On Landis Conjecture for the Fractional Schr\"{o}dinger Equation

In this paper, we study a Landis-type conjecture for the general fractional Schr\"{o}dinger equation $((-P)^{s}+q)u=0$. As a byproduct, we also proved the additivity and boundedness of the linear operator $(-P)^{s}$ for non-smooth coefficents. For differentiable potentials $q$, if a solution decays at a rate $\exp(-|x|^{1+})$, then the solution vanishes identically. For non-differentiable potentials $q$, if a solution decays at a rate $\exp(-|x|^{\frac{4s}{4s-1}+})$, then the solution must again be trivial. The proof relies on delicate Carleman estimates. This study is an extension of the work by R\"{u}land-Wang (2019).


Introduction
In this work, we study a Landis-type conjecture for the fractional Schrödinger equation where {E λ } is the spectral resolution of −P (each {E λ } is a projection in L 2 (R n )) and {e tP } t≥0 is the heat-diffusion semigroup generated by −P , see e.g. [GLX17,ST10].
The Landis conjecture was proposed by E.M. Landis in the 60's [KL88]. He conjectured the following statement: Let |q(x)| ≤ 1 and let u be a solution to (1.1) with P = ∆ and s = 1. If |u(x)| ≤ C 0 and |u(x)| ≤ exp(−C|x| 1+ ), then u ≡ 0. However, this statement is false: In [Mes92], Meshkov constructed a (complex-valued) potential q and a (complex-valued) nontrivial u with |u(x)| ≤ C exp(−C|x| author's knowledge, the real-version of Landis conjecture is still open for n ≥ 3. Here we also refer some related works [Dav20,DKW17,DKW20,DW20,KSW15]. In [RW19], Rüland and Wang consider the Landis conjecture of the fractional Schrödinger equation (1.1) with P = ∆ and 0 < s < 1. For the case when s = 1/2, in [Cas20], we remark that Cassano proved the Landis conjecture for the Dirac equation. In some sense, the Dirac operator is the square root of the Laplacian operator, that is, the phenomena are similar when s = 1/2.
Assume that a jk = a kj ∈ C 0,1 (R n ) for all 1 ≤ j, k ≤ n, and satisfy In this paper, we prove the following Landis-type conjecture for the fractional Schrödinger equations.
We also have the following result for non-differentiable potential q.
Remark 1.3. When s = 1 2 , Theorem 1.1 and Theorem 1.2 still hold without (1.5). Remark 1.4. We prove Theorem 1.2 using the splitting arguments in [RW19]. Therefore, due to the sub-ellipticity nature, we the same restriction s ∈ ( 1 4 , 1). We also see that, as s → 1, the exponent 4s 4s−1 in Theorem 1.2 tends to 4 3 , which is the optimal exponent for the classical Schrödinger equation.
Remark 1.5. The condition (1.4) allows small perturbations of Laplacian only, which works as a sufficient condition in deriving Carleman estimate. In [GFR19], they also imposed similar assumption to prove the strong unique continuation property for (1.1). In contrast to the works [DKW17,Ros21], which studied the real-version of Landis conjecture, such condition is not needed, since their proofs did not involve any Carleman estimate.
1.2. Main ideas. The main method of proving Theorem 1.1 and 1.2 is Carleman estimates. However, due to the non-locality of (−P ) s , the techniques here are much complicated than those for the classical case, i.e., s = 1. One of the major tricks is to localize (−P ) s , which is motivated by Caffarelli-Silvestre's fundamental work [CS07]. Here we will use the Caffarelli-Silvestre type extension of (−P ) s proved in [ST10,Sti10]. After localizing (−P ) s , we will derive a Carleman estimate on R n+1 + mimicking the one proved in [RS20]. This Carleman estimate enables passing of the boundary decay to the bulk decay.
However, extension of these properties to (−P ) s is not trivial. We establish the additivity property of (−P ) s by introducing the Balakrishnan definition of (−P ) s , which is equivalent to (1.2), see e.g. [MCSA01] or [Yos80,Section IX.11]. The continuity of (−P ) s : H 2s (R n ) → L 2 (R n ) can be also obtained by the Balakrishnan operator, as well as the interpolation of the single operator −P . Here, we shall not interpolate on the family of the operator (−P ) s , see also [GM14] for the interpolation theory of the analytic familiy of multilinear operators.
Remark 1.6. In [See67], R.T. Seeley showed that the operator (−P ) s is a pseudo-differential operator of order 2s if a jk are smooth. In this case, we can apply the theory of pseudo-differential operator, see e.g. [Tay74]. As a byproduct, we loosened the smoothness hypothesis that required by theories of the pseudo-differential operator. Moreover, the boundary value theories for the fractional Laplacian have been elaborated in recent years, see e.g. [Gru14,Gru15,Gru16a,Gru16b,Gru20]. In [Gru20], Grubb calculated the first few terms in the symbol of (−P ) s . 1
The proof of Lemma 2.1 is same as in [ST10,Sti10]. The following estimate also holds true: Therefore, by arbitrariness of v ∈ H s (R n ), we conclude the following lemma: Lemma 2.2. Let 0 < s < 1 and a jk given as in Lemma 2.1.
Note that Since a jk is uniformly Lipschitz, then We here also remark that dom (−P ) = H 2 (R n ) is the maximal extension such that −P is selfadjoint and densely defined in L 2 (R n ), see [GLX17, equation (2.8)]. Given any φ ∈ C ∞ c (R n ), we see that [LM72,Remark 7.4]), then we know that We shall prove the followings: Lemma 2.3. Let 0 < s < 1 and a jk given as in Lemma 2.1. We have the inequality Moreover, we have Remark 2.4. Using the duality argument as in (2.7), we know that (2.8) and (2.9) are equivalent.
In order to prove Lemma 2.3, we introduce the Balakrishnan operator as in [MCSA01, Definition 3.1.1 and Definition 5.1.1].

Boundary Decay Implies Bulk Decay
Firstly, we translate the decay behavior on R n to decay behavior which is also holds on R n+1 + .
3.1. Proof of the part (a) of Lemma 3.3 for the case s ∈ [1/2, 1). We first prove the following extension of the Carleman estimate in [RS20, Proposition 5.7].
Lemma 3.4. Let s ∈ [ 1 2 , 1) and let w ∈ H 1 (R n+1 Suppose that We assume that max 1≤j,k≤n for some sufficiently small ǫ > 0. For s = 1 2 , we further assume max 1≤j,k≤n (∇ ′ ) 2 a jk ∞ ≤ C for some positive constant C. Assume additionally that Then there exist τ 0 > 1 and a constant C such that for all τ ≥ τ 0 .
Proof. Now we prove the Carleman estimate for s ∈ ( 1 2 , 1), as the case s = 1 2 is naturally included in our estimates.
Step 2: Estimating the bulk contributions.
Step 2.1.1: Computation the principal term. Note that The following identity can be found in [RS20, equation (5.20) of Proposition 5.7]: For our purpose, we need to refine the estimate [RS20, equation ( Combining (3.1), (3.2) and (3.3), we reach Step 2.1.2: Estimating the remainder. Using integration by parts, we can estimate R from below: Here we would like to highlight some features when estimating the second term of R, that is, So, summing up (3.5) and (3.7), we note that the problematic term τ ∂ 2 n+1 φ∂ j u, (a jk − δ jk )∂ k u is canceled. It problematic because ∂ 2 n+1 φ has singularity x −2s n+1 for s ∈ (1/2, 1). However, when s = 1 2 , ∂ 2 n+1 φ has no singularity. In this case, we consider (3.6) rather than (3.7). This is the reason why we can loosen the second derivative assumption for the case s = 1 2 .
We observe that Using (3.16), we reach Finally, we also have Step 4: Conclusion. Put them together, we reach which is our desired result.
As in [RS20], we introduce the following sets for s ∈ [ 1 2 , 1): With this notation, we infer the following analogous to [RS20, Proposition 5.10]: for some sufficiently small ǫ > 0. For s = 1 2 , we further assume max 1≤j,k≤n Proof. We may assume that x for some sufficiently large constant c 0 > 0. Otherwise the result is trivial. Let η is a smooth cut-off function satisfies Since η and ∇η are bounded, together with |∂ n+1 η| ≤ Cx n+1 , we know that So, by the Carleman estimate in Lemma 3.4, there exists τ 0 > 1 such that for all τ ≥ τ 0 . Then, for large τ 0 , the last term of f was absorbed by the gradient term in the left-hand-side, so we have Hence, Dividing above equation by τ , since τ ≥ 1 and applying Caccioppoli's inequality (Lemma A.6), we obtain in C + s,1/8 , and also since s ≥ 1 2 , for large c 0 , where α ∈ (0, 1) will be chosen later. Note that Finally, choosing α ∈ (0, 1) satisfies α = φ − φ + −φ − (1 − α) will implies our desired result. For our purpose, we only need the following simplified version of the Lemma above: for some sufficiently small ǫ > 0. For s = 1 2 , we further assume max 1≤j,k≤n (∇ ′ ) 2 a jk ∞ ≤ C for some positive constant C. Then there exist α = α(n, s) ∈ (0, 1), c = c(n, s) ∈ (0, 1) and a constant C such that . Now we are ready to prove the part (a) of Lemma 3.3 for the case when s ∈ [ 1 2 , 1).
Proof of the part (a) of Lemma 3.3 for s ∈ [ 1 2 , 1). In order to invoke the estimation from Corollary 3.6, we split our solution u into two partsũ = u 1 + u 2 , where u 1 := E s (ζu) satisfies Since u 2 = 0 on B ′ 8 , by Corollary 3.6, there exist α = α(n, s) ∈ (0, 1), c = c(n, s) ∈ (0, 1) and a constant C such that Let η be a smooth, radial cut-off function with η = 1 in B + 2 and η = 0 outside B + 4 . Plug w = ηx 1−2s n+1 ∂ n+1 u 2 into the trace characterization lemma (Lemma A.5), we reach lim We first control the boundary term of (3.19). Since η is a bounded multiplier on H 2s (R n ), using duality, we have . Applying the Caccioppoli's inequality in Lemma A.6, with zero Dirichlet condition and zero inhomogeneous terms, we have , where the last inequality follows by the boundedness assumptions of a jk . Observe that Applying Caccioppoli's inequality in Lemma A.6 on ∇ ′ u 2 with zero Dirichlet condition and f j = n k=1 ∇ ′ a jk ∂ k u 2 , since ∇ ′ a jk ∞ ≤ ǫ, we have (3.22) x where the second inequality follows by (3.21). Hence, we reach . Plugging this into (3.18) leads to where the second inequality follows by Lemma 2.3.
Let v and f as in (3.28). Letṽ be the Caffarelli-Silvestre type extension of ηf as in (2.1), where η is a cut-off function satisfies with |∂ n+1 η| ≤ Cx n+1 . As consequences, the function v := v −ṽ is the Caffarelli-Silvestre extension of (1 − η)f and solves Hence, by Lemma 3.7 and since s = 1 − s, we have and thus Using where the last inequality follows by Lemma 2.2. Thus, Firstly, we estimate the right hand side of (3.30) by where the second inequality follows by (2.4) and the last one is followed by the Caccioppoli's inequality in Lemma A.6. Similarly, we can estimate the left hand side of (3.30) by , where the last inequality is followed by Poincaré inequality. Thus, (3.30) becomes Next, we estimate the boundary contribution ηf H 1−s (R n ×{0}) . Using the interpolation inequality in Lemma A.4, we have . Using (3.21) and (3.22), we know that . Choosing µ > 0 in (3.32) such that the right contributions become equal, i.e. .
For our purpose, we only need the following version of inequality: Corollary 3.9. Let s ∈ (0, 1/2) and let x 0 ∈ R n × {0}. Suppose with w = 0 on C ′ s,2 . We assume that max 1≤j,k≤n for some sufficiently small ǫ > 0. We further assume max 1≤j,k≤n for some positive constant C. Then there exist C = C(n, s), c = c(n, s) and α = α(n, s) ∈ (0, 1) such that Now, we are ready to proof the part (a) of Lemma 3.3 for the case s ∈ (0, 1/2).
Proof of the part (a) of Lemma 3.3 for s ∈ (0, 1 2 ). The case s ∈ (0, 1/2) is similar as the case s ∈ (1/2, 1). As above, the estimation for u 1 is a direct result of (2.4). For u 2 , we use Corollary 3.9 and the interpolation inequality in Lemma A.5. With this estimation, the analogues of (3.26) and (3.27) are followed by combining the estimates in splitting argument as above. Note that (3.27) becomes which is our desired result.
Finally, combining (3.35) and Lemma A.7, we can immediately obtain the part (b) of Lemma 3.3.

A Carleman estimate with differentiability assumption.
Modifying the arguments in [Reg97], we can proof the following Carleman estimate.
Proof of Theorem 4.1.

Since
(4.1) Since ∂ j and ∂ k commute, then that is, Ω j and Ω k commute up to some lower order terms. Write Also, the vector fields have the following properties n+1 j=1 ω j Ω j = 0 and n+1 j=1 Ω j ω j = n in S n + , n j=1 ω j Ω j = 0 and n j=1 Ω j ω j = n on ∂S n + .
Using this coordinate, Next, let u = e n−2s Step 2: Conjugation. Next, setting v = ω 1−2s 2 n+1 e τ ϕ u, where ϕ(t) = φ(e t ω) = e αt , we reach for some constants C 1 and C 2 . Also, We denote the norm and the scalar product in the bulk and the boundary space by and we omit the notation "lim ω n+1 →0 " in • 0 and •, • 0 .
Step 3: Showing the ellipticity of∆ ω . We need to prove the ellipticity of∆ ω : Lemma 4.2. Suppose (4.4) holds, then Proof. Note that The integration by parts is given by Similar integration by parts formula holds for Ω j for j = 1, · · · , n. Indeed, by (4.1), we know that for j = 1, · · · , n, Ω j and ω n+1 are commute up to some lower order term. So, to estimate the first term, it is suffice to estimate n j=1 Ω 2 j v 2 . Finally, the lower order terms can be easily estimated using integration by parts, Defining L − := S + A + (I) − (II) + (III), Step 4: Estimating the difference D. Observe that D = −4 Sv, Av + R, where By using (4.1) and integration by parts, we can compute by using integration by parts, again we reach Step 5: Estimating the sum S . Note that Observe that For δ ∈ (0, 1), write Hence, by using integration by parts, and apply Lemma 4.2 on the term δ |ϕ ′ | − 1 2∆ ω v 2 , choose δ > 0 small, and then choose ǫ > 0 small, we reach Step 6: Combining the difference D and the sum S . Multiplying (4.5) by τ , and summing with (4.6), we reach Step 7: Obtaining gradient estimates. Note that Step 8: Conclusion. Summing up (4.7) and (4.8), we reach Changing back to the Cartesian coordinate, and we obtain our result.
Proof of Theorem 4.3.
Step 1: Changing the coordinates. As in the proof of Theorem 4.1, firstly, we pass to conformal coordinates. With the notations mentioned before, recall (4.2): Step 2: Splitting u into elliptic and subelliptic parts. We split u into two parts u = u 1 +u 2 . Here u 1 is a solution to We remark that existence of unique energy solution to this problem is followed by the Lax-Milgram theorem in H 1 (S n + × R, ω 1−2s n+1 ).
5. Proofs of Theorem 1.1 and Theorem 1.2 Proof of Theorem 1.1.
Hence, we reach . Multiplying the above inequality by e τ ϕ , and then integrating with respect to the radial variable t, we obtain Similarly, we have So, for large τ , the boundary terms of (5.3) are absorbed, and we reach . Pulling out the exponential weight in the above estimate yields Step 4: Conclusion. Sinceφ(4) ≥φ (2), taking τ → ∞ will leads a contradiction, unlessũ = 0 in B + 6 \ B + 4 . Finally, applying the unique continuation property for classical second order elliptic equations (see e.g. [Reg97, Theorem 1.1]), we conclude thatũ ≡ 0.
Following exactly the arguments in [RW19, Theorem 2], we can obtain Theorem 1.2. For sake of completeness, here we give a sketch of the proof of Theorem 1.2.