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# By localizing the equation with the help of Caarelli-Silvestre extension, we solve the problem by a delicate Carleman inequality in the (n + 1)-dimensional half space

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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 innity. 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 Caarelli-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 innity for the fractional Schrödinger equation with drift term

(1.1) ((−∆)s+ b(x)x · ∇ + q(x))u = 0 in Rn,

where s ∈ (0, 1) and b, q are scalar-valued functions. Precisely, we are interested in investigating the decay rate of u at innity 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 satises |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|43) such that u is a solution of the Schrödinger equation with potential q. He also showed that if |u(x)| ≤ C exp(−C|x|43+), then u ≡ 0.

In view of Meshkov's counterexample, Kenig [Ke06] rened the Landis conjecture and asked whether this conjecture is true for real-valued potentials and solutions. This real version Landis conjecture was conrmed 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 dierentiable and satises

|x · ∇q(x)| ≤ 1,

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

Rn

e|x|α|u|2dx < ∞,

2020 Mathematics Subject Classication. 35R11; 35B40.

Key words and phrases. Fractional Laplacian, Landis-type conjecture, Caarelli-Silvestre extension.

1

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then u ≡ 0. On the other hand, for a non-dierentiable potential q, if s ∈ (1/4, 1), kqkL(Rn) ≤ 1, and u satises the decay behavior: ∃ α > 4s−14s such that

Z

Rn

e|x|α|u|2dx < ∞, 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(Rn) for some r ∈ R and u = (−∆)su = 0 in some open set W ⊂ Rn, 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 satises 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 dierentiable, then any non-trivial solution of the fractional Schrödinger equation does not decay super-exponentially at innity. The detailed statement is described in the following theorem.

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

(s ∈ (12, 1) when n = 1 or 2, s ∈ (34, 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 ∈ Rn and the radial derivatives of q, b satisfy

(1.4) |x · ∇q(x)| ≤ λ and |x · ∇b(x)| ≤ λ|x|−β for all x ∈ Rn for some β > 1. If u ∈ Hs(Rn) is a solution to (1.1) such that

(1.5) Z

Rn

e|x|β



|u(x)|2+ |∇u(x)|2



dx ≤ λ, then u ≡ 0.

For non-dierentiable 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−14s , and both b, q satisfy (1.3). Here we assume the radial derivative of b satises

|x · ∇b(x)| ≤ λ|x|−β for all x ∈ Rn. If u ∈ Hs(Rn) 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 Caarelli-Silvestre extension [CS07] and derive the Carleman estimate for a degenerate elliptic equation in the (n + 1)-dimensional upper half-space Rn+1+ .

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Our strategy in proving both theorems is similar to that of [RW19]. We will mainly handle the CS extended solution ˜u in Rn+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 Rn+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 coecient in (1.1). Such choice is due to the limitation in the Carleman estimate for ˜u in Rn+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 Rd for s ∈ (0, 1), see [Jak08, Sect. 2]. Precisely, let Xt be the 2s-stable Ornstein-Uhlenbeck process in Rd, given in the following stochastic integral:

Xt= e−λtX0+ Z t

0

e−λ(t−s)d ˆXs, Xˆ0 = 0,

where the integral is in the Stieltjes sense, and ( ˆXt, Px) is the isotropic 2s-stable Lévy process in Rd with index of stability s ∈ (0, 1) and characteristic function

E0ei ˆXt·ξ = e−t|ξ|2s for all ξ ∈ Rd and for all t ≥ 0.

Here, Ex denotes the expectation with respect to the distribution Px of the process starting from x ∈ Rd. Indeed, the innitesimal genrator of Xt is equal to (1.6).

This paper is organized as follows. In Section 2, we introduce some notations and state the Caarelli-Silvestre extension. In Section 3, we discuss how the boundary decay of u implies the bulk decay of ˜u in Rn+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 denitions and notations in [RW19]. We now restate what we need. Let Rn+1+ := Rn × R+ =  x = (x0, xn+1) x0 ∈ Rn, xn+1 > 0

and x0 = (x0, 0) ∈ Rn× {0}. For r, R > 0, we denote

Br+(x0) := x ∈ Rn+1+ |x − x0| ≤ r , Br0(x0) := x ∈ Rn× {0} |x − x0| ≤ r ,

B+r := Br+(0), Br0 := Br0(0), A+r,R := x ∈ Rn+1+ r ≤ |x| ≤ R , A0r,R := x ∈ Rn× {0} r ≤ |x| ≤ R .

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For Ω ⊂ Rn+1+ , we dene H˙1(Ω, x1−2sn+1) :=



v : Ω → R Z

x1−2sn+1|∇v|2dx < ∞

 , H1(Ω, x1−2sn+1) :=



v : Ω → R Z

x1−2sn+1(|v|2+ |∇v|2) dx < ∞

 .

Write x = rθ, where r > 0 and θ ∈ S+n, i.e., θ = (θ0, θn+1) ∈ Sn with θn+1 > 0. We also denote

H1(S+n, θn+11−2s) :=



v : S+n → R Z

S+n

θ1−2sn+1 (|v|2+ |∇Snv|2) dθ < ∞

 , where ∇Sn = (Θ1, · · · , Θn+1). Here, Θk are vector elds on Sn.

For s ∈ (12, 1)and u ∈ Hs(Rn), let ˜u ∈ ˙H1(Rn+1+ , x1−2sn+1)be a solution to the degenerate elliptic equation

(2.1)

(∇ · x1−2sn+1∇˜u = 0 in Rn+1+ ,

˜

u = u on Rn× {0},

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

−(−∆)su(x) = cn,s lim

xn+1→0x1−2sn+1n+1u(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)





∇ · x1−2sn+1∇˜u = 0 in Rn+1+ ,

˜

u = u on Rn× {0},

cn,s lim

xn+1→0x1−2sn+1n+1u = bx˜ 0· ∇0u + qu on Rn× {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 Rn+1+ .

Proposition 3.1. Let s be given in (1.2) and ˜u ∈ H1(Rn+1+ , x1−2sn+1) be a solution to (2.2). Assume that there exists a constant λ > 0 such that

(3.1) |q(x0)| ≤ λ and |b(x0)||x0| ≤ λ for all x0 ∈ Rn.

If there exist constants C, β > 1 such that (1.5) holds, then there exist constants C1, c1 >

0 such that

(3.2) |˜u(x)| ≤ C1e−c1|x|β for all x ∈ Rn+1+ . The following lemma can be found in [RW19, Equ.(19)].

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Lemma 3.2. Let s ∈ (12, 1) and ˜u ∈ H1(Rn+1+ , x1−2sn+1) be a solution to (2.1). If x0 ∈ Rn× {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˜ L2(B+cr(x0))

≤ C

 kx

1−2s 2

n+1 uk˜ L2(B16r+ (x0))+ r1−skukL2(B0

16r(x0))

α

×

 rs+1

xn+1lim→0x1−2sn+1n+1

L2(B016r(x0))

+ r1−skukL2(B0

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 ∈ H1(Rn+1+ , x1−2sn+1) be a solution to (2.2).

If q ∈ L(Rn) and b(x0)x0 ∈ (L(Rn))n, then there exist α = α(n, s) ∈ (0, 1) and c = c(n, s) ∈ (0, 1) such that

k˜ukL(B+cr 4)

≤ Crn2



rs−1kxn+11−2s2 uk˜ L2(B+16r)+ kukL2(B0

16r)

α

×



r2skbx0 · ∇0u + qukL2(B0

16r)+ kukL2(B0

16r)

1−α

+ r2skbx0· ∇0u + qukL2(B0

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 Rn+1+ and v(x0) = u(rx0)on Rn× {0}. Note that

cn,s lim

xn+1→0x1−2sn+1n+1v(x) = c˜ n,s lim

xn+1→0x1−2sn+1n+1 ˜u(rx)

= rcn,s lim

xn+1→0x1−2sn+1n+1u(rx)˜

= rcn,s lim

rxn+1→0r−1+2s(rxn+1)1−2sn+1u(rx)˜

= r2scn,s lim

rxn+1→0(rxn+1)1−2sn+1u(rx)˜

= r2sb(rx0)rx0· ∇0u(rx) + q(rx)u(rx)

(By (2.2))

= r2sb(rx0)x0 · ∇0u(rx) + r2sq(rx)u(rx)

= r2sb(rx0)x0 · ∇0v(x) + r2sq(rx)v(x), that is,

(3.5)





∇ · x1−2sn+1∇˜v = 0 in Rn+1+ ,

˜

v = v on Rn× {0},

cn,s lim

xn+1→0x1−2sn+1n+1v = b˜ rx0 · ∇0v + qrv on Rn× {0},

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where br(x0) := r2sb(rx0) and qr(x0) := r2sq(rx). Applying Lemma A.2 to (3.5) with p = 2 (since (1.2)) a1 = 0 and a2 = brx0· ∇0v + qrv, we obtain that

(3.6) k˜vkL(B1/4+ )≤ C



kxn+11−2s2 ˜vkL2(B1+)+ kbrx0 · ∇0v + qrvkL2(B0

1)

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

k˜vkL(B+1/4) = sup

x∈Rn+1+ ,|x|≤14

|˜v(x)| = sup

x∈Rn+1+ ,|rx|≤r4

|˜u(rx)| = k˜ukL(Br/4+ ). On the other hand, we can derive

kx

1−2s 2

n+1vk˜ 2L2(B+1) = Z

x∈Rn+1+ ,|x|≤1

x1−2sn+1|˜v(x)|2dx = r2s−2−nkx

1−2s 2

n+1uk˜ 2L2(B+r)

and

kbrx0 · ∇0v + qrvk2L2(B10)= Z

|x0|<1

|br(x0)x0· ∇0v(x0) + qr(x0)v(x0)|2dx0

= r4s−nkbx0 · ∇0u + quk2L2(Br0). Thus, (3.6) implies

k˜ukL(B+r/4) ≤ Crn2



rs−1kx

1−2s 2

n+1uk˜ L2(Br+)+ r2skbx0· ∇0u + qukL2(B0r)

 .

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˜ukL(Bcr/4+ ) ≤ Crn2



rs−1kxn+11−2s2 uk˜ L2(B+cr)+ r2skbx0· ∇0u + qukL2(B0

cr)

 .

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

cn,s lim

xn+1→0x1−2sn+1n+1u = bx˜ 0· ∇0u + 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 suciently large and x0 ∈ Rn× {0}with |x0| = 32R. From (1.5), it follows that

λ ≥ Z

B16R0 (x0)

e|x0|β



|u(x0)|2+ |∇0u(x0)|2

 dx0

≥ e(16R)β Z

B16R0 (x0)



|u(x0)|2+ |∇0u(x0)|2

 dx0

= e(16R)β



kuk2L2(B16R0 (x0))+ k∇0uk2L2(B16R0 (x0))

 , that is,

(3.7) kukL2(B0

16R(x0))+ k∇0ukL2(B0

16R(x0)) ≤ ˜Ce−˜cRβ

for some constants ˜C and ˜c. Note that the Caarelli-Silvestre extension ˜u satises kx

1−2s 2

n+1∇˜ukL2(Rn+1+ ) ≤ C and k˜u(•, xn+1)kL2(Rn) ≤ kukL2(Rn)

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

kxn+11−2s2 uk˜ 2L2(B+

16R(x0))≤ Z 16R

0

x1−2sn+1k˜u(•, xn+1)k2L2(Rn)dxn+1

≤ Z 16R

0

x1−2sn+1kuk2L2(Rn)dxn+1

= (16R)2−2s

2 − 2s kuk2L2(Rn)

≤ CR2−2s.

Plugging cn,slimxn+1→0x1−2sn+1n+1u = bx˜ 0· ∇0u + qu into (3.3) (with r = R) and using (3.1) yields

kx

1−2s 2

n+1uk˜ L2(B+cR(x0)) ≤ ˜Ce−˜cRβ.

Next, choosing r = cR16 in (3.4) and putting together (3.1) and (1.5), we have k˜ukL(B+

c2R 64

(x0)) ≤ ˜Ce−˜cRβ for |x0| = 32R, which implies

k˜ukL(B+

c ˜R(x0))≤ ˜Ce−˜c ˜Rβ 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 dierentiable and non-dierentiable potentials.

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

Theorem 4.1. Let s ∈ (12, 1) and let ˜u ∈ H1(Rn+1+ , x1−2sn+1) with supp(˜u) ⊂ Rn+1+ \ B1+ be a solution to





∇ · x1−2sn+1∇˜u = f in Rn+1+ ,

˜

u = u on Rn× {0},

xn+1lim→0x1−2sn+1n+1u = W x˜ 0 · ∇0u + V u + g on Rn× {0},

where f ∈ L2(Rn+1+ , x2s−1n+1) is compactly supported in Rn+1+ , g ∈ L2(Rn) is compactly supported in Rn, and x0 · ∇0W, x0· ∇0V exist. Let α > 1 be a real constant and dene φ(x) := |x|α. If

W (x0) ≥ 0 for all x0 ∈ Rn\ B10.

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Then there exists a real number τ0 > 1 such that τ3keτ φ|x|2 −1x

1−2s 2

n+1uk˜ 2L2

(Rn+1+ )+ τ keτ φ|x|α2x

1−2s 2

n+1∇˜uk2L2

(Rn+1+ )

+ τ−1keτ φ|x|α2x

1−2s 2

n+1∇(x · ∇˜u)k2L2

(Rn+1+ )

≤ C



keτ φ|x|x

2s−1 2

n+1f k2L2(Rn+1+ )

+ τ keτ φ|x|α2|V |12uk2L2(Rn×{0})+ τ keτ φ|x|α2|x0 · ∇0V |12uk2L2(Rn×{0})

+ τ2keτ φ|x|α|W |12uk˜ 2L2(Rn×{0})+ τ2keτ φ|x|α|x0· ∇0W |12uk2L2(Rn×{0})

+ τ2keτ φ|x|α2uk2L2(Rn×{0})

+ τ−1keτ φ|x|α2|V |12(x0· ∇0u)k2L2(Rn×{0})

+ τ−1keτ φ|x|α2|x0 · ∇0V |12(x0 · ∇0u)k2L2(Rn×{0})

+ keτ φ|W |12(x0· ∇0u)k2L2(Rn×{0})+ keτ φ|x0· ∇0W |12(x0· ∇0u)k2L2(Rn×{0})

+ keτ φ|x|α2(x0· ∇0u)k2L2(Rn×{0})

+ τ2keτ φ|x|32α|x|1+2s2 gk2L2(Rn×{0})+ keτ φ|x|α2|x|s(x0· ∇0g)k2L2(Rn×{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 = etθ with t ∈ R and θ ∈ S+n. Set u := en−2s2 tu˜, we have

(4.2)











θ1−2sn+12t + ∇Sn· θn+11−2sSn − θ1−2sn+1 (n − 2s)2 4



u = ˜f in S+n× R,

θn+1lim→0θ1−2sn+1ν · ∇Snu =



V −˜ n − 2s 2



u + ˜W ∂tu + ˜g on ∂S+n × R, where ˜f = en+2+2s2 tf, ν = (0, · · · , 0, 1), ˜V = e2stV, ˜W = e2stW, and ˜g = en+2s2 tg.

Next, by setting ˜∆Sn := θ

2s−1 2

n+1Sn·θ1−2sn+1Snθ

2s−1 2

n+1, ϕ(t) = φ(etθ) = eαt, v = eτ ϕθ

1−2s 2

n+1 u, f = eτ ϕf˜, and g = eτ ϕg˜, (4.2) can be written as





Lϕv := (S + A)v = θ

2s−1 2

n+1 f in S+n× R,

lim

θn+1→0θ1−2sn+1ν · ∇Snθ

2s−1 2

n+1 v = I + ˜W θ

2s−1 2

n+1tv + g on ∂S+n× R, where

S := ∂t2+ ˜∆Sn −(n − 2s)2

4 + τ20|2, A := −2τ ϕ0t− τ ϕ00,

I :=



V −˜ n − 2s 2

 θ

2s−1 2

n+1v − τ ϕ0W θ˜

2s−1 2

n+1 v.

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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 • kL2(Sn

+×R), h•, •i := h•, •iL2(Sn

+×R)

k • k0 := k • kL2(∂Sn

+×R), h•, •i0 := h•, •iL2(∂Sn

+×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

kLϕvk2 = kSvk2 + kAvk2+ h[S, A]v, vi + 4τ hθ1−2sn+1ν · ∇Snθ

2s−1 2

n+1 v, ϕ0θ

2s−1 2

n+1tvi0 (4.4)

+ 2τ hθ1−2sn+1ν · ∇Snθ

2s−1 2

n+1 v, ϕ00θ

2s−1 2

n+1 vi0. From ˜W ≥ 0, it follows that h ˜W θ

2s−1 2

n+1tv, ϕ0θ

2s−1 2

n+1tvi0 ≥ 0 and kLϕvk2 ≥ kSvk2+ kAvk2+ h[S, A]v, vi

+ 4τ hI, ϕ0θ

2s−1 2

n+1tvi0+ 4τ hg, ϕ0θ

2s−1 2

n+1tvi0 (4.5)

+ 2τ hθn+11−2sν · ∇Snθ

2s−1 2

n+1v, ϕ00θ

2s−1 2

n+1vi0.

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

[S, A]v = 4τ30|2ϕ00v − 4τ ϕ00t2v − 4τ ϕ000tv − τ ϕ0000v.

Since supp(˜u) ⊂ Rn+1+ \ B+1, we can see that supp(v) ⊂  (θ, t) ∈ S+n× R t ≥ 0 Recall that ϕ(t) = eαt for α > 1. We thus have |ϕ0|2ϕ00 ≥ ϕ0000 in supp(v). By this. inequality, we can estimate

h[S, A]v, vi ≥ 2τ3000)12vk2+ 4τ k(ϕ00)12tvk2. Combining this inequality with (4.5) yields

kLϕvk2 ≥ kSvk2 + kAvk2+ 2τ3000)12vk2+ 4τ k(ϕ00)12tvk2 + 4τ hI, ϕ0θ

2s−1 2

n+1tvi0 + 4τ hg, ϕ0θ

2s−1 2

n+1tvi0 + 2τ hθn+11−2sν · ∇Snθ

2s−1 2

n+1 v, ϕ00θ

2s−1 2

n+1 vi0. (4.6)

Step 3: Derive the full gradient inequality from the symmetric part kSvk2. Let c0 be a positive constant to be determined later. Note that

− ˜∆Sn = −S + ∂t2+ τ20)2− (n − 2s)2

4 .

(10)

Hence, we have

c0τ k(ϕ0)12θ

1−2s 2

n+1Snθ

2s−1 2

n+1 vk2+ c0τ hθ1−2sn+1 ν · ∇Snθ

2s−1 2

n+1 v, ϕ0θ

2s−1 2

n+1 vi0

= −c0τ hϕ0v, ˜∆Snvi

= c0τ



− hϕ0v, Svi − hϕ00v, ∂tvi + k(ϕ0)12tvk2+ τ2k(ϕ0)32vk2

− (n − 2s)2

4 k(ϕ0)12vk2



≤ −c0τ hϕ0v, Svi − c0τ hϕ00v, ∂tvi + c0τ k(ϕ0)12tvk2+ 2c0τ3k(ϕ0)32vk2 (4.7)

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

kSvk2+ kAvk2+ 2τ3000)12vk2+ 4τ k(ϕ00)12tvk2 + c0τ k(ϕ0)12θ

1−2s 2

n+1Snθ

2s−1 2

n+1vk2 + 4τ hI, ϕ0θ

2s−1 2

n+1tvi0+ 4τ hg, ϕ0θ

2s−1 2

n+1tvi0 + 2τ hθ1−2sn+1 ν · ∇Snθ

2s−1 2

n+1v, ϕ00θ

2s−1 2

n+1 vi0 + c0τ hθ1−2sn+1 ν · ∇Snθ

2s−1 2

n+1 v, ϕ0θ

2s−1 2

n+1 vi0

≤ kLϕvk2− c0τ hϕ0v, Svi − c0τ hϕ00v, ∂tvi + c0τ k(ϕ0)12tvk2+ 2c0τ3k(ϕ0)32vk2.

Using ϕ00 = αϕ0 and choosing suciently small c0 > 0, we can derive kSvk2+ kAvk2+ τ3000)12vk2

+ τ k(ϕ00)12tvk2+ τ k(ϕ0)12θ

1−2s 2

n+1Snθ

2s−1 2

n+1vk2

≤ C



kLϕvk2+ τ |hI, ϕ0θ

2s−1 2

n+1tvi0| + τ |hg, ϕ0θ

2s−1 2

n+1tvi0| + τ |hθn+11−2sν · ∇Snθ

2s−1 2

n+1 v, ϕ00θ

2s−1 2

n+1 vi0|

 (4.8)

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

Step 4: Estimate the second derivatives from the symmetric part kSvk2. Let c1 be a positive constant to be determined later. In view of ϕ0 ≥ 1 in supp(v), we obtain

kSvk2 ≥ c1τ−1k(ϕ0)12Svk2 ≥ c1τ−1k(ϕ0)12S0vk2− Cτ−1k(ϕ0)12vk2, where S0 := ∂t2+ ˜∆Sn+ τ20)2. It is clear that

k(ϕ0)12S0vk2 = k(ϕ0)12t2vk2+ k(ϕ0)12∆˜Snvk2+ τ4k(ϕ0)32vk2

+ 2h(ϕ0)−1t2v, ˜∆Snvi + 2τ2h∂t2v, ϕ0vi + 2τ2h ˜∆Snv, ϕ0vi.

Repeating the integration by parts gives

h∂t2v, ϕ0vi = −h∂tv, ϕ00vi − k(ϕ0)12tvk2, h ˜∆Snv, ϕ0vi = −hθn+11−2sν · ∇Snθ

2s−1 2

n+1v, ϕ0θ

2s−1 2

n+1 vi0− k(ϕ0)12θ

1−2s 2

n+1Snθ

2s−1 2

n+1 vk2,

(11)

and

h(ϕ0)−1t2v, ˜∆Snvi = − hθ1−2sn+1ν · ∇Snθ

2s−1 2

n+1 v, (ϕ0)−1θ

2s−1 2

n+1t2vi0

− 1

2kϕ−10)12θ

1−2s 2

n+1Snθ

2s−1 2

n+1 vk2 + k(ϕ0)12θ

1−2s 2

n+1Snθ

2s−1 2

n+1tvk2. Therefore, we can derive

kSvk2 ≥ c1τ−1k(ϕ0)12Svk2

≥ c1τ−1k(ϕ0)12t2vk2+ 2c1τ−1k(ϕ0)12θ

1−2s 2

n+1Snθ

2s−1 2

n+1tvk2 + c1τ3k(ϕ0)32vk2− c1τ−1−10)12θ

1−2s 2

n+1Snθ

2s−1 2

n+1 vk2

− c1τ h∂tv, ϕ00vi − c1τ k(ϕ0)12tvk2

− 2c1τ−1n+11−2sν · ∇Snθ

2s−1 2

n+1v, (ϕ0)−1θ

2s−1 2

n+1t2vi0

− 2c1τ hθn+11−2sν · ∇Snθ

2s−1 2

n+1v, ϕ0θ

2s−1 2

n+1 vi0. (4.9)

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

τ3000)12vk2+ τ k(ϕ00)12tvk2+ τ k(ϕ0)12θ

1−2s 2

n+1Snθ

2s−1 2

n+1 vk2 + τ−1k(ϕ0)12t2vk2 + τ−1k(ϕ0)12θ

1−2s 2

n+1Snθ

2s−1 2

n+1tvk2

≤ C



kLϕvk2 + τ |hI, ϕ0θ

2s−1 2

n+1tvi0| + τ |hg, ϕ0θ

2s−1 2

n+1tvi0| + τ |hθn+11−2sν · ∇Snθ

2s−1 2

n+1v, ϕ00θ

2s−1 2

n+1vi0| + τ−1|hθn+11−2sν · ∇Snθ

2s−1 2

n+1v, (ϕ0)−1θ

2s−1 2

n+1t2vi0|

 (4.10) .

Step 5.1: Estimate the term τ|hI, ϕ0θ

2s−1 2

n+1tvi0| + τ |hg, ϕ0θ

2s−1 2

n+1tvi0|. Write ˜U :=

V −˜ n−2s2 W˜. Performing integration by parts leads to

|τ h ˜U θ

2s−1 2

n+1 v, ϕ0θ

2s−1 2

n+1tvi0|

= τ

− 1 2h ˜U θ

2s−1 2

n+1 v, ϕ00θ

2s−1 2

n+1 vi0− 1 2h∂tU θ˜

2s−1 2

n+1 v, ϕ0θ

2s−1 2

n+1 vi0

≤ τ

2k| ˜U |1200|12θ

2s−1 2

n+1vk20+ τ

2k|∂tU |˜ 120|12θ

2s−1 2

n+1 vk20

≤ C



τ k| ˜V |1200|12θ

2s−1 2

n+1vk20+ τ k|∂tV |˜ 120|12θ

2s−1 2

n+1vk20 + τ k| ˜W |1200|12θ

2s−1 2

n+1vk20+ τ k|∂tW |˜ 120|12θ

2s−1 2

n+1vk20



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