem to obtain,

for som e uniquely determined p ∈ L

(Ω) [respectively, p ∈

(Ω)],

We require the problem in the form of (III.3.65) instead of (III.3.2) so that the

result in the t heorem applies also to unbounded domains.

218 III The Function Spaces of Hydrodynamics

F(v) = L(Av) =

Ω

pAv =

Ω

p∇ · v, v ∈ D

1,q

(Ω),

which completes the proof of the theorem. ut

In view of the results established in Sections III. 3 and III.4, from Theorem

III.5.3 we o bta in the following.

Corollary III.5.1 Let Ω be a bounded or exterior domain of R

, n ≥ 2, that

satisﬁes the cone condition, or Ω = R

. Then, any bounded linear functional

F on D

1,q

(Ω), 1 < q < ∞, identically vanishing on

1,q

(Ω) is of the form

(III.5.3) for some uniq uely determined p ∈ L

(Ω) [respectively, p ∈

(Ω),

if Ω is bounded].

Remark III.5.3 If Ω is an arbitrary domain in R

, n ≥ 2, it is not said that

a representation of the type (III.5.4) holds in Ω. Actually, the assumptions of

Theorem I II.5.3 certainly do not hold if Ω is not smooth enough. However,

since problem (III.3.65) is solvable in every bal l B contained in Ω, we may

use Theorem III.5.3 to show the following result whose proof we leave to the

reader as an exercise.

Corollary III.5.2 Let Ω be an arbitrary domain of R

, n ≥ 2. Suppose F

is a bounded linear functional on D

1,q

(Ω

), 1 < q < ∞, identically vanishing

1,q

(Ω

), where Ω

is any bounded domain of Ω with Ω

⊂ Ω. Then, there

exists p ∈ L

loc

(Ω) such that F admits the following representation:

F(ψ) =

Ω

p∇ · ψ, for all ψ ∈ C

∞

(Ω).



III.6 Approximation Problems in Spaces H

and D

1,q

A problem often encountered in the appli cations is the following. Assume

v ∈ H

(Ω) ∩



∩

i=1

(Ω)



, 1 < q, r

< ∞, i = 1, . . . , k.

Clearly, v can be approximated by a sequence {v

} ⊂ D(Ω) (as a member

of H

(Ω)) and by a sequence {v

} ⊂ C

∞

(Ω) (a s a member of L

(Ω)). The

question now is to establish if v can be approximated by the same sequence

in both spaces or, in other words, taking into account that D(Ω) ⊂ C

∞

(Ω), if

there is a sequence {v

} ⊂ D(Ω) such that as m → ∞

→ v in H

(Ω) ∩



∩

i=1

(Ω)



. (III. 6.1)

Of course this problem admits a trivial positive answer when the domain

Ω is bounded and q, r

are suitably related to each other. For example, take

k = 1 and assume that at least one of the fol lowing conditions is ful ﬁlled

≡ r):

III.6 Approximation Problems in Spaces H

and D

1,q

219

(i) r ≤ q;

(ii) q ≥ n;

(iii) q < n and r ≤ nq/(n − q).

Then (III.6.1) fo llows at once. Actually, denoting by {v

} ⊂ D(Ω) a sequence

converging to v in H

(Ω), in case (i) we have, by the H¨older i nequality,

kv − v

≤ ckv − v

1,q

→ 0.

In case (ii) or (iii) the same conclusion can be drawn by using, instead of the

H¨older inequality, the embedding inequalities of Theorem I I.3.2. Moreover,

the Sobolev inequality (II.3.7) also gives the result for arbitrary Ω, provided

1 < q < n and r = nq/(n −q).

What can b e said in the general case when q, n, and r

are not necessarily

related to each other? The aim of this section is to show that for Ω locally

Lipschitz, i t is always po ssible to ﬁnd a sequence {v

} ⊂ D(Ω) satisfying

(III.6.1). An analogous result holds if we replace H

with D

1,q

The proof will be achieved through several intermediate steps. The ﬁrst

step is to introduce a suitable “cut-oﬀ” function. This function involves the

distance δ(x) of a point x ∈ Ω from the boundary ∂Ω. We need to diﬀerentiate

δ(x) but, in fact, δ(x) is in general not more diﬀerentiable than the obvious

Lipschitz-like condition |δ(x) −δ(y)| ≤ |x −y| . To overcome such a diﬃculty,

we introduce the so-called regularized di stance in the sense of Stein (1970,

p.171). In this respect, we have the f ollowing lemma for whose proof we refer

the reader to Stein (1970, Chapter VI, Theorem 2).

Lemma III.6.1 Let Ω be a domain of R

and set

δ(x) = dist (x, ∂Ω). (III.6.2)

Then there is a function ρ ∈ C

∞

(Ω) such that for al l x ∈ Ω

(i) δ(x) ≤ ρ(x);

(ii) |D

ρ(x)| ≤ κ

|α|+1

[δ(x)]

1−|α|

, |α| ≥ 0,

where κ

|α|+1

depends only on α and n.

Remark III.6.1 A simpl e estimate for the constant κ

is given by Stein

(1970, p.169 and p.171) and one has κ

= (20/3)(12)

. Moreover, if Ω is

suﬃciently smooth (depending on |α|), and x is suﬃciently close to ∂Ω, one

can take ρ = δ and, consequently, κ

= κ

= 1. 

Owing to this result, we can prove the following.

Lemma III.6.2 Let Ω, δ be as in Lemma I II.6.1. For any ε > 0 set

γ(ε) = exp(−1/ε).

Then, there exists a function ψ

∈ C

∞

(Ω) such that

220 III The Function Spaces of Hydrodynamics

(i) |ψ

(x)| ≤ 1, for all x ∈ Ω;

(ii) ψ

(x) = 1, if δ(x) < γ

(ε)/2κ

;

(iii) ψ

(x) = 0, if δ(x) ≥ 2γ(ε);

(iv) |∇ψ

(x)| ≤ κ

ε/δ(x), for all x ∈ Ω,

(v) |D

(x)| ≤ κ ε/δ

|α|

(x) , |α| ≥ 2 , ε ∈ (0, ε

], ε

> 0 ,

where κ

and κ

are given in Lemma III.6.1,

while κ = κ(α, n, ε

) .

Proof. Consider the following function of R into itself:

(t) =











1 if t < γ

(ε)

ε ln(γ(ε)/t) if γ

(ε) < t < γ(ε)

0 if t > γ(ε).

Clearly, choosing η = γ

/2, the molliﬁer Φ

≡ (ϕ

)

of ϕ

satisﬁes Φ

(t) = 1

for t < γ

/2, Φ

(t) = 0 fo r t > 2γ and

|Φ

(t)| ≤ ε/t,

| ≤ c ε/t

, k ≥ 1 , ε ∈ (0, ε

] ,

for all t ∈ R, (III.6.3)

where c = c(k, ε

). In addition, |Φ

(t)| ≤ 1. Setting

(x) ≡ Φ

(ρ(x)),

where ρ is the regularized distance of Lemma III.6.1, and recalling statements

(i) and (ii) of that lemma, we deduce

(x) = 1 if δ(x) < γ

/2κ

(x) = 0 if δ(x) > 2γ.

Moreover, from (III.6.3) and Lemma III.6.1 it follows for all x ∈ Ω

|∇ψ

(x)| ≤ κ

ε/ρ(x) ≤ κ

ε/δ(x)

(x)| ≤ κ ε/δ

|α|

(x) , |α| ≥ 2 ,

whenever ε ≤ ε

, for some ε

> 0 and with κ = κ(α, n, ε

) . The result is

therefore completely proved. ut

Notice that, by Remark III.6.1,

γ(ε) < 2κ

, for all ε > 0.

Of course, w hatever the estimate for κ

, we can always choose a larger value for

such that this latter inequality holds.

III.6 Approximation Problems in Spaces H

and D

1,q

221

Lemma III.6.3 Let Ω be a bounded, locally Lipschitz domain of R

. Then

there exists c = c( Ω, q, n) such that for all u ∈ W

1,q

(Ω), 1 < q < ∞, we have

kuδ

−1

≤ c|u|

1,q

where δ is given in (III.6 .2).

Proof. It is enough to prove the result for u ∈ C

∞

(Ω). By (II.5.5)

kuk

q,Ω

≤ c

|u|

1,q

for any domain Ω

, with Ω

⊂ Ω and where c

= c

(Ω). To show the estimate

“near” the boundary, we recall that ∂Ω can be covered with a ﬁnite number

of cylinders o f the type



∈ D

⊂ R

n−1

: ζ(x

) − σ < x

< ζ(x

) + σ



where ζ is a Lipschitz function locally deﬁning the boundary of Ω. Therefore,

setting V

= Ω ∩ C

, and noting that

− ζ(x

) ≤ c δ(x), x ∈ V

we ﬁnd

|u(x)δ

−1

(x)|

≤ c

(

ζ(x

)+σ

ζ(x

)

|u(x

, x

−ζ(x

−q

)

and the desired estimate follows from the elementary one-dimensional inequal-

ity

∞

|h(t)|

−q

≤



q − 1



∞



, h ∈ C

∞

which can be easily proved by integrating the identity

|h(t)|

−q



1−q

1 − q

|h(t)|



−

1−q

1 − q

|h(t)|

We also have

Lemma III.6.4 Let Ω be as i n Lemma III.6.3. Suppose

u ∈ W

1,q

(Ω) ∩



∩

i=1

(Ω)



for some q, r

∈ (1, ∞), i = 1, . . . , k. Then, for any η > 0 there exists u

∈

∞

(Ω) such that

ku −u

1,q

i=1

ku − u

< η.

222 III The Function Spaces of Hydrodynamics

Proof. For simplicity, we show the result for k = 1 and set r

= r. Given

ε > 0, we set

(x) = 1 − ψ

(x),

where the function ψ

(x) has been introduced in Lemma III.6.2. We then have

that |ϑ

(x)| ≤ 1, ϑ

(x) vanishes in a neighborhood N

of ∂Ω, ϑ

(x) = 1 for

δ(x) ≥ 2 exp(−1/ε) and

|∇ϑ

(x)| ≤ ε/δ(x), for all x ∈ Ω. (III.6.4)

Putting

(x) = ϑ

(x)u(x),

we at once recognize tha t u

(x) is o f compact support in Ω and that

lim

ε→0

− uk

= 0, s = r, q. (III.6.5)

Furthermore, from (III.6.4),

− u|

1,q

≤ k(1 −ϑ

)∇uk

+ εku/δk

and so, by Lemma III.6.3 , we obtain

−u|

1,q

≤ k(1 −ϑ

)∇uk

+ cε| u|

1,q

with c independent of u and ε. Thus

lim

ε→0

− u|

1,q

= 0. (II I.6.6)

However, u

can be approximated by its regularizer (u

)

in both spaces

1,q

(Ω) and L

(Ω) and since, for ρ small enough, (u

)

∈ C

∞

(Ω), the lemma

follows from this and from (III.6.5) and (III.6.6). ut

We are now in a position to prove the main result.

Theorem III.6.1 Let Ω be a locall y Lipschitz domain of R

, n ≥ 2. Assume

v ∈ H

(Ω) ∩



∩

i=1

(Ω)



for some q, r

∈ (1, ∞), i = 1, . . ., k. Then, there exists a sequence {ϕ

} ⊂

D(Ω) such that

lim

m→∞

−vk

1,q

= lim

m→∞

i=1

− vk

= 0.

III.6 Approximation Problems in Spaces H

and D

1,q

223

Proof. Again, for simplicity, we shall treat the case k = 1 and set r

= r.

Let us ﬁrst consider the case Ω bounded. By Lemma III.6.4 there exists a

sequence {ϕ

} ⊂ C

∞

(Ω) satisfying

lim

m→∞

kϕ

−vk

1,q

= lim

m→∞

kϕ

− vk

= 0. (III.6 .7)

Since

v, ∇ · v ∈ L

(Ω),

and v has zero trace at the b oundary, by Theorem III.2.4 we have

−v ∈

0,r

(Ω).

In addition,

∇ · (ϕ

− v) = ∇ · ϕ

∈ C

∞

(Ω),

and so, by Theorem III.3.4 there exists w

∈ C

∞

(Ω) such that, for all m ∈ N,

∇ · w

= −∇ · ϕ

1,q

≤ c k ∇ · ϕ

≤ c kϕ

−vk

where c is independent of w

, ϕ

, and v. Setting

= ϕ

+ w

it follows that

kv − v

1,q

≤ kϕ

− vk

1,q

+ kw

1,q

≤ kϕ

−vk

1,q

+ ck∇ · ϕ

1,q

kv − v

≤ kϕ

− vk

+ kw

≤ (1 + c)kϕ

−vk

which by (III.6.7) completes the proof in the case where Ω is bounded. Assume

now that Ω is an exterior doma in and denote by ζ

∈ C

∞

) a “cut-oﬀ”

function that equals one in Ω

and zero in Ω

with

|∇ζ

| ≤ c

/R, (III.6.8)

with c

independent of R. Let w

be a solution to the problem

∇ · w

= −∇ · (ζ

∈ W

1,q

(Ω

R,2R

) ∩ W

1,r

(Ω

R,2R

)

1,s,Ω

R,2R

≤ c

k∇ζ

· vk

, s = q, r.

(III.6.9)

By Theorem III.3.1 and Theorem III.3.4 such a solution exists and by Lemma

III.3.1 the constant c

entering the estimate can be taken independent of R.

In view of (III. 6.8) and (III.6.9), we also have

224 III The Function Spaces of Hydrodynamics

1,s,Ω

R,2R

≤ c

−1

kvk

s,Ω

R,2R

, s = q, r, (III.6.10)

where, again, c

does not depend o n R. Set

= ζ

v + w

Clearly,

∈ W

1,q

(Ω

) ∩ L

(Ω

), ∇ · v

= 0.

Since Ω

is l ocally Lipschitz, from Subsection 4 (a) it follows that

∈ H

(Ω

) ∩ L

(Ω

By the ﬁrst part of the proof we may then state that for any ε > 0 there is

ε,R

∈ D(Ω) such that

−v

ε,R

1,q

+ kv

− v

ε,R

< ε

and so for s = q, r

kv − v

ε,R

≤ kv

−v

ε,R

+ kv − v

< ε + k(1 −ζ

)vk

+ kw

s,Ω

R,2R

(III.6.11)

Obviously, for all suﬃciently la rge R,

k(1 − ζ

)vk

< ε. (III.6.12)

Moreover, from inequality (II.5.5) and (III.6.10) it follows that

s,Ω

R,2R

≤ c

R|w

1,s,Ω

R,2R

≤ c

kvk

s,Ω

R,2R

with c

independent of R and so, again for all suﬃciently large R,

s,Ω

R,2R

< ε. (III.6.13)

From (III.6.11)–(III.6.13) we thus ﬁnd that v

ε,R

∈ D(Ω) approaches v in

∩L

. It remains to be shown that v

ε,R

also approaches v in D

1,q

. However,

this is obtained at o nce since, as before, we prove

|v − v

ε,R

1,q

< 2ε + |w

1,q,Ω

R,2R

< 2ε + c

−1

kvk

q,Ω

R,2R

, (III. 6.14)

and so for R large enough we deduce

|v − v

ε,R

1,q

< 3ε

and the proof of the theorem is complete. ut

In an analo gous manner, we can prove

III.6 Approximation Problems in Spaces H

and D

1,q

225

Theorem III.6.2 Let Ω be a locall y Lipschitz domain of R

, n ≥ 2. Assume

v ∈ D

1,q

(Ω) ∩



∩

i=1

(Ω)



for some q, r

∈ (1, ∞), i = 1, . . . , k. Then, there exists a sequence {v

} ⊂

D(Ω) such that

lim

m→∞

−v|

1,q

= lim

m→∞

i=1

− vk

= 0.

Proof. The proof goes exactly as in Theorem III.6.1

except for one point

that demands a little more care. Precisely, again for k = 1 and with r

= r,

once we arrive at (II I.6.14), we cannot immediately conclude that for R large

enough

−1

kvk

q,Ω

R,2R

< ε (III.6.15)

because we do not know that v ∈ L

(Ω). To show (III.6.15) we have to argue

somewhat diﬀerently. If q ∈ (1, n), by the H¨older inequality we ﬁnd

kvk

q,Ω

R,2R

≤ c

Rkvk

nq/(n−q),Ω

with c

= c

(n, q) and since, by the Sobolev inequality,

kvk

nq/(n−q),Ω

< ∞,

we conclude (III.6.15). If r ≥ q ≥ n the same conclusion holds since, in such

a case, by the H¨older inequality, it foll ows that

−1

kvk

q,Ω

R,2R

≤ c

n(1/q−1/r−1/n)

kvk

r,Ω

R,2R

and (III.6.15) is recovered. Finally, if q ∈ [n, ∞) ∩(r, ∞), from Lemma II.3.3

(see also Exercise II I.6.3) we ﬁnd that if

u ∈ D

1,q

) ∩ L

), (III.6.16)

then u ∈ L

) and the following inequality holds

kuk

≤ c|u|

1,q

kuk

1−λ

, (I II.6.17)

with

λ =

n(q − r)

r(q − n) + nq

(< 1) (III.6.18)

and c = c(n, q, r). Since v ∈ D

1,q

) ∩ L

), we apply this result to v

to deduce v ∈ L

) and (III.6.15) again follows. The proof is therefore

completed. ut

Recall that for Ω bounded, H

(Ω) = D

1,q

(Ω).