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

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

Proof. By the results of Theorem XI.1.2, we know that v and the a ssoci-

ated pressure p are in C

∞

(Ω). Thus, u and p satisfy (XI.1.5)–(XI.1. 6). Dot-

multiplying both sides of (XI.1.5)

by u, and integrating over B

R,R

∗

, R

∗

> R,

we obtain

R,R

∗

∇u : ∇u =

R,R

∗

T (e

× x · ∇u) · u

∂B

∪∂B

∗



n · ∇u · u +

R|u|

−

R|u|

u · n − p(u · n)



We observe that on ∂B

∪ ∂B

∗

we have n :=

|x|

and thus

R,R

∗

×x · ∇u · u =

∂B

∪∂B

∗

|u|

× x) · n = 0.

We thus conclude that

R,R

∗

∇u : ∇u

∂B

∪∂B

∗



n · ∇u · u +

R|u|

−

R|u|

u ·n − p(u · n)



The rest of the proof follows precisely that of Lemma X.8.2, and will be

therefore omitted. ut

Lemma XI.6.2 Let the assumptions of Lemma XI.6.1 be satisﬁed. Then,

setting u := v + v

∞

, the following properties hold:

(a) ∇u ∈ L

∞

) ;

(b) e

×x · ∇u ∈ L

), for all q ∈ (2, ∞) .

Proof. By Theorem XI.1.2 we know that v, p are in C

∞

). Thus, by The-

orem XI.4.1 (after possibly adding a constant to p), we obtain

(u, ep) ∈ X

) , for all q ∈ (1, 2) , (XI.6.1)

with ep deﬁned in (XI.1.6). Let ψ

be a smooth, nonincreasing function that

is 0 in B

and is 1 i n Ω

2ρ

. We set

w := ψ

(u + Φ σ) −Z , φ := ψ

ep , (XI.6.2)

where

Φ :=

∂B

2ρ

v · n ,

and σ, we recall, is given by (see (XI.3.2))

σ(x) = −∇E(x − x

) , (XI.6.3)

XI.6 On the Asymptotic Structure of Generalized Solutions When v

·ω 6= 0 777

with E the fundamental solution to Laplace’s equation. Moreover, Z ∈

∞

2ρ

) satisﬁes ∇· Z = ∇ψ

·(v + Φ σ). Since

2ρ

∇ψ

· (v + Φ σ) =

∂B

2ρ

(v + Φ σ) ·n = 0 ,

the existence of the function Z is secured by Theorem III.3.3. Using (XI. 3.8)

and (XI.6 .3), we see that (w, φ) satisﬁes the following problem:

∆w + R

∂w

∂x

+ T (e

× x · ∇w − e

× w) −∇d

= R∇ · [(ψ

u) ⊗ (ψ

u)] + F

∇· w = 0











in R

, (XI.6.4)

with d = d(x) := φ(x) − ψ

(x)D

E(x − x

), and F ∈ C

∞

). In view

of (XI.6.1) and the mentioned regularity properties, it is easy to check that

∇ · [(ψ

u) ⊗ (ψ

u)] ∈ L

), for all q ∈ (1, ∞). Thus, by Lemma VIII.8.2,

the uniqueness L emma VIII.7.1, and the fact that ψ

= 1 in B

2ρ

, we conclude

that

u ∈ L

)

× x · ∇u − e

×u) ∈ L

)

for all q ∈ (1, ∞) . (XI.6.5)

However, by (XI.6.1), we have also ∇u ∈ L

) for all s ∈ (4/3, ∞), so that

property (a) follows from (XI.6.5)

and Theorem II.9.1. Furthermore, again

by (XI.6.1), we have e

× u ∈ L

), for all r ∈ (2, ∞), which, by virtue

of (XI.6. 5)

, impl ies e

× x · ∇u ∈ L

), for a ll q ∈ (2, ∞). This proves

property (b) and concludes the proof o f the lemma.

With the help of the previous lemma, we prove the following.

Lemma XI.6.3 Let the assumptions of Lemma XI.6.1 be satisﬁed, and let

Q = Q(t) be the one-parameter family o f proper orthogonal matrices deﬁned

in (VIII.5.10)–(VIII.5.11). Then, setting u := v+v

∞

, the following properties

hold:

(a) For all q ∈ (2, ∞),

Ω



sup

s≥0

|u(Q

(s) · x)|



dx < ∞.

(b) For all ε > 0,

Ω



sup

s≥0

|u(Q

(s) · x)|



dx ≤ c R

where c is independent of R (c → ∞ as ε → 0).

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

Proof. Property (b) is an immediate consequence of property (a). In fact, by

the H¨older inequality, for any q ∈ (2, ∞), we obtain

Ω



sup

s≥0

|u(Q

(s) · x)|



dx ≤ | Ω

(q−2)



Ω



sup

s≥0

|u(Q

(s) · x)|





≤ c R

3(q−2)



Ω



sup

s≥0

|u(Q

(s) ·x)|





and property (b) follows from (a). In order to establish (a), we observe that

since Q is periodic of perio d T := 2π/T (see (VIII.5.12)), it is enough to prove

the result by restricting s to the interval [0, T]. Set w(x, t) := u(Q

(t) · x).

Taking into account that (see also (VIII.5.1 5))



∂w

∂t



= T |e

× (Q

· x) · ∇u|,

and that Q(0) = I, we obtain, for all s ∈ [0, T],

|w(x, s)| ≤ |u(x)|+



∂w

∂t



= |u(x)| +

×(Q

(t) · x) · ∇u(Q

(t) ·x)|dt ,

which implies, by (II.3.3) and H¨older inequality,



sup

s∈[0,T ]

|u(Q

(s) ·x)|



≤ 2

q−1



|u(x)|

+ T

q−1

×(Q

(t) · x) · ∇u(Q

(t) ·x)|



(XI.6.6)

Recalling that Q(t) is proper orthogonal for all t ≥ 0, we have

Ω

×(Q

(t) · x) · ∇u(Q

(t) · x)|

dx =

Ω

× x · ∇u(x)|

dx ,

so that from (XI.6. 6) we get

Ω



sup

s∈[0,T ]

|u(Q

(s) · x)|



dx ≤ 2

q−1



kuk

q,Ω

+ T

× x · ∇uk

q,Ω



(XI.6.7)

From the stated assumptions, Theorem XI.4.1, a nd Lemma XI.6.2, we know

that u, e

, ×x · ∇u ∈ L

(Ω) for all q ∈ (2, ∞), so that claimed property (a)

follows from (XI.6.7). ut

We are now ready to give pointwise estimates for the velocity ﬁeld of a

generalized solution.

XI.6 On the Asymptotic Structure of Generalized Solutions When v

·ω 6= 0 779

Theorem XI.6.1 Let v be a generalized solution to (XI.0.1 0)–(XI.0.11), cor-

responding to f of b ounded support a nd to v

· ω 6= 0. Then, for any δ > 0

and all suﬃciently large |x|,

v(x) + v

∞

(x) = O



|x|

−1

(1 + Rs(x))

−1

+ |x|

−3/2+δ



and where, we recall, s(x) := |x|+ x

Proof. Choose ρ so large that supp (f ) ⊂ B

. By Theorem XI. 1.2, we know

that v, p are in C

∞

(Ω

) ∩ C

∞

(Ω

, r), r > ρ. Moreover, setting u := v + v

∞

by Theorem XI.4.1 (after possibly adding a constant to p), we ﬁnd that (u, ep)

satisﬁes (XI.6. 1). Proceeding exactly as in the proof of Lemma XI.6.2, we

then establish that the functions ψ

, w, and d there deﬁned satisfy problem

(XI.6.4). Let Q = Q(t) be the one-parameter family of proper orthogonal

matrices deﬁned in (VIII.5.10)–(VIII.5.11), and set

y := Q(t) · x ,

S(y, t) := Q(t) · w(Q

(t) · y) , π(y, t) := d(Q

(t) · y) ,

V (y, t) := Q(t) ·[ψ

u](Q

(t) ·y) , H(y, t) := Q(t) · F (Q

(t) ·y) .

(XI.6.8)

From (XI.6.1) and (XI. 6.3) we obtain

w ∈ L

) , for all r > 2, (XI.6.9)

and hence we have

∂S

∂t

= ∆S + R

∂S

∂y

− ∇π − R∇ · [V ⊗ V ] − H

∇ · S = 0











in R

∞

lim

t→0

kS(·, t) − wk

= 0 , all r ∈ (2, ∞) .

(XI.6.10)

Utilizing (XI.6.1), we deduce, in particular,

∇ · [V ⊗ V ] ∈ L

∞,r

∞

) , for all r ∈ (1, 4) .

Moreover,

supp (H (·, t)) ⊂ B

2ρ

, H ∈ L

∞,r

∞

) , for all r ∈ (1, ∞).

In view of all the above, from Theorem VIII.4 .1, Theorem VIII.4.2, and The-

orem VIII.4.3 we can ﬁnd a solution (

S, bπ) to (XI.6.10) such that

(

S, 0) ∈ L

× (ε, T)),

S ∈ L

∞

)

(0, bπ) ∈ L

) , fo r all r ∈ (2, 4), all ε > 0, and all T > ε,

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

having the fol lowing representation:

S(y, t) = (4πt)

−3/2

−|y−z+Rte

/4t

w(z) dz

−

Γ ( y − z, t − τ ) ·



R∇ · [V ⊗V ](z, τ) + H(z, τ )



dz dτ .

(XI.6.11)

Clearly, we have (for exam ple)

∂S

∂t

, D

S , φ , ∇φ ∈ L

loc

((0, T ] × R

)

and, recalling (XI.6.9), also S ∈ L

), for all r ∈ (2, ∞). Thus, by Lemma

VIII.4.2, we conclude, in particular, that S =

S. From (XI.6.11) we can

therefore, derive the following representation for S:

S = S

+ S

, (XI.6.12)

where

(y, t) = R

(y − z, t − τ )V

(z, τ)V

(z, τ) dz dτ , i = 1, 2, 3 ,

(XI.6.13)

(y, t) = −

Γ (y − z, t − τ) · H(z, τ) dz dτ , (XI.6.1 4)

(y, t) = (4πt)

−3/2

−|y−z+Rte

/4t

w(z) dz . (XI.6.15)

We shall now give pointwise estima tes of the functions S

, i = 1, 2, 3. Since

the numerical values of R and T are irrelevant in the proof (provided they

are both positive, of course), in what follows we shall put, for simplicity,

R = T = 1. By Lemma VIII.3.2, we have

(y, t)| ≤ c

(τ + |y − z + τ e

)

−2

|V (z, t − τ)|

dz dτ

= c

(τ + |y − z + τ e

)

−2

|V (z, t − τ)|

dz dτ

(τ + |y − z + τ e

)

−2

|V (z, t − τ)|

dz dτ

: = I

+ I

(XI.6.16)

where R = |y|/3 > 2ρ. Using the H¨older inequality, for any r ∈ (2, ∞) and

with r

:= r/(r − 2) we obtain

For si mplicity, throughout the proof we shall omit, the label R in the notation of

the fundamental tensor Γ .

XI.6 On the Asymptotic Structure of Generalized Solutions When v

·ω 6= 0 781

≤ c

∞



(τ + |y − z + τ e

)

−2r



1/r

kV (t − τ )k

dτ .

From the deﬁnition of V given in (XI.6.8), we obtain kV (t −τ )k

≤ kuk

r,Ω

and so

≤ c

kuk

r,Ω

∞



(τ + |y − z + τ e

)

−2r



1/r

dτ .

Furthermore, putting z

= z − τ e

, we have |z

| ≤ 2R for τ ∈ [ 0, R] and

z ∈ B

. Thus



(τ + |y − z + τ e

)

−2r



1/r

dτ

≤



(τ + |y − z

)

−2r



1/r

dτ .

Also, for |z

| ≤ 2R we have |y − z

| ≥ 3R − 2R = R, and hence



(τ + |y − z + τ e

)

−2r



1/r

dτ

≤



(τ + R

)

−2r



1/r

dτ ≤ c

3/r

(τ + R

)

dτ ≤ c

−2+

Since

∞



(τ + |y − z + τ e

)

−2r



1/r

dτ ≤

∞



−2r



1/r

dτ

≤ c

−1+

we infer, for suﬃciently large |y|,

≤ c

−1+

= c

−1+

3(r−2)

where c

= c

(r). Thus, recalling that r is arbitrary in (2, ∞), we conclude

that

≤ c

|y|

−1+ε

, for all ε > 0 , (XI.6 .17)

with c

→ ∞ as ε → 0. We shall next estimate I

. To this end, we set

δ(ξ) =

+ ξ

, ξ ∈ R

, and write

∩{δ(y−z)<1}

(τ + |y − z + τe

)

−2

|V (z, t −τ)|

dz dτ

∩{δ(y−z)≥1}

(τ + |y − z + τ e

)

−2

|V (z, t − τ )|

dz dτ

: = I

+ I

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

By H¨older’s inequality, we obtain for an arbi trary q

> 6 and with q

/(q

−2) ∈ (1, 3/2):

≤

{δ(y−z)<1}

(τ + |y − z + τ e

)

1/q

kV (t −τ )k

dτ ,

from which, using the result of Exercise VIII.3 .2 and the deﬁnition of V , we

deduce

≤ c

kuk

≤ c

kuk

2(q

−6)

∞,Ω

kuk

2(6−q

)

6,B

. (XI.6.18)

We shall now use in the above relation the inequality

kuk

6,B

R ≤ c

|u|

1,2,B

R , (XI.6 .19)

where, as we know from Exercise II.6.5, the constant c

is independent of R.

Combining (XI.6.19) with Lemma XI.6.1 and placing this information back

into (XI.6.18), it follows that

≤ c

−1+ε

= c

|y|

−1+ε

, for all ε > 0 , (XI. 6.20)

where c

→ ∞ as ε → 0. In order to estimate I

we choose arbitrary

∈ (2, 6) and set q

:= q

/(q

− 2) ∈ (3/2, ∞). We thus obtain

≤

{δ(y−z)≥1}

(τ + |y − z + τ e

)

1/q

kV (t −τ )k

dτ ,

from which, using again the result of Exercise VIII.3.2, the deﬁnition of V ,

and the interpol ation inequality (II.2.7) we deduce

≤ c

kuk

≤ c



kuk

1−θ

6,B

kuk



, (XI.6.21)

where θ → 0 and c

→ ∞ as q

→ 6. Combining (XI.6.21), (XI. 6.19) and

Lemma XI.6.1, we deduce that

≤ c

−1+ε

= c

|y|

−1+ε

, for all ε > 0 , (XI. 6.22)

where c

→ ∞ as ε → 0. Thus, (XI.6.16), (XI.6.17), (XI.6.20), and (XI.6.22)

allow us to conclude that

(y, t)| ≤ c

|y|

−1+ε

, for all ε > 0 , (XI.6.2 3)

with c

→ ∞ as ε → 0. Concerning w

, from Theorem VIII.4.4 (see

(VIII.4.48)) and the deﬁnition of H, we immediately deduce for all s > 3

(y, t)| ≤ C

kF k

(1 + |y|)(1 + s(y))

, (XI.6.24)

XI.6 On the Asymptotic Structure of Generalized Solutions When v

·ω 6= 0 783

so that in particular,

(y, t)| ≤ c

|y|

−1

. (XI.6.25)

We also notice that from Theorem VIII.4.3 (see (VIII.4.42)), it follows that

(y, t)| ≤ c

−

kwk

. (XI.6.26)

Combining (XI.6.2 ), (XI.6.8), and (XI.6.12 ) together with (XI.6.24)–(XI.6.26),

we thus conclude, for suﬃciently large |x|, that

|u(x)| ≤ |w(x)| + |Φ||σ(x)| ≤ |S(Q(t) · x)| + c

|x|

−2

≤ c



|x|

−1+ε

+ t

−

kwk



, for all ε > 0 a nd all t > 0 ,

(XI.6.27)

where c

depends o n ε, but not on t. Thus, letting t → ∞, from this last

inequality we obta in

|u(x)| ≤ c

|x|

−1+ε

(XI.6.28)

With the estimate (XI.6. 28) in hand, we will give an im proved bound on the

term S

given i n (XI.6.13). We begin by observing that, clearly, from the

deﬁnition of V and (XI.6.28), it follows that

|V (z, t)| ≤ c

(1 + |z|)

−1+ε

sup

s≥0

|V (z, s)| ≤ sup

s≥0

|u(Q

(s) ·z)|.

(XI.6.29)

We next partition R

into three regions:

, B

R,4R

, B

and denote the corresponding contributions of S

over these spatial re-

gions by I

, I

, and I

, respectively. Thus, employing Lemma VIII.3.3 and

(XI.6.29)

, we may increase I

as follows

| ≤ c



sup

s≥0

|V (z, s)|





∞

|∇Γ (y − z, t)|dt



≤ c

Ω



sup

s≥0

|u(Q

(s) · z)|



|y −z|

3/2

dz ≤

|y|

Ω



sup

s≥0

|u(Q

(s) · z)|



Using the result of Lemma XI.6.3 in this l atter, we then conclude that

| ≤ c

|y|

−3/2

= c

|y|

−3/2+δ

(XI.6.30)

for a rbitrary δ > 0. In a similar fashion, we establish

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

| ≤ c



sup

s≥0

|V (z, s)|





∞

|∇Γ (y − z, t)|dt



≤ c

Ω



sup

s≥0

|u(Q

(s) · z)|



|y −z|

3/2

dz .

Therefore, since for z ∈ Ω

we have |y − z| ≥

|z| + 3R − |y| =

|z|, with

the help of the H¨ol der inequality we obtain for arbitrary r ∈ (1, 2)

| ≤ c

Ω



sup

s≥0

|u(Q

(s) · z)|





∞

−

2(r−1)

dρ



r−1

which, in turn, by Lemma XI.6.3, furnishes

| ≤ c

|R|

−3/2+δ

= c

|y|

−3/2+δ

, (XI.6.31)

where δ := 3(r − 1)/r is positive and arbitrarily clo se to zero, since r can be

chosen arbitrarily close to 1. In order to increase I

, we notice that

(ξ) :=

∞

|∇Γ (ξ, t)|dt

satisﬁes the foll owing property:

∂B

)

(x − x

) ≤ M ρ

−1/2

, (XI.6.32)

for a ll x

∈ R

and ρ > 0. The proof of (XI.6.32) i s entirely analogous to the

analogous property proved in Exercise VII.3.1 for the fundamental tensor E,

once we take into account the estimate fo r Γ

furnished in Lemma VIII.3.3.

Taking into account (XI.6.29)

, we derive

| ≤ c

R,4R

(1 + |z|)

−2+2ε



∞

|∇Γ (y −z, t)|dt



≤ c

−2+2ε

R,4R

Γ (y − z) dz ≤ c

−2+2ε

|Γ |

1,B

(y )

Consequently, by (XI.6.32),

| ≤ c

|R|

−2+2ε

1/2

= c

|y|

−3/2+δ

(XI.6.33)

for arbi trary δ > 0. Recalling that S

i=1

, from (XI.6.30), (XI. 6.31),

and (XI.6 .33) we deduce

(y, t)| ≤ c

|y|

−3/2+δ

, for all δ > 0. (XI.6.34)

XI.6 On the Asymptotic Structure of Generalized Solutions When v

·ω 6= 0 785

Putting together (XI.6.9), (XI.6.12), (XI. 6.25), (XI.6.26), and (XI.6.34), ar-

guing as in (XI.6.27), and letting t → ∞, we conclude, for suﬃciently large

|x|, that

|u(x)| ≤ c



(1 + |x|)

−1

(1 + s(x))

−1

+ |x|

−3/2+δ



, for all δ > 0 ,

which completes the proof of the theorem. ut

Our next result concerns the asymptotic behavior of ∇v.

Theorem XI.6.2 Let the assumptions o f Theorem XI.6.1 be satisﬁed. Then,

for a ny η > 0 and all suﬃciently large |x|,

∇(v(x) + v

∞

(x)) = O



|x|

−3/2

(1 + Rs(x))

−3/2

+ |x|

−2+η



Proof. As in the proof of the previo us theorem, we set u := v +v

∞

. We begin

by recalling that, by Lemma XI.6.2, ∇u is bounded for a ll la rge |x|. From the

deﬁnition (XI. 6.8) of V , we thus obtain

|∇V (y, t)| ≤ c

(XI.6.35)

with c

independent of y and t. We also recall that by the same token, from

Theorem XI.6.1 we have, fo r all large |y|,

|V (y, t)| ≤ c

|y|

−1

(XI.6.36)

with c

independent of y and t. Our next step is to prove that |∇u(x)| decays

at least like |x|

−1

. Our starting point is again (XI.6.12)–(XI.6.15) for large

values of |y|, which, as immediately proved, yields for k, i = 1, 2, 3,

(y, t) = R

(y −z, t −τ )D

(z, τ)V

(z, τ)] dz dτ ,

(y, t) = −

Γ (y − z, t − τ ) · H(z, τ) dz dτ ,

(y, t) = (4πt)

−3/2



−|y−z+Rte

/4t

w(z) dz



(XI.6.37)

As done previously, we set, for simplicity, R = T = 1. We notice that from

Theorem VIII.4.4 (see, in particular, the argument after (VIII.4.66)), we get

(y, t)| ≤ c

|y|

−3/2

(1 + s(y))

−3/2

, (XI.6.38)

while from Theorem VIII. 4.3 it follows that

(y, t)| ≤ c

−3/4

kwk

. (XI.6.39)