# Optimal three-ball inequalities and

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## quantitative uniqueness for the Stokes system

### Jenn-Nan Wang

Abstract

In this paper we study the local behavior of a solution to the Stokes system with singular coefficients. One of the main results is the bound on the vanishing order of a nontrivial solution to the Stokes system, which is a quantitative version of the strong unique continuation prop- erty. Our proof relies on some delicate Carleman-type estimates. We first use these estimates to derive crucial optimal three-ball inequal- ities. Taking advantage of the optimality, we then derive an upper bound on the vanishing order of any nontrivial solution to the Stokes system from those three-ball inequalities.

### 1 Introduction

Assume that Ω is a connected open set containing 0 in Rnwith n ≥ 2. In this paper we are interested in the local behavior of (u, p) satisfying the following Stokes system:

( ∆u + A(x) · ∇u + ∇p = 0

divu = 0, (1.1)

where A is measurable satisfying

|A(x)| ≤ C0| log |x||−3|x|−1 ∀ x ∈ Ω (1.2)

Department of Mathematics, National Cheng Kung University, Tainan 701, Taiwan.

Email:cllin2@mail.ncku.edu.tw

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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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 . On the other hand, when A(x) satisfies |A(x)| = O(|x|−1+) with  >

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

The results in  and  concern only the qualitative unique continuation theorem. In this work we aim to derive a quantitative estimate of the strong unique continuation for (1.1).

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 , , , , . For the power of Laplacian, a quanti- tative estimate was obtained in . We refer to  and references therein for the development of this investigation.

Since there is no equation for p in the Stokes system (1.1), to prove the unique continuation theorem for (1.1), one usually apply the divergence on the first equation and obtain

∆p + div(A(x) · ∇u) = 0. (1.3)

However, the first equation of (1.1) and (1.3) 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 . Carleman-type estimates derived in  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 . It is tempting to derive doubling inequalities for (1.1) by (2.4) and (2.15) using the ideas in  or . But this seems hard to reach with estimates (2.4), (2.15). One of the difficulties is the appearance of the parameter β on the right hand side of (2.15).

Even though we are not able to prove doubling inequalities for (1.1), we can derive certain three-ball inequalities which are optimal in the sense explained in  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 inequality

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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 vanishing order of a nontrivial solution from the optimal three-ball inequality. A recent result by Bourgain and Kenig  (more precisely, Kenig’s lecture notes for 2006 CNA Summer School ) indicates that this is indeed possible, 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 coefficients.

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 , , or the inverse problem .

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. Also, we let U (x) = [|x|4|∇p|2+ |x|2|p|2+ |u|2]1/2.

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

τZ

|x|<R3

|U |2dx

1−τ

(1.4)

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 .

Remark 1.3 We want to say a few words about the appearance of ∇p term in U . In the derivation of the three-ball inequality (1.4), it is crucial to control ∇u in a smaller region by quantities of u and p in a bigger region (see (3.1)). Roughly speaking, this is an interior estimate for ∇u. In view of the first equation of (1.1), ∇p needs be included in this estimate.

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Theorem 1.4 Let (u, p) ∈ (H1(BR0))n+1 be a nontrivial solution to (1.1), i.e, (u, p) 6= (0, 0), then there exist positive constants K and m, depending on n and (u, p), such that

Z

|x|<R

|U |2dx ≥ KRm (1.5)

for all R sufficiently small.

Remark 1.5 Based on Theorem 1.1, the constants K and m in (1.5) 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 n and R2/R3.

Corollary 1.6 Let (u, p) ∈ (Hloc1 (Ω))n× L2loc(Ω) be a solution of (1.1) with A satisfying (1.2). Assume that (u, p) vanishes of infinite order at the origin, i.e., for all N > 0,

Z

|x|<R

(|u|2+ |p|2)dx = O(RN) as R → 0. (1.6) Then (u, p) ≡ 0 in Ω.

This corollary is a small improvement of the strong unique continuation prop- erty for the Stokes system proved in  where |A(x)| = O(|x|−1+) with

 > 0.

This paper is organized as follows. In Section 2, we derive suitable Carleman-type estimates. A technical interior estimate is proved in Section 3.

Section 4 is devoted to the proofs of Theorem 1.1, 1.4, and Corollary 1.6.

### 2 Carleman estimates

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

∂xj = e−tjt+ Ωj), 1 ≤ j ≤ n,

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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,

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+.

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Motivated by the ideas in , 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|α|

Z Z

|t|−2|∂tjαv|2dtdω ≤ ˜C1 Z Z

|Pβv|2dtdω, (2.7)

where ˜C1 is a positive constant depending on n.

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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 the pressure p, we need another Carleman estimate. The derivation here follows the line in .

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 . 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 + 2 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

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).

### 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|2|∇u|2dx ≤ C Z

a3r<|x|<a4r

(|x|4|∇p|2+ |u|2)dx, (3.1) where the constant C is independent of r and (u, p).

Proof. 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 → B−1x in (3.2), we will have

P

|α|≤1B2−|α|kDαvkL2(Rn) . (k∆vkL2(Rn)+ B2kvkL2(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 (1.1) to get that

B2 Z

|x−y|≤d(y)/4

|∇u|2dx .

Z

|x−y|≤d(y)/2

(|A|2+ d(y)−2)|∇u|2dx + Z

|x−y|≤d(y)/2

|∇p|2dx +(B4+ d(y)−4)

Z

|x−y|≤d(y)/2

|u|2dx. (3.4)

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Now taking B = 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 .

Z

|x−y|≤d(y)/2

(d(y)4|A|2+ d(y)2)|∇u|2dx +

Z

|x−y|≤d(y)/2

d(y)4|∇p|2dx + (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 .

Z

X

Z

|x−y|≤d(y)/2

(d(y)2+ d(y)4|A|2)|∇u(x)|2d(y)−ndydx +

Z

X

Z

|x−y|≤d(y)/2

d(y)4−n|∇p|2dydx +M4

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)

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Combining (3.6)–(3.9), we obtain M2

Z

X

d(x)2|∇u|2dx .

Z

X

(d(x)2+ d(x)4|A|2)|∇u(x)|2dx + Z

X

d(x)4|∇p|2dx +M4

Z

X

|u|2dx. (3.10)

In view of (1.2), we can take M large enough to absorb the first term on the right hand side of (3.10). Thus we conclude that

Z

X

d(x)2|∇u|2dx . Z

X

(d(x)4|∇p|2+ |u|2)dx. (3.11) 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 (3.11).

### 4 Proof of Theorem 1.1 and Theorem 1.4

This section is devoted to the proofs of Theorem 1.1 and Theorem 1.4. To begin, we first consider the case where 0 < R1 < R2 < R < 1 and BR ⊂ Ω.

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)

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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 = χp and f =

|x|χA · ∇u, we get that C2

Z

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

≤ β Z

ϕ2β(log |x|)4|x|2−n[|x|2∆(χp) + |x|div(|x|χA · ∇u)]2dx +β

Z

ϕ2β(log |x|)4|x|2−nk|x|χA · ∇uk2dx. (4.3) 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|∇p|2+ |x|2|p|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|∇(χp)|2+ |x|2|χp|2)dx

≤ M1C3 Z

ϕ2β|x|−n|x|4|∆(χu)|2dx +βC3

Z

(log |x|)4ϕ2β|x|−n[|x|3∆(χp) + |x|2div(|x|χA · ∇u)]2dx +βC3

Z

(log |x|)4ϕ2β|x|−nk|x|2χA · ∇uk2dx. (4.4)

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By (1.1), (1.2), (1.3), 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|∇p|2+ |x|2|p|2)dx

≤ C4M1

Z

R1/2<|x|<eR2

(log |x|)−2ϕ2β|x|−n|x|2|∇u|2dx +C4M1

Z

R1/2<|x|<eR2

ϕ2β|x|−n|x|4|∇p|2dx +C4β

Z

R1/2<|x|<eR2

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

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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| 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)

Recall that |U (x)|2 = |x|4|∇p|2+ |x|2|p|2+ |u|2. Replacing 2β + n by β, (4.7) becomes

R2−β(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)

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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.

|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)

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),

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(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)

IfR

|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

| bU (y)|2dy ≤ C11( Z

|y|<R1/R3

| bU (y)|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))},

and bU (y) = |y|4|∇p(y)|b 2+ |y|2|p(y)|b 2+ |bu(y)|2. Note that C11 is 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 1.4. We fix R2, R3 in Theorem 1.1 and define





eu(x) := u(x)/qR

|x|<R2|U |2dx, p(x) := p(x)/e qR

|x|<R2|U |2dx,

V (x) = |x|4|∇p(x)|e 2+ |x|2|p(x)|e 2+ |eu(x)|2.

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Note that R

|x|<R2|V |2dx = 1. From the three-ball inequality (1.4), we have that

1 ≤ C(

Z

|x|<R1

|V |2dx)τ( Z

|x|<R3

|V |2dx)1−τ. (4.16) Raising both sides by 1/τ yields that

Z

|x|<R3

|V |2dx ≤ ( Z

|x|<R1

|V |2dx)(C Z

|x|<R3

|V |2dx)1/τ. (4.17)

In view of the formula for τ , we can deduce from (4.17) that Z

|x|<R3

|V |2dx ≤ ( Z

|x|<R1

|V |2dx)(1/R1)C log(˜

R

|x|<R3|V |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 1.4.

Finally, we come to the proof of Corollary 1.6. In view of Theorem 1.4, it is enough to show that

p ∈ Hloc1 (Ω) (4.19)

and for all N > 0 Z

|x|<R

|∇p|2dx = O(RN) as R → 0. (4.20)

It is only a matter of checking that the arguments used in [18, page 1898-1899]

can be applied to prove (4.19) and (4.20). To avoid unnecessary repetition, we only sketch the main steps here. By virtue of (1.3), u ∈ Hloc1 (Ω), and (1.2), we get that p ∈ Hloc1 (Ω \ {0}) by elliptic regularity. Using elliptic

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regularity and the first equation of (1.1), we have u ∈ Hloc2 (Ω \ {0}). By the vanishing assumption (1.6), we can derive that

Z

R<|x|<2R

|∇p|2dx = O(RN) as R → 0

for all N > 0. It follows that p is the sum of a function in Hloc1 (Ω) and a distribution supported at 0. But no distribution supported at 0 is in L2loc(Ω).

Thus, p ∈ Hloc1 (Ω) and (4.20) holds.

### Acknowledgements

The authors were supported in part by the National Science Council of Tai- wan.

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