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Optimal three-ball inequalities and

quantitative uniqueness for the Stokes system

Ching-Lung Lin

Gunther Uhlmann

Jenn-Nan Wang

Abstract

In this paper we study the local behavior of a solution to the Stokes system with singular coefficients in Rnwith n = 2, 3. One of the main results is the bound on the vanishing order of a nontrivial solution u satisfying the Stokes system, which is a quantitative version of the strong unique continuation property for u. Different from the previous known results, our strong unique continuation result only involves the velocity field u. Our proof relies on some delicate Carleman-type esti- mates. We first use these estimates to derive crucial optimal three-ball inequalities for u. Taking advantage of the optimality, we then derive an upper bound on the vanishing order of any nontrivial solution u to the Stokes system from those three-ball inequalities. As an applica- tion, we derive a minimal decaying rate at infinity of any nontrivial u satisfying the Stokes equation under some a priori assumptions.

1 Introduction

Assume that Ω is a connected open set containing 0 in Rn with n = 2, 3. In this paper we are interested in the local behavior of u satisfying the following

Department of Mathematics, NCTS, National Cheng Kung University, Tainan 701, Taiwan. Email:cllin2@mail.ncku.edu.tw

Department of Mathematics, University of Washington, Box 354350, Seattle 98195- 4350, USA. Email:gunther@math.washington.edu

Department of Mathematics, Taida Institute of Mathematical Sciences, NCTS (Taipei), National Taiwan University, Taipei 106, Taiwan. Email: jn- wang@math.ntu.edu.tw

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Stokes system:

( ∆u + A(x) · ∇u + B(x)u + ∇p = 0 in Ω,

∇ · u = 0 in Ω, (1.1)

where A and B are measurable satisfying

|A(x)| ≤ λ1|x|−1| log |x||−3, |B(x)| ≤ λ1|x|−2| log |x||−3 ∀ x ∈ Ω (1.2) and A · ∇u = (A · ∇u1, · · · , A · ∇un).

For the Stokes system (1.1) with essentially bounded coefficients A(x), the weak unique continuation property has been shown by Fabre and Lebeau [6]. On the other hand, when A(x) satisfies |A(x)| = O(|x|−1+) with  >

0, the strong unique continuation property was proved by Regbaoui [19].

The results in [6] and [19] concern only the qualitative unique continuation theorem and both results require the vanishing property for u and p. In this work we aim to derive a quantitative estimate of the strong unique continuation for u satisfying (1.1) with an appropriate p.

For the second order elliptic operator, using Carleman or frequency func- tions methods, quantitative estimates of the strong unique continuation (in the form of doubling inequality) under different assumptions on coefficients were derived in [4], [7], [8], [14], [16]. For the power of Laplacian, a quanti- tative estimate was obtained in [17]. We refer to [16] and references therein for the development of this investigation.

Since there is no equation for p in the Stokes system (1.1), we apply the curl operator ∇× on the first equation and obtain

∆q + ∇ · F = 0, (1.3)

where q = ∇ × u and for n = 2, ∇ × u = ∂1u2 − ∂2u1. For n = 3, ∇ · F is a vector function defined by (∇ · F )i = P3

j=1jFij, i = 1, 2, 3, where Fij = P3

k,`=1ijk`(x)∂ku` +P3

k=1ijk(x)uk with appropriate ˜Aijk`(x) and B˜ijk(x) satisfying

| ˜Aijk`(x)| ≤ C0| log |x||−3|x|−1, | ˜Bijk(x)| ≤ C0| log |x||−3|x|−2 ∀ x ∈ Ω.

(1.4) When n = 2, ∇ · F is a scalar and we simply drop i in the definition above.

Now we define ∇× G = ∇ × G for any three-dimensional vector function

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G and ∇× g = (∂2g, −∂1g) for a scalar function g if n = 2. It is easy to check that ∆u = ∇(∇ · u) − ∇× (∇ × u) and thus we have

∆u + ∇× q = 0 (1.5)

if ∇·u = 0. However, the equations (1.3) and (1.5) do not give us a decoupled system. The frequency functions method does not seem to work in this case.

So we prove our results along the line of Carleman’s method. On the other hand, since the coefficient A(x) is more singular than the one considered in [19]. Carleman-type estimates derived in [19] can not be applied to the case here. Hence we need to derive new Carleman-type estimates for our purpose. The key is to use weights which are slightly less singular than the negative powers of |x| (see estimates (2.4) and (2.15)). The estimate (2.15) is to handle (1.3) and the idea is due to Fabre and Lebeau [6].

We can derive certain three-ball inequalities which are optimal in the sense explained in [5] using (2.4) and (2.15). We would like to remark that usually the three-ball inequality can be regarded as the quantitative estimate of the weak unique continuation property. However, when the three-ball in- equality is optimal, one is able to deduce the strong unique continuation from it. It seems reasonable to expect that one could derive a bound on the van- ishing order of a nontrivial solution from the optimal three-ball inequality.

A recent result by Bourgain and Kenig [3] (more precisely, Kenig’s lecture notes for 2006 CNA Summer School [13]) indicates that this is indeed possi- ble, at least for the Schr¨odinger operator. In this paper, we show that by the optimal three-ball inequality, we can obtain a bound on the vanishing order of a nontrivial solution to (1.1) containing ”nearly” optimal singular coeffi- cients. Finally, we would like to mention that quantitative estimates of the strong unique continuation are useful in studying the nodal sets of solutions for elliptic or parabolic equations [4], [9], [15], or the inverse problem [1].

We now state main results of this paper. Their proofs will be given in the subsequent sections. Assume that there exists 0 < R0 ≤ 1 such that BR0 ⊂ Ω. Hereafter Br denotes an open ball of radius r > 0 centered at the origin.

Theorem 1.1 There exists a positive number ˜R < 1, depending only on n, such that if 0 < R1 < R2 < R3 ≤ R0 and R1/R3 < R2/R3 < ˜R, then

Z

|x|<R2

|u|2dx ≤ C

Z

|x|<R1

|u|2dx

τZ

|x|<R3

|u|2dx

1−τ

(1.6)

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for (u, p) ∈ (H1(BR0))n+1 satisfying (1.1) in BR0, where the constant C de- pends on R2/R3, n, and 0 < τ < 1 depends on R1/R3, R2/R3, n. Moreover, for fixed R2 and R3, the exponent τ behaves like 1/(− log R1) when R1 is sufficiently small.

Remark 1.2 It is important to emphasize that C is independent of R1 and τ has the asymptotic (− log R1)−1. These facts are crucial in deriving an vanishing order of a nontrivial (u, p) to (1.1). Due to the behavior of τ , the three-ball inequality is called optimal [5].

It should be emphasized that three-ball inequalities (1.6) involve only the velocity field u. This is important in the application to inverse problems for the Stokes system, for example, see [2]. Using (1.6), we can also derive an upper bound of the vanishing order for any nontrivial u satisfying (1.1), which is a quantitative form of the strong unique continuation property for u. Let us now pick any R2 < R3 such that R3 ≤ R0 and R2/R3 < ˜R.

Theorem 1.3 Let (u, p) ∈ (H1(BR0))n+1 be a nontrivial solution to (1.1), then there exist positive constants K and m, depending on n and u, such that

Z

|x|<R

|u|2dx ≥ KRm (1.7)

for all R with R < R2.

Remark 1.4 Based on Theorem 1.1, the constants K and m in (1.7) are given by

K = Z

|x|<R3

|u|2dx and

m = ˜C log R

|x|<R3|u|2dx R

|x|<R2|u|2dx

 ,

where ˜C is a positive constant depending on λ1, n and R2/R3.

From Theorem 1.3, we immediately conclude that if (u, p) ∈ (Hloc1 (Ω))n+1 satisfies (1.1) and for any N ∈ N, there exists CN > 0 such that

Z

|x|<r

|u|2dx ≤ CNrN,

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then u vanishes identically in Ω. Consequently, p is a constant in Ω. This is a new strong unique continuation result for the Stokes system with singular coefficients.

By three-ball inequalities (1.6), one can also study the minimal decaying rate of any nontrivial velocity u to (1.1) with a suitable assumption on co- efficients A and B (see [3] for a related result for the Schr¨odinger equation).

Consider (u, p) satisfying (1.1) with Ω = Rn, n = 2, 3. Assume here that kukL(Rn)+ kAkL(Rn)+ kBkL(Rn) ≤ λ2. (1.8) Denote

Mr(t) = inf

|x|=t

Z

|y−x|<r

|u(y)|2dy.

Then we can prove that

Theorem 1.5 Let (u, p) ∈ (Hloc1 (Rn))n+1 be a nontrivial solution to (1.1).

Assume that (1.8) holds. Then for any r < 1, there exists c > 0 such that Mr(t) ≥ r(1+ tr ),

where c depends on λ2, n, R

|x|<r|u|2dx and ζ = 1 + 2 ˜C log(1/r) with ˜C given in Remark 1.4.

We can apply Theorem 1.5 to the stationary Navier-Stokes equation.

Corollary 1.6 Let (u, p) ∈ (Hloc1 (Rn))n+1 be a nontrivial solution of the stationary Navier-Stokes equation:

−∇u + u · ∇u + ρu + ∇p = 0, ∇ · u = 0, in Rn with n = 2, 3. Assume that

kukL(Rn)+ kρkL(Rn) ≤ λ3. Then for any r < 1, there exists ˜c > 0 such that

Mr(t) ≥ r˜(1+ tr ), where ˜c depends on λ3, n, andR

|x|<r|u|2dx.

This paper is organized as follows. In Section 2, we derive suitable Carleman-type estimates. A technical interior estimate is proved in Sec- tion 3. Section 4 is devoted to the proofs of Theorem 1.1, 1.3. The proof of Theorem 1.5 is given in Section 5.

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2 Carleman estimates

Similar to the arguments used in [10], we introduce polar coordinates in Rn\{0} by setting x = rω, with r = |x|, ω = (ω1, · · · , ωn) ∈ Sn−1. Further- more, using new coordinate t = log r, we can see that

∂xj = e−tjt+ Ωj), 1 ≤ j ≤ n,

where Ωj is a vector field in Sn−1. We could check that the vector fields Ωj satisfy

n

X

j=1

ωjj = 0 and

n

X

j=1

jωj = n − 1.

Since r → 0 iff t → −∞, we are mainly interested in values of t near −∞.

It is easy to see that

2

∂xj∂x` = e−2tjt− ωj+ Ωj)(ω`t+ Ω`), 1 ≤ j, ` ≤ n.

and, therefore, the Laplacian becomes

e2t∆ = ∂t2+ (n − 2)∂t+ ∆ω, (2.1) where ∆ω = Σnj=12j denotes the Laplace-Beltrami operator on Sn−1. We recall that the eigenvalues of −∆ω are k(k + n − 2), k ∈ N, and the corre- sponding eigenspaces are Ek, where Ek is the space of spherical harmonics of degree k. It follows that

Z Z

|∆ωv|2dtdω =X

k≥0

k2(k + n − 2)2 Z Z

|vk|2dtdω (2.2)

and

X

j

Z Z

|Ωjv|2dtdω =X

k≥0

k(k + n − 2) Z Z

|vk|2dtdω, (2.3) where vk is the projection of v onto Ek. Let

Λ =

r(n − 2)2

4 − ∆ω,

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then Λ is an elliptic first-order positive pseudodifferential operator in L2(Sn−1).

The eigenvalues of Λ are k + n−22 and the corresponding eigenspaces are Ek. Denote

L±= ∂t+ n − 2 2 ± Λ.

Then it follows from (2.1) that

e2t∆ = L+L= LL+.

Motivated by the ideas in [18], we will derive Carleman-type estimates with weights ϕβ = ϕβ(x) = exp(−β ˜ψ(x)), where β > 0 and ˜ψ(x) = log |x| + log((log |x|)2). Note that ϕβ is less singular than |x|−β, For simplicity, we denote ψ(t) = t + log t2, i.e., ˜ψ(x) = ψ(log |x|). From now on, the notation X . Y or X & Y means that X ≤ CY or X ≥ CY with some constant C depending only on n.

Lemma 2.1 There exist a sufficiently small r0 > 0 depending on n and a sufficiently large β0 > 1 depending on n such that for all u ∈ Ur0 and β ≥ β0, we have that

β Z

ϕ2β(log |x|)−2|x|−n(|x|2|∇u|2+ |u|2)dx . Z

ϕ2β|x|−n|x|4|∆u|2dx, (2.4) where Ur0 = {u ∈ C0(Rn\ {0}) : supp(u) ⊂ Br0}.

Proof. By the polar coordinate system described above, we have Z

ϕ2β|x|4−n|∆u|2dx

= Z Z

e−2βψ(t)e4t|∆u|2dtdω

= Z Z

|e−βψ(t)e2t∆u|2dtdω. (2.5)

If we set u = eβψ(t)v and use (2.1), then

e−βψ(t)e2t∆u = ∂t2v + b∂tv + av + ∆ωv =: Pβv, (2.6) where a = (1 + 2t−1)2β2+ (n − 2)β + 2(n − 2)t−1β − 2t−2β and b = n − 2 + 2β + 4t−1β. By (2.5) and (2.6), (2.4) holds if for t near −∞ we have

X

j+|α|≤1

β3−2(j+|α|) Z Z

|t|−2|∂tjαv|2dtdω ≤ ˜C1 Z Z

|Pβv|2dtdω, (2.7)

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where ˜C1 is a positive constant depending on n.

From (2.6), using the integration by parts, for t < t0 and β > β0, where t0 < −1 and β0 > 0 depend on n, we have that

Z Z

|Pβv|2dtdω

= Z Z

|∂t2v|2dtdω + Z Z

|b∂tv|2dtdω + Z Z

|av|2dtdω + Z Z

|∆ωv|2dtdω

− Z Z

tb|∂tv|2dtdω − 2 Z Z

a|∂tv|2dtdω + Z Z

t2a|v|2dtdω

− Z Z

t(ab)|v|2dtdω + 2X

j

Z Z

|∂tjv|2dtdω

+X

j

Z Z

tb|Ωjv|2dtdω − 2X

j

Z Z

a|Ωjv|2dtdω

≥ Z Z

|∆ωv|2dtdω + Z Z

{b2− ∂tb − 2a}|∂tv|2dtdω

+X

j

Z Z

{∂tb − 2a}|Ωjv|2dtdω + Z Z

{a2+ ∂t2a − ∂t(ab)}|v|2dtdω

≥ Z Z

|∆ωv|2dtdω +X

j

Z Z

{−4t−2β − 2a}|Ωjv|2dtdω

+ Z Z

{a2+ 11t−2β3}|v|2dtdω + Z Z

β2|∂tv|2dtdω. (2.8) In view of (2.8), using (2.2),(2.3), we see that

Z Z

|∆ωv|2dtdω − 2X

j

Z Z

a|Ωjv|2dtdω + Z Z

a2|v|2dtdω

= X

k≥0

Z Z

[a − k(k + n − 2)]2|vk|2dtdω. (2.9)

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Substituting (2.9) into (2.8) yields Z Z

|Pβv|2dtdω

≥ X

k≥0

Z Z

{11t−2β3 − 4t−2βk(k + n − 2) + [a − k(k + n − 2)]2}|vk|2dtdω

+ Z Z

β2|∂tv|2dtdω

= X

k,k(k+n−2)≥2β2

+ X

k,k(k+n−2)<2β2

 Z Z

{11t−2β3− 4t−2βk(k + n − 2)

+[a − k(k + n − 2)]2}|vk|2dtdω + Z Z

β2|∂tv|2dtdω. (2.10) For k such that k(k + n − 2) < 2β2, we have

11t−2β3− 4t−2βk(k + n − 2) ≥ t−2β3+ t−2βk(k + n − 2). (2.11) On the other hand, if 2β2 < k(k + n − 2), then, by taking t even smaller, if necessary, we get that

−4t−2βk(k + n − 2) + [a − k(k + n − 2)]2 & t−2βk(k + n − 2). (2.12) Finally, using formula (2.3) and estimates (2.11), (2.12) in (2.10), we imme- diately obtain (2.7) and the proof of the lemma is complete.

2

To handle the auxiliary equation corresponding to q, we need another Carleman estimate. The derivation here follows the line in [19].

Lemma 2.2 There exists a sufficiently small number t0 < 0 depending on n such that for all u ∈ Vt0, β > 1, we have that

X

j+|α|≤1

β1−2(j+|α|) Z Z

t−2ϕ2β|∂tjαu|2dtdω . Z Z

ϕ2β|Lu|2dtdω, (2.13)

where Vt0 = {u(t, ω) ∈ C0((−∞, t0) × Sn−1)}.

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Proof. If we set u = eβψ(t)v, then simple integration by parts implies Z Z

ϕ2β|Lu|2dtdω

= Z Z

|∂tv − Λv + βv + 2βt−1v + (n − 2)v/2|2dtdω

= Z Z

|∂tv|2dtdω + Z Z

| − Λv + βv + 2βt−1v + (n − 2)v/2|2dtdω +β

Z Z

t−2|v|2dtdω.

By the definition of Λ, we have Z Z

| − Λv + βv + 2βt−1v + (n − 2)v/2|2dtdω

= X

k≥0

Z Z

| − kvk+ βvk+ 2βt−1vk|2dtdω

= X

k≥0

Z Z

(−k + β + 2βt−1)2|vk|2dtdω,

where, as before, vk is the projection of v on Ek. Note that (−k + β + 2βt−1)2+ βt−2 ≥ 1

8β(2βt−1)2+ 1

16β(β − k)2. Considering β > (1/2)k and β ≤ (1/2)k, we can get that

Z Z

ϕ2β|Lu|2dtdω

= Z Z

|∂tv|2dtdω + Σk≥0 Z Z

[(−k + β + 2βt−1)2+ βt−2]|vk|2dtdω

&

Z Z

|∂tv|2dtdω + Σk≥0

Z Z

−1t−2k(k + n − 2) + βt−2)|vk|2dtdω.

(2.14)

The estimate (2.13) then follows from (2.3).

2

Next we need a technical lemma. We then use this lemma to derive another Carleman estimate.

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Lemma 2.3 There exists a sufficiently small number t1 < −2 depending on n such that for all u ∈ Vt1, g = (g0, g1, · · · , gn) ∈ (Vt1)n+1 and β > 0, we have that

Z Z

ϕ2β|u|2dtdω . Z Z

ϕ2β(|L+u + ∂tg0+

n

X

j=1

jgj|2+ kgk2)dtdω.

Proof. This lemma can be proved by exactly the same arguments used in

Lemma 2.2 of [19]. So we omit the proof here.

2

Lemma 2.4 There exist a sufficiently small number r1 > 0 depending on n and a sufficiently large number β1 > 2 depending on n such that for all w ∈ Ur1 and f = (f1, · · · , fn) ∈ (Ur1)n, β ≥ β1, we have that

Z

ϕ2β(log |x|)2(|x|4−n|∇w|2+ |x|2−n|w|2)dx . β

Z

ϕ2β(log |x|)4|x|2−n[(|x|2∆w + |x|divf )2+ kf k2]dx, (2.15) where Ur1 is defined as in Lemma 2.1.

Proof. Replacing β by β + 1 in (2.15), we see that it suffices to prove Z

ϕ2β(log |x|)−2(|x|2|∇w|2+ |w|2)|x|−ndx . β

Z

ϕ2β[(|x|2∆w + |x|divf )2+ kf k2]|x|−ndx. (2.16) Working in polar coordinates and using the relation e2t∆ = L+L, (2.16) is equivalent to

X

j+|α|≤1

Z Z

β2−2(j+|α|)t−2ϕ2β|∂tjαu|2dtdω

. β Z Z

ϕ2β(|L+Lw + ∂t(

n

X

j=1

ωjfj) +

n

X

j=1

jfj|2+ kf k2)dtdω.(2.17) Applying Lemma 2.3 to u = Lw and g = (Pn

j=1ωjfj, f1, · · · , fn) yields β

Z Z

ϕ2β|Lw|2dtdω . β

Z Z

ϕ2β(|L+Lw + ∂t(

n

X

j=1

ωjfj) +

n

X

j=1

jfj|2+ kf k2)dtdω.(2.18)

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Now (2.17) is an easy consequence of (2.13) and (2.18).

2

3 Interior estimates

To establish the three-ball inequality for (1.1), the following interior estimate is useful.

Lemma 3.1 Let (u, p) ∈ (Hloc1 (Ω))n+1 be a solution of (1.1). Then for any 0 < a3 < a1 < a2 < a4 such that Ba4r ⊂ Ω and |a4r| < 1, we have

Z

a1r<|x|<a2r

|x|4|∇q|2+ |x|2|q|2+ |x|2|∇u|2dx ≤ C0 Z

a3r<|x|<a4r

|u|2dx (3.1) where the constant C0 is independent of r and u. Here q = ∇ × u.

Proof. The proof of this lemma is motivated by ideas used in [11]. Let X = Ba4r\ ¯Ba3r and d(x) be the distant from x ∈ X to Rn\X. By the elliptic regularity, we obtain from (1.1) that u ∈ Hloc2 (Ω\{0}). It is trivial that

kvkH1(Rn) . k∆vkL2(Rn)+ kvkL2(Rn) (3.2) for all v ∈ H2(Rn). By changing variables x → E−1x in (3.2), we will have

P

|α|≤1E2−|α|kDαvkL2(Rn) . (k∆vkL2(Rn)+ E2kvkL2(Rn)) (3.3) for all v ∈ H2(Rn). To apply (3.3) on u, we need to cut-off u. So let ξ(x) ∈ C0(Rn) satisfy 0 ≤ ξ(x) ≤ 1 and

ξ(x) =

( 1, |x| < 1/4, 0, |x| ≥ 1/2.

Let us denote ξy(x) = ξ((x−y)/d(y)). For y ∈ X, we apply (3.3) to ξy(x)u(x) and use equation (1.5) to get that

E2 Z

|x−y|≤d(y)/4

|∇u|2dx

≤ C10 Z

|x−y|≤d(y)/2

|∇q|2dx + C10 Z

|x−y|≤d(y)/2

d(y)−2|∇u|2dx +C10(E4+ d(y)−4)

Z

|x−y|≤d(y)/2

|u|2dx. (3.4)

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Now taking E = M d(y)−1 for some positive constant M and multiplying d(y)4 on both sides of (3.4), we have

M2d(y)2 Z

|x−y|≤d(y)/4

|∇u|2dx

≤ C10 Z

|x−y|≤d(y)/2

d(y)4|∇q|2dx + C10 Z

|x−y|≤d(y)/2

d(y)2|∇u|2dx +C10(M4+ 1)

Z

|x−y|≤d(y)/2

|u|2dx. (3.5)

Integrating d(y)−ndy over X on both sides of (3.5) and using Fubini’s Theorem, we get that

M2 Z

X

Z

|x−y|≤d(y)/4

d(y)2−n|∇u|2dydx

≤ C10 Z

X

Z

|x−y|≤d(y)/2

d(y)4|∇q(x)|2d(y)−ndydx +C10

Z

X

Z

|x−y|≤d(y)/2

d(y)2−n|∇u|2dydx +2C10M4

Z

X

Z

|x−y|≤d(y)/2

|u|2d(y)−ndydx. (3.6) Note that |d(x) − d(y)| ≤ |x − y|. If |x − y| ≤ d(x)/3, then

2d(x)/3 ≤ d(y) ≤ 4d(x)/3. (3.7)

On the other hand, if |x − y| ≤ d(y)/2, then

d(x)/2 ≤ d(y) ≤ 3d(x)/2. (3.8)

By (3.7) and (3.8), we have ( R

|x−y|≤d(y)/4d(y)−ndy ≥ (3/4)nR

|x−y|≤d(x)/6d(x)−ndy ≥ 8−nR

|y|≤1dy, R

|x−y|≤d(y)/2d(y)−ndy ≤ 2nR

|x−y|≤3d(x)/4d(x)−ndy ≤ (3/2)nR

|y|≤1dy (3.9) Combining (3.6)–(3.9), we obtain

M2 Z

X

d(x)2|∇u|2dx

≤ C20 Z

X

d(x)2|∇u(x)|2dx + C20 Z

X

d(x)4|∇q|2dx + C20M4 Z

X

|u|2dx.

(3.10)

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On the other hand, we have from (1.3) that

n

X

i=1

Z

y(x)∇qi|2dx =

n

X

i=1

Z

∇qi· ∇(ξy2(x)¯qi)dx −

n

X

i=1

2 Z

ξy∇qi· ¯qi∇ξydx

≤ C30

n

X

i=1

| Z

(divF )iξy2qidx| +

n

X

i=1

2 Z

y∇qi· ¯qi∇ξy|dx

≤ C30

n

X

i=1

| Z n

X

j=1

Fij · ∂jy2qi)dx| +1 4

n

X

i=1

Z

y∇qi|2dx + 4 Z

|x−y|≤d(y)/2

d(y)−2|q|2dx

≤ C40 Z

|x−y|≤d(y)/2

|F |2dx +1 4

n

X

i=1

Z

y∇qi|2dx + C40 Z

|x−y|≤d(y)/2

d(y)−2|q|2dx

+1 4

n

X

i=1

Z

y∇qi|2dx + C40 Z

|x−y|≤d(y)/2

d(y)−2|q|2dx.

(3.11) Therefore, we get that

Z

|x−y|≤d(y)/4

|∇q|2dx

≤ Z

y(x)∇q|2dx

≤ C50 Z

|x−y|≤d(y)/2

|F |2dx + C50 Z

|x−y|≤d(y)/2

d(y)−2|q|2dx.

(3.12) Multiply d(y)4 on both sides of (3.12), we obtain that

Z

|x−y|≤d(y)/4

d(y)4|∇q|2dx

≤ C60 Z

|x−y|≤d(y)/2

d(y)4| ˜A|2|∇u|2dx + C60 Z

|x−y|≤d(y)/2

d(y)4| ˜B|2|u|2dx +C60

Z

|x−y|≤d(y)/2

d(y)2|q|2dx. (3.13)

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Repeat (3.6)∼(3.10), we have that Z

X

d(x)4|∇q|2dx

≤ C70 Z

X

d(x)4| ˜A|2|∇u|2dx + C70 Z

X

d(x)4| ˜B|2|u|2dx +C70

Z

X

d(x)2|q|2dx. (3.14)

Combining K×(3.14), (3.10) and R

Xd(x)2|q|2dx, we obtain that M2

Z

X

d(x)2|∇u|2dx + K Z

X

d(x)4|∇q|2dx + Z

X

d(x)2|q|2dx

≤ Z

X

(C20d(x)2 + C70Kd(x)4| ˜A|2)|∇u(x)|2dx + C70K Z

X

d(x)4| ˜B|2|u|2dx +C20M4

Z

X

|u|2dx + C20 Z

X

d(x)4|∇q|2dx + (C70K + 1) Z

X

d(x)2|q|2dx.

(3.15) Taking K = 2C20, one can eliminate R

Xd(x)4|∇q|2dx on the right hand side of (3.15). Observe that

Z

X

d(x)2|q|2dx ≤ C80 Z

X

d(x)2|∇u(x)|2dx.

So, by choosing M large enough, we can ignore R

Xd(x)2|∇u(x)|2dx on the right hand side of (3.15). Finally, we get that

M2 Z

X

d(x)2|∇u|2dx + K Z

X

d(x)4|∇p|2dx + Z

X

d(x)2|q|2dx

≤ C90 Z

X

|u|2dx. (3.16)

We recall that X = Ba4r\ ¯Ba3r and note that d(x) ≥ ˜Cr if x ∈ Ba2r\ ¯Ba1r, where ˜C is independent of r. Hence, (3.1) is an easy consequence of (3.16).

2

4 Proof of Theorem 1.1 and Theorem 1.3

This section is devoted to the proofs of Theorem 1.1 and Theorem 1.3. To begin, we first consider the case where 0 < R1 < R2 < R < 1 and BR ⊂ Ω.

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The small constant R will be determined later. Since (u, p) ∈ (H1(BR0))n+1, the elliptic regularity theorem implies u ∈ Hloc2 (BR0\ {0}). Therefore, to use estimate (2.4), we simply cut-off u. So let χ(x) ∈ C0(Rn) satisfy 0 ≤ χ(x) ≤ 1 and

χ(x) =





0, |x| ≤ R1/e,

1, R1/2 < |x| < eR2, 0, |x| ≥ 3R2,

where e = exp(1). We remark that we first choose a small R such that R ≤ min{r0, r1}/3 = ˜R0, where r0 and r1 are constants appeared in (2.4) and (2.15). Hence ˜R0 depends on n. It is easy to see that for any multiindex

α (

|Dαχ| = O(R−|α|1 ) for all R1/e ≤ |x| ≤ R1/2

|Dαχ| = O(R−|α|2 ) for all eR2 ≤ |x| ≤ 3R2. (4.1) Applying (2.4) to χu gives

C1β Z

(log |x|)−2ϕ2β|x|−n(|x|2|∇(χu)|2+|χu|2)dx ≤ Z

ϕ2β|x|−n|x|4|∆(χu)|2dx.

(4.2) From now on, C1, C2, · · · denote general constants whose dependence will be specified whenever necessary. Next applying (2.15) to w = χq and f = |x|χF , we get that

C2 Z

ϕ2β(log |x|)2(|x|4−n|∇(χq)|2+ |x|2−n|χq|2)dx

≤ β Z

ϕ2β(log |x|)4|x|2−n[|x|2∆(χq) + |x|div(|x|χF )]2dx +β

Z

ϕ2β(log |x|)4|x|2−nk|x|χF k2dx. (4.3)

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Multiplying by M1 on (4.2) and combining (4.3), we obtain that M1β

Z

R1/2<|x|<eR2

(log |x|)−2ϕ2β|x|−n(|x|2|∇u|2+ |u|2)dx +

Z

R1/2<|x|<eR2

(log |x|)2ϕ2β|x|−n(|x|4|∇q|2+ |x|2|q|2)dx

≤ M1β Z

ϕ2β(log |x|)−2|x|−n(|x|2∇(χu)|2+ |χu|2)dx +

Z

(log |x|)2ϕ2β|x|−n(|x|4|∇(χq)|2+ |x|2|χq|2)dx

≤ M1C3 Z

ϕ2β|x|−n|x|4|∆(χu)|2dx +βC3

Z

(log |x|)4ϕ2β|x|−n[|x|3∆(χq) + |x|2div(|x|χF )]2dx +βC3

Z

(log |x|)4ϕ2β|x|−nk|x|2χF k2dx. (4.4) By (1.2), (1.3), (1.4), and estimates (4.1), we deduce from (4.4) that

M1β Z

R1/2<|x|<eR2

(log |x|)−2ϕ2β|x|−n(|x|2|∇u|2+ |u|2)dx +

Z

R1/2<|x|<eR2

(log |x|)2ϕ2β|x|−n(|x|4|∇q|2+ |x|2|q|2)dx

≤ C4M1 Z

R1/2<|x|<eR2

ϕ2β|x|−n|x|4|∇q|2dx +C4β

Z

R1/2<|x|<eR2

(log |x|)−2ϕ2β|x|−n(|x|2|∇u|2+ |u|2)dx +C4M1

Z

{R1/e≤|x|≤R1/2}∪{eR2≤|x|≤3R2}

ϕ2β|x|−n| ˜U |2dx +C4β

Z

{R1/e≤|x|≤R1/2}∪{eR2≤|x|≤3R2}

(log |x|)4ϕ2β|x|−n| ˜U |2dx, (4.5)

where | ˜U (x)|2 = |x|4|∇q|2+|x|2|q|2+|x|2|∇u|2+|u|2 and the positive constant C4 only depends on n.

Now letting M1 = 2 + 2C4, β ≥ 2 + 2C4, and R small enough such that (log(eR))2 ≥ 2C4M1, then the first three terms on the right hand

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side of (4.5) can be absorbed by the left hand side of (4.5). Also, it is easy to check that there exists ˜R1 > 0, depending on n, such that for all β > 0, both (log |x|)−2|x|−nϕ2β(|x|) and (log |x|)4|x|−nϕ2β(|x|) are decreas- ing functions in 0 < |x| < ˜R1. So we choose a small R < ˜R2, where R˜2 = min{exp(−2√

2C4M1− 1), ˜R1/3, ˜R0}. It is clear that ˜R2 depends on n.

With the choices described above, we obtain from (4.5) that

R−n2 (log R2)−2ϕ2β(R2) Z

R1/2<|x|<R2

|u|2dx

≤ Z

R1/2<|x|<eR2

(log |x|)−2ϕ2β|x|−n|u|2dx

≤ C5β Z

{R1/e≤|x|≤R1/2}∪{eR2≤|x|≤3R2}

(log |x|)4ϕ2β|x|−n| ˜U |2dx

≤ C5β(log(R1/e))4(R1/e)−nϕ2β(R1/e) Z

{R1/e≤|x|≤R1/2}

| ˜U |2dx +C5β(log(eR2))4(eR2)−nϕ2β(eR2)

Z

{eR2≤|x|≤3R2}

| ˜U |2dx. (4.6)

Using (3.1), we can control | ˜U |2 terms on the right hand side of (4.6). In other words, it follows from (3.1) that

R−2β−n2 (log R2)−4β−2 Z

R1/2<|x|<R2

|u|2dx

≤ C622β+n(log(R1/e))4(R1/e)−nϕ2β(R1/e) Z

{R1/4≤|x|≤R1}

|u|2dx +C622β+n(log(eR2))4(eR2)−nϕ2β(eR2)

Z

{2R2≤|x|≤4R2}

|u|2dx

= C622β+n(log(R1/e))−4β+4(R1/e)−2β−n Z

{R1/4≤|x|≤R1}

|u|2dx +C622β+n(log(eR2))−4β+4(eR2)−2β−n

Z

{2R2≤|x|≤4R2}

|u|2dx. (4.7)

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Replacing 2β + n by β, (4.7) becomes R−β2 (log R2)−2β+2n−2

Z

R1/2<|x|<R2

|u|2dx

≤ C72β(log(R1/e))−2β+2n+4(R1/e)−β Z

{R1/4≤|x|≤R1}

|u|2dx +C72β(log(eR2))−2β+2n+4(eR2)−β

Z

{2R2≤|x|≤4R2}

|u|2dx. (4.8) Dividing R−β2 (log R2)−2β+2n−2 on the both sides of (4.8) and providing β ≥ n + 2, we have that

Z

R1/2<|x|<R2

|u|2dx

≤ C8(log R2)6(2eR2/R1)β Z

{R1/4≤|x|≤R1}

|u|2dx +C8(log R2)6(2/e)β[(log R2/ log(eR2))2]β−n−2

Z

{2R2≤|x|≤4R2}

|u|2dx

≤ C8(log R2)6(2eR2/R1)β Z

{R1/4≤|x|≤R1}

|u|2dx +C8(log R2)6(4/5)β

Z

{2R2≤|x|≤4R2}

|u|2dx. (4.9)

In deriving the second inequality above, we use the fact that log R2

log(eR2) → 1 as R2 → 0, and thus

2

e · log R2 log(eR2) < 4

5

for all R2 < ˜R3, where ˜R3is sufficiently small. We now take ˜R = min{ ˜R2, ˜R3}, which depends on n.

Adding R

|x|<R1/2|u|2dx to both sides of (4.9) leads to Z

|x|<R2

|u|2dx ≤ C9(log R2)6(2eR2/R1)β Z

|x|≤R1

|u|2dx +C9(log R2)6(4/5)β

Z

|x|≤1

|u|2dx. (4.10)

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It should be noted that (4.10) holds for all β ≥ ˜β with ˜β depending only on n. For simplicity, by denoting

E(R1, R2) = log(2eR2/R1), B = log(5/4), (4.10) becomes

Z

|x|<R2

|u|2dx

≤ C9(log R2)6n

exp(Eβ) Z

|x|<R1

|u|2dx + exp(−Bβ) Z

|x|<1

|u|2dxo . (4.11) To further simplify the terms on the right hand side of (4.11), we consider two cases. If R

|x|<R1|u|2dx 6= 0 and exp (E ˜β)

Z

|x|<R1

|u|2dx < exp (−B ˜β) Z

|x|<1

|u|2dx,

then we can pick a β > ˜β such that exp (Eβ)

Z

|x|<R1

|u|2dx = exp (−Bβ) Z

|x|<1

|u|2dx.

Using such β, we obtain from (4.11) that Z

|x|<R2

|u|2dx

≤ 2C9(log R2)6exp (Eβ) Z

|x|<R1

|u|2dx

= 2C9(log R2)6

Z

|x|<R1

|u|2dx

E+BB Z

|x|<1

|u|2dx

E+BE

. (4.12) If R

|x|<R1|u|2dx = 0, then letting β → ∞ in (4.11) we haveR

|x|<R2|u|2dx = 0 as well. The three-ball inequality obviously holds.

On the other hand, if exp (−B ˜β)

Z

|x|<1

|u|2dx ≤ exp (E ˜β) Z

|x|<R1

|u|2dx,

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then we have Z

|x|<R2

|u|2dx

Z

|x|<1

|u|2dx

E+BB Z

|x|<1

|u|2dx

E+BE

≤ exp (B ˜β)

Z

|x|<R1

|u|2dx

E+BB Z

|x|<1

|u|2dx

E+BE

. (4.13)

Putting together (4.12), (4.13), and setting C10= max{2C9(log R2)6, exp ( ˜β log(5/4))}, we arrive at

Z

|x|<R2

|u|2dx ≤ C10

Z

|x|<R1

|u|2dx

E+BB Z

|x|<1

|u|2dx

E+BE

. (4.14) It is readily seen that E+BB ≈ (log(1/R1))−1 when R1 tends to 0.

Now for the general case, we consider 0 < R1 < R2 < R3 < 1 with R1/R3 < R2/R3 ≤ ˜R, where ˜R is given as above. By scaling, i.e. defining u(y) := u(Rb 3y), p(y) := Rb 3p(R3y) and bA(y) = A(R3y), (4.14) becomes

Z

|y|<R2/R3

|u(y)|b 2dy ≤ C11( Z

|y|<R1/R3

|u(y)|b 2dy)τ( Z

|y|<1

|bu(y)|2dy)1−τ, (4.15) where

τ = B/[E(R1/R3, R2/R3) + B],

C11 = max{2C9(log R2/R3)6, exp ( ˜β log(5/4))}.

Note that C11is independent of R1. Restoring the variable x = R3y in (4.15) gives

Z

|x|<R2

|u|2dx ≤ C11( Z

|x|<R1

|u|2dx)τ( Z

|x|<R3

|u|2dx)1−τ. The proof of Theorem 1.1 is complete.

We now turn to the proof of Theorem 1.3. We fix R2, R3 in Theorem 1.1.

By dividing R

|x|<R2|u|2dx on the three-ball inequality (1.5), we have that 1 ≤ C(

Z

|x|<R1

|u|2dx/

Z

|x|<R2

|u|2dx)τ( Z

|x|<R3

|u|2dx/

Z

|x|<R2

|u|2dx)1−τ. (4.16)

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Raising both sides by 1/τ yields that Z

|x|<R3

|u|2dx ≤ ( Z

|x|<R1

|u|2dx)(C Z

|x|<R3

|u|2dx/

Z

|x|<R2

|u|2dx)1/τ. (4.17) In view of the formula for τ , we can deduce from (4.17) that

Z

|x|<R3

|u|2dx ≤ ( Z

|x|<R1

|u|2dx)(1/R1)C log(˜

R

|x|<R3|u|2dx/R

|x|<R2|u|2dx)

, (4.18) where ˜C is a positive constant depending on n and R2/R3. Consequently, (4.18) is equivalent to

( Z

|x|<R3

|u|2dx)Rm1 ≤ Z

|x|<R1

|u|2dx for all R1 sufficiently small, where

m = ˜C log R

|x|<R3|u|2dx R

|x|<R2|u|2dx

 . We now end the proof of Theorem 1.3.

5 Proof of Theorem 1.5

We prove Theorem 1.5 in this section. Let us first choose a > max{2, ˜R−1}, where ˜R is given in Theorem 1.1. By doing so, we can see that if we set R2 = ar and R3 = a2r, then R2/R3 < ˜R for r > 0. Now let 0 < r < 1 and define R2, R3 accordingly. Let |˜x| = t. We pick a sequence of points 0 = x0, x1, · · · , xN = ˜x such that |xj+1 − xj| ≤ r. We shall prove the desired estimate iteratively. To see how the iteration goes, let us assume that R

|x−xl|<r|u|2dx ≥ rml for some ml > 0 since u is nontrivial. By Theorem 1.3 and Remark 1.4, we have that

Z

|x−xl+1|<r

|u|2dx ≥ Z

|x−xl+1|<R3

|u|2dx · rm, (5.1) where

m = ˜C log R

|x−xl+1|<R3|u|2dx R

|x−xl+1|<R2|u|2dx

 .

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