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

Подождите немного. Документ загружается.

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

(∇(v

+ v

−v), ∇ψ) = (p

+ p

− p, ∇ ·ψ) , ψ ∈ C

∞

(Ω

) ,

and so, from Theorem IV.4.3 we ﬁnd that V := v

−v and P := p

−p

satisfy the following properties

∆V = ∇P in Ω

, V , P ∈ C

∞

(Ω

) . (X.1.8)

Suppose, at ﬁrst, q ≥ 2. By the H¨ol der inequality, Lemma II.6. 3, and Exercise

II.6.3 we ﬁnd (with |x| = r, x ∈ Ω

)

(r) :=

|V (r, ω)|

dω ≤ c



3,S

+ kv

q,S

+ kvk

2,S



≤ c



(ln r)

4/3

+ g(r) + r

−1

+ 1



where g(r) = (ln r)

4/3

if q = 3, while g(r) = r

2−6/q

, if q 6= 3. If q ∈ (3/2, 2),

by the same token, we estimate as follows

(r) :=

|V (r, ω)|

dω ≤ c



3,S

+ kv

q,S

+ kvk

2,S



≤ c



(ln r)

2q/3

+ r

q−3

+ r

−q/2

+ 1



As a consequence, we have, for all ρ > R,

∞

(τ)

(1 + τ)

dτ < ∞, if q ∈ [2, ∞)

∞

(τ)

(1 + τ)

3+q

dτ < ∞, if q ∈ (3/2, 2) ,

so that V satisﬁes the assumption of Theorem V.3. 2, from which we derive,

in particular, P ∈ L

(Ω

), for all ρ > R, and all s > 3/2. The proof of the

theorem is then completed. ut

Remark X.1.2 If n = 2, the proof just g iven fails, because Theorem II.6.1

does not ensure any summability for v + v

∞

on Ω

. Consequently, for plane

motions, a generalized solution has a priori no summability at all in the neigh-

borhood of inﬁnity. However, by a more detailed study performed in Chapter

XII, where we will show that if v

∞

6= 0 the pressure ﬁeld of any weak solution

satisfying some mild extra property and corresponding to suitable b ody force

is i n L

(Ω

) for all q > 2 . If v

∞

= 0, the question is completely open. 

Remark X.1.3 Lemma X. 1.1 is readily extended to dimension n ≥ 4. In

fact, under the assumption

f ∈ D

−1,q

(Ω

) , q ∈ (n/(n −1), ∞)

one can show that

X.1 Generalized Solutions and Related Properties 657

p = p

+ p

, p

∈ L

n/(n−2)

(Ω

) , p

∈ L

(Ω

This result is obtained by the same technique used in Lemma X.1.1, observing

that, by Theorem II.6.1, one has

(v + v

∞

) · ∇v ∈ D

−1,n /(n−2)

(Ω

In particular, if Ω is locally Lipschitz and q = n/(n − 2), we have

p ∈ L

n/(n−2)

(Ω),

thus recovering for p the same property proved in the case Ω bounded;

cf. Remark IX.1.5. It is interesting to observe that if the asymptotic properties

of v merely reduce to v ∈ D

1,2

(Ω), the exponent n/(n − 2) is sharp, even if

f ∈ C

∞

(Ω), as the f ollowing argument shows. Consider (X.0.8), (X.0.4) in

Ω ≡ R

, n ≥ 3, with v

∞

= 0 and assume v, p to be a smooth solution cor-

responding to f = ∇ · F , where F ≡ {F

} ∈ C

∞

). (As will be proved in

Theorem X.4.1, such a solution does exist.) By taking the divergence operator

of both sides of (X.0.8) we then recover that the pressure ﬁeld satisﬁes the

following equation

∆p =

i,j=1

+ F

Starting with this relation, we can then gi ve an explicit representation f or p.

Actually, if p tends to a limit p

at i nﬁnity, employing classical arguments we

can easily show (with c

= c

(n), i = 1, 2) that

p(x) = p

+ R{c

i=1

(x) + F

(x)]

+ c

i,j=1

lim

ε→0

−B

(x)

(y) + F

(y))[D

|x −y|

2−n

]dy}

≡ p

+ p

(x) + p

(x).

v ∈ D

1,2

), (X.1.9)

by the Sobolev inequality it follows that

∈ L

n/(n−2)

) (X.1.10)

and so

∈ L

n/(n−2)

Clearly,

∈ L

) f or all q > 1.

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

Moreover, the quantity

|x − y|

2−n

is a singular kernel, and so, by the Calder´on–Zygmund theorem, by (X.1.9)

and the properties of F we have

∈ L

n/(n−2)

)

∈ L

) f or all q > 1.

We then conclude that

p ∈ L

n/(n−2)

). (X.1.11)

Since (X.1.10 ) cannot be improved under the sole assumption (X.1.9), the

estimates on p

and p

are sharp and therefore (X.1 .11) is sharp too. 

Exercise X.1.1 Let Ω be a exterior domain of R

, n ≥ 3, of class C

. By the same

argument used in the proof of Lemma X.1.1, show t hat if v is a weak solution to

(X.0.8), (X.0.4) with

∗

∈ W

1−1/q,q

(∂Ω), f ∈ D

−1,q

(Ω) , q ∈ (n/(n − 1), ∞)

then p = p

+ p

with p

∈ L

n/(n−2)

(Ω), p

∈ L

(Ω), where p is the pressure ﬁeld

associated to v by Lemma X.1.1.

In the ﬁnal part of this section we shall discuss the diﬀerentiability prop-

erties of a generalized solution. This will be done with the help of Theorem

IX.5. 1 and Theorem IX.5.2. Speciﬁcally, we have the following

Theorem X.1.1 Let v be a generalized solution to (X.0.8), (X.0.4 ) in an

exterior domain Ω ⊆ R

, n ≥ 2 with v ∈ L

loc

(Ω).

Then, if

f ∈ W

m,q

loc

(Ω), m ≥ 0 , (X.1.12)

where q ∈ (1, ∞) if either m = 0 or n = 2, while q ∈ [n/2, ∞) if m > 0 and

n ≥ 3, it fol lows that

v ∈ W

m+2,q

loc

(Ω), p ∈ W

m+1,q

loc

(Ω),

where p is the pressure associated to v by Lemma X.1.1.

Thus, in particular,

These assumptions on on v can be replaced by the weaker requirements (i)–(iii)

of Theorem IX.5.1; see also Remark IX.5.1.

Observe that if f satisﬁes (XI.1.23) for the speciﬁed values of m and q, t hen f or

all bounded Ω

with Ω

⊂ Ω it holds that

f ∈ W

−1,2

(Ω

), if n = 2, 3

while

f ∈ W

−1,n/(n−2)

(Ω

), if n ≥ 4.

X.2 On the Validity of the Energy Equation for Generalized Solutions 659

f ∈ C

∞

(Ω),

then

v, p ∈ C

∞

(Ω).

Assume, further, tha t v ∈ L

(Ω

), for some R > δ(Ω

). Then if Ω is of class

m+2

and

∗

∈ W

m+2−1/q,q

(∂Ω), f ∈ W

m,q

(Ω

) (X.1.13)

with the values of m and q speciﬁed earlier, we have

v ∈ W

m+2,q

(Ω

), p ∈ W

m+1,q

(Ω

). (X.1.14 )

Thus, i n particular, if Ω is of class C

∞

and

∗

∈ C

∞

(∂Ω), f ∈ C

∞

(Ω

it follows that

v, p ∈ C

∞

(Ω

Proof. The ﬁrst part of the theorem (local regularity) has already been shown

in Theorem IX.5.1. To show the second, from the hypothesis on f made in

(X.1.13) we o bta in, again by Theorem IX.5.1,

v ∈ W

m+2,q

loc

(Ω),

and, as a consequence,

v ∈ W

m+2−1/q,q

(∂B

Therefore, v is a generalized solution to the Navier–Stokes problem in the

bounded domain Ω

with v ∈ L

(Ω

) and corresponding to data satisfying

the assumption of Theorem IX.5.2. Therefore, (X.1.14) foll ows and the result

is completely proved. ut

X.2 On the Validity of the Energy Equation for

Generalized Soluti ons

When we give a mathematical deﬁnition of solution to a system of equations

relating to physics, one of the ﬁrst questions that arises is that of establishing

if these solutions meet all basic physical principles such as conservation laws

or energy balance. Thus, in the present situa tion, a ﬁrst issue to investigate is

to ascertain if a generalized solutio n satisﬁes the energy equation. In the case

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

when the velocity at the bo unda ry vanishes identically

this equatio n in its

classical f ormulation reads:

Ω

D(v) : D(v) −v

∞

∂Ω

T (v, p) ·n + R

Ω

f · (v + v

∞

) = 0 (X.2.1)

where D and T are stretching and stress tensors deﬁned in (IV.8.6). The

meaning of (X.2.1) is quite clear: it represents the balance between the dissi-

pation due to the viscosity, the power of external body force and, for v

∞

6= 0,

the tota l force exerted by the l iquid on the bodies. If Ω is bounded, the proof

of the energy equation (i.e., the relation formally obtained from (X.2.1) with

∞

= 0) depends solely on the regularity in Ω of the solution and one readily

shows that, if f has a mild degree of smoothness, every corresponding gen-

eralized solution satisﬁes the energy equation; see Exercise IX.3.1. However,

if Ω is an exterior domain, the pro of of (X.2.1) demands not o nly certain

regularity in Ω

, for all R > δ(Ω

), but it also depends in an essential way on

the asymptotic properties of the solution. To see this, assume v, p is a classical

solution to (X.0.8), (X.0 .4) with v

∗

≡ 0. Multipl ying (X.0.8) by u := v + v

∞

integrating by parts over Ω

and taking into account (X.0.8)

, we deduce

Ω

D(v) : D(v) −v

∞

∂Ω

T ( v, p) · n + R

Ω

f · (v + v

∞

)

= R

∂B

u · T (v, p) · n −

∂B

v · n.

(X.2.2)

If v is a generalized solution,

v ∈ D

1,2

(Ω) . (X.2.3)

So, if f is “well-behaved” at inﬁnity, letting R → ∞ into (X.2.2), the l eft-

hand side of (X. 2.2) tends to the left-hand side of (X.2.1); nevertheless the

sole condition (X.2.3), together with condition (X.1.6) o n the pressure,

is not

enough to ensure that the surface integral on the right-hand side of (X.2. 2)

tends to zero as R → ∞, even along a sequence. Thus, we are not able to

conclude the validity o f (X.2.1), unless we have further information on the

behavior of v a t large distances.

Another (equivalent) way of looking a t the problem is the following one.

Assume, for simpli ci ty, v

∗

≡ v

∞

≡ 0. Then, ta king into account that | v|

1,2

2kD(v)k

, in our standard no tation, the energy equation (X.2.1) becomes

|v|

1,2

= −R[f, v] . (X.2.4)

This assumption i s made here for the sake of simplicity. The case v

∗

6= 0 will be

considered later in this section.

Recall that, by Remark X.1.3, (X.1.6) is sharp for a generalized solution. More-

over, (X.1.6) holds if Ω ⊂ R

, otherwise, if Ω ⊂ R

, we do not even know if

p ∈ L

(Ω

) for some q ∈ [1, ∞).

X.2 On the Validity of the Energy Equation for Generalized Solutions 661

From the deﬁnitio n of g eneralized solution we know that

(∇v, ∇ϕ) + R(v · ∇v, ϕ) = −R[f, ϕ] (X.2.5)

for all ϕ ∈ D(Ω). The natural step would then be to put v into (X.2.5) in

place of ϕ. However, unlike the case where Ω is bounded (see Exercise IX.3.1),

we are not allowed a priori to do this, since v does not have the appropriate

summability prop erties at large distances. This fact is intimately related to

the continuity properties of the trilinear form

a(u, v, w) ≡ (u · ∇v, w) (X.2.6)

which, for an arbitrary domain Ω are speciﬁed in the following.

Lemma X.2.1 Let Ω be an arbitrary domain in R

, n ≥ 2. Then the trilinear

form (X.2.6) is continuous in the space

(Ω) ×

1,r

(Ω) × L

(Ω), q

−1

+ r

−1

+ s

−1

= 1.

If n ≥ 3, it is al so continuous in the space

1,2

(Ω) ×

1,2

(Ω) × L

(Ω).

Proof. The ﬁrst assertion is a trivial consequence of the H¨older inequality,

while the second is proved by choosing q = 2n/(n − 2), r = n and using the

Sobolev inequality (II.3.7):

kuk

2n/(n−2)

≤

q(n − 1)

2(n − q)

√

|u|

1,2

In view of this lemma we can thus for m ally

replace ϕ in (X. 2.5) with the

generalized solution v if, for instance, v ∈ L

(Ω) (by cho osing q = s = 4, r =

2) or, if n ≥ 3, v ∈ L

(Ω).

The reasonings just developed explain why, in the case of an exterior do-

main Ω, a solution should still be considered a “generalized” solution even

though it is smooth in Ω

for any R > δ(Ω

). Actually, unlike the case where

Ω is bounded, where the word “generalized” expresses only a possible lack

of diﬀerentiability, for Ω exterior it is also to mean the possibility that the

solution need not be as “regular” in the neighborhood of i nﬁnity as expected

from the physical (and intuitive) point of view.

Our objective in this section is to investigate what regularity must be

imposed on the velocity ﬁeld at large distances in order that it satisﬁes equa-

tion (X. 2.1). Successively (cf. Section X.7 and Section X.9), we shall analyze

We have, in fact, to ascertain that v can be suitably approximated by functions

from D(Ω).

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

whether such a regularity can be deduced by maki ng suitable assumptions on

the data (on f in particular). Speciﬁcally, when the region of ﬂow is three-

dimensional, we shall prove that if f enjoys certain summability conditions,

every generalized solution corresponding to suitable f and v

∞

6= 0 obeys the

energy equation. If v

∞

= 0, however, we are able to show the same result

only f or small Reynolds number R. Thus, if v

∞

= 0 the validity of (X.2.1) for

arbitrary R remains open. If the region of ﬂow is two-dimensional the picture

is even less clear and wi ll be treated i n detail in Chapter XI. Speciﬁcall y, for

plane ﬂows one shows that, if f is suitable and v

∞

6= 0, every generalized so-

lution satisfying a mild extra property obeys (X.2.1) (even though existence

is only known for small R) while, if v

∞

= 0, the question remains entirely

open.

Before performing the above investigation, however, there is a formal as-

pect that must be ﬁxed. If v

∞

6= 0 and v merely belo ngs to W

1,2

loc

(Ω

) for all

R > δ(Ω

), as prescribed by Deﬁnition X.1.1, the second integral in (X.2.1)

need not be meaningful. Thus, in such a case, we have to introduce a suitable

generalization of (X.2.1). Let us denote by a any vector ﬁeld in Ω verifying

(i) a ∈ C

∞

(Ω);

(ii) ∇ · a = 0;

(iii) a = 0 in Ω

and a + v

∞

≡ 0 in Ω

for som e d > δ(Ω

(X.2.7)

For instance, we may take a as in (V.2.5). If v

∞

= 0 we take a ≡ 0. The

ﬁeld a will be called an extension of v

∞

. We then give the following.

Deﬁnition X.2.1. Let v be a generalized solution to (X.0.8), (X.0.4) corre-

sponding to v

∗

≡ 0 and let a be a n extension of v

∞

. The relation

|v|

1,2

+ R[f, v + a] = (∇v, ∇a) −R(v ·∇a, v + a) (X.2.8)

is called generalized energy equation.

To justify the preceding deﬁnition, we notice that, if v

∞

= 0, (X.2.8)

coincides with (X.2.1), while it reduces to (X.2.1) for v

∞

6= 0, provided Ω, f ,

v, and p are suﬃciently smooth. For example, if Ω is of class C

and

v ∈ C

(Ω

) ∩ C

(Ω

), p ∈ C(Ω

) ∩ C

(Ω

), f ∈ C(Ω),

multiplying (X.0.8) by a + v

∞

, integrating by parts over Ω

and observing

that supp (∇a) ⊂ Ω

d,2d

it follows that

∂Ω

n·T (v, p)·v

∞

= R[f , a+v

∞

]+ 2(D(v), D(a))−R(v·∇a, v−a). (X.2. 9)

Taking into account that

Let u

, u

∈ D

1,2

(Ω), with u

−u

∈ D

1,2

(Ω). Then, the following identity holds

X.2 On the Validity of the Energy Equation for Generalized Solutions 663

|v|

1,2

−(∇v, ∇a) = 2[kD(v)k

−(D(v), D(a))],

(X.2.9) along with (X.2.8), impl ies (X.2.1).

Exercise X.2.1 The validity of (X.2. 9) can be shown under very mild regularity

assumptions on f . For example, suppose that, for some R > δ(Ω

), f ∈ L

(Ω

where r is arbitrary in (1, ∞). Assume, further, v

∗

≡ 0 and Ω is of class C

. Show

that every generalized solution v corresponding to these data satisﬁes (X.2.9) with

p pressure ﬁeld associated to v by Lemma X.1.1. Hint: Use Theorem X.1.1 and

Exercise II.4.3.

Remark X.2.1 In Section X.4 we shall prove that, regardless of the dimen-

sion n ≥ 2, for any f ∈ D

−1,2

(Ω) there exists a corresponding generalized

solution that (for v

∗

≡ 0) satisﬁes the generalized energy inequality, that is, re-

lation (X.2.8) with “=” replaced by “ ≤ ”. However, the case Ω = R

appears

to be open; see Remark X.4.4 

The main results of this section are contained i n Theorem X.2.1 and The-

orem X.2.2, which we are now go ing to prove.

Theorem X.2.1 Let v be a generalized solution to the Navier–Stokes prob-

lem (X.0.8), (X.0.4) in an exterior locally Lipschitz domain of R

, n = 2, 3,

corresponding to f ∈ D

−1,2

(Ω), v

∗

= 0 and to some v

∞

∈ R

. Then if

∞

= 0, a suﬃcient condition in order that v satisﬁes the generalized energy

equation (X.2.8) with a ≡ 0 is

v ∈ L

(Ω) (X.2.10)

while, if v

∞

6= 0, a suﬃcient condition in order that v satisﬁes (X.2.8) for any

extension a of v

∞

v + v

∞

∈ L

(Ω) ∩ L

(Ω)

∞

· ∇v ∈ L

(Ω)

(X.2.11)

for som e q ∈ (1, ∞), q

= q/(q − 1).

Proof. Let us ﬁrst consider the case where v

∞

= 0. By Deﬁnition X.1.1 and

Remark XI.1.1 we have

v ∈

1,2

(Ω)

and, since Ω is locally Lipschitz, by the results of Section III.5,

v ∈ D

1,2

(Ω).

1,2

− (∇u

, ∇u

) = 2[kD(u

− (D(u

), D(u

))] .

In fact, let {w

} ⊂ D(Ω) be such that |w

− (u

− u

1,2

→ 0 as k → ∞.

Multiplying both sides of the identity 2∇ · D(w

) = ∆w

by u

, integrating by

parts over Ω and letting k → ∞ proves the result.

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

Denote by {v

} ⊂ D(Ω) a sequence of functions converging to v in the space

1,2

(Ω) ∩ L

(Ω). In view of (X.2.10) and Theorem III.6.2 this sequence cer-

tainly exists. Replacing ϕ w ith v

into (X.1.2) furnishes

(∇v, ∇v

) + R(v · ∇v, v

) = −R[f, v

]. (X.2.12)

Clearly, as k → ∞,

(∇v, ∇v

) → |v|

1,2

[f , v

] → [f , v] .

(X.2.13)

Furthermore, by Lemma X.2.1, we have

|(v ·∇v, v

) − (v · ∇v, v)| ≤ kvk

|v|

1,2

kv − v

and so, in virtue of (X.2.10),

(v ·∇v, v

) → (v · ∇v, v). (X.2.14)

However, it is immediately seen that

(v · ∇v, v) = 0. (X.2.15)

Actually, (X.2.15) is a consequence of Lemma X.2.1 and the fact that, by

Lemma IX.2.1,

·∇v

, v

) = 0, for all k ∈ N.

The proo f of the theorem when v

∞

= 0 then follows from (X.2. 12)–(X.2.15).

Consider next the case where v

∞

6= 0. To this end, let a be any extension of

∞

and set

u = v + a.

Since Ω is locally Lipschitz, by (i)-(iv) of Deﬁnition X.1.1, Theorem II.6.3,

and the results of Section III.5 we o bta in, as before,

u ∈ D

1,2

(Ω). (X.2.16)

Furthermore, from (X.2.11 )

we also have

u ∈ L

(Ω) ∩ L

(Ω). (X.2.17 )

In view of (X.2.16) and (X.2 .17), Theorem III.6.2 implies the existence of a

sequence {u

} ⊂ D(Ω) converging to u in the space

1,2

(Ω) ∩ L

(Ω).

Setting ϕ = u

into (X.1.2 ) we thus ﬁnd

(∇v, ∇u

) + R((v + v

∞

) ·∇v, u

) −R(v

∞

·∇v, u

) = −R[f , u

] . (X.2.18)

X.2 On the Validity of the Energy Equation for Generalized Solutions 665

Taking into account Lemma X.2.1, (X.2.11), and (X.2.17), we recover, in the

limit k → ∞

(∇v, ∇u

) → (∇v, ∇u)

((v + v

∞

) ·∇v, u

) → ((v + v

∞

) ·∇v, u)

∞

·∇v, u

) → (v

∞

· ∇v, u).

(X.2.19)

Moreover, since for all k ∈ N

((v + v

∞

) · ∇u

, u

) = 0,

we have, again by Lemma X.2.1,

((v + v

∞

) · ∇u, u) = 0

and (X.2 .18), (X.2.19) furnish

|v|

1,2

+ R[f , v −a] = (∇v, ∇a) − R(v · ∇a, v − a) − R(v

∞

·∇u, u) .

(X.2.20)

We have now to show that the last term on the right-hand side o f (X.2.20)

vanishes identically. First of all, we observe that it is clearly well-deﬁned since,

by (X.2.11)

and (X.2.17) it follows that

∞

· ∇u · u ∈ L

(Ω). (X.2.21)

Next, taking v

∞

= e

, for ρ > δ(Ω

) we let {C

k,ρ

}, k ∈ N, denote a fam ily of

cylinder-like domains of the form

k,ρ

= {x ∈ Ω : |x

| < k, |x

| < ρ},

where x

= (x

, . . . , x

). For suﬃciently large k and ρ it is C

k,ρ

⊃ Ω

and,

correspondingly, we ﬁnd

k,ρ

∞

· ∇u · u =

k,ρ

∂u

∂x

· u =

(k,ρ)

−

(k,ρ)

, (X.2.22)

where

(k, ρ) = {x ∈ R : x

= ρ, |x

| ≤ k},

(k, ρ) = {x ∈ R : x

= −ρ, |x

| ≤ k}.

By (X.2.11)

, it i s easy to show that, for each ﬁxed k,

lim

ρ→∞

(k,ρ)

= 0, i = 1, 2. (X.2.23 )

In fact, setting x

= (ρ, 0, . . ., 0), by the boundary inequality (II.4.1), it follows

that