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

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396 VI Steady Stokes Flow in Domains with U nbounded Boundaries

Clearly, ζ(x) satisﬁes estimates of the type (VI.4 .9) and so, using the identity

∇(ζv) = ∇ζ × ω + ζ∇ ×ω + v∆ζ

together with (VI.4.12) it follows that

k∆(ζv)k

2,Ω

≤ c



ωk

2,Ω

+ kf

r−1

2,Ω



≤ c

(VI.4.13)

where, in the last step, we made use of the assumption on v. By considering ζv

as a function deﬁned in the whole of R

, (VI.4.13) furnishes ∆(ζv) ∈ L

)

and, by the Calder´on–Zygmund Theorem II.11.4 (see Exercise II.11.9 and

Exercise II.11. 11) we have D

(ζv) ∈ R

. Since

ζD

v = −D

ζD

v − D

ζD

v −vD

ζ + D

(ζv)

we prove, a s before,

ζD

v ∈ L

Recalling the relations

β,γ

(x) = 1, δ(x

) = f(x

) in ω

and the properties of ω

, from this latter inf ormation we conclude

1+r

v ∈ L

(Ω

), (VI.4.14)

where γ

can be taken arbitrarily close to γ/2. The lemma is thus shown for

|α| = 2 and q = 2. Still assuming q = 2, the case |α| > 2 can be easily proved

by iteration. Actually, setting ω

≡ D

ω, j = 1, . . ., n, we obtain that ω

satisﬁes (VI.4.7) with ω

in lieu of ω. Replace now Ψ with

= δ

2+r

β,γ

,R,R

(x),

so that

|∇Ψ

(x)| ≤ c

1+r

)

(x)| ≤ c

)

and, consequently,



Ω

(Ψ

∆Ψ

+ |∇Ψ

)ω

+ (∇Ψ

⊗ ω

, ∇(Ψ

))



≤ c

1+r

2,Ω

+ (1/2)k∇(Ψ

2,Ω

In two space dimension this identity is replaced by

∆(ζv) = 2∇ζ · ∇v + (1/2)ζϑ + v∆ζ

with ϑ = (−∂ω/∂x

, ∂ω/∂x

VI.4 Decay of Flow in Channels with Unbounded Cross Section 397

Therefore, from (VI.4.7) and (VI. 4.8) with ω ≡ ω

, Ψ ≡ Ψ

and from (VI.4.1 4)

we deduce

k∇(Ψ

2,Ω

≤ c

Repeati ng step by step the arguments previously used for the case |α| = 2 we

may thus conclude for all |`| = 3

2+r

v ∈ L

(Ω

), (VI.4.15)

where γ

< γ

can b e ta ken arbitrarily close to γ

, i.e., to γ/2. Iterating this

procedure as many times as we please we ﬁnally prove the lemma for q = 2.

In showing the result for arbitrary q > 2 we shall conﬁne ourselves to proving

it for |α| = 2; the general case |α| > 2 can be obtained by the same iterating

procedure just used. Consider (VI.4.7) in three space dimensions. By the L

theory for the Laplace operator in the whole space (see Exercise II.1 1.9) it

follows that

k∇(Ψω)k

q,R

≤ c|f|

−1,q,R

, (VI.4.16)

where f denotes the right-hand side of (VI.4 .7). If ϕ indicates an arbitrary

vector function from C

∞

) we have

|(f, ϕ)| = |(∇Ψ · ∇ω, ϕ) −(∇Ψ ⊗ ω, ∇ϕ)| ≡ |F

+ F

Use of (VI.4.9) then gives

| ≤ c

ωk

q,Ω

k∇ϕk

3 ; (VI.4.17)

furthermore,

| = |(ω∆Ψ, ϕ) − (∇Ψ · ∇ϕ, ω)|. (VI.4.18)

Since ϕ ∈ C

∞

) and q > 2, we ﬁnd that ϕ obeys inequality (II.6.10), i.e.,

|ϕ|

+ x

)

≤

2 − q

|∇ϕ|

(VI.4.19)

and so, by (VI.4.9) and (VI.4.19), by recalling that |x

| ≤ f(x

), for all x ∈ Ω,

we deduce

|(ω∆Ψ, ϕ)| ≤ c

Ω

|ω||ϕ|

≤ c

ωk

q,Ω

(

∞

|ϕ|

+ x

)

≤ c

ωk

q,Ω

k∇ϕk

3 .

(VI.4.20)

On the other hand, again by (VI.4.19), i t is clear that

|(∇Ψ · ∇ϕ, ω)| ≤ c

ωk

q,Ω

k∇ϕk

(VI.4.21)

398 VI Steady Stokes Flow in Domains with U nbounded Boundaries

and therefore, collecting (VI.4.16)–(VI.4. 18), (VI.4.20), and (VI.4.21) we re-

cover

k∇(Ψ ω)k

q,Ω

≤ c

. (VI.4.22)

By reasoning as in the case where q = 2, from (VI.4.22) we obtain

1+r

v ∈ L

(Ω

)

and the proof of the lemma is then completed. ut

Remark VI.4.1 The only p oint in the proof just g iven where we need the

assumption q > 2 is in increasing the ﬁrst integral on the right-hand side of

(VI.4.18), by means of (VI.4.19 ) (see (VI.4.20)). However, if

f(t) = f

+ Mt, (VI.4. 23)

i.e., Ω is an inﬁnite portion o f a straight cone, we have |x| ≤ cf(x

) for all

x ∈ Ω and so if q

< n, that is, q > n/(n − 1), using (II.6.10 ) we have

|(∆Ψ ω, ϕ)| ≤ c

ωk

q,Ω

kϕ/|x|k

≤ c

ωk

q,Ω

k∇ϕk

Therefore, if f sati sﬁes (VI.4.23), Lemma VI.4.1 is valid for all q ≥ 3/2 if

n = 3 and for all q ≥ 2 if n = 2. 

Remark VI.4.2 If f satisﬁes |f

(t)f(t)| ≤ c, Lemma VI.4.1 can be proved

with γ

arbitrarily close to one. This is easily seen by using, throughout the

proof, f(t) in place of δ(t). 

The theorem to foll ow gives pointwise decay f or a solution v ∈ D

1,q

(Ω).

Theorem VI.4.1 Assume v is a regular soluti on to (VI.4.1) such that v ∈

1,q

(Ω), for some q > 1. Then for all |α| ≥ 0 and all γ ∈ (0, 1)

lim

|x|→∞

∇v(x)| = 0, uniformly in Ω

(VI.4.24)

and, if 1 < q ≤ n,

lim

|x|→∞

|v(x)| = 0, unifo rmly in Ω

, (VI.4.25)

where γ

∈ (0, 1) [respectively, γ

∈ (0, 1 /2)] can b e taken arbitrarily clo se to

1 [respectively, to 1/2] if 1 < q < n [respectively, q = n] .

Proof. From Lemma V.3.1 we have for all x ∈ Ω

(x) =

(d)

(x − y)v

(y)dy, (VI.4.26)

where d < (1 − γ)f

. Diﬀerentiating once (VI.4.26), and then diﬀerentiating

it a gain |α| times, after using the H¨older inequality we obtain

VI.4 Decay of Flow in Channels with Unbounded Cross Section 399

(x)| ≤ kD

(d)

(x − y)k

q,B

≤ ckD

q,B

which proves (VI.4.24). Assume now 1 < q < n. By the Sobolev inequality we

have kvk

s,Ω

< ∞, s = nq/(n − q) (see Exercise VI.4.1) and (VI. 4.26) gi ves

(x)| ≤ kH

(d)

(x − y)k

s,B

showing (VI.4.25) if 1 < q < n. To prove the theorem completely, it remains

to prove (VI.4.25 ) for q = n. To this end, setting for a ∈ (0, 1]

) = {x ∈ Σ(x

) : |x

| < af(x

)} (VI.4. 27)

we denote by ψ = ψ(|x

|) a smooth function that is one in Σ

) and zero

outside Σ

σ+ε

), ε > 0. Moreover, we take

|∇ψ(|x

|)| ≤

f(x

)

with c independent of x. Since q > n − 1 and ψv ∈ W

1,n

(Σ), we may apply

inequality (II.11.14) to ψv and use the latter to obtain for all x = (x

, x

) ∈ Ω

|v(x

, x

≤ c

n−1

)

σ+ε

)

(ξ

, x

)dξ

+f(x

)

σ+ε

)

|∇v(ξ

, x

dξ

≡ I

) + I

From (VI.3.5) and (VI. 4.24) it follows that

lim

|x|→∞

) = 0 in Ω

and so it remains to show

lim

|x|→∞

) = 0 in Ω

. (VI.4.28)

By a simple calculation we derive



k∇vk

n,Σ

σ+ε

(t)



≤ c



−1

(t)k∇vk

n,Σ

σ+ε

(t)

+ k∇vk

n−1

n,Σ

σ+ε

(t)

+kD

n,Σ

σ+ε

(t)



and so, by employing inequality (II.2. 7), it f ollows that

400 VI Steady Stokes Flow in Domains with U nbounded Boundaries



≤ c

k∇vk

n,Σ

σ+ε

(t)

+ f

(t)kD

n,Σ

σ+ε

(t)

By assumption and Lemma VI.4.1 with |α| = 2, r = 0 we have, for all σ < 1/2,

∈ L

(0, ∞)

and therefore ` ∈ R

exists such that

lim

|t|→∞

(t) = `.

However, again by assumption, there exi sts at least a sequence {t

} tending

to i nﬁnity along which i t holds

|v|

1,n,Σ

σ+ε

)

= o(1/t

)

while, by (VI.4.3)

f(t

) = O(t

As a consequence,

lim

→∞

I(t

) = 0,

which shows (VI.4.28 ). The proof of the theorem is therefore completed. ut

In the following theorem we establish the decay rate.

Theorem VI.4.2 Let v satisfy the hypotheses of Theorem VI.4.1. If n = 3,

then for al l x ∈ Ω

and all |α| ≥ 0

v(x)| ≤

|α|+ϑ

)

, 1 < q < 3

∇v(x)| ≤

|α|

)

, q ≥ 3

(VI.4.29)

where ϑ = 1/2 if 1 < q ≤ 2, while if 2 < q < 3, ϑ = (3 − q)/q for α = 0 and

ϑ = 0 for |α| ≥ 1. If n = 2, then for all x ∈ Ω

and |α| ≥ 0

∇v(x)| ≤

|α|

)

, 1 < q ≤ 2. (VI.4. 30)

Here γ

is any number less than 1/4; moreover, the constants c

depend on

n, q, α, M, and on the norm | v|

1,q

Proof. We begin to prove (VI.4.29)

. By assumption

v ∈ D

1,q

(Ω

), γ < 1,

and, by (VI.3.6), v/f ∈ L

(Ω

). Without loss we may assume q ≥ 2 since, by

Theorem VI.4.1 and the regularity hypothesis on v, if

VI.4 Decay of Flow in Channels with Unbounded Cross Section 401

v ∈ D

1,q

(Ω), for some q > 1,

then

v ∈ D

1,r

(Ω

), v/f ∈ L

(Ω

) f or all r ≥ q.

Set

w = δu

β,γ,R,R

where δ is the function constructed in Lemma III.4.2 and u

β,R,R

is the “cut-

oﬀ” function introduced in the proof of Lemm a VI.4.1. From (VI.4.5) and the

properties of δ we derive

q,Ω

≤ c



kv/fk

q,Ω

+ k∇vk

q,Ω

+ kfD

q,Ω



with c

independent of R and γ < 1/2 arbitrari ly close to 1/2. From Lemma

VI.4. 1 it follows that

q,Ω

≤ c

. (VI.4.3 1)

Considering the function w in the whole of R

and recalling that q < 3, from

the Sobolev inequality and (VI.4.3 1) we have

k∇wk

s,R

≤ κk D

q,R

≤ c

, s = 3q/(3 −q)

with c

independent o f R. By taking into account the properties of the function

δ it is not hard to show that the preceding inequality i n the limit R → ∞

furnishes

kf∇vk

s,Ω

≤ c

, (VI.4.32)

where γ

< γ can b e taken arbitrarily close to γ/2, i.e., to 1/4. Since q < 3

we may a pply to v the Sobolev inequality to deduce v ∈ L

(Ω) (see Exercise

VI.4. 1). Thus, from (VI.4.32) it follows that v satisﬁes the assumption of

Lemma VI.4.1 and for all |α| ≥ 1 we conclude

|α|

s,Ω

≤ c

, (VI.4.33)

where γ

can b e taken close to 1/4. Take x ∈ Ω

, d < di st (x, ∂Ω

) and set

= δ

|α|

with δ ≡ f in Ω

. Applying Theorem II.3.2 to θu

with θ = 1 in B

d/2

(x)

and θ = 0 outside B

(3/4 ) d

(x) we obtain

|α|

v(x)| ≤ cku

m,s,B

(x)

(VI.4.34)

whenever m ≥ s/3. However, from the properties of the function δ we easily

deduce

m,s,B

(x)

≤ kf

|α|

s,B

(x)

|`|=1

|α|

s,B

(x)

(VI.4.35)

402 VI Steady Stokes Flow in Domains with U nbounded Boundaries

and so by (VI.4.33)–(VI.4.35) we recover (VI.4.29)

. Using Theorem II.11. 2

and inequality (II.11.11), and proceeding as in the proof of Theorem VI.4.1,

we establish the estimate

)|v(x

, x

≤ c

f(x

)kvk

s,Σ

σ+ε

)

+ f

s+1

)k∇vk

s,Σ

σ+ε

)

≡ I(x

(VI.4.36)

Denoting, temporarily, x

by t, we want to show that the function I = I(t) is

bounded for all suﬃciently large t. To this end, it is suﬃcient to show that

dI/dt ∈ L

(0, ∞).

We have



≤ c

kvk

s,Σ

σ+ε

(t)

+ f(t)kvk

s−1

s,Σ

σ+ε

(t)

k∇vk

s,Σ

σ+ε

(t)

(t)k∇vk

s,Σ

σ+ε

(t)

+ f

s+1

(t)k∇vk

s−1

s,Σ

σ+ε

(t)

s,Σ

σ+ε

(t)

≤ c

kvk

s,Σ

σ+ε

(t)

+ f

(t)k∇vk

s,Σ

σ+ε

(t)

+ f

(t)kD

s,Σ

σ+ε

(t)

(VI.4.37)

and since, by the Sobolev inequality (see Exercise VI.4.1)

k∇vk

s,Σ(t)

∈ L

(0, ∞),

from (VI.4.33), (VI.4.36) and (VI.4.37) we obtain (VI.4.29)

also in the case

where α = 0. In order to show (VI.4.29)

we set

= δ

|α|

∇v,

with δ ≡ f in Ω

. Proceeding as in the proof of (VI.4 .29)

we may establish

the inequality (see (VI.4.34))

|α|

∇v(x)| ≤ cku

2,q,B

(x)

, (VI.4.38)

where x ∈ Ω

, d < dist (x, ∂Ω

). Therefore, since

2,q,B

(x)

≤ c





|α|

∇vk

q,B

(x)

|`|=1

|α|

∇vk

q,B

(x)





(VI.4.39)

(VI.4.29)

follows from (VI.4.38), (VI.4.3 9) and Lemma VI.4 .1. The proof of

(VI.4.30) is entirely identical to the one just given for (VI.4.29), if one recalls

that, as already observed, v ∈ D

1,q

(Ω) for some q < 2 implies

v ∈ D

1,2

(Ω

), v/f ∈ L

(Ω

) f or all σ < 1.

The proo f of the theorem is therefore completed. ut

VI.4 Decay of Flow in Channels with Unbounded Cross Section 403

Remark VI.4.3 If f(t) veriﬁes (VI.4.23), then estimate (VI.4.30) holds for

any q ∈ (1, ∞). This is a consequence of Remark VI.4.1. 

Remark VI.4.4 If |f

(t)f(t)| ≤ c, then all conclusions in Theorem VI. 4.2

remain valid for γ, γ

< 1. 

Exercise VI.4.1 Let Ω be a domain of the type introduced at the beginning of this

section and let u be a smooth function in Ω, vanishing on Γ and with u ∈ D

1,q

(Ω),

1 < q < n. Show that kuk

< ∞, s = nq/(n −q) (Sobolev inequality). Hint: Extend

u to zero outside Ω. The function ψ(x/R

)u(x), with ψ given in the proof of Lemma

VI.4.1 belongs to D

1,q

We now turn our attention to the behavior of the pressure. First of all,

from Theorem VI.4.1, Theorem VI.4.2, and (VI.4.1)

we at once obtain

Theorem VI.4.3 Let v satisfy the assumptions of Theorem VI.4.1. Then for

all |α| ≥ 0

lim

|x|→∞

∇p(x) = 0

uniform ly in Ω

, γ < 1. Moreover if n = 3, then for all x ∈ Ω

∇p(x)| ≤

|α|+1

)

1 < q < 3

∇p(x)| ≤

|α|

)

q ≥ 3

while, if n = 2,

∇p(x)| ≤

|α|

)

1 < q ≤ 2,

where γ

< 1/4 a nd, f or ﬁxed α, it can be taken arbitrarily close to 1/4.

Furthermore, using Theorem VI.4.2 we can prove the following result.

Theorem VI.4.4 Let v ∈ D

1,2

(Ω). Then, there exists a constant p

∈ R

such that

lim

|x|→∞

p(x) = p

uniform ly in Ω

, for any γ < 1/4.

Proof. Denote by p(x

) the mean of the pressure p over Σ(x

), i.e.,

p(x

) =

|Σ

)

p(x

, x

)dx

where Σ

is deﬁned in (VI.4.27). The following inequality holds:

404 VI Steady Stokes Flow in Domains with U nbounded Boundaries

|p(x

, x

) − p(x

≤ cf(x

)

|∇p(x

, x

, (VI.4.40 )

with x

∈, and γ an arbitrary number less than σ. Actually, a pplying inequality

(II.11.14) to ψ(p − p), with ψ the “cut-oﬀ” function introduced in the proof

of Theorem VI.4.1, we readily deduce

|p(x

, x

) −p(x

≤ c

1−n

)

|p(x

, x

) − p(x

+f(x

)

|∇p(x

, x

(VI.4.41)

Increasing the ﬁrst integral on the ri ght-hand side of (VI.4.41) by means of

inequality (II.5.10), we recover (VI.4.40). Denoting temporarily x

by t and

the ri ght-hand side of (VI.4.40) by I = I(t), we also have



≤ c

k∇pk

n,Σ

(t)

+ f(t)k∇pk

n,Σ

(t)

n,Σ

(t)

which, by Lemma VI.4.1 and (VI.4.1)

, in turn implies dI/dt ∈ L

(0, ∞) for

all σ < 1/4. Thus,

lim

t→∞

I(t)

exists and since

lim

→∞

I(t

) = 0

at least along a suitable sequence {t

} (as a consequence of the summability

of ∇p in L

(Ω

), which follows from (VI.4.1)

, Lemma VI.4.1 and Theorem

VI.4. 3, and of the fact that f(t) = O(t)) we may conclude

lim

|x|→∞

|p(x

, x

) − p(x

)| = 0 (VI.4.42)

uniform ly in Ω

. Next, we shall show that p tends to a prescrib ed limi t as

|x| → ∞ . From the nth component of (VI.4.1) we derive

∂

∂x

p(x

) =

|Σ

)

∆v

which in turn furnishes

|p(t

) − p(t

)| ≤ c

1−n

(t)

v(x

, x

)|dx

for a rbitrary t

, t

> 0. Use of the Schwarz inequality gives

VI.4 Decay of Flow in Channels with Unbounded Cross Section 405

|p(t

) − p(t

)| ≤ c

−(n+1)/2

(t)



f(t)kD

2,Σ

(t)



≤ c



−n−1

(t) + f

(t)kD

2,Σ

(t)



≡ I

(t) + I

(t).

Now, as t

, t

→ ∞, I

(t) tends to zero because the assumption v ∈ D

1,2

(Ω)

implies (VI.4.4) with q = 2, whil e I

tends to zero by L emma VI.4.1 and we

conclude

lim

→∞

p(x

) = p

for some constant p

. The theorem then follows from this latter condition and

(VI.4.42). ut

Remark VI.4.5 If |f

(t)f(t)| ≤ c, in the previous theorem we can take

γ < 1. 

Remark VI.4.6 If f does not satisfy (VI.4.4) with q ≤ 2 and, consequently,

if v 6∈ D

1,2

(Ω), it is not expected that the pressure ﬁeld tends to a constant a t

large distances in the exit. Actually, if n = 2, Amick & Fraenkel (198 0) prove

that p diverges. However, using the following heuristic argument, it is easy

to convince oneself that the same property should hold a lso for n = 3 . Since

f(x

) is the only “natural length” of the problem, by the incompressibility

condition ∇ · v = 0 the longitudinal velocity component v

has to have the

asymptotic form g(|x

|/f(x

))/f

). Now, if we take the third component

of the Stokes equations we ﬁnd ∂p/∂x

= ∂

/∂x

+ ∂

/∂x

≈ f

−4

As a consequence, if f does not verify (VI.4. 4) with some q ≤ 2, p diverges;

see also Pileckas (1996a, 1996b, 1996c). 

Remark VI.4.7 The physical meaning of the constant to which the pres-

sure tends in the exits is very important and it is tightly related to the ﬂux

Φ. Speciﬁcally, for solutions w hose existence has been established in Theorem

VI.3. 1, to prescribe the ﬂux is equivalent to prescribing the diﬀerence p

be-

tween the constant values p

to which the pressure ﬁeld p(x) tends i n the exits

Ω

(i = 1, 2). This property, which was originally di scovered for a particular

region of ﬂow by Heywood (19 76, Section 6), is simply proved in our case as

follows. Since p is deﬁned up to a constant, we can always adjust i t in such a

way that

lim

|x|→∞

p(x) = 0 in (Ω

)

≡ Ω

1γ

lim

|x|→∞

p(x) = p

in (Ω

)

≡ Ω

2γ

By the linearity of the problem (VI.4.1), it is suﬃcient to show that if p

= 0

then the solution v determined in Theorem VI.3.1 is identically zero. On the

other hand, this will follow if we show