FRACTIONAL LAPLACIAN WITH A DRIFT TERM

PU-ZHAO KOW AND JENN-NAN WANG

Abstract. In this paper, we study a Landis-type conjecture for the fractional Schrödinger equation with a drift term. The Landis-type conjecture is a question of unique continuation property from the innity. More precisely, we would like to prove the following: if any solution of the fractional Schrödinger equation with a drift term decays at a certain exponential rate, then such solution must be trivial. By localizing the equation with the help of Caarelli-Silvestre extension, we solve the problem by a delicate Carleman inequality in the (n + 1)-dimensional half space.

1. Introduction

In this work, we consider the unique continuation from the innity for the fractional Schrödinger equation with drift term

(1.1) ((−∆)^{s}+ b(x)x · ∇ + q(x))u = 0 in R^{n},

where s ∈ (0, 1) and b, q are scalar-valued functions. Precisely, we are interested in
investigating the decay rate of u at innity that implies the solution u is trivial. This
problem is closely related to the Landis conjecture [KL88]. For s = 1, b = 0, Landis
conjectured that if if |q(x)| ≤ 1 and |u(x)| ≤ C0 satises |u(x)| ≤ exp(−C|x|^{1+}), then
u ≡ 0. The Landis conjecture was disproved by Meshkov [Me91], who constructed a
complex-valued potential q and a nontrivial complex-valued u with |u(x)| ≤ C exp(−C|x|^{4}^{3})
such that u is a solution of the Schrödinger equation with potential q. He also showed
that if |u(x)| ≤ C exp(−C|x|^{4}^{3}^{+}), then u ≡ 0.

In view of Meshkov's counterexample, Kenig [Ke06] rened the Landis conjecture and asked whether this conjecture is true for real-valued potentials and solutions. This real version Landis conjecture was conrmed partially in [KSW15] where n = 2 and q ≥ 0.

This result was later extended to the more general situation with ∆ being replaced by any second order elliptic operator [DKW17]. The Landis conjecture in the real case with n = 1 was studied in [Ro18]. Recently, the real version Landis conjecture in the plane case was resolved by Logunov, Malinnikova, Nadirashvili, Nazarov [LMNN20].

A Landis-type conjecture was considered in [RW19] for the fractional Laplacian with b = 0. Both qualitative and quantitative estimates were proved in [RW19]. For example, when q is dierentiable and satises

|x · ∇q(x)| ≤ 1,

if u satises the following decay behavior: ∃ α > 1 such that Z

R^{n}

e^{|x|}^{α}|u|^{2}dx < ∞,

2020 Mathematics Subject Classication. 35R11; 35B40.

Key words and phrases. Fractional Laplacian, Landis-type conjecture, Caarelli-Silvestre extension.

1

then u ≡ 0. On the other hand, for a non-dierentiable potential q, if s ∈ (1/4, 1),
kqk_{L}^{∞}_{(R}^{n}_{)} ≤ 1, and u satises the decay behavior: ∃ α > _{4s−1}^{4s} such that

Z

R^{n}

e^{|x|}^{α}|u|^{2}dx < ∞,
then u ≡ 0.

The main theme of this work is to extend the results in [RW19] to the fractional
Schrödinger equation with a drift term (1.1). The unique continuation property established
in [GSU20] states that if u ∈ H^{−r}(R^{n}) for some r ∈ R and u = (−∆)^{s}u = 0 in
some open set W ⊂ R^{n}, then u ≡ 0. It follows from this property that the unique
continuation property holds for (1.1). However, to our best knowledge, the strong
unique continuation property for the fractional Laplacian with a drift term remains
open. Our work is the rst attempt to study the unique continuation property for
any solution of the fractional Laplacian with a drift term that satises some decaying
condition. Motivated by the result of [Ro18], we consider the drift term only involving
the radial derivative. Inspired by the ideas in [Me89] and [RW19], we show that if both
b and q are dierentiable, then any non-trivial solution of the fractional Schrödinger
equation does not decay super-exponentially at innity. The detailed statement is
described in the following theorem.

Theorem 1.1. Let n = 1, 2, 3 and (1.2)

(s ∈ (^{1}_{2}, 1) when n = 1 or 2,
s ∈ (^{3}_{4}, 1) when n = 3.

We also assume that there exists a constant λ such that

(1.3) |q(x)| ≤ λ and 0 ≤ b(x) ≤ λ|x|^{−β} for all x ∈ R^{n}
and the radial derivatives of q, b satisfy

(1.4) |x · ∇q(x)| ≤ λ and |x · ∇b(x)| ≤ λ|x|^{−β} for all x ∈ R^{n}
for some β > 1. If u ∈ H^{s}(R^{n}) is a solution to (1.1) such that

(1.5) Z

R^{n}

e^{|x|}^{β}

|u(x)|^{2}+ |∇u(x)|^{2}

dx ≤ λ, then u ≡ 0.

For non-dierentiable potential q, we also can prove the following result.

Theorem 1.2. Let n = 1, 2, 3 and s given in (1.2). Let β > _{4s−1}^{4s} , and both b, q satisfy
(1.3). Here we assume the radial derivative of b satises

|x · ∇b(x)| ≤ λ|x|^{−β} for all x ∈ R^{n}.
If u ∈ H^{s}(R^{n}) is a solution to (1.1) satisfying (1.5), then u ≡ 0.

Remark 1.3. Since we treat the drift term as a lower order addition, it is reasonable to expect that s > 1/2 in Theorem 1.1 and 1.2.

Like several existing results, we want to prove Theorem1.1and1.2by an appropriate
Carleman estimate. Since such estimate is a local estimate, we rst localize the equation
(1.1) by the Caarelli-Silvestre extension [CS07] and derive the Carleman estimate
for a degenerate elliptic equation in the (n + 1)-dimensional upper half-space R^{n+1}+ .

Our strategy in proving both theorems is similar to that of [RW19]. We will mainly
handle the CS extended solution ˜u in R^{n+1}+ . The condition (1.5) can be considered as a
boundary decay for ˜u. We then pass the boundary decay to the bulk decay of ˜u in R^{n+1}+ .
We would like to point out that unlike the pure potential case considered in [RW19],
here, in order to guarantee the bulk decay of ˜u, we also need the boundary decay of ∇u
due to the addition of the drift term.

We now comment on the form of drift coecient in (1.1). Such choice is due to the
limitation in the Carleman estimate for ˜u in R^{n+1}+ . Since the boundary term contains
the rst derivatives of u, to bound this boundary term, we need to include the second
derivatives of ˜u in the Carleman estimate in view of the trace inequality (LemmaA.3).

However, the parameter appears in the rst derivatives of u is τ, while the parameter
in the second derivatives of ˜u is τ^{−1} (see (4.4)). Without further restrictions, this
boundary term can not be removed. We also remark that the operator

(1.6) − ((−∆)^{s}+ λx · ∇) (where λ is a positive constant)

is related to a 2s-stable Ornstein-Uhlenbeck process in R^{d} for s ∈ (0, 1), see [Jak08,
Sect. 2]. Precisely, let Xt be the 2s-stable Ornstein-Uhlenbeck process in R^{d}, given in
the following stochastic integral:

X_{t}= e^{−λt}X_{0}+
Z t

0

e^{−λ(t−s)}d ˆX_{s}, Xˆ_{0} = 0,

where the integral is in the Stieltjes sense, and ( ˆXt, P^{x}) is the isotropic 2s-stable Lévy
process in R^{d} with index of stability s ∈ (0, 1) and characteristic function

E^{0}e^{i ˆ}^{X}^{t}^{·ξ} = e^{−t|ξ|}^{2s} for all ξ ∈ R^{d} and for all t ≥ 0.

Here, E^{x} denotes the expectation with respect to the distribution P^{x} of the process
starting from x ∈ R^{d}. Indeed, the innitesimal genrator of Xt is equal to (1.6).

This paper is organized as follows. In Section 2, we introduce some notations and
state the Caarelli-Silvestre extension. In Section 3, we discuss how the boundary
decay of u implies the bulk decay of ˜u in R^{n+1}+ . The derivation of the needed Carleman
estimate is discussed in great detail in Section4. Section 5 is devoted to the proofs of
main results.

2. The Caffarelli-Silvestre extension

Basically, we shall follow the denitions and notations in [RW19]. We now restate
what we need. Let R^{n+1}+ := R^{n} × R+ = x = (x^{0}, x_{n+1}) x^{0} ∈ R^{n}, x_{n+1} > 0

and
x_{0} = (x^{0}, 0) ∈ R^{n}× {0}. For r, R > 0, we denote

B_{r}^{+}(x_{0}) := x ∈ R^{n+1}+ |x − x_{0}| ≤ r ,
B_{r}^{0}(x_{0}) := x ∈ R^{n}× {0} |x − x0| ≤ r ,

B^{+}_{r} := B_{r}^{+}(0), B_{r}^{0} := B_{r}^{0}(0),
A^{+}_{r,R} := x ∈ R^{n+1}+ r ≤ |x| ≤ R ,
A^{0}_{r,R} := x ∈ R^{n}× {0} r ≤ |x| ≤ R .

For Ω ⊂ R^{n+1}+ , we dene
H˙^{1}(Ω, x^{1−2s}_{n+1}) :=

v : Ω → R Z

Ω

x^{1−2s}_{n+1}|∇v|^{2}dx < ∞

,
H^{1}(Ω, x^{1−2s}_{n+1}) :=

v : Ω → R Z

Ω

x^{1−2s}_{n+1}(|v|^{2}+ |∇v|^{2}) dx < ∞

.

Write x = rθ, where r > 0 and θ ∈ S+^{n}, i.e., θ = (θ^{0}, θn+1) ∈ S^{n} with θn+1 > 0. We also
denote

H^{1}(S_{+}^{n}, θ_{n+1}^{1−2s}) :=

v : S_{+}^{n} → R
Z

S_{+}^{n}

θ^{1−2s}_{n+1} (|v|^{2}+ |∇S^{n}v|^{2}) dθ < ∞

,
where ∇^{S}^{n} = (Θ_{1}, · · · , Θ_{n+1}). Here, Θk are vector elds on S^{n}.

For s ∈ (^{1}_{2}, 1)and u ∈ H^{s}(R^{n}), let ˜u ∈ ˙H^{1}(R^{n+1}+ , x^{1−2s}_{n+1})be a solution to the degenerate
elliptic equation

(2.1)

(∇ · x^{1−2s}_{n+1}∇˜u = 0 in R^{n+1}+ ,

˜

u = u on R^{n}× {0},

where ∇ = (∇^{0}, ∂n+1) = (∂1, · · · , ∂n, ∂n+1). It was established in [CS07] that

−(−∆)^{s}u(x) = c_{n,s} lim

xn+1→0x^{1−2s}_{n+1}∂_{n+1}u(x)˜

for some constant cn,s > 0. In view of this observation, (1.1) can be reformulated as the local, degenerate elliptic equation

(2.2)

∇ · x^{1−2s}_{n+1}∇˜u = 0 in R^{n+1}+ ,

˜

u = u on R^{n}× {0},

cn,s lim

xn+1→0x^{1−2s}_{n+1} ∂n+1u = bx˜ ^{0}· ∇^{0}u + qu on R^{n}× {0}.

3. Boundary decay implies bulk decay

In this section, we will show that the boundary decay (1.5) implies the bulk decay of

˜

uin R^{n+1}+ .

Proposition 3.1. Let s be given in (1.2) and ˜u ∈ H^{1}(R^{n+1}+ , x^{1−2s}_{n+1}) be a solution to
(2.2). Assume that there exists a constant λ > 0 such that

(3.1) |q(x^{0})| ≤ λ and |b(x^{0})||x^{0}| ≤ λ for all x^{0} ∈ R^{n}.

If there exist constants C, β > 1 such that (1.5) holds, then there exist constants C1, c_{1} >

0 such that

(3.2) |˜u(x)| ≤ C_{1}e^{−c}^{1}^{|x|}^{β} for all x ∈ R^{n+1}+ .
The following lemma can be found in [RW19, Equ.(19)].

Lemma 3.2. Let s ∈ (^{1}_{2}, 1) and ˜u ∈ H^{1}(R^{n+1}+ , x^{1−2s}_{n+1}) be a solution to (2.1). If x0 ∈
R^{n}× {0}, then there exist α = α(n, s) ∈ (0, 1) and c = c(n, s) ∈ (0, 1) such that

kx

1−2s 2

n+1uk˜ _{L}2(B^{+}cr(x0))

≤ C

kx

1−2s 2

n+1 uk˜ _{L}2(B_{16r}^{+} (x0))+ r^{1−s}kuk_{L}^{2}_{(B}^{0}

16r(x0))

α

×

r^{s+1}

xn+1lim→0x^{1−2s}_{n+1}∂_{n+1}u˜

L^{2}(B^{0}_{16r}(x0))

+ r^{1−s}kuk_{L}^{2}_{(B}^{0}

16r(x0))

1−α

(3.3)

for some positive constant C.

Combining Lemma 3.2 and LemmaA.2 implies the following lemma.

Lemma 3.3. Let s be given by (1.2) and ˜u ∈ H^{1}(R^{n+1}+ , x^{1−2s}_{n+1}) be a solution to (2.2).

If q ∈ L^{∞}(R^{n}) and b(x^{0})x^{0} ∈ (L^{∞}(R^{n}))^{n}, then there exist α = α(n, s) ∈ (0, 1) and
c = c(n, s) ∈ (0, 1) such that

k˜uk_{L}∞(B^{+}cr
4)

≤ Cr^{−}^{n}^{2}

r^{s−1}kx_{n+1}^{1−2s}^{2} uk˜ _{L}2(B^{+}_{16r})+ kuk_{L}^{2}_{(B}^{0}

16r)

α

×

r^{2s}kbx^{0} · ∇^{0}u + quk_{L}^{2}_{(B}^{0}

16r)+ kuk_{L}^{2}_{(B}^{0}

16r)

1−α

+ r^{2s}kbx^{0}· ∇^{0}u + quk_{L}^{2}_{(B}^{0}

16r)

(3.4)

for all r > 1, where the constant C is independent to r.

Proof. Given any r > 1, let ˜v(x) = ˜u(rx) in R^{n+1}+ and v(x^{0}) = u(rx^{0})on R^{n}× {0}. Note
that

c_{n,s} lim

xn+1→0x^{1−2s}_{n+1}∂_{n+1}v(x) = c˜ _{n,s} lim

xn+1→0x^{1−2s}_{n+1} ∂_{n+1} ˜u(rx)

= rcn,s lim

xn+1→0x^{1−2s}_{n+1}∂n+1u(rx)˜

= rc_{n,s} lim

rxn+1→0r^{−1+2s}(rx_{n+1})^{1−2s}∂_{n+1}u(rx)˜

= r^{2s}c_{n,s} lim

rxn+1→0(rx_{n+1})^{1−2s}∂_{n+1}u(rx)˜

= r^{2s}b(rx^{0})rx^{0}· ∇^{0}u(rx) + q(rx)u(rx)

(By (2.2))

= r^{2s}b(rx^{0})x^{0} · ∇^{0}u(rx) + r^{2s}q(rx)u(rx)

= r^{2s}b(rx^{0})x^{0} · ∇^{0}v(x) + r^{2s}q(rx)v(x),
that is,

(3.5)

∇ · x^{1−2s}_{n+1}∇˜v = 0 in R^{n+1}+ ,

˜

v = v on R^{n}× {0},

cn,s lim

xn+1→0x^{1−2s}_{n+1}∂n+1v = b˜ rx^{0} · ∇^{0}v + qrv on R^{n}× {0},

where br(x^{0}) := r^{2s}b(rx^{0}) and qr(x^{0}) := r^{2s}q(rx). Applying Lemma A.2 to (3.5) with
p = 2 (since (1.2)) a1 = 0 and a2 = b_{r}x^{0}· ∇^{0}v + q_{r}v, we obtain that

(3.6) k˜vk_{L}∞(B_{1/4}^{+} )≤ C

kx_{n+1}^{1−2s}^{2} ˜vk_{L}2(B_{1}^{+})+ kb_{r}x^{0} · ∇^{0}v + q_{r}vk_{L}^{2}_{(B}^{0}

1)

, where C = C(n, s). Since ˜v(x) = ˜u(rx), we have

k˜vk_{L}∞(B^{+}_{1/4}) = sup

x∈R^{n+1}_{+} ,|x|≤^{1}_{4}

|˜v(x)| = sup

x∈R^{n+1}_{+} ,|rx|≤^{r}_{4}

|˜u(rx)| = k˜uk_{L}∞(B_{r/4}^{+} ).
On the other hand, we can derive

kx

1−2s 2

n+1vk˜ ^{2}_{L}2(B^{+}_{1}) =
Z

x∈R^{n+1}_{+} ,|x|≤1

x^{1−2s}_{n+1}|˜v(x)|^{2}dx = r^{2s−2−n}kx

1−2s 2

n+1uk˜ ^{2}_{L}2(B^{+}r)

and

kb_{r}x^{0} · ∇^{0}v + q_{r}vk^{2}_{L}2(B_{1}^{0})=
Z

|x^{0}|<1

|b_{r}(x^{0})x^{0}· ∇^{0}v(x^{0}) + q_{r}(x^{0})v(x^{0})|^{2}dx^{0}

= r^{4s−n}kbx^{0} · ∇^{0}u + quk^{2}_{L}2(B_{r}^{0}).
Thus, (3.6) implies

k˜uk_{L}∞(B^{+}_{r/4}) ≤ Cr^{−}^{n}^{2}

r^{s−1}kx

1−2s 2

n+1uk˜ _{L}2(Br^{+})+ r^{2s}kbx^{0}· ∇^{0}u + quk_{L}^{2}_{(B}^{0}_{r}_{)}

.

Here, r > 1 is arbitrary and C is independent of r. Replacing r by cr, where c is the constant given in Lemma3.2, we have

k˜uk_{L}∞(B_{cr/4}^{+} ) ≤ Cr^{−}^{n}^{2}

r^{s−1}kx_{n+1}^{1−2s}^{2} uk˜ _{L}2(B^{+}cr)+ r^{2s}kbx^{0}· ∇^{0}u + quk_{L}^{2}_{(B}^{0}

cr)

.

Here, C is another constant independent of r. Combining this inequality with (3.3) and using

c_{n,s} lim

xn+1→0x^{1−2s}_{n+1}∂_{n+1}u = bx˜ ^{0}· ∇^{0}u + qu,

we obtain our desired result.

We are ready to prove the main result of this section.

Proof of Proposition3.1. Let R > 1 be suciently large and x0 ∈ R^{n}× {0}with |x0| =
32R. From (1.5), it follows that

λ ≥ Z

B_{16R}^{0} (x0)

e^{|x}^{0}^{|}^{β}

|u(x^{0})|^{2}+ |∇^{0}u(x^{0})|^{2}

dx^{0}

≥ e^{(16R)}^{β}
Z

B_{16R}^{0} (x0)

|u(x^{0})|^{2}+ |∇^{0}u(x^{0})|^{2}

dx^{0}

= e^{(16R)}^{β}

kuk^{2}_{L}2(B_{16R}^{0} (x0))+ k∇^{0}uk^{2}_{L}2(B_{16R}^{0} (x0))

, that is,

(3.7) kuk_{L}^{2}_{(B}^{0}

16R(x0))+ k∇^{0}uk_{L}^{2}_{(B}^{0}

16R(x0)) ≤ ˜Ce^{−˜}^{cR}^{β}

for some constants ˜C and ˜c. Note that the Caarelli-Silvestre extension ˜u satises kx

1−2s 2

n+1∇˜uk_{L}2(R^{n+1}+ ) ≤ C and k˜u(•, xn+1)k_{L}^{2}_{(R}^{n}_{)} ≤ kuk_{L}^{2}_{(R}^{n}_{)}

(see [St10, page 48-49]). Thus, we have

kx_{n+1}^{1−2s}^{2} uk˜ ^{2}_{L}_{2}_{(B}+

16R(x0))≤ Z 16R

0

x^{1−2s}_{n+1}k˜u(•, x_{n+1})k^{2}_{L}2(R^{n})dx_{n+1}

≤ Z 16R

0

x^{1−2s}_{n+1}kuk^{2}_{L}2(R^{n})dx_{n+1}

= (16R)^{2−2s}

2 − 2s kuk^{2}_{L}2(R^{n})

≤ CR^{2−2s}.

Plugging cn,slim_{x}_{n+1}→0x^{1−2s}_{n+1}∂_{n+1}u = bx˜ ^{0}· ∇^{0}u + qu into (3.3) (with r = R) and using
(3.1) yields

kx

1−2s 2

n+1uk˜ _{L}2(B^{+}_{cR}(x0)) ≤ ˜Ce^{−˜}^{cR}^{β}.

Next, choosing r = ^{cR}_{16} in (3.4) and putting together (3.1) and (1.5), we have
k˜uk_{L}∞(B^{+}

c2R 64

(x0)) ≤ ˜Ce^{−˜}^{cR}^{β} for |x0| = 32R,
which implies

k˜uk_{L}∞(B^{+}

c ˜R(x0))≤ ˜Ce^{−˜}^{c ˜}^{R}^{β} for all large ˜R 1.

The decay estimate (3.2) then follows from the chain-of-balls argument described in

[RW19, Proposition 2.2, Step 2].

4. Carleman estimates

This section is mainly devoted to the derivations of Carleman estimates. We will discuss the estimates corresponding to both dierentiable and non-dierentiable potentials.

4.1. Carleman estimate with dierentiable potential. The proof of the Carleman estimate below follows from the argument in [RW19].

Theorem 4.1. Let s ∈ (^{1}_{2}, 1) and let ˜u ∈ H^{1}(R^{n+1}+ , x^{1−2s}_{n+1}) with supp(˜u) ⊂ R^{n+1}+ \ B_{1}^{+}
be a solution to

∇ · x^{1−2s}_{n+1}∇˜u = f in R^{n+1}+ ,

˜

u = u on R^{n}× {0},

xn+1lim→0x^{1−2s}_{n+1}∂_{n+1}u = W x˜ ^{0} · ∇^{0}u + V u + g on R^{n}× {0},

where f ∈ L^{2}(R^{n+1}+ , x^{2s−1}_{n+1}) is compactly supported in R^{n+1}+ , g ∈ L^{2}(R^{n}) is compactly
supported in R^{n}, and x^{0} · ∇^{0}W, x^{0}· ∇^{0}V exist. Let α > 1 be a real constant and dene
φ(x) := |x|^{α}. If

W (x^{0}) ≥ 0 for all x^{0} ∈ R^{n}\ B_{1}^{0}.

Then there exists a real number τ0 > 1 such that
τ^{3}ke^{τ φ}|x|^{3α}^{2} ^{−1}x

1−2s 2

n+1uk˜ ^{2}_{L}_{2}

(R^{n+1}_{+} )+ τ ke^{τ φ}|x|^{α}^{2}x

1−2s 2

n+1∇˜uk^{2}_{L}_{2}

(R^{n+1}_{+} )

+ τ^{−1}ke^{τ φ}|x|^{−}^{α}^{2}x

1−2s 2

n+1∇(x · ∇˜u)k^{2}_{L}_{2}

(R^{n+1}_{+} )

≤ C

ke^{τ φ}|x|x

2s−1 2

n+1f k^{2}_{L}2(R^{n+1}_{+} )

+ τ ke^{τ φ}|x|^{α}^{2}|V |^{1}^{2}uk^{2}_{L}2(R^{n}×{0})+ τ ke^{τ φ}|x|^{α}^{2}|x^{0} · ∇^{0}V |^{1}^{2}uk^{2}_{L}2(R^{n}×{0})

+ τ^{2}ke^{τ φ}|x|^{α}|W |^{1}^{2}uk˜ ^{2}_{L}2(R^{n}×{0})+ τ^{2}ke^{τ φ}|x|^{α}|x^{0}· ∇^{0}W |^{1}^{2}uk^{2}_{L}2(R^{n}×{0})

+ τ^{2}ke^{τ φ}|x|^{α}^{2}uk^{2}_{L}2(R^{n}×{0})

+ τ^{−1}ke^{τ φ}|x|^{−}^{α}^{2}|V |^{1}^{2}(x^{0}· ∇^{0}u)k^{2}_{L}2(R^{n}×{0})

+ τ^{−1}ke^{τ φ}|x|^{−}^{α}^{2}|x^{0} · ∇^{0}V |^{1}^{2}(x^{0} · ∇^{0}u)k^{2}_{L}2(R^{n}×{0})

+ ke^{τ φ}|W |^{1}^{2}(x^{0}· ∇^{0}u)k^{2}_{L}2(R^{n}×{0})+ ke^{τ φ}|x^{0}· ∇^{0}W |^{1}^{2}(x^{0}· ∇^{0}u)k^{2}_{L}2(R^{n}×{0})

+ ke^{τ φ}|x|^{−}^{α}^{2}(x^{0}· ∇^{0}u)k^{2}_{L}2(R^{n}×{0})

+ τ^{2}ke^{τ φ}|x|^{3}^{2}^{α}|x|^{1+2s}^{2} gk^{2}_{L}2(R^{n}×{0})+ ke^{τ φ}|x|^{−}^{α}^{2}|x|^{s}(x^{0}· ∇^{0}g)k^{2}_{L}2(R^{n}×{0})

(4.1)

for all τ ≥ τ0. Here, the positive constant C is independent of τ.

Proof. Step 1: Pass to conformal polar coordinates. Let x = e^{t}θ with t ∈ R and
θ ∈ S_{+}^{n}. Set u := e^{n−2s}^{2} ^{t}u˜, we have

(4.2)

θ^{1−2s}_{n+1}∂^{2}_{t} + ∇S^{n}· θ_{n+1}^{1−2s}∇S^{n} − θ^{1−2s}_{n+1} (n − 2s)^{2}
4

u = ˜f in S+^{n}× R,

θn+1lim→0θ^{1−2s}_{n+1}ν · ∇S^{n}u =

V −˜ n − 2s 2

W˜

u + ˜W ∂_{t}u + ˜g on ∂S+^{n} × R,
where ˜f = e^{n+2+2s}^{2} ^{t}f, ν = (0, · · · , 0, 1), ˜V = e^{2st}V, ˜W = e^{2st}W, and ˜g = e^{n+2s}^{2} ^{t}g.

Next, by setting ˜∆_{S}^{n} := θ

2s−1 2

n+1 ∇_{S}^{n}·θ^{1−2s}_{n+1}∇_{S}^{n}θ

2s−1 2

n+1, ϕ(t) = φ(e^{t}θ) = e^{αt}, v = e^{τ ϕ}θ

1−2s 2

n+1 u,
f = e^{τ ϕ}f˜, and g = e^{τ ϕ}g˜, (4.2) can be written as

L_{ϕ}v := (S + A)v = θ

2s−1 2

n+1 f in S+^{n}× R,

lim

θn+1→0θ^{1−2s}_{n+1}ν · ∇S^{n}θ

2s−1 2

n+1 v = I + ˜W θ

2s−1 2

n+1∂_{t}v + g on ∂S+^{n}× R,
where

S := ∂_{t}^{2}+ ˜∆S^{n} −(n − 2s)^{2}

4 + τ^{2}|ϕ^{0}|^{2},
A := −2τ ϕ^{0}∂_{t}− τ ϕ^{00},

I :=

V −˜ n − 2s 2

W˜

θ

2s−1 2

n+1v − τ ϕ^{0}W θ˜

2s−1 2

n+1 v.

To shortened the notations, we denote the norm and the scalar product in the bulk and the boundary space by

(4.3)

(k • k := k • k_{L}^{2}_{(S}^{n}

+×R), h•, •i := h•, •i_{L}^{2}_{(S}^{n}

+×R)

k • k_{0} := k • k_{L}^{2}_{(∂S}^{n}

+×R), h•, •i_{0} := h•, •i_{L}^{2}_{(∂S}^{n}

+×R)

and we omit the notation limθn+1→0 in k • k0 and h•, •i0.

Since S, A are only symmetric and anti-symmetric up to boundary contributions, we can see that

kL_{ϕ}vk^{2} = kSvk^{2} + kAvk^{2}+ h[S, A]v, vi
+ 4τ hθ^{1−2s}_{n+1}ν · ∇_{S}^{n}θ

2s−1 2

n+1 v, ϕ^{0}θ

2s−1 2

n+1∂_{t}vi_{0}
(4.4)

+ 2τ hθ^{1−2s}_{n+1}ν · ∇S^{n}θ

2s−1 2

n+1 v, ϕ^{00}θ

2s−1 2

n+1 vi_{0}.
From ˜W ≥ 0, it follows that h ˜W θ

2s−1 2

n+1 ∂_{t}v, ϕ^{0}θ

2s−1 2

n+1∂_{t}vi_{0} ≥ 0 and
kL_{ϕ}vk^{2} ≥ kSvk^{2}+ kAvk^{2}+ h[S, A]v, vi

+ 4τ hI, ϕ^{0}θ

2s−1 2

n+1 ∂_{t}vi_{0}+ 4τ hg, ϕ^{0}θ

2s−1 2

n+1 ∂_{t}vi_{0}
(4.5)

+ 2τ hθ_{n+1}^{1−2s}ν · ∇S^{n}θ

2s−1 2

n+1v, ϕ^{00}θ

2s−1 2

n+1vi_{0}.

Step 2: Estimate the commutator term h[S, A]v, vi. The commutator term can be computed explicitly

[S, A]v = 4τ^{3}|ϕ^{0}|^{2}ϕ^{00}v − 4τ ϕ^{00}∂_{t}^{2}v − 4τ ϕ^{000}∂_{t}v − τ ϕ^{0000}v.

Since supp(˜u) ⊂ R^{n+1}+ \ B^{+}_{1}, we can see that supp(v) ⊂ (θ, t) ∈ S+^{n}× R t ≥ 0
Recall that ϕ(t) = e^{αt} for α > 1. We thus have |ϕ^{0}|^{2}ϕ^{00} ≥ ϕ^{0000} in supp(v). By this.
inequality, we can estimate

h[S, A]v, vi ≥ 2τ^{3}kϕ^{0}(ϕ^{00})^{1}^{2}vk^{2}+ 4τ k(ϕ^{00})^{1}^{2}∂_{t}vk^{2}.
Combining this inequality with (4.5) yields

kL_{ϕ}vk^{2} ≥ kSvk^{2} + kAvk^{2}+ 2τ^{3}kϕ^{0}(ϕ^{00})^{1}^{2}vk^{2}+ 4τ k(ϕ^{00})^{1}^{2}∂_{t}vk^{2}
+ 4τ hI, ϕ^{0}θ

2s−1 2

n+1 ∂_{t}vi_{0} + 4τ hg, ϕ^{0}θ

2s−1 2

n+1 ∂_{t}vi_{0}
+ 2τ hθ_{n+1}^{1−2s}ν · ∇S^{n}θ

2s−1 2

n+1 v, ϕ^{00}θ

2s−1 2

n+1 vi0. (4.6)

Step 3: Derive the full gradient inequality from the symmetric part kSvk^{2}.
Let c0 be a positive constant to be determined later. Note that

− ˜∆S^{n} = −S + ∂_{t}^{2}+ τ^{2}(ϕ^{0})^{2}− (n − 2s)^{2}

4 .

Hence, we have

c_{0}τ k(ϕ^{0})^{1}^{2}θ

1−2s 2

n+1 ∇S^{n}θ

2s−1 2

n+1 vk^{2}+ c_{0}τ hθ^{1−2s}_{n+1} ν · ∇S^{n}θ

2s−1 2

n+1 v, ϕ^{0}θ

2s−1 2

n+1 vi_{0}

= −c_{0}τ hϕ^{0}v, ˜∆S^{n}vi

= c_{0}τ

− hϕ^{0}v, Svi − hϕ^{00}v, ∂_{t}vi + k(ϕ^{0})^{1}^{2}∂_{t}vk^{2}+ τ^{2}k(ϕ^{0})^{3}^{2}vk^{2}

− (n − 2s)^{2}

4 k(ϕ^{0})^{1}^{2}vk^{2}

≤ −c_{0}τ hϕ^{0}v, Svi − c_{0}τ hϕ^{00}v, ∂_{t}vi + c_{0}τ k(ϕ^{0})^{1}^{2}∂_{t}vk^{2}+ 2c_{0}τ^{3}k(ϕ^{0})^{3}^{2}vk^{2}
(4.7)

for all τ 1. Combining (4.6) and (4.7) gives

kSvk^{2}+ kAvk^{2}+ 2τ^{3}kϕ^{0}(ϕ^{00})^{1}^{2}vk^{2}+ 4τ k(ϕ^{00})^{1}^{2}∂tvk^{2}
+ c_{0}τ k(ϕ^{0})^{1}^{2}θ

1−2s 2

n+1 ∇S^{n}θ

2s−1 2

n+1vk^{2}
+ 4τ hI, ϕ^{0}θ

2s−1 2

n+1∂_{t}vi_{0}+ 4τ hg, ϕ^{0}θ

2s−1 2

n+1 ∂_{t}vi_{0}
+ 2τ hθ^{1−2s}_{n+1} ν · ∇S^{n}θ

2s−1 2

n+1v, ϕ^{00}θ

2s−1 2

n+1 vi_{0}
+ c_{0}τ hθ^{1−2s}_{n+1} ν · ∇S^{n}θ

2s−1 2

n+1 v, ϕ^{0}θ

2s−1 2

n+1 vi_{0}

≤ kL_{ϕ}vk^{2}− c_{0}τ hϕ^{0}v, Svi − c_{0}τ hϕ^{00}v, ∂_{t}vi
+ c_{0}τ k(ϕ^{0})^{1}^{2}∂_{t}vk^{2}+ 2c_{0}τ^{3}k(ϕ^{0})^{3}^{2}vk^{2}.

Using ϕ^{00} = αϕ^{0} and choosing suciently small c0 > 0, we can derive
kSvk^{2}+ kAvk^{2}+ τ^{3}kϕ^{0}(ϕ^{00})^{1}^{2}vk^{2}

+ τ k(ϕ^{00})^{1}^{2}∂_{t}vk^{2}+ τ k(ϕ^{0})^{1}^{2}θ

1−2s 2

n+1 ∇S^{n}θ

2s−1 2

n+1vk^{2}

≤ C

kLϕvk^{2}+ τ |hI, ϕ^{0}θ

2s−1 2

n+1∂tvi0| + τ |hg, ϕ^{0}θ

2s−1 2

n+1 ∂tvi0|
+ τ |hθ_{n+1}^{1−2s}ν · ∇S^{n}θ

2s−1 2

n+1 v, ϕ^{00}θ

2s−1 2

n+1 vi_{0}|

(4.8)

for some positive constant C, which is independent of τ.

Step 4: Estimate the second derivatives from the symmetric part kSvk^{2}.
Let c1 be a positive constant to be determined later. In view of ϕ^{0} ≥ 1 in supp(v), we
obtain

kSvk^{2} ≥ c1τ^{−1}k(ϕ^{0})^{−}^{1}^{2}Svk^{2} ≥ c1τ^{−1}k(ϕ^{0})^{−}^{1}^{2}S^{0}vk^{2}− Cτ^{−1}k(ϕ^{0})^{−}^{1}^{2}vk^{2},
where S^{0} := ∂_{t}^{2}+ ˜∆S^{n}+ τ^{2}(ϕ^{0})^{2}. It is clear that

k(ϕ^{0})^{−}^{1}^{2}S^{0}vk^{2} = k(ϕ^{0})^{−}^{1}^{2}∂_{t}^{2}vk^{2}+ k(ϕ^{0})^{−}^{1}^{2}∆˜S^{n}vk^{2}+ τ^{4}k(ϕ^{0})^{3}^{2}vk^{2}

+ 2h(ϕ^{0})^{−1}∂_{t}^{2}v, ˜∆S^{n}vi + 2τ^{2}h∂_{t}^{2}v, ϕ^{0}vi + 2τ^{2}h ˜∆S^{n}v, ϕ^{0}vi.

Repeating the integration by parts gives

h∂_{t}^{2}v, ϕ^{0}vi = −h∂tv, ϕ^{00}vi − k(ϕ^{0})^{1}^{2}∂tvk^{2},
h ˜∆S^{n}v, ϕ^{0}vi = −hθ_{n+1}^{1−2s}ν · ∇S^{n}θ

2s−1 2

n+1v, ϕ^{0}θ

2s−1 2

n+1 vi0− k(ϕ^{0})^{1}^{2}θ

1−2s 2

n+1 ∇S^{n}θ

2s−1 2

n+1 vk^{2},

and

h(ϕ^{0})^{−1}∂_{t}^{2}v, ˜∆S^{n}vi = − hθ^{1−2s}_{n+1}ν · ∇S^{n}θ

2s−1 2

n+1 v, (ϕ^{0})^{−1}θ

2s−1 2

n+1 ∂_{t}^{2}vi0

− 1

2kϕ^{−1}(ϕ^{0})^{1}^{2}θ

1−2s 2

n+1 ∇S^{n}θ

2s−1 2

n+1 vk^{2}
+ k(ϕ^{0})^{−}^{1}^{2}θ

1−2s 2

n+1 ∇S^{n}θ

2s−1 2

n+1∂_{t}vk^{2}.
Therefore, we can derive

kSvk^{2} ≥ c1τ^{−1}k(ϕ^{0})^{−}^{1}^{2}Svk^{2}

≥ c_{1}τ^{−1}k(ϕ^{0})^{−}^{1}^{2}∂_{t}^{2}vk^{2}+ 2c_{1}τ^{−1}k(ϕ^{0})^{−}^{1}^{2}θ

1−2s 2

n+1 ∇S^{n}θ

2s−1 2

n+1 ∂_{t}vk^{2}
+ c_{1}τ^{3}k(ϕ^{0})^{3}^{2}vk^{2}− c_{1}τ^{−1}kϕ^{−1}(ϕ^{0})^{1}^{2}θ

1−2s 2

n+1 ∇_{S}^{n}θ

2s−1 2

n+1 vk^{2}

− c_{1}τ h∂_{t}v, ϕ^{00}vi − c_{1}τ k(ϕ^{0})^{1}^{2}∂_{t}vk^{2}

− 2c_{1}τ^{−1}hθ_{n+1}^{1−2s}ν · ∇S^{n}θ

2s−1 2

n+1v, (ϕ^{0})^{−1}θ

2s−1 2

n+1 ∂_{t}^{2}vi_{0}

− 2c_{1}τ hθ_{n+1}^{1−2s}ν · ∇S^{n}θ

2s−1 2

n+1v, ϕ^{0}θ

2s−1 2

n+1 vi_{0}.
(4.9)

Putting this inequality and (4.8) together and choosing suciently small c1 > 0, we have

τ^{3}kϕ^{0}(ϕ^{00})^{1}^{2}vk^{2}+ τ k(ϕ^{00})^{1}^{2}∂_{t}vk^{2}+ τ k(ϕ^{0})^{1}^{2}θ

1−2s 2

n+1 ∇S^{n}θ

2s−1 2

n+1 vk^{2}
+ τ^{−1}k(ϕ^{0})^{−}^{1}^{2}∂_{t}^{2}vk^{2} + τ^{−1}k(ϕ^{0})^{−}^{1}^{2}θ

1−2s 2

n+1 ∇S^{n}θ

2s−1 2

n+1∂tvk^{2}

≤ C

kL_{ϕ}vk^{2} + τ |hI, ϕ^{0}θ

2s−1 2

n+1∂_{t}vi_{0}| + τ |hg, ϕ^{0}θ

2s−1 2

n+1 ∂_{t}vi_{0}|
+ τ |hθ_{n+1}^{1−2s}ν · ∇S^{n}θ

2s−1 2

n+1v, ϕ^{00}θ

2s−1 2

n+1vi_{0}|
+ τ^{−1}|hθ_{n+1}^{1−2s}ν · ∇_{S}^{n}θ

2s−1 2

n+1v, (ϕ^{0})^{−1}θ

2s−1 2

n+1 ∂_{t}^{2}vi_{0}|

(4.10) .

Step 5.1: Estimate the term τ|hI, ϕ^{0}θ

2s−1 2

n+1∂tvi0| + τ |hg, ϕ^{0}θ

2s−1 2

n+1∂tvi0|. Write ˜U :=

V −˜ ^{n−2s}_{2} W˜. Performing integration by parts leads to

|τ h ˜U θ

2s−1 2

n+1 v, ϕ^{0}θ

2s−1 2

n+1∂_{t}vi_{0}|

= τ

− 1 2h ˜U θ

2s−1 2

n+1 v, ϕ^{00}θ

2s−1 2

n+1 vi_{0}− 1
2h∂_{t}U θ˜

2s−1 2

n+1 v, ϕ^{0}θ

2s−1 2

n+1 vi_{0}

≤ τ

2k| ˜U |^{1}^{2}|ϕ^{00}|^{1}^{2}θ

2s−1 2

n+1vk^{2}_{0}+ τ

2k|∂_{t}U |˜ ^{1}^{2}|ϕ^{0}|^{1}^{2}θ

2s−1 2

n+1 vk^{2}_{0}

≤ C

τ k| ˜V |^{1}^{2}|ϕ^{00}|^{1}^{2}θ

2s−1 2

n+1vk^{2}_{0}+ τ k|∂_{t}V |˜ ^{1}^{2}|ϕ^{0}|^{1}^{2}θ

2s−1 2

n+1vk^{2}_{0}
+ τ k| ˜W |^{1}^{2}|ϕ^{00}|^{1}^{2}θ

2s−1 2

n+1vk^{2}_{0}+ τ k|∂_{t}W |˜ ^{1}^{2}|ϕ^{0}|^{1}^{2}θ

2s−1 2

n+1vk^{2}_{0}