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

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386 VI Steady Stokes Flow in Domains with U nbounded Boundaries

Proof. By Theorem VI. 2.1 and what we have observed at the beginning of this

section, we have to show only the validity of estimate (VI.2.17). For the sake

of sim plicity we shall restrict ourselves to treat the case n = 3; also, Cartesian

coordinates will be denoted by x

, x

and x, y, z, indiﬀerently. Proceeding

as in the proof of the previous theorem we may write the identity

−

Σ(ζ)

∇u : ∇ udΣ



dζ

Σ(z

)



τu

−

∂u

∂z



−

Σ(z)



τu

−

∂u

∂z



(VI.2.18)

From Theorem VI.2.1 and Lemma VI.1.2 we know that u, ∇τ ∈ W

m,2

(Ω) for

all m ≥ 0 and so, in parti cular, it easily follows that

ı(z

) ≡

Σ(z

)

τu

, z

)dx

= o(1) as z

→ ∞. (VI.2.19)

In fact, setting

τ(z

) =

|Σ|

τ(x

, z

)dx

from (VI.2.1)

and Poincar´e’s inequality (II.5.10) it follows that

|ı(z

)| =



Σ(z

)

(τ −τ)u

, z

)dx



≤ c|τ|

1,2,Σ

kuk

2,Σ

and (VI.2.19) becomes a consequence of (VI.1.21) and (VI.1.22). Thus, from

(VI.2.19) and (VI.1. 21), by letting z

→ ∞ into (VI.2.18), we ﬁnd

H(z) ≡

∞

Σ(ζ)

∇u : ∇udΣ

dζ =

Σ(z)



τu

−

∂u

∂z



. (VI.2.20)

We next integrate both sides of (VI.2.20) between t + ` a nd t + ` + 1 with ` a

non-negative integer to obtain

t+`+1

t+`

H(z) =

t+`+1

t+`

Σ(z)

τu

−

Σ(t+`+1)

Σ(t+`)

. (VI.2.21)

By writing u

= ∇ · ω, with ω a solution to (VI.2.14) and by argui ng as in

the proof of Theorem VI.2.1, from (VI.2.21) it follows that

t+`+1

t+`

H(z) ≤ c

√

µ|u|

1,2,Ω

t+`,t+`+1

−

Σ(t+`+1)

Σ(t+`)

Summing both sides of this relation from ` = 0 to ` = ∞ and observing that,

by R emark VI.2.1,

VI.3 Existence of Flow in Channels with Unbounded Cross Sections 387

lim

z→∞

Σ(z)

, z) = 0,

leads to

∞

H(z) ≤ c

√

µH(t) +

Σ(t)

. (VI.2.22)

Since, by (VI.2.15), we have

Σ(t)

≤ µ

Σ(t)

∇u : ∇u = −µH

(t),

inequality (VI.2.22) ﬁnally gives

(t) +

∞

H ≤

√

H(t)

which, by Lemma VI.2.2, completes the proof o f the theorem. ut

VI.3 Flow in Unbounded Channels with Unbounded

Cross Sections. Existence , Uniqueness, and Regularity

In the present section we shall investigate existence and uniqueness of steady,

slow motions of a viscous liquid in a domain Ω ⊂ R

, n = 2, 3, with m ≥ 2

“exits” to inﬁnity Ω

, i = 1, . . ., m, whose cross sections become suitably

unbounded at large distances. To start, we shall assume m = 2 and that the

domains Ω

are bodies of rotation (see Section III.4.3). Several generalizations

will be considered in Exercise VI.3. 1-Exercise VI.3.4. Speciﬁcally, we take

Ω =

[

i=0

Ω

with Ω

a compact subset of R

and Ω

, i = 1, 2, disjoint domains which, i n

possibly diﬀerent coordinate systems, are given by

Ω

= {x ∈ R

: x

> 0, |x

| < f

)},

where the functions f

satisfy, for all t, t

, t

> 0 and i = 1, 2,

(i) f

(t) ≥ f

= const. > 0;

(ii) |f

) −f

)| ≤ M |t

−t

with M a positive constant. Furthermore, we shal l assume f

(and therefore Ω)

of class C

∞

We shall also suppose that the planar cross section Σ

= Σ

)

See footnote 1 in Section VI. 1.

388 VI Steady Stokes Flow in Domains with U nbounded Boundaries

of Ω

, i = 1, 2, perpendicular to the axis x

= 0 and passing through the point

(0, x

) tends to inﬁnity as |x| → ∞ in such a way that

∞

−(n+1)

(t)dt < ∞, i = 1, 2. (VI.3.1)

This condition i s equivalent to condition (i) stated in the introduction to the

chapter.

We denote, as usual, by Σ a cross section of Ω, that is, any b ounded

intersection o f Ω with an (n − 1)-dimensional plane that in Ω

, i = 1, 2,

reduces to Σ

and by n a unit vector normal to Σ and oriented from Ω

toward Ω

(say). We want to study the following problem: given Φ ∈ R, to

determine a pair v, p such that

∆v = ∇p

∇ · v = 0

)

in Ω

v = 0 at ∂Ω

v ·n = Φ

lim

|x|→∞

v(x) = 0 in Ω

, i = 1, 2.

(VI.3.2)

To sol ve this problem, analogously to what we di d in Section VI.1, we

shall put it into an equivalent form that can be handled by the technique of

generalized solutions. To this end, by multiplying (VI.3.1) by ϕ ∈ D(Ω) and

integrating by parts we obtain fo rmally

(∇v, ∇ϕ) = 0, for all ϕ ∈ D(Ω). (VI.3.3)

We then give the foll owing.

Deﬁnition VI.3.1. A ﬁeld v : Ω → R

is called a weak (or generalized )

solution to problem (VI.3.2) if and only if

(i) v ∈

1,2

(Ω);

(ii) v satisﬁes (VI.3.3 );

(iii) v satisﬁes (VI.3.2)

in the trace sense.

The meaning of conditions (ii) and (iii) is quite obvious. Let us comment

on condition (i). First, we observe that it requires v ∈

1,2

(Ω) and not v ∈

1,2

(Ω). As already remarked, thi s allows for a nonzero ﬂux Φ of v through Σ;

actually, if we required, instead, v ∈ D

1,2

(Ω) we would automa tically impose

Φ = 0. Moreover, (i) ensures a certain degree of regularity for v and that v

vanishes on the boundary in the trace sense. Finally, as we are going to show,

condition (i) implies

VI.3 Existence of Flow in Channels with Unbounded Cross Sections 389

|Σ(x

Σ(x

)

(x) → 0 as |x| → ∞ in Ω

, i = 1, 2, (VI.3.4)

which represents the generalized form of (VI.3 .2)

. Property (VI.3.4) is a

consequence of the general inequality

Σ(t)

|w|

, t)dx

≤ cf

q−1

(t)

Ω(t)

|∇w|

, (VI.3.5)

where

Ω(t) = {x ∈ R

: x

> t ≥ 0, |x

| < f(x

)}

Σ(t) =



∈ R

n−1

: |x

| ≤ f(t)



f sati sﬁes assumptions of the type (i), (ii) stated for f

and, ﬁnally, w is an

arbitrary member from D

1,q

(Ω(t)), 1 < q < ∞, vanishing at the lateral surface

| = f(x

), x

> t. To show (VI.3.5) we observe that (II.5.5) furnishes for

almost all t ≥ 0

Σ(t)

|w|

, t)dx

≤ c

(t)

Σ(t)

|∇w(x

, t)|

, 1 < q < ∞. (VI.3.6)

Consider, next, the function g(x

) ≡ f

−q+1

with e

unit vector in the

direction x

. Integrating the identity

∇ · (g|w|

) = |w|

∇ · g + g · ∇|w|

over Ω

t,t

, in virtue of assumption (ii) on f we deduce for a ll t

> t

q−1

(t)

Σ(t)

|w|

≤

q−1

)

Σ(t

)

|w|

q−1

(τ)

Σ(τ)

|w|

q−1

|∇w|dΣ

dτ

Ω

t,t

|w|

and so, from (VI.3.6) and the H¨older inequali ty, it follows that

q−1

(t)

Σ(t)

|w|

≤

q−1

)

Σ(t

)

|w|

+ c

Ω(t)

|∇w|

However, again by (VI.3.6) and assumption (ii) on f, since w ∈ D

1,q

(Ω(t)),

we have that the ﬁrst term o n the ri ght-hand side of this latter inequality

must tend to zero as t

tends to inﬁnity, at least along a sequence of values.

Therefore, (VI.3 .5) is proved.

390 VI Steady Stokes Flow in Domains with U nbounded Boundaries

Remark VI.3.1 By using Lemma IV.1.1 and Theorem VI.1.1, one can show

that to every weak solution there corresponds a pressure ﬁeld p satisfying the

identity (VI.1.5) and that the pair v, p is inﬁnitely diﬀerentiable up to the

boundary. 

Our next objective is to show exi stence and uniqueness of a generalized

solution. To this end, we observe that in view of hypothesis (i) on f

, from

the results of the ﬁrst part of Section III.5 and, i n particular, from Theorem

III.5.2, a vector ﬁeld a ∈ C

∞

(Ω) vanishing near ∂Ω exists such that

a ∈

1,2

(Ω)

a · n = 1.

(VI.3.7)

We then look for a generalized solution to (VI.3.2) of the form v = u + Φa,

with u ∈ D

1,2

(Ω) obeying the identity

(∇u, ∇ϕ) = −Φ(∇a, ∇ϕ), for all ϕ ∈ D(Ω). (VI.3.8)

Clearly, if such a u exists, the vector ﬁeld v satisﬁes all requirements of Def-

inition VI.3.1. On the other hand, the existence of u is at once established

by means of the Riesz theorem for, evidently, the right-hand side of (VI.3.8)

deﬁnes a linear functional in D

1,2

(Ω). Let us next establish uniqueness. Let

be another generalized solution to (VI.3.2) corresponding to the same ﬂux

Φ. Letting w = v − v

we have w ∈

(Ω) and the ﬂux of w through Σ is

zero. Therefore, by Theorem III. 5.2, w ∈ D

1,2

(Ω). However, from (VI.3.3) we

also have

(∇w, ∇ϕ) = 0, for all ϕ ∈ D

1,2

(Ω),

so that we conclude w ≡ 0.

Finally, we wish to give an estimate for weak solutions. We recall that, by

Theorem I II.5.2, any v ∈

1,2

(Ω) can be written as v = u + Φa, where a

is a given vector in

1,2

(Ω) (independent of v) and u is a vector in D

1,2

(Ω)

uniquely related to v. Writing (VI.3.3) for ϕ ∈ D

1,2

(Ω) and ta king ϕ = u we

thus readily recover

|v|

1,2

≤ |Φ||a|

1,2

≤ c|Φ|,

where c depends only on Ω and n.

Results obtained so far can then be summa rized in the following.

Theorem VI.3.1 For any Φ ∈ R there exists one and only one generalized

solution v to problem (VI.3.2). Such a solution satisﬁes the estimate

|v|

1,2

≤ c|Φ|,

where c depends onl y on Ω and n. Furthermore, denoting by p the correspond-

ing pressure ﬁeld, v, p are inﬁnitely diﬀerentiable in Ω

, with Ω

any bounded

domain contained in Ω , and satisfy (VI.3.1)–(VI.3.4) in the ordinary sense.

VI.3 Existence of Flow in Channels with Unbounded Cross Sections 391

Remark VI.3.2 If the cross sections widen at such a rate that condition

(VI.3.1) on f

is violated, then the unique solvability of problem (VI.3.2)

becomes much more complicated. In this respect, we quote the results of

Amick & Fraenkel (1980), when Ω is two-dimensional with two exits to inﬁnity,

and the more recent ones of Pileckas (1996a, 1996b, 1996c) valid for two-

dimensional and three-dimensional domai ns a s well, under the assumption

that f

satisﬁes (i), (ii) and the following condi tion

∞

−(n−1)(q−1)−q

(t)dt < ∞, for some q ∈ (1, ∞). (VI.3.9)

A typical result proved there is the following one.

Theorem VI.3.2 Let Ω ⊂ R

, n = 2, 3, with f

satisfying (i ), (ii) and

(VI.3.9). Then, for any Φ ∈ R there exists a unique solution v, p to (VI.3.2)

with v ∈

1,q

(Ω), and satisfying the estimate

|v|

1,q

≤ c|Φ|

where c = c(Ω, n, q).



Remark VI.3.3 Another approach to uniq ue solvability that holds in both

two and three dimensions and for cross sections that need not verify (VI.3.1)

is that proposed by Ladyzhenskaya and Sol onnikov (1980) and Solonnikov

(1983). However, it is not known if their solutions, which have a ﬁnite Dirichlet

integral only on bounded subdomains of Ω, verify the condition at inﬁnity

(VI.3.2)

. 

Exercise VI.3.1 Assume that a body force is acting on the liquid and add its

contribution f to the right-hand side of (VI.3.2)

. Denoting by [F, ϕ] the value of a

linear functional F on D

1,2

(Ω) at ϕ, show that, given any bounded linear functional

f on D

1,2

(Ω) and any Φ ∈ R, there exists one and only one vector ﬁeld v satisfying

(i) and (ii) of Deﬁnition VI. 3.1 and the identity

(∇v, ∇ϕ) = −[f , ϕ], for all ϕ ∈ D

1,2

(Ω). (VI.3.10)

Moreover, prove that v satisﬁes the estimate

|v|

1,2

≤ c (|Φ|+ |f|

−1,2

) (V I.3.11)

with |f |

−1,2

denoting the norm of f . Finally, show that if f is inﬁnitely diﬀerentiable

up to the boundary, the same holds for v and for the corresponding pressure p.

Exercise VI.3.2 Assume Ω has m > 2 exits Ω

, i = 1, . . . , m, all of the form speci-

ﬁed at the beginning of this section and verifying (VI.3.1). Given m real numbers Φ

subject to the restriction

i=1

= 0, we shall say that v is a generalized solution

to the Stokes problem in Ω corresponding to the ﬂuxes Φ

if v satisﬁes (i) and (ii)

of Deﬁnition VI.3.1 and if

392 VI Steady Stokes Flow in Domains with U nbounded Boundaries

v · n = Φ

, in the trace sense.

Show existence and uniqueness of this generalized solution. Hint: Use Lemma III.4.3.

Exercise VI.3.3 Extend the results of Exercise VI.3.2 to the case when Ω

are not

necessarily bodies of rotation but rather verify the more general conditions:

(a) D

⊂ Ω

⊂ D

where

= {x ∈ R

: x

> 0, | x

| < f

)}

= {x ∈ R

: x

> 0, | x

| < a

), a

> 1}

and

(b) In the domains

{x ∈ Ω : R

< x

< R + f(R

)}, i = 1, .., m,

problem (III.4.19) is solvable with a constant c independent of R. Hint: See Remark

III.5.1.

Exercise VI.3.4 Let Ω be a C

∞

-smooth domain with m ≥ 2 exits to inﬁnity

Ω

. Suppose that the ﬁrst ` (≤ m) exits satisfy the condition stated in Exercise

VI.3.3, while the remaining m −` are cylindrical. Show that, given m real numbers

subject to the condition

i=1

= 0, there exists one and only one pair v, p

inﬁnitely diﬀerentiable up to the boundary such that

∆v = ∇p

∇· v = 0

)

in Ω

v = 0 at ∂Ω

v · n = Φ

, i = 1, . . . m,

v ∈ D

1,2

(Ω

), i = 1, . . . `,

v − v

(i)

∈ W

1,2

(Ω

), i = ` + 1 . . . m,

where v

(i)

are the Poiseuille velocity ﬁelds associated to Φ

. Hint: Construct suitable

extensions of the Poiseuille velocity ﬁelds and vectors having prescribed ﬂux in Ω

by means of the method used in Section V I.2 and Lemma III.4.3.

Exercise VI.3.5 (Flow through an aperture, Heywood 1976). Let Ω be the domain

(VI.0.7), with S containing the unit disk {|x

| < 1}. Show that, given any bounded

linear functional f on D

1,2

(Ω) and any Φ ∈ R, there exists one and only one vector

ﬁeld v satisfying (i) and (ii) of Deﬁnition V I.3.1 and identity (VI.3.10) where, in this

case, Φ is the ﬂux of v through S. Furthermore, show that such a solution satisﬁes

inequality (VI.3.11).

VI.4 Decay of Flow in Channels with Unbounded Cross Section 393

VI.4 Pointwise Decay of Flows in Channels with

Unbounded Cross Section

To complete the study of problem (VI.3 .2), it remains to investigate pointwise

decay to zero of the velocity and pressure ﬁelds at l arge distances in the exits

Ω

. Seemingly, this study is not easily performed by a simple modiﬁcation of

the methods used in the case of channels with bounded cross sections and we

shall employ a diﬀerent technique.

Let v, p be a pair of smooth functions

satisfying the system

∆v = ∇p

∇ · v = 0

)

in Ω

v = 0 at Γ

v · n = Φ,

(VI.4.1)

where Ω is a semi-inﬁnite channel with unbounded cross section, Γ its lateral

surface, Σ its cross section, and Φ a prescribed number. For simplicity, we

shall assume hereafter tha t Ω is a body of rotation. However, the proofs we

give apply unchanged to the more general case where Ω contains a body of

rotation

Ω. In such a case, the results we ﬁnd remain valid in

Ω. We thus take

for n = 2, 3,

Ω = {x ∈ R

: x

> 0, |x

| < f(x

)}

Σ = Σ(x

) = {x

∈ Ω : |x

| = f(x

)}

(VI.4.2)

with f ∈ C

∞

) verifying the assumptions (i) and (ii) of the previous sec-

tion, i.e.,

f(t) ≥ f

= const. > 0

|f(t

) − f(t

)| ≤ M |t

− t

(VI.4.3)

for a ll t, t

, t

> 0 and with M a positive constant. The aim of this section is

to investigate decay as |x| → ∞ of solutions to (VI.4.1) having v ∈ D

1,q

(Ω).

Speciﬁcally, we show that if 1 < q ≤ n then v and all its derivatives of arbitrary

order tend to zero pointwise; moreover, if 1 < q < n, we are also able to give

the decay rate. Of course, for such solutions to exist, by the conservation o f

the ﬂow through Σ, it i s necessary that f satisﬁes the condition

∞

−(n−1)(q−1)−q

(t)dt < ∞ (VI.4.4)

Alternatively, as suggested by Pileckas (1996a, 1996b, 1996c), one may use a

“weighted de Saint-Venant principle” in conjunction with local estimates for the

Stokes problem, to obtain sharper results than those obtained here.

For example, v ∈ C

(Ω

), p ∈ C

(Ω

) for all bounded subdomains Ω

⊂ Ω.

Some of the results we show, such as those of Theorem VI.4.1, hold in any space

dimension n ≥ 2.

394 VI Steady Stokes Flow in Domains with U nbounded Boundaries

These results furnish, in the particular case where q = 2, pointwise decay of

solutions whose existence has been established in Theorem VI.3.1. Also, in

such a case, we prove that the pressure ﬁeld tends to a certain constant value

at l arge distances.

To show all the a bove, we need some preliminary considerations. For γ ∈

(0, 1], set

Ω

= {x ∈ Ω : |x

| < γf(x

)}, Ω

≡ Ω.

The following result gives a basic a priori estimate for solutions to (VI.4 .1).

Lemma VI.4.1 Let Ω ⊂ R

. Assume that for some γ ∈ (0, 1], q ≥ 2 and

r ≥ 0

the following conditions hold:

(i) f

∇v ∈ L

(Ω

(ii) f

r−1

v ∈ L

(Ω

where assumption (ii) is needed only i f γ < 1. Then, for every |α| ≥ 2 we

have

|α|+r−1

v ∈ L

(Ω

where γ

is any positive number less than γ/2. If Ω ⊂ R

the same conclusion

holds with q = 2.

Proof. First of all we notice that hypothesis (ii) follows from (i) if γ = 1,

as a consequence of inequality (VI.3.6). To show the theorem we need a

suitable“cut-oﬀ” function. Denote by ψ ∈ C

∞

(R) a function such that

ψ(t) = 1 if t ≤ 1 and ψ(t) = 0 if t ≥ 2 and set for β ∈ (0, γ), R

≥ 0

and all R > 2R

β,γ

(x) = ψ



− β



)

+ γ

− 2β



R,R

(x) = ψ



|x|



1 − ψ



|x|



where the function δ(t) has been introduced in Lemma III.4.2. Clearly, the

function

β,γ,R,R

≡ χ

β,γ

R,R

vanishes in the set

{x ∈ Ω : |x

| ≥ γf(x

)}

and for |x| ≥ 2R, while

β,γ,R,R

(x) = 1 for x ∈



[ω

∩B

) − B



Evidently, if γ = 1, for such solutions to exist the conservation of ﬂux requires

∞

−(n−1)(q−1)+q(r−1)

(t)dt < ∞.

VI.4 Decay of Flow in Channels with Unbounded Cross Section 395

where

= {x ∈ Ω : |x

| ≤ βδ(x

)}.

By the properties of the function δ(t), ω

can be taken arbitraril y close to

Ω

γ/2

by picking β suﬃciently close to γ. Using Lemma III.4.2, it is not diﬃcult

to see that

β,R,R

(x)| ≤

|`|

)

, for all | `| ≥ 0, (VI.4 .5)

with c = c(`, β, ψ, R

). We begin to show the lemma for |α| = 2, the general

case being obtained by iteration. To this end, letting

Ψ(x) = δ

1+r

β,R,R

(x) (VI.4.6)

taking the curl of both sides o f (VI.4.1)

, we obtain with ω = ∇ × v

∆(Ψω) = ∇Ψ · ∇ ω + ∇ · [ ∇Ψ ⊗ ω]. (VI.4.7)

Multiplying both sides of (VI.4.7) by Ψ ω furnishes

(∇(Ψω), ∇(Ψω)) =

Ω

(Ψ∆Ψ + |∇Ψ |

)ω

+ (∇Ψ ⊗ ω, ∇(Ψω))

≡ I

+ I

(VI.4.8)

From (VI.4.6), (VI.4.5), and Lemma III. 4.2 it follows that

|∇Ψ (x)| ≤ c

)

Ψ(x)| ≤ c

r−1

)

(VI.4.9)

and, consequently, by (VI.4.3)

and the Yo ung inequality (II.2.5),

| ≤ c

ωk

2,Ω

| ≤ c

ωk

2,Ω

k∇(Ψω)k

2,Ω

2,γ

(VI.4.10)

Thus, (VI.4.8) and (VI.4.10) along with the assumption on ∇v give

k∇(Ψ ω)k

2,Ω

≤ c

(VI.4.11)

with c

independent of R. Letting R → ∞ into (VI.4.11) yields

k∇(ζω)k

2,Ω

≤ c

(VI.4.12)

with

ζ(x) = δ

1+r

)χ

β,γ

(x)ψ



|x|



In two space dimension ω has only one nonzero component gi ven by ∂v

/∂x

−

∂v

/∂x