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

Подождите немного. Документ загружается.

756 XI Three-Dimensional Flow in Exterior Domains. Rotational Case

surface integrals on ∂B

converge to zero a s R → ∞. We also notice that,

up to the pressure term, (XI.2.1) coi ncides with the energy equality (X.2.29)

proved for the irrotational case. The reason i s that the total power of the

rotational contribution vanishes identically.

We shall deal directly with the validity of (XI.2.1), leaving the ana logous

proof of the “generalized energy equality” (in the sense of Deﬁnition X.2.2) as

an exercise to the interested reader. We recall that this latter coincides with

(XI.2.1) for suﬃciently smooth domains (for exam ple, of class C

) and data.

We thus have the following.

Theorem XI.2.1 Let Ω be of cla ss C

, and assume

f ∈ L

4/3

(Ω) , v

∗

∈ W

5/4,4/3

(∂Ω) . (XI.2.2)

Then, any generalized solution corresponding to the above data that in addi-

tion satisﬁes

(v + v

∞

) ∈ L

(Ω) (XI.2.3)

satisﬁes the energy equality (XI.2.1).

Proof. Set u := v + v

∞

. From (IV.8.9), (XI.2.2), and Theorem XI.1.2 we

obtain

∇· T (u, ep) + λ

∂u

∂x

+ T (e

× x · ∇u − e

× u) = λ

u · ∇u + f

∇· u = 0











a.e. in Ω ,

(XI.2.4)

with ep deﬁned in (XI.1.6), and where λ

= λ

= R if v

·ω 6= 0 (with v

∞

as in

(XI.0.11)

), while λ

= 0 and λ

= T if v

·ω = 0 (with v

∞

as in (XI.0.11 )

Since v is a weak solution, we have (see (XI.1.9))

u ∈ D

1,2

(Ω) ∩ L

(Ω) . (XI.2.5)

We analyze ﬁrst the case v

· ω 6= 0 (see (XI.0.11)

), and notice that in

particular, by (XI.2.2), (XI.2.3), and the H¨older inequality,

u · ∇u + f ∈ L

4/3

(Ω) . (XI.2.6)

Consequently, from this property, (XI.2.2), (XI.2.5), a nd Theorem VIII.8 .1, it

follows that

∂u

∂x

∈ L

4/3

(Ω) , ep ∈ L

12/5

(Ω) . (XI.2.7)

Let ψ

be the “cut-oﬀ” function used in the proof of Theorem X.3.2, and let us

dot-multiply both sides of (XI.2.4) by ψ

u. By an easily justiﬁed integration

by parts on the resulting equation, and taking into account that u = v

∗

+ v

∞

at ∂Ω (in the trace sense), we obtain

XI.2 On the Energy Equation and the Uniqueness of Generalized Solutions 757

2(D(u), D(ψ

u)) −

∂Ω



∞

+ v

∗

) · T (v, ep) −

∗

+ v

∞

)

∗



· n



∂u

∂x

, ψ



R(|u|

∇ψ

, u) − (epu, ∇ψ

) + (f, ψ

u) = 0 ,

(XI.2.8)

where we have used ∇ψ

·(e

×x) = 0. We now recall the f ollowing properties

of the function ψ

lim

R→∞

(x) = 1 , for all x ∈ Ω ,

supp (ψ

) ⊂ Ω

R,2R

, |∇ψ

| ≤ C/R

(XI.2.9)

for a constant C independent of x and R. Thus, using this la tter along with

(XI.2.3), (XI.2.5), and (XI.2.7), the reader will have no diﬃculty in proving

the following:

lim

R→∞

(D(u), D(ψ

u)) = (D(u), D(u)) , lim

R→∞

(f, ψ

u) = (f, u) ,

lim

R→∞

(|u|

∇ψ

, u) = 0 , lim

R→∞



∂u

∂x

, ψ





∂u

∂x

, u



(XI.2.10)

However, as shown in the proof of Theorem X.2.1, we obtain



∂u

∂x

, u



= 0 . (XI.2.11)

Finally, by the H¨older inequality and by (XI.2.7), we have

|(epu, ∇ψ

)| ≤ kepk

12/5,Ω

R,2R

kuk

4,Ω

R,2R

k∇ψ

3,Ω

R,2R

≤ C kepk

12/5,Ω

R,2R

kuk

4,Ω

R,2R

where in the last inequality, we have used (XI.2.9)

. Thus,

lim

R→∞

(epu, ∇ψ

) = 0 . (XI.2.12)

Consequently, letting R → ∞ in (XI.2.8) and employing (XI.2.10)–(XI.2.12),

we recover (XI.2.1), which is therefore proved in the case (XI.0.11)

. The proof

in the case (XI.0.11)

is entirely analogous. (We recall that in this situation,

in (XI.2.4 ) we must take λ

= 0 and λ

= T .) It suﬃces to observe that from

(XI.2.3), (XI.2.6), and Theorem VIII.7.2 , it follows that ep (up to an inessential

constant) satisﬁes (XI.2.7). C onsequently, following step by step the argument

previously used for the case (XI.0.11)

, we prove the desired property also i n

case (XI.0.11)

. The theorem is completely proved. ut

Our next objective is to furnish suﬃcient conditions for uniqueness of

generalized solutions. We have the foll owing.

758 XI Three-Dimensional Flow in Exterior Domains. Rotational Case

Theorem XI.2.2 Let Ω, f, and v

∗

satisfy the assumptions of Theorem

XI.2. 1, and let v, v

be two corresponding generalized solutions satisfyi ng,

in addition, (XI.2.3).

Then, if (v + v

∞

) ∈ L

(Ω) with

kv + v

∞

√

, (XI.2.13)

necessarily v(x) = v

(x), a.e. in Ω.

Proof. We shall give the proof w hen v

· ω 6= 0, so that v

∞

is given in

(XI.0.11)

. In fact, as the reader will immediately realize, it remains basically

unchanged in the case v

·ω = 0. Set u := v

−v, φ := ep

− ep, w := v + v

∞

:= v

∞

, where ep and ep

are the (modiﬁed, according to (XI.1.6 )) pres-

sure ﬁelds associated to v and v

by Lemma XI.1.1. From (IV.8.9), (XI.2.2),

and Theorem XI.1.2 we thus obtain

∇ · T (u, φ) + R

∂u

∂x

+T (e

×x · ∇u −e

×u)

= R(u · ∇u + w · ∇u + u · ∇w)

∇ · u = 0











a.e. in Ω ,

u = 0 at ∂Ω .

(XI.2.14)

We now foll ow exactly the same pro cedure used in the proof of Theorem

XI.2. 1, that is, after dot-multiply ing bo th sides of (XI.2.14 )

by ψ

u, inte-

grating by parts over Ω, and tak ing into account the boundary conditions, we

let R → ∞. Using the fact that both v and v

satisfy (XI.2 .3) we then show

that u obeys the following “perturbed” energy equality:

|u|

1,2

= R(u · ∇u, w) , (XI.2.15)

where we have used the relation |u|

1,2

√

2kD(u)k

By (XI.2.15), the

H¨older inequality, and the Sobolev inequality (II.3.11), we thus obtain

|u|

1,2



1 − R

√

kwk



≤ 0 ,

which, in turn, proves the result under the assumption (XI.2. 13). ut

For future use, we need another uniqueness result that generalizes to the

rotational case the one proved in Theorem X.3.2. We shall limit ourselves to

the case v

·ω = 0, namely, when v

∞

satisﬁes (XI.0.11)

, in that it wi ll suﬃce

for o ur purposes.

Thus, let C be the class of general ized solutions v to (XI.0.10), (XI.0.11)

with v

·ω = 0 satisfying the energy inequality

The assumption on v

can be somehow weakened. See Exercise XI.2.1.

See Footnote 4 in Chapter X.

XI.2 On the Energy Equation and the Uniqueness of Generalized Solutions 759

Ω

D(v) : D(v) −

∂Ω



∞

+ v

∗

) · T (v, ep) −

∗

+ v

∞

)

∗



· n

Ω

f · (v + v

∞

) ≤ 0 .

(XI.2.16)

As we shall prove in the following section, under suitable regularity assump-

tions on Ω, f , and v

∗

, the class C is not empty.

We have the foll owing.

Theorem XI.2.3 Let Ω be of cla ss C

, and let

f ∈ L

(Ω) ∩ L

6/5

(Ω) , v

∗

∈ W

3/2,2

(∂Ω) , v

·ω = 0.

Suppose v i s a corresponding generalized solution such that for all x ∈ Ω,

(1 + |x|)|v(x) + v

∞

| ≤ M , M <

. (XI.2.17)

Then v is unique in the class C.

Proof. Let v

be any element in C corresponding to the given data, let p

the associated pressure, and set w := v

+ v

∞

. From (IV.8.9), (XI.2.2), and

Theorem XI.1.2 we deduce that w satisﬁes

∇ · T (w, ep

) + T (e

× x · ∇w − e

× w) = T w ·∇w + f

∇ · w = 0

)

a.e. in Ω ,

w = v

∗

+ v

∞

≡ d

∗

at ∂Ω,

(XI.2.18)

with ep

the modiﬁed pressure as in (XI.1.6). Dot-multiplying both sides of

(XI.2.18)

by u := v + v

∞

and integrating by parts over Ω

, we obtain

−2

Ω

D(w) : D(u) + 2

∂B

n · D(w) · u −

∂B

(u · n)

−T

Ω

w · ∇w · u + T

Ω

× x · ∇w − e

× w) ·u

= −

∂Ω

n · T (w, ep

) · d

∗

Ω

f · u .

(XI.2.19)

Likewise, interchanging the roles of u and w, we get

−2

Ω

D(w) : D(u) + 2

∂B

n · D(u) · w −

∂B

φ(w ·n)

−T

Ω

u · ∇u · w + T

Ω

×x · ∇u −e

× u) · w

= −

∂Ω

n · T (u,

φ) · d

∗

Ω

f · w ,

(XI.2.20)

760 XI Three-Dimensional Flow in Exterior Domains. Rotational Case

where

φ is the modiﬁed pressure, according to (XI.1.6), associated to v. Our

next task is to show that all integrals on ∂B

in (XI.2 .19) and (XI.2.20)

converge to zero as R → ∞, along a sequence at least. To this end, we notice

that in (XI.2.18), written with w ≡ u and ep

≡

φ, its right-hand side can be

put in the form T ∇·(u⊗u)+f . Since u decays pointwise in the way speciﬁed

in (XI.2.17), and u ∈ D

1,2

(Ω), it follows that u ⊗ u ∈ L

(Ω) and ∇ · (u ⊗

u) ∈ L

(Ω). Moreover, by assumption, f ∈ L

6/5

(Ω). As a consequence, f rom

Lemma VIII.2 .2 we deduce

φ ∈ L

(Ω

), for suﬃciently large R. Moreover, by

Theorem XI.1.2, ep

∈ L

(Ω

). Thus, recall ing al so that w ∈ L

(Ω), we can

ﬁnd an unbounded sequence {R

} such that

lim

k→∞

∂B



|ep

+ |∇w|

+ |w|

+ |

φ|

+ |∇u|

+ |u|



= 0 . (XI. 2.21)

Using this property, we shall now show that all surface integrals over ∂B

(XI.2.19), (XI.2.20) tend to zero as R → ∞, at least along the sequence {R

In fact, setting

[]u[]

:= sup

x∈Ω

(1 + |x|)|u(x)|,

we have, as k → ∞,



∂B

n · D(w) · u



≤ c

[]u[]

∂B

|∇w|

≤ c

[]u[]

k∇wk

2,∂B

→ 0 ,



∂B

(u · n)



≤ c

[]u[]

∂B

|ep

≤ c

[]u[]

kep

3,∂B

→ 0 ,



∂B

n · D(u) ·w



≤ c

k∇uk

2,∂B

kwk

6,∂B

→ 0 ,



∂B

φ(w · n)



≤ c

φk

2,∂B

kwk

6,∂B

→ 0 .

(XI.2.22)

We next investigate the convergence of the volume integrals in (XI.2.19),

(XI.2.20). To this end, we begin by observing that in view of the a ssumptions

made on u a nd Theorem II.6.1(i), we obtain



Ω

w · ∇w · u



≤ c

[]u[]

|w|

1,2

so that

lim

k→∞

Ω

w · ∇w · u =

Ω

w · ∇w · u . (XI.2.2 3)

By the same token,

XI.2 On the Energy Equation and the Uniqueness of Generalized Solutions 761

lim

k→∞

Ω

u · ∇u · w =

Ω

u · ∇u · w . (XI.2.24)

Likewise, since f ∈ L

6/5

(Ω) and u, w ∈ L

(Ω), we deduce

lim

k→∞

Ω

f · w =

Ω

f · w , lim

k→∞

Ω

f · u =

Ω

f · u . (XI.2.25)

We now observe that by an integration by parts, we can show for all R > δ(Ω

)

that

Ω

× x · ∇w− e

× w) ·u +

Ω

×x ·∇u −e

×u) · w

∂B

× x) · n w ·u +

∂Ω

× x) · n d

∂Ω

×x) · n d

∗

(XI.2.26)

where in the last step, we have taken into account that on ∂B

, n and x

are parallel. Adding side by side (XI.2.18) and (XI.2.19), using (XI.2.26), and

then passing to the limit k → ∞, with the help o f (XI.2.23)–(XI.2.25) we

recover

−4

Ω

D(w) : D(u) = T

Ω

(w · ∇w · u + u ·∇u · w) +

Ω

(f · w + f · u)

∂Ω



n · T (w, ep

) + n · T (u,

φ)



· d

∗

−T

∂Ω

×x) · n d

∗

(XI.2.27)

However, by (XI.2.17) and Theorem XI.2.1, (u,

φ) satisﬁes the energy equality

(XI.2.1), while by assumption, (w, ep

) satisﬁes the energy inequality (XI.2.16).

Thus adding, side by side, these two relations and (XI.2.27), and observing

that

kD(w − u)k

= kD(w)k

+ kD(u)k

−2(D(w), D(u)) ,

we easily deduce

kD(z)k

≤ T ((w · ∇w, u) + (u · ∇ u, w)) − T

∂Ω

∗

·n , (XI.2.28)

with z := w − u. However, by a slight modiﬁcation of the reasoning used in

the proof of Theorem X.3.2

we show that

Speciﬁcally, see the proof of properties (i) and (ii) after (X.3.29) and t he argument

that follows.

762 XI Three-Dimensional Flow in Exterior Domains. Rotational Case

(w · ∇w, u) + (u · ∇u, w) = (z · ∇z, u) +

∂Ω

∗

· n ,

and so, placing this latter into (XI.2.27), we conclude that

2kD(z)k

≤ T (z ·∇z, u) .

Using in this relation the assumption (XI.2.17) along with the Schwarz in-

equality, we obtain

2kD(z)k

≤ T M



Ω

|z|

|x|



1/2

|z|

1,2

. (XI.2.29)

We now observe that for ϕ ∈ D(Ω), by dot-multiplying both sides of the

identity ∆ϕ = 2 ∇ · D(ϕ) by ϕ and integrating by parts, we have

2kD(ϕ)k

= |ϕ|

1,2

. (XI.2.30)

Moreover, by (II. 6.10), we also have that

Ω

|ϕ|

|x|

≤ 4|ϕ|

1,2

. (XI.2.3 1)

Now, z ∈ D

1,2

(Ω), with ∇ · z = 0. Furthermore, z has zero trace at the

boundary, and so by Theorem III.5.1, z ∈ D

1,2

(Ω). Then, by a simple density

argument we prove that (XI.2.30) and (XI. 2.31) continue to hold for z too.

Thus, placing (XI. 2.30) and (XI.2.31) (with ϕ ≡ z) into (XI.2.29), we obtain

|z|

1,2

(1 − 2T M) ≤ 0 ,

from which, if M < 1/(2T ), uniqueness follows. ut

Exercise XI.2.1 Show that in the uniqueness Theorem XI.2.2 the hypothesis v

∈

(Ω) can be replaced by the assumption that v

and the associated pressure p

satisfy the energy inequality (XI.2.16). As shown in the next section, the class of

such solutions is not empty.

XI.3 Existence of Generalized Solutions

The o bjective of this section is to prove the existence of a generali zed so-

lution to problem (XI.0.10 ), (XI.0.11). Thi s will be achieved by a suitable

generalization of the arguments used i n the proof o f Theorem X.4.1.

We begin with the following result.

Lemma XI.3.1 Let Ω be locally Lipschitz and let

∗

∈ W

1/2,2

(∂Ω) , v

∞

be given in (XI.0.11) .

XI.3 Existence of Generalized Solutions 763

Then, for any η > 0 there exist ε = ε(η, v

∗

, Ω) > 0 and V = V (ε) : Ω → R

satisfying properties (i)–(iv) of Lemma X.4.1 with v

∞

≡ v

∞

. The ﬁeld V can

be written as follows:

V (x) = V

(x) + Φσ(x) −v

∞

(x) , (XI.3.1)

where V

(x) is o f bounded support in Ω, and

Φ :=

∂Ω

∗

·n , σ(x) =

4π

∇



|x − x



, (XI.3.2)

with x

∈

◦

Ω

. Moreover, for all u ∈ D

1,2

(Ω), we have

|(u · ∇(V + v

∞

), u)| ≤



η +

|Φ|

4πr



|u|

1,2

, (XI.3.3)

where r

= dist (x

, ∂Ω). Finally, if kv

∗

1/2,2(∂Ω)

≤ M, for some M > 0, then

V satisﬁes the i nequalities given in (X. 4.6), with v

∞

≡ v

∞

Proof. The proof is a direct consequence of that of Lemma X.4.1 (with v

∞

≡

∞

), a nd Remark X.4.1.

We are now in a position to prove the main result o f this section.

Theorem XI.3.1 Let Ω be a locally Lipschitz domain of R

, with a con-

nected boundary. Moreover, let

f ∈ D

−1,2

(Ω), v

∗

∈ W

1/2

(∂Ω), v

∞

be given in (XI.0.11) .

The following prop erties hold, with Φ deﬁned in (XI.3.2).

(i) Existence. If |Φ| < 4πr

/R, there is at least one generalized solution v to

the Navier–Stokes problem (XI.0.10), (XI.0.11). Such a solution satisﬁes

the conditions:

|v(x) + v

∞

| = O(1/

|x|) as |x| → ∞,

kpk

2,Ω

≤ c (|v|

1,2,Ω

+ Rkvk

1,2Ω

+ T kvk

2,Ω

+ |f|

1,2

) ,

(XI.3.4)

for all R > δ(Ω

), where p is the pressure ﬁeld associated to v by Lemma

XI.1. 1, while c = c(Ω, R) with c → ∞ as R → ∞.

Furthermore, if Ω is of class C

, f ∈ L

(Ω), and v

∗

∈ W

3/2,2

(∂Ω),

then v and the corresponding pressure ﬁeld p (see Theorem XI.1.2) satisfy

the energy inequali ty

We recall that throughout this chapter, we are assuming for simplicity that

◦

Ω

is connected.

764 XI Three-Dimensional Flow in Exterior Domains. Rotational Case

Ω

D(v) : D(v) −

∂Ω

[(v

∞

+ v

∗

) · T (v, ep) −

∗

+ v

∞

)

∗

] · n

+[f, v + v

∞

] ≤ 0 ,

(XI.3.5)

where ep is deﬁned in (XI.1.6), and T the Cauchy stress tensor (IV.8.6).

(ii) Estimate by the data. If v

∗

∈ M

1/2,2

(∂Ω) (deﬁned in (IX.4.52)) and

|Φ| ≤ 2πr

/R, then the generalized solution determined in (i) satisﬁes

the following estimate:

|v + v

∞

1,2

≤ 4|f|

−1,2

+ C kv

∗

1/2,2(∂Ω)



1 + R



1 + kv

∗

1/2,2(∂Ω)



+ T



(XI.3.6)

where C = C(Ω, R, M ) .

Proof. We look for a solution of the form v = u+V , where V is the extension

constructed in Lemma XI.3.1. Placing this latter into (XI.1.2), after a simple

mani pulatio n we obtain, fo r all ϕ ∈ D(Ω),

(∇u, ∇ϕ) +(∇(V + v

∞

), ∇ ϕ) + R[(u · ∇u, ϕ) + (u · ∇(V + v

∞

), ϕ)

−(u · ∇v

∞

, ϕ) + ((V + v

∞

) · ∇u, ϕ) − (v

∞

· ∇u, ϕ)

+((V + v

∞

) · ∇(V + v

∞

), ϕ) − ((V + v

∞

) · ∇v

∞

, ϕ)

−(v

∞

· ∇(V + v

∞

), ϕ)] + 2T [(e

×u, ϕ) + (e

×(V + v

∞

), ϕ)]

= −[f , ϕ] ,

(XI.3.7)

where we have used (∇v

∞

, ∇ϕ) = (e

×v

∞

, ϕ) = 0. Taking into a ccount the

identity Ra · ∇v

∞

= T e

× a, and that by a cal culation entirely analo gous

to that leading to (VIII.1.8),

×x · ∇σ − e

×σ = 0, (XI.3.8)

with the help of (XI.3.1) we deduce that (XI.3.7) reduces to

(∇u, ∇ϕ) + (∇(V + v

∞

), ∇ ϕ) + R[(u · ∇u, ϕ) + (u ·∇(V + v

∞

), ϕ)

+ ((V + v

∞

) · ∇u, ϕ) − (v

∞

·∇u, ϕ) + ((V + v

∞

) · ∇(V + v

∞

), ϕ)

− (e

·∇(V + v

∞

), ϕ)] + T [(e

× u, ϕ) + (e

× V

−e

× x · ∇V

, ϕ)]

= −[f , ϕ] .

(XI.3.9)

It is clear that if we ﬁnd u ∈ D

1,2

(Ω) satisfying (XI.3 .9) for all ϕ ∈ D(Ω),

then v := u + V is a generalized solution to (XI.0. 10), (XI.0.11). In order

to construct such a u we use the Galerkin method. Let {ϕ

} ⊂ D(Ω) be the

basis of D

1,2

(Ω), introduced in Lemma VII.2.1. A sequence of approximating

solutions {u

} is then sought of the form

XI.3 Existence of Generalized Solutions 765

k=1

(∇u

, ∇ϕ

) = −(∇(V + v

∞

), ∇ ϕ

) −R[(u

· ∇u

, ϕ

)

+(u

· ∇(V + v

∞

), ϕ

) + ((V + v

∞

) ·∇u

, ϕ

) − (v

∞

· ∇u

, ϕ

)

+((V + v

∞

) · ∇(V + v

∞

), ϕ

) −(e

· ∇(V + v

∞

), ϕ)]

−T [(e

× u, ϕ) + (e

×V

−e

× x · ∇V

, ϕ)] − [f , ϕ

] := F

(ξ) ,

(XI.3.10)

k = 1, . . . , m. For each m ∈ N, we may establish existence of a solution

ξ = (ξ

, . . . , ξ

) to (XI.3.10) by means of Lemma IX.3.2. In fact, since ∇·v

∞

∇· (V + v

∞

) = 0 and u

∈ D(Ω), by Lemma IX.2.1 we obtai n

·∇u

, u

) = (v

∞

· ∇u

, u

) = ((V + v

∞

) · ∇u

, u

) = 0.

We thus deduce

F · ξ = −(∇(V + v

∞

), ∇ u

) − R[(u

· ∇(V + v

∞

), u

)

+((V + v

∞

) · ∇(V + v

∞

), u

) − (e

· ∇(V + v

∞

), u

)]

−T [(e

×V

−e

× x · ∇V

, u

)] − [f, u

] .

(XI.3.11)

Using the inequalities of Schwarz, H¨older, and Sobolev (see (II.3.11)), and

recalling that V

is of bounded support, we deduce

−(∇(V + v

∞

), ∇ u

) ≤ |V + v

∞

1,2

−((V + v

∞

) · ∇(V + v

∞

), u

) ≤ kV + v

∞

|V + v

∞

1,2

≤ c

kV + v

∞

|V + v

∞

1,2

· ∇(V + v

∞

), u

) ≤ c

|V + v

∞

1,6/5

≤ c

|V + v

∞

1,6/5

1,2

−(e

×V

, u

) ≤ kV

6/5

≤ c

6/5

1,2

×x · ∇V

, u

) ≤ c

1,6/5

≤ c

1,6/5

1,2

−[f, u

] ≤ |f|

−1,2

1,2

(XI.3.12)

Furthermore, by Lemma XI.3.1, for any given η > 0 we can choose V such

that

−(u

· ∇(V + v

∞

), u

)| ≤



η +

|Φ|

4πr



1,2

. (XI.3.13)

Therefore, i f

|Φ| <

4πr

, (XI.3.14)