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 Br(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 Qs,τx,R = {(t, y) : y ∈ BR(x) ⊂ Rn, t ∈ (−τ + s, τ + s)} Let u ∈ H1((−1, 1); Hloc2 (Ω)) be a nontrivial solution of (1.1) with an appropriate pressure function p ∈ L2((−1, 1); Hloc1 (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
Q0,T −t00,R2
|u|2dxdt ≤ ˜C Z Z
Q0,T0,R1
|u|2dxdt
!κ
Z Z
Q0,T0,R3
|u|2dxdt
!1−κ
, (1.3)
where ˜C depends on , λ, n, T, t0, R2/R3 and κ = log(2R3R3
2) 4 log(4RR2
1 ) + log(2R3R3
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
Q0,T −t00,R2
|u|2dxdt ≤ Ce4(log4R2R1)m1 Z Z
Q0,T0,R1
|u|2dxdt, (1.4)
where
m1 := 64 +1 2 + 64
log 2 RR
Q0,T
0,R3
|u|2dxdt RR
Q0,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
Q0,T0,R3
|u|2dxdt
Rm1 ≤
Z Z
Q0,T0,R1
|u|2dxdt for all R1 sufficiently small, where
m = C1+ C1log RR
Q0,T0,R3|u|2dxdt RR
Q0,T −t00,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); Hloc2 (Ω))×
L2((−1, 1); Hloc1 (Ω)) satisfies (1.1) and Z Z
Q0,T0,r
|u(t, x)|2dxdt = O(rN) ∀ 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); Hloc2 (Ω)) × L2((−1, 1); Hloc1 (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(vij − vji), where v = (vij)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∂jFij, i = 1, 2, · · · , n, where Fij = Pn
k,`=1A˜ijk`(x, t)∂ku` +Pn
k=1B˜ijk(x, t)uk with appropriate ˜Aijk`(x) and ˜Bijk(x) satisfying
| ˜Aijk`(x, t)| ≤ λ|x|−1+, | ˜Bijk(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 64β ∈ 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, · · · , ωn) ∈ Sn−1. Using new coordinate y = − log r, we obtain that
∂
∂xj = ey(−ωj∂y+ 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
ωjΩj = 0 and
n
X
j=1
Ωjωj = n − 1.
It is easy to see that
∂2
∂xj∂x` = e2y(−ωj∂y− ωj+ Ωj)(−ω`∂y+ Ω`), 1 ≤ j, ` ≤ n.
and, therefore, the Laplacian becomes
e−2y∆ = ∂y2 − (n − 2)∂y+ ∆ω,
where ∆ω = Σnj=1Ω2j 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 Ek, where Ek is the space of spherical harmonics of degree k. We remark that
X
j
Z Z
|Ωjv|2dydω =X
k≥0
k(k + n − 2) Z
|vk|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 Ekwhich represents the space of spherical harmonics of degree k. Hence
Λ = Σk≥0(k + n − 2
2 )πk, (2.7)
where πk is the orthogonal projector on Ek. 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 = eψ(y)L±(e−ψ(y)v) and e−2yeψ(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
β2−2(j+|α|) Z
(1 + ψ00)|∂yjΩα(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]
=Σk≥0χ(∂y2vk− ˜b∂yvk+ ˜avk), where
(˜a = (ψ0− k)(ψ0+ k + n) − ψ00
˜b = 2ψ0 + n.
By abusing the notation v = vk, it is enough to prove that X
j≤1
Z
(1 + ψ00)|(β2−2j+ k2−2j)∂yj(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)∂2y(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 ∂2y[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 ≤ 2log βσ, 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 ≥ 2log βσ and 64β ∈ 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, · · · , gn) ∈ (C0(R; C0∞(R × Sn−1)))n+1 and β > β10 with 64β ∈ N + 2561 , we have that
β2 Z
(1 + ψ00)|χu|2 . Z
|χ(L−ψL+ψu − e−2y∂tu) + ∂yg0+
n
X
j=1
Ωjgj|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
Tψv(t, y, ω) = (2π)−1/2X
k≥0
Z ∞
−∞
eiηy(iη − ψ0− k − (n − 2))πkv(t, η, ω)dη.ˆ
(2.22) Similarly, we define
T˜ψv(t, y, ω) = (2π)−1/2X
k≥0
Z ∞
−∞
eiηy(iη − β − k − (n − 2))πkv(t, η, ω)dη.ˆ
(2.23) Note that we have Tψv = L−ψv = ∂yv − ψ0v −(n−2)2 v − Λv = ˜Tψv + (β − ψ0)v. It is clear that ˜Tψ is invertible whose inverse is given by
T˜ψ−1v(t, y, ω) = (2π)−1/2X
k≥0
Z ∞
−∞
eiηy(iη − β − k − (n − 2))−1πkv(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 ˜TψvkL2(R2×Sn−1) ≥ βkvkL2(R2×Sn−1), βk ˜Tψ−1vkL2(R2×Sn−1)≤ kvkL2(R2×Sn−1), k ˜Tψ−1( X
j+|α|≤1
∂yjΩαv)kL2(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 Tψ−1 = (I + ˜Tψ−1(β − ψ0)I)−1T˜ψ−1.
Consequently, it follows from (2.25) that for v ∈ L2(R; C0∞(R × Sn−1)) we have
kTψvkL2(R2×Sn−1) ≥ β
2kvkL2(R2×Sn−1), βkTψ−1vkL2(R2×Sn−1) . kvkL2(R2×Sn−1), kTψ−1( X
j+|α|≤1
∂yjΩαv)kL2(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) = (Tψ−1u)(t, y, ω), i.e., u(t, y, ω) = Tψv(t, y, ω). Let uk(t, y, ω) = πku(t, y, ω) and vk(t, y, ω) = πkv(t, y, ω).
From the definition of Tβ, we have that X
k≥0
uk = u = Tψv
= (2π)−1/2X
k≥0
Z ∞
−∞
eiηy(iη − ψ0− k − (n − 2))πkv(t, η, ω)dηˆ
= X
k≥0
(∂yvk− (ψ0+ k + n − 2)vk). (2.27)
We would like to find vk satisfying lim|y|→∞vk(t, y, ω) = 0. Thus, from (2.27), we have that
vk(t, y, ω) =
0, y ≥ y1,
− Z y1
y
e−ψ(ξ)−kξ−(n−2)ξ
uk(t, ξ, ω)dξ · eψ(y)+ky+(n−2)y, y < y1. (2.28) Note that for y < y0, vk 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
uk(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 e2yv(t, y, ω) =P
k≥0e2yvk(t, y, ω) with vk given in (2.28).
Using Tψ = L−ψ and estimates in (2.26), we obtain that
kχL−ψL+ψu − χe−2y∂tu + ∂yg0+
n
X
j=1
ΩjgjkL2(R2×Sn−1)
= kTψ L+ψ(χu) + Tψ−1 −χe−2y∂tu + ∂yg0+
n
X
j=1
ΩjgjkL2(R2×Sn−1)
≥ βkL+ψ(χu) + Tψ−1 −χe−2y∂tu + ∂yg0 +
n
X
j=1
ΩjgjkL2(R2×Sn−1)
≥ βkL+ψ(χu) − ∂t(Tψ−1(χe−2yu)) + Tψ−1(χ0e−2yu)kL2(R2×Sn−1)
−βkTψ−1 ∂yg0+
n
X
j=1
ΩjgjkL2(R2×Sn−1)
& βkL+ψ(χu) − ∂t(Tψ−1(χe−2yu)) + Tψ−1(χ0e−2yu)kL2(R2×Sn−1)
−βkgkL2(R2×Sn−1). (2.29)
Now, we let v = Tψ−1(e−2yu) and hence Tψv = e−2yu and χv = Tψ−1(e−2yχu). We immediately obtain from (2.8) that
kL+ψ(χu) − ∂t(Tψ−1(χe−2yu)) + Tψ−1(χ0e−2yu)kL2(R2×Sn−1)
=kχ L+ψL−ψ(e2yv) − 2L+ψ(e2yv) − e−2y∂t(e2yv)kL2(R2×Sn−1)
& X
j+|α|≤1
β1−(j+|α|)k(1 + ψ00)1/2∂yjΩα(χe2yv)kL2(R2×Sn−1)− (χ0v)
L2(R2×Sn−1). (2.30) From the substitution u = e2yTβv, the definitions of Tβ, Λ, and the first inequality of (2.26), it is not difficult to show that
k(1 + ψ00)1/2χukL2(R2×Sn−1)
. X
j+|α|≤1
β1−(j+|α|)k(1 + ψ00)1/2∂yjΩα(χe2yv)kL2(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(Tψ−1(χe−2yu)) + Tψ−1(χ0e−2yu)kL2(R2×Sn−1)
& βk(1 + ψ00)1/2χukL2(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, · · · , fn) ∈ (C0(R; C0∞(Rn\ {0})))n and β ≥ β2 with 64β ∈ 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
β2−2(j+|α|) Z
(1 + ψ00)|∂yjΩα(χu)|2
. Z
|χL−ψL+ψu − e−2y∂t(χu) + ∂yg0+
n
X
j=1
Ωjgj|2
+ β2kgk2L2(R2×Sn−1)+ Z
|e−2yχ0u|2,
(2.34)
where g0 = hω, e−ψf i, gj = e−ψfj, j = 1, 2, · · · , n. Denote J (u) = χL−ψL+ψu − e−2y∂t(χu) + ∂yg0+Pn
j=1Ωjgj. 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
Ωjgj)
= −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
Ωjgj)
≥ Z
(1 + ψ00)|∂y(χu)|2− Cβ2 Z
(1 + ψ00)|χu|2+ Z
(1 + ψ00)
n
X
j=1
|Ωj(χu)|2
− Cβkgk2L2(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 = Ba2r\ ¯Ba1r, 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 ξY(Z) = ξY(t, x) = ξ((Z−Y )/d(Y )) for Y ∈ X . Multiplying ξ2Y(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 ξY2(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)