Quantitative uniqueness estimates for the generalized non-stationary Stokes system

Quantitative uniqueness estimates for the generalized non-stationary Stokes system

Ching-Lung Lin

Jenn-Nan Wang

Abstract

We study the local behavior of a solution to a generalized non-stationary Stokes system with singular coefficients in Rⁿ with n ≥ 2. One of our main results is a bound on the vanishing order of a nontrivial solution u satisfying the generalized non-stationary Stokes system, which is a quantitative version of the (strong) unique continuation property for u. Different from the previous known results, our unique continuation result only involves the velocity field u.

Our proof relies on some delicate Carleman-type estimates. We first use these estimates to derive crucial optimal three-cylinder inequalities for u.

Keywords: Stokes system, Quantitative uniqueness estimates, Carleman estimates

1 Introduction

In this work we study the local behavior of the solution to a generalized non-stationary Stokes system. This generalized Stokes system includes the usual Navier-Stokes equa- tions with velocity that can become singular at one point. Let Ω be a connected bounded domain in Rⁿ with n ≥ 2. Without loss of generality, we assume 0 is in the interior of Ω and define B_r(x) = {y : |y − x| < r}. We consider the following time-dependent Stokes systems

( ∂_tu − ∆u + A(t, x) · ∇u + B(t, x)u + ∇p = 0 in (−1, 1) × Ω,

∇ · u = 0 in (−1, 1) × Ω. (1.1)

We assume that coefficients A(t, x) and B(t, x) satisfy ( |A(t, x)| ≤ λ|x|^−1+, t ∈ (−1, 1),

|B(t, x)| ≤ λ|x|^−2+, t ∈ (−1, 1) (1.2)

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

Email:cllin2@mail.ncku.edu.tw

†Institute of Applied Mathematical Sciences, NCTS, National Taiwan University, Taipei 106, Taiwan. Email: jnwang@math.ntu.edu.tw

for some λ > 0 and 0 <  < 1. Our main concern is the local vanishing behavior of u(t, x) in (−1, 1) × Ω.

We now describe our main results. For s ∈ (−1, 1), let Q^s,τ_x,R = {(t, y) : y ∈ B_R(x) ⊂ Rⁿ, t ∈ (−τ + s, τ + s)} Let u ∈ H¹((−1, 1); H_loc² (Ω)) be a nontrivial solution of (1.1) with an appropriate pressure function p ∈ L²((−1, 1); H_loc¹ (Ω)). For the local case, we derive the following optimal three cylinder inequality and doubling cylinder inequality.

Theorem 1.1 Given T and t₀ such that 0 < t₀ < T ≤ 1. For any ˆR < 1 such that if 0 < R₁ < R₂ < R₃/3 < ˆR then

Z Z

Q^{0,T −t0}_0,R2

|u|²dxdt ≤ ˜C Z Z

Q^0,T_0,R1

|u|²dxdt

!κ

Z Z

Q^0,T_0,R3

|u|²dxdt

!1−κ

, (1.3)

where ˜C depends on , λ, n, T, t₀, R₂/R₃ and κ = log(^2R_3R³

2) 4 log(^4R_R²

1 ) + log(^2R_3R³

2).

We remark that the optimality of (1.3) is due to the fact that κ ∼ −1/ log R₁. From Theorem 1.1, we can get the following double cylinder inequality.

Theorem 1.2 Let , n, λ, R₂, T, t₀ and R₃ be as in Theorem 1.2. Then we have Z Z

Q^{0,T −t0}_0,R2

|u|²dxdt ≤ Ce^{4(log4R2R1)m1} Z Z

Q^0,T_0,R1

|u|²dxdt, (1.4)

where

m₁ := 64 +1 2 + 64





log 2 RR

Q^0,T

0,R3

|u|²dxdt RR

Q^{0,T −t0}

0,R3 2

|u|²dxdt







 and [x] is the largest integer small than x.

We also obtain a vanishing order of the velocity over one cylinder.

Corollary 1.3 Let , n, λ, R₂, T, t₀ and R₃ in Theorem 1.2 be fixed. Then we have

Z Z

Q^0,T_0,R3

|u|²dxdt

 R^m₁ ≤

Z Z

Q^0,T_0,R1

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

m = C₁+ C₁log RR

Q^0,T_0,R3|u|²dxdt RR

Q^{0,T −t0}_0,R2 |u|²dxdt



and C₁ is a positive constant depending on , n, λ, T, t₀ and R₂/R₃.

Corollary 1.3 and Theorem 1.1 immediately imply that if (u, p) ∈ H¹((−1, 1); H_loc² (Ω))×

L²((−1, 1); H_loc¹ (Ω)) satisfies (1.1) and Z Z

Q^0,T_0,r

|u(t, x)|²dxdt = O(r^N) ∀ N ∈ N, (1.5)

then u is zero and p is a constant in (−T, T ) × Ω. It is clear that our theorems imply the following unique continuation property of the Navier-Stokes equation. Assume that (u, p) ∈ H¹((−1, 1); H_loc² (Ω)) × L²((−1, 1); H_loc¹ (Ω)) satisfies

( ∂_tu − ∆u + u · ∇u + ∇p = 0 in (−1, 1) × Ω,

∇ · u = 0 in (−1, 1) × Ω.

If the velocity field u satisfies the first condition of (1.2) and (1.5), then u is trivial and p is a constant.

The main tool in the derivation of (1.3) and (1.4) is Carleman type estimates. We derive such estimates with a weight function depending only on the spatial variable x. To take care of the time variable t, we introduce a cut-off function χ(t) into the Carleman estimates. The main difficulty comes from the integral over the set where 0 < χ(t) < 1. Fortunately, we can overcome the difficulty by carefully choosing the cut-function χ(t) (see the definition of χ in (4.1)).

For parabolic equations with time-independent or time-dependent coefficients, optimal three cylinder inequalities similar to (1.3) are derived in Vessella [13] and Escauriaza-Vessella [2]. These quantitative estimates are useful in the study of the stability for various types of inverse parabolic problems with unknown boundaries (see Vessella’s review article [14]). Our approach share the same spirit as in Ves- sella’s works. We refer readers to Yamamoto’s review article [15] for other types of Carleman estimates for the parabolic equation and their applications to inverse problems. In the consideration of the inverse source problem for the Navier-Stokes equations [1], a Carleman estimate for the linearized Navier-Stokes equations is de- rived. We would like to point out that the weight function in our Carleman estimate is time-independent and singular at the origin, while the weight function used in the Carleman estimate of [1] is regular in the spatial variable x while blows up at end points of time. Other Carleman estimates for the Navier-Stokes equations can also be found in the references in [15]. For the stationary Stokes or Navier-Stokes equations, unique continuation and quantitative estimates (local or at infinity) are proved in [3, 7, 8, 12].

The singular behavior (1.2) of A(t, x) and B(t, x) is motivated by the study of the strong unique continuation property for the elliptic equations [4]. For the ellip- tic equation, the critical case (when  = 0) are considered in [9, 11]. This paper is organized as follows. In Section 2, we derive suitable Carleman estimates. A tech- nical interior estimate is proved in Section 3. Section 4 is devoted to the proofs of Theorem 1.1, Corollary 1.3, and Theorem 1.2.

2 Reduced system and Carleman estimates

To study the Stokes type equation, it is useful to consider the vorticity equation. Let us now define the vorticity q of the velocity u by

q = curl u := 1

√2(∂iuj− ∂_jui)1≤i,j≤n.

Note that here q is a matrix-valued function. The formal transpose of curl is given by

(curl^>v)_1≤i≤n := 1

√2 X

1≤j≤n

∂_j(v_ij − v_ji), where v = (v_ij)_1≤i,j≤n. It is easy to see that

∆u = ∇(∇ · u) − curl^>curlu (see, for example, [10] for a proof), which implies

∆u + curl^>q = 0. (2.1)

Since there is no equation for p in the Stokes system (1.1), we apply the curl operator

∇× on the first equation and obtain

∂_tq − ∆q + ∇ · F = 0, (2.2)

where q = ∇ × u is the vorticity and ∇ · F is a vector function defined by (∇ · F )_i = Pn

j=1∂_jF_ij, i = 1, 2, · · · , n, where F_ij = Pn

k,`=1A˜<sub>ijk`</sub>(x, t)∂_ku_` +Pn

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

| ˜A_ijk`(x, t)| ≤ λ|x|^−1+, | ˜B_ijk(x, t)| ≤ λ|x|^−2+, ∀ (x, t) ∈ Ω × (−1, 1). (2.3) In summary, to study our problem it suffices to consider equations (2.1), (2.2), which is a system of elliptic and parabolic equations.

The proof of main theorem relies heavily on suitable Carleman estimates. In the rest of the section, we will derive two needed Carleman estimates. We begin with the delicate construction of the weight function ϕ = ϕ(x) = exp(ψ(y)), where β > 0, y = − log |x| and

ψ(y) = βy + 1

16y tan⁻¹y − 1

32ln(1 + y²) + ψ0(y).

To define ψ₀(y), we let µ(y) ∈ C^∞(R) satisfy 0 ≤ µ(y) ≤ 1 and

µ(y) =

(0, y ≤ 0, 1, y ≥ 1.

We now set

ψ⁰⁰₀(y) = β

128σµ(y)e^−y/2 and

ψ⁰₀(y) = Z y

0

ψ₀⁰⁰(s)ds − Z 1

0

ψ₀⁰⁰(s)ds − β

64σe^−/2, ψ₀(y) =

Z y 0

ψ₀⁰(s)ds − Z 1

0

ψ₀⁰(s)ds + β

32σe^−/2 with

σ = Z 1

0



2µ(y)e^−y/2+ e^−/2> 1 2. It should be noted that

ψ₀⁰(y) =







− β

64, y ≤ 0,

− β

64σe^−y/2, y ≥ 1 and

ψ₀(y) =











− β 64y −

Z 1 0

ψ⁰₀(s)ds + β

32σe^−/2, y ≤ 0, β

32σe^−y/2, y ≥ 1.

Moreover, the function ψ(y) for y > 0 is a convex function satisfying for ₆₄^β ∈ N +₂₅₆¹

that 



 1

2β ≤ ψ⁰ ≤ 2β, dist(2ψ⁰, Z) + ψ⁰⁰& 1

(2.4) and for any C > 0 there exists R_ > 0 such that

C|x|^β ≤ (1 + ψ⁰⁰(− log |x|)) (2.5) for all β and |x| ≤ R_. From now on, the notation X . Y or X & Y means that X ≤ CY or X ≥ CY with some constant C depending only on a priori given constants. To check the second inequality of (2.4), we note that







ψ⁰⁰ ≥ 

128, y ≤ 2

 logβ σ, dist(2ψ⁰, Z) ≥ 1

4, y ≥ 2

log β σ.

We now introduce polar coordinates in Rⁿ\{0} by setting x = rω, with r = |x|, ω = (ω₁, · · · , ω_n) ∈ Sⁿ⁻¹. Using new coordinate y = − log r, we obtain that

∂

∂x_j = e^y(−ω_j∂_y+ 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

ωjΩj = 0 and

n

X

j=1

Ωjωj = n − 1.

It is easy to see that

∂²

∂x_j∂x<sub>`</sub> = e<sup>2y</sup>(−ωj∂y− ωj+ Ωj)(−ω`∂y+ Ω`), 1 ≤ j, ` ≤ n.

and, therefore, the Laplacian becomes

e^−2y∆ = ∂_y² − (n − 2)∂_y+ ∆_ω,

where ∆_ω = Σⁿ_j=1Ω²_j denotes the Laplace-Beltrami operator on Sⁿ⁻¹. We recall that the eigenvalues of −∆_ωare k(k+n−2), k ∈ N∪{0}, and the corresponding eigenspaces are E_k, where E_k is the space of spherical harmonics of degree k. We remark that

X

j

Z Z

|Ω_jv|²dydω =X

k≥0

k(k + n − 2) Z

|v_k|²dy, (2.6)

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 E_kwhich represents the space of spherical harmonics of degree k. Hence

Λ = Σ_k≥0(k + n − 2

2 )π_k, (2.7)

where π_k is the orthogonal projector on E_k. Let L^± = ∂y− n − 2

2 ± Λ, then it follows that

e^−2y∆ = L⁺L⁻= L⁻L⁺.

Denote L^±_ψ = ∂_y −ⁿ⁻²₂ ± Λ − ψ⁰(y). Then we have that L^±_ψv = e^ψ(y)L^±(e^−ψ(y)v) and e^−2ye^ψ(y)∆(e^−ψ(y)v) = L⁺_ψL⁻_ψv = L⁻_ψL⁺_ψv.

Lemma 2.1 Let χ(t) ∈ C₀²(R). There exists a sufficiently large constant β₁, de- pending on n, such that for all v(t, y, ω) ∈ C¹(R; C₀^∞(R × Sⁿ⁻¹)) and β ≥ β₁ with

β

64 ∈ N + ₂₅₆¹ , we have that Z

|χ(L⁺_ψL⁻_ψv − 2L⁺_ψv − e^−2y∂_tv)|²+ Z

|χ⁰e^−2yv|²

& X

j+|α|≤1

β^{2−2(j+|α|)} Z

(1 + ψ⁰⁰)|∂_y^jΩ^α(χv)|²,

(2.8)

where we denote R f = R f dtdydω for any function f (t, y, ω).

Proof. From v = Σkvk, we can compute χ(L⁺_ψL⁻_ψv − 2L⁺_ψv)

=χ[∂_y²v − 2ψ⁰∂_yv − n∂_yv + (n − 2 + (ψ⁰)²+ nψ⁰ − ψ⁰⁰)v − 2Λv + ∆_ωv]

=Σ_k≥0χ(∂_y²v_k− ˜b∂_yv_k+ ˜av_k), where

(˜a = (ψ⁰− k)(ψ⁰+ k + n) − ψ⁰⁰

˜b = 2ψ⁰ + n.

By abusing the notation v = v_k, it is enough to prove that X

j≤1

Z

(1 + ψ⁰⁰)|(β^2−2j+ k^2−2j)∂_y^j(χv)|²

. Z

|χ(∂_y²v − ˜b∂_yv + ˜av − e^−2t∂_tv)|²+ Z

|χ⁰e^−2yv|².

(2.9)

It is helpful to note that ψ⁰ = β + 1

16tan⁻¹y + ψ₀⁰(y) ≥ 31β 32 , 0 < ψ⁰⁰= 1

16(1 + y²)+ β

128σµ(y)e^−y/2 ≤ 1 16+ β

64 ≤ ψ⁰

32, β ≥ 4,

|ψ⁰⁰⁰| = | −y

8(1 + y²)² − β

256σ²µ(y)e^−y/2+ β

128σµ⁰(y)e^−y/2| ≤ 2ψ⁰⁰, where we choose |µ⁰| ≤ 4 in the last inequality. Observe that

2|χ(∂_y²v − ˜b∂_yv + ˜av − e^−2y∂_tv)|²+ 4|χ⁰e^−2yv|²

≥|χ(∂_y²v − ˜b∂_yv + ˜av) − e^−2y∂_t(χv)|²

=|H(v)|²− 2˜b∂_y(χv)H(v) − 2e^−2y∂_t(χv)H(v) + |˜b∂_y(χv) + e^−2y∂_t(χv)|²,

(2.10)

where H(v) := χ(∂_y²v + ˜av). Now we write











−2 Z

˜b∂_y(χv)H(v) = −2 Z

˜b∂_y(χv)∂_y²(χv) − 2 Z

˜

a˜b(χv)∂_y(χv)

−2 Z

e^−2y∂_t(χv)H(v) = −2 Z

e^−2y∂_t(χv)∂_y²(χv) − 2 Z

˜

a(χv)e^−2y∂_t(χv).

(2.11)

Straightforward computations imply that











− 2 Z

˜b∂_y(χv)∂²_y(χv) = 2 Z

ψ⁰⁰|∂_y(χv)|²,

− 2 Z

˜

a˜b(χv)∂y(χv) = Z

∂y(˜a˜b)|χv|²,

(2.12)

− 2 Z

e^−2y∂t(χv)∂_y²(χv) = −4 Z

e^−2y∂t(χv)∂y(χv), (2.13)

−2 Z

˜

a(χv)e^−2y∂_t(χv) = 0, (2.14) Note that here ˜a is independent of t. Combining (2.10) to (2.14) yields

2 Z

|χ(∂_y²v − ˜b∂_yv + ˜av − e^−2t∂_tv)|²+ 4 Z

|χ⁰e^−2yv|²

≥ Z 

|H(v)|²+ |˜b∂y(χv) + e^−2y∂t(χv)|²

 + 2

Z

ψ⁰⁰|∂y(χv)|²

− 4 Z

e^−2y∂_t(χv)∂_y(χv) + 17 3

Z

(ψ⁰)²ψ⁰⁰|χv|²− 5 2

Z

k²ψ⁰⁰|χv|²

(2.15)

for β ≥ β₀. In deriving (2.15), we have used estimates (ψ⁰⁰)² < ₃₂¹ψ⁰ψ⁰⁰ < _β¹(ψ⁰)²ψ⁰⁰ and |ψ⁰ψ⁰⁰⁰| < 2ψ⁰ψ⁰⁰ < _β¹(ψ⁰)²ψ⁰⁰.

Likewise, we write















|˜b∂_y(χv) + e^−2y∂_t(χv)|²

= |(2ψ⁰+ n − 2)∂_y(χv) + e^−2y∂_t(χv)|²+ 4(2ψ⁰+ n − 1)|∂_y(χv)|² + 4e^−2y∂_t(χv)∂_y(χv).

1

2|H(v)|² = 1

2|H(v) + 3ψ⁰⁰χv|²− 3ψ⁰⁰χvH(v) −9

2(ψ⁰⁰)²|χv|².

(2.16)

It is easy to check that

− 3 Z

ψ⁰⁰χvH(v)

= −3 Z

ψ⁰⁰χ²v(∂_y²v + ˜av)

≥ 3 Z

ψ⁰⁰|∂_y(χv)|²− 10 3

Z

(ψ⁰)²ψ⁰⁰|χv|²+ 3 Z

k²ψ⁰⁰|(χv)|² (2.17)

for all β ≥ β₀. From (2.15)-(2.17), we have that for β ≥ β₀ 2

Z

|χ(∂_y²v − ˜b∂yv + ˜av − e^−2t∂tv)|²+ 4 Z

|χ⁰e^−2yv|²

≥ 8 Z

ψ⁰|∂_y(χv)|²+ 2 Z

(ψ⁰)²ψ⁰⁰|χv|²+1 2

Z

k²ψ⁰⁰|χv|² + 1

2 Z

|H(v)|². (2.18)

Our next strategy is to divide y into three region. Before doing so, we need to further simplify (2.18). Note that

1 4

Z

|H(v)|² =1 4

Z

|H(v) −β(ψ⁰− k)χv 100|β − k| |²+

Z β(ψ⁰− k)

200|β − k|χvH(v)

−

Z β²(ψ⁰ − k)² 40000|β − k|²|χv|²

≥

Z β(ψ⁰− k)

200|β − k|χvH(v) −

Z β²(ψ⁰− k)² 40000|β − k|²|χv|² and note

Z β(ψ⁰− k)

200|β − k|χ²v∂_y²v = −

Z β(ψ⁰− k)

200|β − k||∂_y(χv)|²+

Z βψ⁰⁰⁰

400|β − k||χv|²

= −

Z β(β − k)

200|β − k||∂_y(χv)|²+

Z β(−ψ₀⁰(y) − ₁₆¹ tan⁻¹y)

200|β − k| |∂_y(χv)|² +

Z βψ⁰⁰⁰

400|β − k||χv|² with

Z β(ψ⁰ − k)

200|β − k|χ²v˜av =

Z β(ψ⁰− k)²(ψ⁰+ k + n)

200|β − k| |χv|²−

Z β(ψ⁰ − k)ψ⁰⁰ 200|β − k| |χv|². Combining (2.18), we have that

2 Z

|χ(∂_y²v − ˜b∂_yv + ˜av − e^−2t∂_tv)|²+ 4 Z

|χ⁰e^−2yv|²

≥ 7 Z

ψ⁰|∂_y(χv)|²+ 3 2

Z

(ψ⁰)²ψ⁰⁰|χv|²+1 2

Z

k²ψ⁰⁰|χv|² +

Z β(ψ⁰− k)²(ψ⁰ + k + n)

400|β − k| |χv|²+1 4

Z

|H(v)|². (2.19) We need further simplification. Let ˜µ(y) = µ(y + 1), then

1 − ˜µ(y) =

( 1, y ≤ −1 0, y ≥ 0.

For a small positive constant ρ < ₁₀¹ which will be determined later, we can estimate 1

4 Z

|H(v)|² =1 4

Z

|H(v) −ρβ(1 − ˜µ)(ψ⁰− k)χv

|^63β₆₄ − k| |²+

Z ρβ(1 − ˜µ)(ψ⁰− k)

2|^63β₆₄ − k| χvH(v)

−

Z ρ²β²(1 − ˜µ)²(ψ⁰ − k)² 4|^63β₆₄ − k|² |χv|²

≥

Z ρβ(1 − ˜µ)(ψ⁰− k)

2|^63β₆₄ − k| χvH(v) −

Z ρ²β²(1 − ˜µ)²(ψ⁰− k)² 4|^63β₆₄ − k|² |χv|². Observe that

Z ρβ(1 − ˜µ)(ψ⁰− k)

2|^63β₆₄ − k| χ²v∂_y²v

= −

Z ρβ(1 − ˜µ)(ψ⁰− k)

2|^63β₆₄ − k| |∂_y(χv)|²+

Z ∂_y²[ρβ(1 − ˜µ)(ψ⁰ − k)]

4|^63β₆₄ − k| |χv|²

= −

Z ρβ(1 − ˜µ)(^63β₆₄ − k)

2|^63β₆₄ − k| |∂_y(χv)|²−

Z ρβ(1 − ˜µ) tan⁻¹y

32|^63β₆₄ − k| |∂_y(χv)|² +

Z ∂_y²[ρβ(1 − ˜µ)(ψ⁰− k)]

4|^63β₆₄ − k| |χv|²

≥ −

Z ρβ(1 − ˜µ)(^63β₆₄ − k)

2|^63β₆₄ − k| |∂_y(χv)|²+

Z ∂_y²[ρβ(1 − ˜µ)(ψ⁰− k)]

4|^63β₆₄ − k| |χv|² and

Z ρβ(1 − ˜µ)(ψ⁰ − k) 2|^63β₆₄ − k| χ²v˜av

=

Z ρβ(1 − ˜µ)(ψ⁰ − k)²(ψ⁰+ k + n)

2|^63β₆₄ − k| |χv|²−

Z ρβ(1 − ˜µ)(ψ⁰− k)ψ⁰⁰ 2|^63β₆₄ − k| |χv|². To estimate the termR ∂²_y[ρβ(1−˜µ)(ψ⁰−k)]

4|^63β₆₄−k| |χv|², we compute that

|∂_y²[(1 − ˜µ)(ψ⁰− k)]| = | − ˜µ⁰(ψ⁰− k) − 2˜µ⁰ψ⁰⁰+ (1 − ˜µ)ψ⁰⁰⁰|

≤ |˜µ⁰⁰||ψ⁰− k| + 2|˜µ⁰|ψ⁰⁰+ (1 − ˜µ)|ψ⁰⁰⁰|

It should be noted that the supports of ˜µ⁰ and ˜µ⁰⁰ are [−1, 0]. We assume ||˜µ⁰||∞+

||˜µ⁰⁰||∞≤ C2, then we can choose ρ < _20000C¹

2. Recall that for y ≤ 0 ψ⁰ = 63

64β + 1

16tan⁻¹y, ψ⁰⁰= 1

16(1 + y²), ψ⁰⁰⁰ = − y 8(1 + y²)².

Therefore, combining (2.19), we have that 2

Z

|χ(∂_y²v − ˜b∂_yv + ˜av − e^−2t∂_tv)|²+ 4 Z

|χ⁰e^−2yv|²

≥ 6 Z

ψ⁰|∂_y(χv)|²+ Z

(ψ⁰)²ψ⁰⁰|χv|²+1 4

Z

k²ψ⁰⁰|χv|² +

Z β(ψ⁰− k)²(ψ⁰+ k + n)

400|β − k| |χv|²+

Z ρβ(1 − ˜µ)(ψ⁰ − k)²(ψ⁰+ k + n) 4|^63β₆₄ − k| |χv|².

(2.20) We now use (2.20) to deduce (2.9). The trick is to divide the domain of integration into three regions.

(i) For −1 ≤ y ≤ ²_log ^β_σ, we have ψ⁰⁰ ≥ ₂₅₆^ (1 + ψ⁰⁰) and (2.9) follows immediately.

Recall that ψ⁰⁰≤ ^ψ₃₂⁰ for all y ∈ R provided β ≥ 4.

(ii) For y ≥ ²_log ^β_σ and ₆₄^β ∈ N + ₂₅₆¹ , we can see that

|ψ⁰− k| ≥ |β − k| − 1 16, which implies

|ψ⁰− k|² ≥ |β − k|²− 1

8|β − k| ≥ 1

2|β − k|². The fourth term on the right hand side of (2.20) now satisfies Z β(ψ⁰− k)²(ψ⁰+ k + n)

400|β − k| |χv|² ≥

Z β|β − k|(ψ⁰ + k + n)

800 |χv|² &

Z

(β²+k²)|χv|². Consequently, (2.9) holds over this domain of integration.

(iii) For y ≤ −1, we obtain from the last term on the right hand side of (2.20) Z ρβ(1 − ˜µ)(ψ⁰− k)²(ψ⁰+ k + n)

4|^63β₆₄ − k| |χv|² ≥ Z

{y≤−1}

ρβ(ψ⁰− k)²(ψ⁰+ k + n) 4|^63β₆₄ − k| |χv|². That the estimate (2.9) is satisfied over y ≤ −1 follows from the same arguments as in (ii).

2

We need another Carleman estimate to handle the divergence terms in the reduced system.

Lemma 2.2 Let χ(t) ∈ C₀²(R). There exists a sufficiently large number β₁⁰ depending on n such that for all u(t, y, ω) ∈ C¹(R; C₀^∞(R × Sⁿ⁻¹)), g = (g₀, g₁, · · · , g_n) ∈ (C₀(R; C₀^∞(R × Sⁿ⁻¹)))ⁿ⁺¹ and β > β₁⁰ with ₆₄^β ∈ N + ₂₅₆¹ , we have that

β² Z

(1 + ψ⁰⁰)|χu|² . Z

|χ(L⁻_ψL⁺_ψu − e^−2y∂_tu) + ∂_yg₀+

n

X

j=1

Ω_jg_j|²

+ β²kgk²+ Z

|χ⁰e^−2yu|².

(2.21)

Proof. For fixed t ∈ R, we consider v(t, y, ω) ∈ C₀^∞(R × Sⁿ⁻¹). We denote ˆv(t, η, ω) its Fourier transformation with respect to y, and define

T_ψv(t, y, ω) = (2π)^−1/2X

k≥0

Z ∞

−∞

e^iηy(iη − ψ⁰− k − (n − 2))π_kv(t, η, ω)dη.ˆ

(2.22) Similarly, we define

T˜_ψv(t, y, ω) = (2π)^−1/2X

k≥0

Z ∞

−∞

e^iηy(iη − β − k − (n − 2))π_kv(t, η, ω)dη.ˆ

(2.23) Note that we have T_ψv = L⁻_ψv = ∂_yv − ψ⁰v −⁽ⁿ⁻²⁾₂ v − Λv = ˜T_ψv + (β − ψ⁰)v. It is clear that ˜T_ψ is invertible whose inverse is given by

T˜_ψ⁻¹v(t, y, ω) = (2π)^−1/2X

k≥0

Z ∞

−∞

e^iηy(iη − β − k − (n − 2))⁻¹π_kv(t, η, ω)dη. (2.24)ˆ From (2.23), (2.24), Plancherel’s theorem, and integrating in t, we have for v ∈ L²(R; C₀^∞(R × Sⁿ⁻¹)) that











k ˜T_ψvk_L²_(R²×Sⁿ⁻¹) ≥ βkvk_L²_(R²×Sⁿ⁻¹), βk ˜T_ψ⁻¹vk_L²_(R²×Sⁿ⁻¹)≤ kvk_L²_(R²×Sⁿ⁻¹), k ˜T_ψ⁻¹( X

j+|α|≤1

∂_y^jΩ^αv)k_L²_(R²×Sⁿ⁻¹) . kvkL²(R²×Sⁿ⁻¹)

(2.25)

(see [12, p.1896] for the proofs). From (2.22), (2.23) and (2.24), we get that T_ψ⁻¹ = (I + ˜T_ψ⁻¹(β − ψ⁰)I)⁻¹T˜_ψ⁻¹.

Consequently, it follows from (2.25) that for v ∈ L²(R; C₀^∞(R × Sⁿ⁻¹)) we have















kT_ψvk_L²_(R²_×Sⁿ⁻¹₎ ≥ β

2kvk_L²_(R²_×Sⁿ⁻¹₎, βkT_ψ⁻¹vk_L²_(R²×Sⁿ⁻¹) . kvkL²(R²×Sⁿ⁻¹), kT_ψ⁻¹( X

j+|α|≤1

∂_y^jΩ^αv)k_L²_(R²×Sⁿ⁻¹). kvkL²(R²×Sⁿ⁻¹).

(2.26)

Now for u(t, y, ω) ∈ C₀¹(R; C₀^∞((y₀, y₁)×Sⁿ⁻¹)), we define v(t, y, ω) = (T_ψ⁻¹u)(t, y, ω), i.e., u(t, y, ω) = T_ψv(t, y, ω). Let u_k(t, y, ω) = π_ku(t, y, ω) and v_k(t, y, ω) = π_kv(t, y, ω).

From the definition of T_β, we have that X

k≥0

u_k = u = T_ψv

= (2π)^−1/2X

k≥0

Z ∞

−∞

e^iηy(iη − ψ⁰− k − (n − 2))π_kv(t, η, ω)dηˆ

= X

k≥0

(∂_yv_k− (ψ⁰+ k + n − 2)v_k). (2.27)

We would like to find v_k satisfying lim|y|→∞v_k(t, y, ω) = 0. Thus, from (2.27), we have that

v_k(t, y, ω) =







0, y ≥ y₁,

− Z y1

y

e−ψ(ξ)−kξ−(n−2)ξ

u_k(t, ξ, ω)dξ · eψ(y)+ky+(n−2)y, y < y₁. (2.28) Note that for y < y₀, v_k of (2.28) is equivalent to

vk(t, y, ω) =







0, y ≥ y₁,

− eψ(y)+ky+(n−2)y

Z y1

y0

e−ψ(ξ)−kξ−(n−2)ξ

u_k(t, ξ, ω)dξ, y < y₀. We remark that although estimate (2.8) is proved for v ∈ C¹(R; C₀^∞(R × Sⁿ⁻¹)), it remains valid for e^2yv(t, y, ω) =P

k≥0e^2yv_k(t, y, ω) with v_k given in (2.28).

Using T_ψ = L⁻_ψ and estimates in (2.26), we obtain that

kχL⁻_ψL⁺_ψu − χe^−2y∂_tu + ∂_yg₀+

n

X

j=1

Ω_jg_jk_L²_(R²_×Sⁿ⁻¹₎

= kTψ L⁺_ψ(χu) + T_ψ⁻¹ −χe^−2y∂tu + ∂yg0+

n

X

j=1

Ωjgj

k_L²_(R²×Sⁿ⁻¹)

≥ βkL⁺_ψ(χu) + T_ψ⁻¹ −χe^−2y∂_tu + ∂_yg₀ +

n

X

j=1

Ω_jg_j
k_L²_(R²×Sⁿ⁻¹)

≥ βkL⁺_ψ(χu) − ∂_t(T_ψ⁻¹(χe^−2yu)) + T_ψ⁻¹(χ⁰e^−2yu)k_L²_(R²×Sⁿ⁻¹)

−βkT_ψ⁻¹ ∂_yg₀+

n

X

j=1

Ω_jg_j
k_L²_(R²×Sⁿ⁻¹)

& βkL⁺_ψ(χu) − ∂_t(T_ψ⁻¹(χe^−2yu)) + T_ψ⁻¹(χ⁰e^−2yu)k_L²_(R²_×Sⁿ⁻¹₎

−βkgk_L²_(R²×Sⁿ⁻¹). (2.29)

Now, we let v = T_ψ⁻¹(e^−2yu) and hence T_ψv = e^−2yu and χv = T_ψ⁻¹(e^−2yχu). We immediately obtain from (2.8) that

kL⁺_ψ(χu) − ∂t(T_ψ⁻¹(χe^−2yu)) + T_ψ⁻¹(χ⁰e^−2yu)k_L²_(R²×Sⁿ⁻¹)

=kχ L⁺_ψL⁻_ψ(e^2yv) − 2L⁺_ψ(e^2yv) − e^−2y∂_t(e^2yv)
k_L²_(R²_×Sⁿ⁻¹₎

& X

j+|α|≤1

β^1−(j+|α|)k(1 + ψ⁰⁰)^1/2∂_y^jΩ^α(χe^2yv)k_L²_(R²_×Sⁿ⁻¹₎− (χ⁰v)

L²(R²×Sⁿ⁻¹). (2.30) From the substitution u = e^2yT_βv, the definitions of T_β, Λ, and the first inequality of (2.26), it is not difficult to show that















k(1 + ψ⁰⁰)^1/2χuk_L²_(R²_×Sⁿ⁻¹₎

. X

j+|α|≤1

β^1−(j+|α|)k(1 + ψ⁰⁰)^1/2∂_y^jΩ^α(χe^2yv)k_L²_(R²×Sⁿ⁻¹), β

χ⁰v

L²(R²×Sⁿ⁻¹) .

χ⁰e^−2yu

L²(R²×Sⁿ⁻¹).

(2.31)

By (2.30) and (2.31), we obtain that

βkL⁺_ψ(χu) − ∂t(T_ψ⁻¹(χe^−2yu)) + T_ψ⁻¹(χ⁰e^−2yu)k_L²_(R²×Sⁿ⁻¹)

& βk(1 + ψ⁰⁰)^1/2χuk_L²_(R²_×Sⁿ⁻¹₎−

χ⁰e^−2yu

L²(R²×Sⁿ⁻¹). (2.32) Finally, combining (2.29) and (2.32) yields (2.21).

2

Now we are ready to prove our main Carleman estimate.

Lemma 2.3 Let χ(t) ∈ C₀²(R). There exists a sufficiently large number β₂ de- pending on n such that for all w(t, x) ∈ C¹(R; C₀^∞(Rⁿ\ {0})), f = (f₁, · · · , f_n) ∈ (C₀(R; C₀^∞(Rⁿ\ {0})))ⁿ and β ≥ β₂ with ₆₄^β ∈ N + ₂₅₆¹ , we have that

Z Z

ϕ²(1 + ψ⁰⁰)(|x|⁴|∇(χw)|²+ β²|x|²|χw|²)dxdt .

Z Z

ϕ²|x|²(χ|x|²∆w − χ|x|²∂_tw + |x|divf )²dxdt +β²

Z Z

ϕ²|x|²kf k²dxdt + Z Z

ϕ²|x|⁶|χ⁰w|²dxdt. (2.33) Proof. The estimate (2.33) remains valid if we replace ψ by ψ + ⁿ₂y − y. We first set w = e^−ψu and g = e^−ψf . Working in polar coordinates and using the relation e^−2y∆ = L⁺L⁻, it suffices to prove that

X

j+|α|≤1

β^{2−2(j+|α|)} Z

(1 + ψ⁰⁰)|∂_y^jΩ^α(χu)|²

. Z

|χL⁻_ψL⁺_ψu − e^−2y∂t(χu) + ∂yg0+

n

X

j=1

Ωjgj|²

+ β²kgk²_L2(R²×Sⁿ⁻¹)+ Z

|e^−2yχ⁰u|²,

(2.34)

where g₀ = hω, e^−ψf i, g_j = e^−ψf_j, j = 1, 2, · · · , n. Denote J (u) = χL⁻_ψL⁺_ψu − e^−2y∂_t(χu) + ∂_yg₀+Pn

j=1Ω_jg_j. We now write

|J(u)|² = |J (u) + (1 + ψ⁰⁰)χu|²− 2(1 + ψ⁰⁰)χuJ (u) − (1 + ψ⁰⁰)²|χu|². (2.35) Denote that

( ˜A = (ψ⁰)²+ (n − 2)ψ⁰− ψ⁰⁰

˜b = −2ψ⁰− (n − 2).

We obtain that for β ≥ β₂ Z

|J(u)|²+ (1 + ψ⁰⁰)²|χu|²

≥ −2 Z

(1 + ψ⁰⁰)χuJ (u)

≥ −2 Z

(1 + ψ⁰⁰)χu(χL⁻_ψL⁺_ψu − e^−2y∂_t(χu) + ∂_yg₀ +

n

X

j=1

Ω_jg_j)

= −2 Z

(1 + ψ⁰⁰)χu ∂_y²(χu) + ˜b∂_y(χu) + ˜A(χu) + ∆_ω(χu)
+ 2

Z

(1 + ψ⁰⁰)χue^−2y∂_t(χu) − 2 Z

(1 + ψ⁰⁰)χu(∂_yg₀ +

n

X

j=1

Ω_jg_j)

≥ Z

(1 + ψ⁰⁰)|∂_y(χu)|²− Cβ² Z

(1 + ψ⁰⁰)|χu|²+ Z

(1 + ψ⁰⁰)

n

X

j=1

|Ω_j(χu)|²

− Cβkgk²_L2(R²×Sⁿ⁻¹),

(2.36)

where C is an absolute constant. By multiplying a large constant K on both side of (2.21), if necessary, and adding this new estimate to (2.36), we obtain the desired

estimate (2.34).

2

We will apply Carleman estimate (2.33) to the parabolic equation (2.2). To treat the elliptic equation (2.1), we use the Carleman estimate below, which can be proved by following the same lines above while ignoring the t derivative. Alternatively, we can simply recall Theorem 3.1 of [6]. By replacing ψ by ψ + y +¹₂log(1 + ψ⁰⁰), which also satisfies (2.4), we can derive the following Carleman estimate.

Lemma 2.4 Let χ(t) ∈ C₀²(R). There exists a sufficiently large number β₃ with

β

64 ∈ N + ₂₅₆¹ depending on n such that for all w(t, x) ∈ C¹(R; C₀^∞(Rⁿ\ {0})) and β ≥ β₂, we have that

Z Z

(1 + ψ⁰⁰)²ϕ²(|x|²|∇(χw)|²+ β²|χw|²)dxdt .

Z Z

(1 + ψ⁰⁰)ϕ²|x|⁴|∆(χw)|²dxdt. (2.37)

3 Interior estimate

Due to the use of cut-off functions, in addition to Carleman estimates, we also need the following interior estimate.

Lemma 3.1 Let X = {(t, x) : t₁ < t < t₂, x ∈ ω ⊂ Ω} and (u, p) be a solution of (1.1). Assume that

diam X < 2

Denote d(t, x) the distance from (t, x) ∈ X to Rⁿ⁺¹\X . Then for any 0 < a₁ < a₂ such that ω = B_a2r\ ¯B_a1r, we have that

Z Z

X

d(t, x)²|∇u|²dxdt + Z Z

X

d(t, x)²|q|²dxdt + Z Z

X

d(t, x)⁴|∇q|²dxdt .

Z Z

X

|u|²dxdt. (3.1)

Proof. The method used here is motivated by the proof of Theorem 17.1.3 in [5].

We apply a suitable cut-off function on u. Take ξ(X) ∈ C₀^∞(Rⁿ⁺¹) satisfying 0 ≤ ξ(X) ≤ 1 and

ξ(X) =

( 1, |X| < 1/4, 0, |X| ≥ 1/2.

Let us denote ξ_Y(Z) = ξ_Y(t, x) = ξ((Z−Y )/d(Y )) for Y ∈ X . Multiplying ξ²_Y(Z)u(Z) on (2.1) yields that

Z Z

|Z−Y |≤d(Y )/4

|∇u|²dZ

≤ δ Z Z

|Z−Y |≤d(Y )/2

d(Y )²|∇q|²dZ + δ Z Z

|Z−Y |≤d(Y )/2

|∇u|²dZ +C₁

δ Z Z

|Z−Y |≤d(Y )/2

d(Y )⁻²|u|²dZ (3.2)

for some absolute constant C₁. Now multiplying d(Y )² on both sides of (3.2), we obtain

Z Z

|Z−Y |≤d(Y )/4

d(Y )²|∇u|²dZ

≤ δ Z Z

|Z−Y |≤d(Y )/2

d(Y )⁴|∇q|²dZ + δ Z Z

|Z−Y |≤d(Y )/2

d(Y )²|∇u|²dZ +C₁

δ Z Z

|Z−Y |≤d(Y )/2

|u|²dZ. (3.3)

Integrating d(Y )⁻ⁿ⁻¹dY over X on both sides of (3.3) and using Fubini’s Theorem, we get that

Z Z

X

Z Z

|Z−Y |≤d(Y )/4

d(Y )¹⁻ⁿ|∇u|²dY dZ

≤ δ Z Z

X

Z Z

|Z−Y |≤d(Y )/2

d(Y )³⁻ⁿ|∇q|²dY dZ +δ

Z Z

X

Z Z

|Z−Y |≤d(Y )/2

d(Y )¹⁻ⁿ|∇u|²dY dZ +C₁

δ Z Z

X

Z Z

|Z−Y |≤d(Y )/2

d(Y )⁻¹⁻ⁿ|u|²dY dZ. (3.4)

Note that |d(Z) − d(Y )| ≤ |Z − Y |. If |Z − Y | ≤ d(Z)/3, then

2d(Z)/3 ≤ d(Y ) ≤ 4d(Z)/3. (3.5)

If |Z − Y | ≤ d(Y )/2, then

d(Z)/2 ≤ d(Y ) ≤ 3d(Z)/2. (3.6)

By (3.5) and (3.6), we have









 Z

|Z−Y |≤d(Y )/4

d(Y )⁻ⁿ⁻¹dY ≥ (3/4)ⁿ⁺¹ Z

|Z−Y |≤d(Z)/6

d(Z)⁻ⁿ⁻¹dy ≥ 8⁻ⁿ⁻¹ Z

|Y |≤1

dY, Z

|Z−Y |≤d(Y )/2

d(Y )⁻ⁿ⁻¹dY ≤ 2ⁿ⁺¹ Z

|Z−Y |≤3d(Z)/4

d(Z)⁻ⁿ⁻¹dy ≤ (3/2)ⁿ⁺¹ Z

|Y |≤1

dY.

(3.7) Combining (3.4)–(3.7) and taking δ small, we obtain

Z Z

X

d(Z)²|∇u|²dZ ≤ C₂δ Z Z

X

d(Z)⁴|∇q|²dZ + C₂ δ

Z Z

X

|u|²dZ, (3.8) where C₂ is another absolute constant.

On the other hand, if we multiply ξ_Y²(Z)q(Z) on both sides of (2.2), we have that Z Z

|Z−Y |≤d(Y )/4

|∇q|²dZ ≤ C₃ Z Z

|Z−Y |≤d(Y )/2

|F |²dZ + 1

2 Z Z

|Z−Y |≤d(Y )/2

|∇q|²dZ + C₃

Z Z

|Z−Y |≤d(Y )/2

d(Y )⁻²|q|²dZ

(3.9)

since d(Y ) < 1 for all Y ∈ X . Now multiplying d(Y )⁴ on both sides of (3.9) and repeating the argument (3.4)–(3.8), we have for ν small that

ν Z Z

X

d(Z)⁴|∇q|²dZ

≤C₄ν Z Z

X

d(Z)²|∇u|²dZ + C₄ν Z Z

X

|u|²dZ + C₄ν Z Z

X

d(Z)²|q|²dZ

≤C5ν Z Z

X

d(Z)²|∇u|²dZ + C5ν Z Z

X

|u|²dZ.

(3.10)

Taking ν = _2C¹

5 and δ = _4C¹

2C5, we obtain Z Z

X

d(Z)²|∇u|²dZ + Z Z

X

d(Z)²|q|²dZ + Z Z

X

d(Z)⁴|∇q|²dZ

≤C6

Z Z

X

|u|²dZ

by adding (3.8) and (3.10).

2

4 Proofs of main results

This section is devoted to the proofs of Theorem 1.1, Corollary 1.3, and Theorem 1.2.

We choose the cut-off function χ as follows

χ(t) =











1, |t| ≤ T₄, 0, |t| ≥ T₃,

exp



−( T

T₃− |t|)³(|t| − T4

T₃− T₄)⁴



, T₄ < |t| < T₃,

(4.1)

where T₃ = T −^t₂⁰, T₄ = T − t₀. Moreover, we let θ(x) ∈ C₀^∞(Rⁿ) satisfy 0 ≤ θ(x) ≤ 1 and

θ(x) =















0, |x| ≤ R₁ e , 1, R1

2 < |x| < 2 ˜R₃, 0, |x| ≥ 3 ˜R₃

with ˜R₃ < 1.

Applying (2.37) to θu gives Z Z

(1 + ψ⁰⁰)²ϕ²(|x|²|∇(χθu)|²+ β²|χθu|²)dxdt .

Z Z

(1 + ψ⁰⁰)ϕ²|x|⁴|∆(χθu)|²dxdt. (4.2)