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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 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 Rn,

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|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] 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

Rn

e|x|α|u|2dx < ∞,

2020 Mathematics Subject Classication. 35R11; 35B40.

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

1

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then u ≡ 0. On the other hand, for a non-dierentiable potential q, if s ∈ (1/4, 1), kqkL(Rn) ≤ 1, and u satises 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 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 ∈ (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-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−14s , 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 ∈ 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 Caarelli-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 coecient 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 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 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 denitions 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 dene 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)˜

= r2sb(rx0)rx0· ∇0u(rx) + q(rx)u(rx)

(By (2.2))

= r2sb(rx0)x0 · ∇0u(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 suciently 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 Caarelli-Silvestre extension ˜u satises 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 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 ∈ (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 dene 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 suciently 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 suciently 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



參考文獻

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