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

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536 VIII Steady Generalized Oseen Flow in Exterior Domains

Lemma VIII.4.2 Suppose u, φ are such that

∂u

∂t

, D

u , φ , ∇φ ∈ L

loc

((0, T ] × R

) , for some s ∈ (1, ∞) ,

and satisfy a.e. the system

∂u

∂t

= ∆u + R

∂u

∂x

+ ∇φ

∇· u = 0











in R

× (0, T )

(VIII.4.32)

along with the initial condition

lim

t→0

ku(t)k

r,B

= 0 , for all ρ > 0 and some r ∈ (1, ∞) . (VIII.4.33)

Then, if

u =

i=1

, u

∈ L

) , for some q

∈ (1, ∞) , i = 1, . . ., N ,

it follows that u = 0 and ∇φ = 0 a.e. in R

Proof. For simplicity, we consider the case N = 2, the general case being

treated in an entirely similar way. L et ψ

= ψ

(x) be a “cut-oﬀ” function

that is 1 for |x| < R, is 0 for |x| > 2R, and satisﬁes |D

| ≤ M R

−|α|

|α| = 1, 2, for some M independent of R. Moreover, let V , Z be the solution

to (VIII.4.1 6), (VIII.4.17), and denote by w = w(x, t) a vector ﬁeld satisfying

the following properties for a.a. t ∈ [0, T ]:

∇ · w = −∇ψ

· V (x, t) in B

R,2R

w(t) ∈ W

3,q

R,2R

) ,

∂w

∂t

∈ W

1,q

R,2R

) ,

k∇w(t)k

q,B

R,2R

≤ C

(k∇ψ

· V (t)k

q,B

R,2R

w(t)k

q,B

R,2R

≤ C

(k∇ψ

· V (t)k

q,B

R,2R

+ k∇(∇ψ

·V (t))k

q,B

R,2R

) ,



∇



∂w

∂t





≤ C



∇ψ

∂V (t)

∂t



q,B

R,2R

(VIII.4.34)

Obviously, f ≡ −∇ψ

·V (x, t) bel ongs to W

2,q

R,2R

), while

∂f

∂t

∈ L

R,2r

for a .a. t ∈ [0, T ]. Furthermore, tak ing into account that ψ

(x) = 0 for |x| =

2R, and that ∇ · V = 0 in R

, we have

R,2R

f = −

R,2R

∇· (ψ

V ) =

∂B

V ·x = 0 .

Thus, from Exercise III.3.7 and from the fact that V is in the class (VII I.4.17),

we deduce the existence of a ﬁeld w satisfying all the properties listed in

VIII.4 On the Unique Solvability of the Oseen Initial-Value Problem 537

(VIII.4.34), for all q ∈ (1, ∞). Furthermore, again by Exercise III.3.7 and by

the properties of V , we ﬁnd that w(t) is continuous, in the L

-norm, for all

t ≥ 0, and in particular, w(· , T ) = 0. We next observe that since B

R,2R

homothetic to B

1,2

via the transforma tion Φ

(x) = x

/R, i = 1, 2, 3, from

Lemma I II.3.3 we obtain that C

and C

are independent of R. By a similar

argument, we can show that C

≤ k(1 + 1/R), with k independent of R. We

then conclude that for R large enough, all constants C

, i = 1, 2, 3, can be

taken independent of R. Consequently, bearing in mind the properties of ψ

from (VIII.4.3 4)

3,4

, we deduce in particular

k∇w(t)k

+ kD

w(t)k

≤ k

kV (t)k

1,q,B

R,2R

, (VIII.4.35)

with k

independent of R. Next, from Exercise II.5.4 we obtain



∂w

∂t



q,B

R,2R

≤ C



∇



∂w

∂t





q,B

R,2R

with C

independent of R. Consequently, again from the properties of ψ

and

(VIII.4.34)

we obtain



∂w

∂t



q,B

R,2R

≤ k



∂V

∂t



q,B

R,2R

, (VIII.4.36)

with k

independent of R. Let us extend w to zero outside B

R,2R

and con-

tinue to denote by w the extension. Thus, if we dot-multiply (VIII .4.32)

(x, t) ≡ ψ

V (x, t) + w(x, t), then integrate over [η, T ] × R

, η > 0, and

recall that V , Z satisfy (VII I.4.16), we obtain

(u(η) , V

(η))

− 2∇ψ

·∇V −V ∆ψ

+ RV

∂ψ

∂x

− ψ

∇Z

+ψ

H −

∂w

∂t

− ∆w + R

∂w

∂x



dx dt .

(VIII.4.37)

We now pass to the limit η → 0 and use (VIII.4.33) along with the continuity

of V

at t = 0 in the L

-norm and the hypothesis on u to deduce

0 =

+ u

) ·

−2∇ψ

· ∇V −V ∆ψ

+ RV

∂ψ

∂x

− ψ

∇Z

+ψ

H −

∂w

∂t

−∆w + R

∂w

∂x



dx dt .

(VIII.4.38)

Next, we choose q = q

, i = 1, 2, and recall the prop erties of V given in

(VIII.4.17) and those of w given in (VIII.4.35), (VIII.4.36). As a consequence,

by means of the H¨older inequality, the properties of ψ

, and the Lebesgue

domi nated converge theorem (Lemma II.2.1), it is easy to show that

538 VIII Steady Generalized Oseen Flow in Exterior Domains

lim

R→∞



−2∇ψ

· ∇V − V ∆ψ

+ RV

∂ψ

∂x

−

∂w

∂t

− ∆w + R

∂w

∂x



dx dt = 0 , i = 1, 2.

(VIII.4.39)

Furthermore, we observe tha t, f or i = 1, 2,



· ∇Z −

· ∇Z



≤

|1 − ψ

||u

||∇Z|,

and since |u

||∇Z| ∈ L

), we may use a gain the Lebesgue dominated

convergence theorem along with the property of ψ

to show that

lim

R→∞

·∇Z =

· ∇Z .

However, from the assumption and (VIII.4.32)

, we have that u

(t) ∈ H

i = 1, 2, for a.a. t ∈ [0 , T ] (see Theorem III.2.3), and since ∇Z(t) ∈ L

for a .a. t ∈ [0 , T ], by Lemma III.2.1 we conclude that

lim

R→∞

u · ∇Z = 0 . (VIII.4.40)

Passing to the limi t R → ∞ in (VIII.4.39), utilizing (VIII.4.3 9) and (VIII. 4.40),

and recalling that H (x, t) = f(x, T − t), we ﬁnally deduce

u(x, t) · f (x, T − t)dx dt , for all f ∈ C

∞

) ,

which entails u = 0 a.e. in R

. The proof of the lemma is complete.

We now turn to the well-posedness of problem (VIII.4.2). Speciﬁcally, we

have the fol lowing.

Theorem VIII.4.3 Let u

∈ H

), 1 ≤ q ≤ ∞. Then there exists v such

that

ess sup

t∈[0,T ]

kv(t)k

< ∞,

v ,

∂v

∂t

, D

v ∈ L

([ε, T ] × R

) , for all ε > 0 and all r ≥ q ,

(VIII.4.41)

and satisfying (VIII.4.2)

1,2

a.e. in R

, for any T > 0. Moreover, for arbitrary

t > 0, we have

v(t)k

≤ c t

−(µ+(j+|α|)/2)

, j = 0, 1 ; 0 ≤ |α| ≤ 2 ,

lim

t→0

kv(t) − u

= 0 ,

(VIII.4.42)

VIII.4 On the Unique Solvability of the Oseen Initial-Value Problem 539

where µ = 3(1/q − 1 /r)/2 and c = c(r, q, j, α, R). The function v has the

following representation

v(x, t) =



4π t



3/2

−|x+R te

−y|

/4t

(y)dy .

Finally, let v

be such that

∈ L

) ,

∂v

∂t

, D

∈ L

loc

((0, T] × R

) , some s ∈ (1, ∞) ,

lim

t→0

(t) −u

= 0 ,

and satisfying (VIII.4.2)

1,2

a.e. in R

. Then v = v

, a.e. in R

Proof. The uniqueness part is a direct consequence of Lemma VIII.4.2. We

shall now prove the existence part. Using the properties of the fundamental

solution Γ (see, in particular, (VI II.3.9)), it is immediately veriﬁed that a

solution to (VIII.4.2)

1,2

is given by the following volume potential:

(x, t) =

(x − y, t; R)u

(y)dy , i = 1, 2, 3 , t > 0 . (VIII.4.43)

However, from (VIII.3.7), Lemma VIII.3.2, and the fact that u

∈ H

), it

follows that for i = 1, 2, 3 and all t > 0,

∂

∂x

Ψ(|x + Rte

−y|, t)u

(y)dy

∇



∂

∂x

Ψ(|x + Rte

− y|, t)



· u

(y)dy = 0 .

As a consequence, taking into account (VIII.3.7) and (VIII.3.4), (VIII.3.5 ),

the volume potential (VIII.4.43) reduces to

v(x, t) =



4π t



3/2

−|x+R te

−y|

/4t

(y)dy .

From this relation we easily deduce, for any multi-index β with | β| ≥ 0,

v(x, t) =





3/2



√



|β|



−|z|



(x −Rte

− 2z

√

t)dz ,

(VIII.4.44)

which, in turn, with the help of Young’s inequality (II.11.2), furnishes, in

particular, (VII I.4.42)

with j = 0. The case j = 1 follows from (VIII.4.44)

and (VIII. 4.2)

. To complete the existence proof, it remains to check the

validity o f (VIII.4.42)

. To this end, we observe that

540 VIII Steady Generalized Oseen Flow in Exterior Domains

−|y|

dy = π

3/2

and thus from (VIII.4.44) with β = 0, we obtain

v(x, t) −u

(x) = π

−3/2

−|z|

(x −Rt e

− 2z

√

t) − u(x)

dz .

Therefore, using the generalized Minkowski inequality (II.2.8), we infer

kv(t) − u

≤ π

−3/2



−|z|

(· − Rt e

− 2

√

tz) − u



dz .

(VIII.4.45)

In view of Exercise II.2.8, we have, for each ﬁxed z ∈ R

lim

t→0

(· − Rt e

− 2

√

tz) − u

= 0 ,

and consequently, passing to the limit t → 0

on both si des of (VI II.4.45)

and using the Lebesgue dominated convergence theorem (see Lemma II.2.1),

we establish (VIII.4.42)

. The proof of the theorem is complete. ut

The last part of this section is devoted to the solvability of (VIII.4.1) in

a space of functions that possess a suitable asympto tic behavior in space,

uniform ly in time. To this end, we introduce the following notation.

If U is a vector or a second-order tensor ﬁeld, α a nonnegative integer, A

a domain in R

, and R ≥ 0, we set

[]U []

α,R,A

= sup

x∈A

[(1 + |x|)

(1 + 2Rs(x))

|U (x)|] , (VIII.4.46)

where, we recall, s(x) is deﬁned in (VIII.3.11).

If R = 0, we shall simply wri te []U[]

α,A

instead of []U[]

α,0,A

. Furthermore,

whenever confusi on does not arise, we shall omit the subscript A.

We have the foll owing.

Theorem VIII.4.4 Let G be a second-order tensor ﬁeld in R

×(0, ∞) such

that

ess sup

t≥0

[](G(t)[]

2,R

+ ess sup

t≥0

k∇· G(t)k

< ∞,

and let h ∈ L

∞,q

× (0, ∞)), q ∈ (3, ∞), with spatial support contained in

, for some ρ > 0. Then, the problem (VIII.4.1) with f ≡ ∇·G + h has one

and only one solution such that f or all T > 0,

(w, φ) ∈ L

) , φ ∈ L

2,6

). (VIII.4.47)

Moreover,

ess sup

t≥0

[]w(t)[]

1,R

+ ess sup

t≥0

kφ(t)k

< ∞,

VIII.4 On the Unique Solvability of the Oseen Initial-Value Problem 541

for a rbitrary r >

, and the following inequality holds:

ess sup

t≥0

[]w(t)[]

1,R

+ ess sup

t≥0

kφ(t)k

≤ C ess sup

t≥0

([]G(t)[]

2,R

+ kh(t)k

)

(VIII.4.48)

with C = C(r, q, ρ, B), whenever R ∈ [0 , B], for some B > 0.

Finally, assume G = 0 and that h satisﬁes the further assumption

H := khk

∞,R

∞

+ k∇hk

∞,R

∞

< ∞. (VIII.4.49)

Then,

ess sup

t≥0



[]∇w(t)[]

+ []D

w(t)[]



< ∞, if R = 0 ,

ess sup

t≥0



[]∇w(t)[]

3/2,R

+ []D

w(t)[]

2,R



< ∞, if R 6= 0

(VIII.4.50)

and there exist constants C1 = C

(ρ) and C

= C

(ρ, B), whenever R ∈

(0, B], such that

ess sup

t≥0



[]∇w(t)[]

+ []D

w(t)[]



≤ C H , if R = 0 ,

ess sup

t≥0



[]∇w(t)[]

3/2,R

+ []D

w(t)[]

2,R



≤ C H , if R 6= 0

(VIII.4.51)

Proof. The existence of a uni que solution satisfying (VIII.4.47) is an i mme-

diate consequence of Theorem VIII.4.1, and the assumption on G and h. In

order to prove the remaining properties, we observe that, by Theorem VIII.4.2

the vector ﬁeld w admits the following representation:

w(x, t) =

Γ (x − y, t − τ) · [(∇ · G)(y, τ) + h(y, τ)] dy dτ

≡ w

(x, t) + w

(x, t) .

By an integration by parts we ﬁnd that

(x, t) = −

∂Γ

∂x

(x − y, τ )G

(y, t − τ )e

dy dτ,

and so by Fubini’s theorem and by Lemma VIII.3.4, Lemma VIII.3.5, we

deduce that

To alleviate the notation, and unless otherwise speciﬁed, we temporarily suppress

the dependence of Γ on R.

542 VIII Steady Generalized Oseen Flow in Exterior Domains

(x, t)| ≤

|∇Γ (x − y, τ)||G(y, t − τ )|dyds

|∇Γ (x − y, τ)||G(y, t − τ )|dτ dy

≤ ess sup

t≥0

[]G(t)[]

2,R

∞

|∇Γ (χ −y, τ)|

[(1 + |y|)(1 + Rs(y))]

dτ dy

≤ C max{1, R}

ess sup

t≥0

[]G(t)[]

2,R

(1 + |x|)(1 + 2 Rs(x))

for all x ∈ R

and all t ≥ 0. We now prove a n analogous estimate for w

We begin by observing that from Exercise III.3.9 and the hypothesis on h, it

follows that there is a second-order tensor ﬁeld H ∈ L

∞

) with ∇H ∈

), for all T > 0, such that

∇ · H (t) = h(t) , a.a. t ∈ [0, ∞) ,

ess sup

t≥0

kH(t)k

∞

≤ c ess sup

t≥0

kh(t)k

≡ h

(VIII.4.52)

Replacing ∇ · H for h in the expression of w

, integrating by parts, and

recalling that supp (h(t)) ⊂ B

, for all t ≥ 0, we obtain

(x − y, τ )D

(y, t − τ )dy dτ

= −

(x − y, τ ) H

(y, t − τ )dy dτ

∂B

(x − y, τ )H

(y, t − τ )n

(y)dσ

dτ

≡ I

+ I

(VIII.4.53)

From (VIII.4.52 ), Lemma VIII.3.6, a nd Fubini ’s theorem, we have

| ≤ c h



|x − y|

−2

dy + R

1/2

[|x − y|(1 + 2Rs(x − y))]

−3/2



≡ c h



(1)

+ I

(2)



(VIII.4.54)

If |x| ≤ 2ρ, we obta in, obviously,

(1)

≤

3ρ

dr ≤

(ρ) K

(1 + |x|)(1 + 2Rs(x))

whereas since

VIII.4 On the Unique Solvability of the Oseen Initial-Value Problem 543

|x| ≥ 2ρ =⇒ |x − y| ≥ |x|/2 for any y ∈ B

, (VIII.4.55)

we deduce

(1)

≤

(ρ)

(1 + |x|)

≤

(ρ) K

(1 + |x|)(1 + 2Rs(x))

Thus, we obtain

(1)

≤

(ρ) K

(1 + |x|)(1 + 2Rs(x))

, x ∈ R

. (VIII.4.56)

We next observe that for |x| ≤ 2ρ,

(2)

≤

3ρ

−1/2

dr ≤

(ρ) K

(1 + |x|)(1 + 2Rs(x))

, (VIII.4.57)

whereas if |x| ≥ 2ρ, from (VIII.4.55) and the fact that fo r any y ∈ B

1 + 2Rs(x) ≤ 1 + 2Rs(x − y) + 2Rs(y)

≤ (1 + 4Rρ)(1 + 2Rs(x −y))

≤ c

K (1 + 2Rs(x − y)) ,

(VIII.4.58)

we infer

(2)

≤

(ρ) K

(1 + |x|)(1 + 2Rs(x))

. (VIII.4.59)

By (VIII.4.5 4)–(VIII.4.59) we conclude that

≤

(ρ) K

(1 + |x|)(1 + 2Rs(x))

, x ∈ R

. (VI II.4. 60)

Furthermore, by (VIII.4.52), Lemma VIII. 3.6, and Fubini’s theorem, we have

≤ c

∞

∂B

|Γ (x − y, s)|ds

≤ c

∂B

(1 + |x − y|)(1 + 2Rs(x − y))

If we integrate both sides of this inequality over ρ between ρ

and 2ρ

, we

deduce

≤

2ρ

(1 + |x − y|)(1 + 2Rs(x − y))

Therefore, proceeding exactly as in the proof of the estimates (VIII.4.57) and

(VIII.4.59), we conclude that

≤

(ρ) K

(1 + |x|)(1 + 2Rs(x))

, x ∈ R

. (VI II.4. 61)

544 VIII Steady Generalized Oseen Flow in Exterior Domains

The desired estimate for w

is then a consequence of (VIII.4.5 3), (VIII.4.60),

and (VIII.4.61). Concerning the stated properties of the pressure, we notice

that from (VIII.4 .47), (VIII.4.1) and from the Helmholtz–Weyl decomposition

theorem, Theorem III.1.2, it follows that

(∇φ(t), ∇χ) = (∇ · G(t), ∇χ) + (h(t), ∇χ) ,

for a.a. t ∈ (0, ∞), and all χ ∈ D

1,2

(VIII.4.62)

We then choose in (VIII.4.62) the function χ as a solution to the Poisson

problem ∆χ = ψ, where ψ is arbitrary from C

∞

). Recalling that from the

representation χ = E ∗ ψ, we have D

χ = O(|x|

1−|β|

), |β| ≥ 0, and using the

properties of φ and G, by (VIII.4.62) we easily obtain

(φ, ψ) = (G, ∇∇χ) −(h, ∇χ) . (VIII.4.63)

By Exercise II. 11.9, we have kD

χk

≤ ckψk

, for all r

∈ (1, ∞), and

k∇χk

/(3−r

)

≤ ckD

χk

, for all r

∈ (1, 3). Thus, with the help of the

H¨older inequality, we obtain, on the one hand,

(G(t), ∇∇χ) ≤ c []G(t)[]

2,R

kψk

, for a.a. t ∈ (0, ∞), all r

∈ (1, ∞) .

(VIII.4.64)

On the other hand, since q > 3, we have q

< 3r

/(3 − r

) for all r

∈ (1, 3),

and consequently,

−(h(t), ∇χ) ≤ kh(t)k

k∇χk

q,B

≤ c

(ρ)kh(t)k

k∇χk

/(3−r

)

≤ c

(ρ)kh(t)k

χk

≤ c

(ρ)kh(t)k

kψk

for a .a. t ∈ (0, ∞), and all r

∈ (1, 3) .

(VIII.4.65)

Collecting (VIII.4.63)–(VIII.4.65) we conclude that

|(φ(t), ψ)| ≤ C

([]G(t)[]

2,R

+ kh(t)k

) kψk

, for a.a. t ∈ (0, ∞), all r

∈ (1, 3) ,

with C

depending o n r, q, and ρ. However, by Exercise II.2.12, this latter

inequality implies

ess sup

t≥0

kφ(t)k

≤ C

ess sup

t≥0

([]G(t)[]

2,R

+ kh(t)k

) ,

with C = C(r, q, ρ), and the proof of the ﬁrst part of the theorem i s complete.

Assume next that G = 0 and h satisﬁes the further assumptions (VI II.4.49).

We then have w(x, t) ≡ w

(x, t), where w

satisﬁes (VIII.4.53), namely, re-

calling (VIII.4.52)

, we deduce

(x, t) =

(x − y, τ ; R)h

(y, t − τ )dy dτ . (VIII.4.66)

We ﬁrst consider the case R = 0. Diﬀerentiating both sides with respect to x,

and taking into account estimate (VIII.3.10) of Lemma VIII.3.3 , we obtain

VIII.4 On the Unique Solvability of the Oseen Initial-Value Problem 545

w(x, t)| ≤ khk

∞,R

∞

Γ (x − y, τ)|dτ dy

≤ c

khk

∞,R

∞

|x − y|

from which, by distinguishing the two cases |x| ≤ 2ρ and |x| > 2ρ, we easily

prove the desired estimate for ∇w. We now diﬀerentiate (VIII.4.66) two times.

Taking into account that

(x, t) =

(y, τ)h

(x − y, t −τ)dy dτ

and the assumptions on h, we readily prove that

(x, t) =

(x − y, τ )D

(y, t − τ )dy dτ , (VIII.4.67 )

so that again by Lemma VIII.3 .3, we obtain

(x, t)| ≤ c

k∇hk

∞,R

∞

|x −y|

From this relation, i t easily follows that

w(x, t)| ≤ c

k∇hk

∞,R

∞

, for all (x, t) ∈ B

2ρ

×R

. (VII I.4.68)

Take now x outside B

2ρ

, and observe that, for such x and for y ∈ B

, Γ (x−y, t)

is a smooth function of x. Therefore, we may diﬀerentiate (VIII.4.66 ) twice

with respect to x, and use the estimates given in part (i) of Lemma VIII.3 .3

to obtain

(x, t)| ≤ c

khk

∞,R

∞

Γ ( x −y, τ)|dt dy

≤ c

khk

∞,R

∞

|x − y|

≤ c

khk

∞,R

∞

|x|

for a ll (x, t) ∈ B

2ρ

×R

(VIII.4.69)

The claimed estimates on the second derivatives of w follow from (VIII.4.68),

(VIII.4.69). The proof of (VIII.4.50) and (VIII.4.51) in the case R 6= 0 is very

similar, once we use the estimates (VIII.3.12), a nd will be only sketched here.

We take ﬁrst |x| ≥ 2ρ, a nd in (VIII.3.12)

we choose β = ρR. Consequently,

since |x − y| ≥ ρ = β/R, diﬀerentiating (VIII.4.66) once i n x and then using

(VIII.3.12)

, we obtain

w(x, t)| ≤ c

khk

∞,R

∞

|x −y|

3/2

(1 + 2Rs(x − y))

3/2

, |x| ≥ 2ρ ,