Optimal three-ball inequalities and quantitative uniqueness for the Stokes system

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

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 · ∇u_n).

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=1∂jFij, i = 1, 2, 3, where F_ij = P3

k,`=1A˜<sub>ijk`</sub>(x)∂_ku_` +P3

k=1B˜_ijk(x)u_k with appropriate ˜A_ijk`(x) and B˜ijk(x) satisfying

| ˜A_ijk`(x)| ≤ C₀| log |x||⁻³|x|⁻¹, | ˜B_ijk(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

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)

for (u, p) ∈ (H¹(B_R0))ⁿ⁺¹ satisfying (1.1) in B_R0, 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(B_R0))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 ≥ KR^m (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) ∈ (H_loc¹ (Ω))ⁿ⁺¹ satisfies (1.1) and for any N ∈ N, there exists CN > 0 such that

Z

|x|<r

|u|²dx ≤ C_Nr^N,

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 kuk_L^∞_(Rⁿ₎+ kAk_L^∞_(Rⁿ₎+ kBk_L^∞_(Rⁿ₎ ≤ λ₂. (1.8) Denote

M_r(t) = inf

|x|=t

Z

|y−x|<r

|u(y)|²dy.

Then we can prove that

Theorem 1.5 Let (u, p) ∈ (H_loc¹ (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 M_r(t) ≥ r^{cζ(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) ∈ (H_loc¹ (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

kuk_L^∞_(Rⁿ₎+ kρk_L^∞_(Rⁿ₎ ≤ λ₃. Then for any r < 1, there exists ˜c > 0 such that

M_r(t) ≥ r^{˜cζ(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.

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, · · · , ω_n) ∈ Sⁿ⁻¹. Further- more, using new coordinate t = log r, we can see that

∂

∂x_j = e^−t(ω_j∂_t+ Ω_j), 1 ≤ j ≤ n,

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

n

X

j=1

ω_jΩ_j = 0 and

n

X

j=1

Ω_jω_j = n − 1.

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

It is easy to see that

∂²

∂x_j∂x_{`</sub> = e<sup>−2t</sup>(ω<sub>j</sub>∂<sub>t</sub>− ω<sub>j</sub>+ Ω<sub>j</sub>)(ω<sub>`}∂_t+ Ω<sub>`</sub>), 1 ≤ j, ` ≤ n.

and, therefore, the Laplacian becomes

e^2t∆ = ∂_t²+ (n − 2)∂_t+ ∆_ω, (2.1) where ∆_ω = Σⁿ_j=1Ω²_j 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

|Ω_jv|²dtdω =X

k≥0

k(k + n − 2) Z Z

|v_k|²dtdω, (2.3) where v_k is the projection of v onto E_k. 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 E_k. Denote

L^±= ∂_t+ n − 2 2 ± Λ.

Then it follows from (2.1) that

e^2t∆ = L⁺L⁻= L⁻L⁺.

Motivated by the ideas in [18], we will derive Carleman-type estimates with weights ϕ_β = ϕ_β(x) = exp(−β ˜ψ(x)), where β > 0 and ˜ψ(x) = log |x| + log((log |x|)²). Note that ϕ_β 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 ∈ U_r0 and β ≥ β₀, we have that

β Z

ϕ²_β(log |x|)⁻²|x|⁻ⁿ(|x|²|∇u|²+ |u|²)dx . Z

ϕ²_β|x|⁻ⁿ|x|⁴|∆u|²dx, (2.4) where U_r0 = {u ∈ C₀^∞(Rⁿ\ {0}) : supp(u) ⊂ B_r0}.

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

ϕ²_β|x|⁴⁻ⁿ|∆u|²dx

= Z Z

e^−2βψ(t)e^4t|∆u|²dtdω

= Z Z

|e^−βψ(t)e^2t∆u|²dtdω. (2.5)

If we set u = e^βψ(t)v and use (2.1), then

e^−βψ(t)e^2t∆u = ∂_t²v + b∂_tv + av + ∆_ωv =: P_βv, (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

β^{3−2(j+|α|)} Z Z

|t|⁻²|∂_t^jΩ^αv|²dtdω ≤ ˜C₁ Z Z

|P_βv|²dtdω, (2.7)

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

|P_βv|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

|∂_tΩ_jv|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}|Ω_jv|²dtdω + Z Z

{a²+ ∂_t²a − ∂_t(ab)}|v|²dtdω

≥ Z Z

|∆_ωv|²dtdω +X

j

Z Z

{−4t⁻²β − 2a}|Ω_jv|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|Ω_jv|²dtdω + Z Z

a²|v|²dtdω

= X

k≥0

Z Z

[a − k(k + n − 2)]²|v_k|²dtdω. (2.9)

Substituting (2.9) into (2.8) yields Z Z

|P_βv|²dtdω

≥ X

k≥0

Z Z

{11t⁻²β³ − 4t⁻²βk(k + n − 2) + [a − k(k + n − 2)]²}|v_k|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}|v_k|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 ∈ V_t0, β > 1, we have that

X

j+|α|≤1

β^{1−2(j+|α|)} Z Z

t⁻²ϕ²_β|∂_t^jΩ^αu|²dtdω . Z Z

ϕ²_β|L⁻u|²dtdω, (2.13)

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

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

ϕ²_β|L⁻u|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

| − kv_k+ βv_k+ 2βt⁻¹v_k|²dtdω

= X

k≥0

Z Z

(−k + β + 2βt⁻¹)²|v_k|²dtdω,

where, as before, v_k is the projection of v on E_k. 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

ϕ²_β|L⁻u|²dtdω

= Z Z

|∂_tv|²dtdω + Σ_k≥0 Z Z

[(−k + β + 2βt⁻¹)²+ βt⁻²]|v_k|²dtdω

&

Z Z

|∂tv|²dtdω + Σk≥0

Z Z

(β⁻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.

Lemma 2.3 There exists a sufficiently small number t₁ < −2 depending on n such that for all u ∈ V_t1, g = (g₀, g₁, · · · , g_n) ∈ (V_t1)ⁿ⁺¹ and β > 0, we have that

Z Z

ϕ²_β|u|²dtdω . Z Z

ϕ²_β(|L⁺u + ∂_tg₀+

n

X

j=1

Ω_jg_j|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 ∈ U_r1 and f = (f₁, · · · , f_n) ∈ (U_r1)ⁿ, β ≥ β₁, we have that

Z

ϕ²_β(log |x|)²(|x|⁴⁻ⁿ|∇w|²+ |x|²⁻ⁿ|w|²)dx . β

Z

ϕ²_β(log |x|)⁴|x|²⁻ⁿ[(|x|²∆w + |x|divf )²+ kf k²]dx, (2.15) where U_r1 is defined as in Lemma 2.1.

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

ϕ²_β(log |x|)⁻²(|x|²|∇w|²+ |w|²)|x|⁻ⁿdx . β

Z

ϕ²_β[(|x|²∆w + |x|divf )²+ kf k²]|x|⁻ⁿdx. (2.16) Working in polar coordinates and using the relation e^2t∆ = L⁺L⁻, (2.16) is equivalent to

X

j+|α|≤1

Z Z

β^{2−2(j+|α|)}t⁻²ϕ²_β|∂_t^jΩ^αu|²dtdω

. β Z Z

ϕ²_β(|L⁺L⁻w + ∂_t(

n

X

j=1

ω_jf_j) +

n

X

j=1

Ω_jf_j|²+ kf k²)dtdω.(2.17) Applying Lemma 2.3 to u = L⁻w and g = (Pn

j=1ω_jf_j, f₁, · · · , f_n) yields β

Z Z

ϕ²_β|L⁻w|²dtdω . β

Z Z

ϕ²_β(|L⁺L⁻w + ∂_t(

n

X

j=1

ω_jf_j) +

n

X

j=1

Ω_jf_j|²+ kf k²)dtdω.(2.18)

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) ∈ (H_loc¹ (Ω))ⁿ⁺¹ be a solution of (1.1). Then for any 0 < a₃ < a₁ < a₂ < a₄ such that B_a4r ⊂ 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 = B_a4r\ ¯B_a3r and d(x) be the distant from x ∈ X to Rⁿ\X. By the elliptic regularity, we obtain from (1.1) that u ∈ H_loc² (Ω\{0}). It is trivial that

kvk_H¹_(Rⁿ₎ . k∆vkL²(Rⁿ)+ kvk_L²_(Rⁿ₎ (3.2) for all v ∈ H²(Rⁿ). By changing variables x → E⁻¹x in (3.2), we will have

P

|α|≤1E^2−|α|kD^αvk_L²_(Rⁿ₎ . (k∆vkL²(Rⁿ)+ E²kvk_L²_(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 ξ_y(x) = ξ((x−y)/d(y)). For y ∈ X, we apply (3.3) to ξ_y(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)

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)

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

n

X

i=1

Z

|ξ_y(x)∇q_i|²dx =

n

X

i=1

Z

∇q_i· ∇(ξ_y²(x)¯q_i)dx −

n

X

i=1

2 Z

ξ_y∇q_i· ¯q_i∇ξ_ydx

≤ C₃⁰

n

X

i=1

| Z

(divF )_iξ_y²q_idx| +

n

X

i=1

2 Z

|ξ_y∇q_i· ¯q_i∇ξ_y|dx

≤ C₃⁰

n

X

i=1

| Z ⁿ

X

j=1

F_ij · ∂_j(ξ_y²q_i)dx| +1 4

n

X

i=1

Z

|ξ_y∇q_i|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

|ξ_y∇q_i|²dx + C₄⁰ Z

|x−y|≤d(y)/2

d(y)⁻²|q|²dx

+1 4

n

X

i=1

Z

|ξ_y∇q_i|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

|ξ_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)

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 = B_a4r\ ¯B_a3r and note that d(x) ≥ ˜Cr if x ∈ B_a2r\ ¯B_a1r, 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 B_R ⊂ Ω.

The small constant R will be determined later. Since (u, p) ∈ (H¹(B_R0))ⁿ⁺¹, the elliptic regularity theorem implies u ∈ H_loc² (B_R0\ {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 (

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

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

C1β Z

(log |x|)⁻²ϕ²_β|x|⁻ⁿ(|x|²|∇(χu)|²+|χu|²)dx ≤ Z

ϕ²_β|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

ϕ²_β(log |x|)²(|x|⁴⁻ⁿ|∇(χq)|²+ |x|²⁻ⁿ|χq|²)dx

≤ β Z

ϕ²_β(log |x|)⁴|x|²⁻ⁿ[|x|²∆(χq) + |x|div(|x|χF )]²dx +β

Z

ϕ²_β(log |x|)⁴|x|²⁻ⁿk|x|χF k²dx. (4.3)

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

Z

R1/2<|x|<eR2

(log |x|)⁻²ϕ²_β|x|⁻ⁿ(|x|²|∇u|²+ |u|²)dx +

Z

R1/2<|x|<eR2

(log |x|)²ϕ²_β|x|⁻ⁿ(|x|⁴|∇q|²+ |x|²|q|²)dx

≤ M1β Z

ϕ²_β(log |x|)⁻²|x|⁻ⁿ(|x|²∇(χu)|²+ |χu|²)dx +

Z

(log |x|)²ϕ²_β|x|⁻ⁿ(|x|⁴|∇(χq)|²+ |x|²|χq|²)dx

≤ M₁C₃ Z

ϕ²_β|x|⁻ⁿ|x|⁴|∆(χu)|²dx +βC₃

Z

(log |x|)⁴ϕ²_β|x|⁻ⁿ[|x|³∆(χq) + |x|²div(|x|χF )]²dx +βC₃

Z

(log |x|)⁴ϕ²_β|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|)⁻²ϕ²_β|x|⁻ⁿ(|x|²|∇u|²+ |u|²)dx +

Z

R1/2<|x|<eR2

(log |x|)²ϕ²_β|x|⁻ⁿ(|x|⁴|∇q|²+ |x|²|q|²)dx

≤ C₄M₁ Z

R1/2<|x|<eR2

ϕ²_β|x|⁻ⁿ|x|⁴|∇q|²dx +C₄β

Z

R1/2<|x|<eR2

(log |x|)⁻²ϕ²_β|x|⁻ⁿ(|x|²|∇u|²+ |u|²)dx +C₄M₁

Z

{R1/e≤|x|≤R1/2}∪{eR2≤|x|≤3R2}

ϕ²_β|x|⁻ⁿ| ˜U |²dx +C₄β

Z

{R1/e≤|x|≤R1/2}∪{eR2≤|x|≤3R2}

(log |x|)⁴ϕ²_β|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

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|⁻ⁿϕ²_β(|x|) and (log |x|)⁴|x|⁻ⁿϕ²_β(|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)⁻²ϕ²_β(R₂) Z

R1/2<|x|<R2

|u|²dx

≤ Z

R1/2<|x|<eR2

(log |x|)⁻²ϕ²_β|x|⁻ⁿ|u|²dx

≤ C₅β Z

{R1/e≤|x|≤R1/2}∪{eR2≤|x|≤3R2}

(log |x|)⁴ϕ²_β|x|⁻ⁿ| ˜U |²dx

≤ C₅β(log(R₁/e))⁴(R₁/e)⁻ⁿϕ²_β(R₁/e) Z

{R1/e≤|x|≤R1/2}

| ˜U |²dx +C5β(log(eR2))⁴(eR2)⁻ⁿϕ²_β(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

≤ C₆2^2β+n(log(R₁/e))⁴(R₁/e)⁻ⁿϕ²_β(R₁/e) Z

{R1/4≤|x|≤R1}

|u|²dx +C₆2^2β+n(log(eR₂))⁴(eR₂)⁻ⁿϕ²_β(eR₂)

Z

{2R₂≤|x|≤4R₂}

|u|²dx

= C₆2^2β+n(log(R₁/e))^−4β+4(R₁/e)^−2β−n Z

{R1/4≤|x|≤R1}

|u|²dx +C₆2^2β+n(log(eR₂))^−4β+4(eR₂)^−2β−n

Z

{2R₂≤|x|≤4R₂}

|u|²dx. (4.7)

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

Z

R1/2<|x|<R2

|u|²dx

≤ C₇2^β(log(R₁/e))^−2β+2n+4(R₁/e)^−β Z

{R1/4≤|x|≤R1}

|u|²dx +C₇2^β(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)

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+B^B Z

|x|<1

|u|²dx

_E+B^E

. (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,

then we have Z

|x|<R2

|u|²dx

≤

Z

|x|<1

|u|²dx

_E+B^B Z

|x|<1

|u|²dx

_E+B^E

≤ exp (B ˜β)

Z

|x|<R1

|u|²dx

_E+B^B Z

|x|<1

|u|²dx

_E+B^E

. (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+B^B Z

|x|<1

|u|²dx

_E+B^E

. (4.14) It is readily seen that _E+B^B ≈ (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)

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)R^m₁ ≤ 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, · · · , x_N = ˜x such that |x_j+1 − x_j| ≤ r. We shall prove the desired estimate iteratively. To see how the iteration goes, let us assume that R

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

Z

|x−x_l+1|<r

|u|²dx ≥ Z

|x−x_l+1|<R₃

|u|²dx · r^m, (5.1) where

m = ˜C log R

|x−x_l+1|<R3|u|²dx R

|x−x_l+1|<R2|u|²dx

 .