On Landis’ conjecture in the plane

On Landis’ conjecture in the plane

Carlos Kenig

∗

Luis Silvestre

†

Jenn-Nan Wang

‡

Abstract

In this paper we prove a quantitative form of Landis’ conjecture in the plane. Precisely, let W (z) be a measurable real vector-valued function and V (z) ≥ 0 be a real measurable scalar function, satisfying kW kL∞_(R2₎ ≤ 1 and kV k_L∞_(R2₎ ≤ 1. Let

u be a real solution of ∆u − ∇(W u) − V u = 0 in R2. Assume that u(0) = 1 and |u(z)| ≤ exp(C0|z|). Then u satisfies inf

|z0|=R

sup

|z−z0|<1

|u(z)| ≥ exp(−CR log R), where C depends on C0. In addition to the case of the whole plane, we also establish a

quantitative form of Landis’ conjecture defined in an exterior domain.

1

Introduction

In the late 60’s (see [KL88]), E.M. Landis conjectured that if ∆u + V u = 0 in Rn _with

kV kL∞_(Rn₎ ≤ 1 and kuk_L∞_(Rn₎ ≤ C₀ satisfying |u(x)| ≤ C exp(−C|x|1+), then u ≡ 0.

Landis’ conjecture was disproved by Meshkov [Me92] who constructed such V and nontrivial u satisfying |u(x)| ≤ C exp(−C|x|43). He also showed that if |u(x)| ≤ C exp(−C|x|

4

3+), then

u ≡ 0. A quantitative form of Meshkov’s result was derived by J. Bourgain and the first author [BK05] in their resolution of Anderson localization for the Bernoulli model [An58] in higher dimensions. It should be noted that both V and u constructed by Meshkov are complex-valued functions. It remains an open question whether Landis’ conjecture is true for real-valued V and u. In this paper, we confirmed Landis’ conjecture for any real solution u of

∆u − ∇(W u) − V u = 0 in R2, (1.1) where W (x, y) = (W1(x, y), W2(x, y)) and V (x, y) are measurable, real-valued, and V (x, y) ≥

0 a.e. In view of the scaling argument in [BK05], Landis’ conjecture is an easy consequence

∗_{Department of Mathematics, University of Chicago, Chicago, IL 60637, USA. Email:}

[email protected]. Supported in part by NSF Grant DMS-1265249.

†_{Department of Mathematics, University of Chicago, Chicago, IL 60637, USA. Email:}

[email protected]. Supported in part by NSF grants DMS-1254332 and DMS-1065979.

‡_{Institute of Applied Mathematical Sciences, NCTS (Taipei), National Taiwan University, Taipei 106,}

of the estimate for the maximal vanishing order of u satisfying (1.1) in a bounded domain (see also [Ke07]). To be precise, we prove that

Theorem 1.1 Assume that W1(x, y), W2(x, y) and V (x, y) are measurable, real-valued, and

V (x, y) ≥ 0 a.e. in B2, moreover, there exist K ≥ 1, M ≥ 1 such that

kW kL∞_(B

2) ≤ K, kV kL∞(B2) ≤ M.

Let u be a real solution to

∆u − ∇(W u) − V u = 0 in B2. (1.2)

Assume that kukL∞_(B

2)≤ exp(C0( √ M + K)) and kukL∞_(B 1) ≥ 1. Then kukL∞_(B r)≥ r C(√M +K) _(1.3)

for all sufficiently small r, where C depends on C0.

Hereafter, we denote Br(a) an open disc of radius r centered at a. In the case when

a = 0, we simply denote Br(0) = Br. Having proved Theorem 1.1, Landis’ conjecture in

quantitative form follows by a scaling argument [BK05].

Theorem 1.2 Assume that W1(x, y), W2(x, y) and V (x, y) are measurable, real-valued, and

V (x, y) ≥ 0 a.e. in R2, furthermore,

kW kL∞_(R2₎ ≤ 1, kV k_L∞_(R2₎≤ 1. (1.4)

Let u be a real solution to (1.1). Assume that |u(z)| ≤ exp(C0|z|) and u(0) = 1, where

z = (x, y). Let z0 = (x0, y0). Then we have that

inf

|z0|=R

sup

|z−z0|<1

|u(z)| ≥ exp(−CR log R) for R  1 (1.5)

where C depends on C0.

Using the same techniques, we can also confirm Landis’ conjecture for

∆u + W · ∇u − V u = 0 in R2 (1.6) with V ≥ 0. As before, we first prove an estimate of the maximal vanishing order of u in B2.

Theorem 1.3 Assume that W and V satisfy the same assumptions as in Theorem 1.1. Let u be a real solution to

∆u + W · ∇u − V u = 0 in B2.

Assume that kukL∞_(B

2)≤ exp(C0( √ M + K)) and kukL∞_(B 1) ≥ 1. Then kukL∞_(B r)≥ r C(√M +K) _(1.7)

for all sufficiently small r, where C depends on C0.

By Bourgain-Kenig’s scaling argument, we then show that

Theorem 1.4 Assume that W, V satisfies the assumptions described in Theorem 1.2. Let u be a real solution to (1.6). Assume that |u(z)| ≤ exp(C0|z|) and u(0) = 1. Then we have

that

inf

|z0|=R

sup

|z−z0|<1

|u(z)| ≥ exp(−CR log R) for R  1 (1.8)

where C depends on C0.

In this paper we also establish a quantitative form of Landis’ conjecture in an exterior domain. Precisely, we show that

Theorem 1.5 Let u be a real solution of

∆u − V (x, y)u = 0 in B₁c, where V (x, y) ≥ 0 a.e. is measurable, real-valued, and satisfies

kV kL∞_(Bc 1) ≤ 1.

Assume that kukL∞_(Bc

1)≤ 1 and there exists C0 > 0 such that

inf

|z0|=5₂

Z

B1(z0)

|u|2 ≥ C0. (1.9)

Then we have that inf

|z0|=R

sup

|z−z0|<1

|u(z)| ≥ C exp(−C0R(log R)2) for R  1,

Before outlining the ideas of our proof, we remark on some related works. The exponent √

M + K in (1.3) is known to be optimal. For the case where u is a λ-eigenfunction of the Laplace-Beltrami operator in a smooth compact Riemannian manifold without boundary, the maximal vanishing order of u is less than C√λ proved by Donnelly and Fefferman in [DF88]. Donnelly and Fefferman’s proof was based on the Carleman method. Using the method of the frequency function developed by Garofalo and Lin [GL86, GL87], Kukavica [Ku98] proved the maximal vanishing order of u solving Lu + V u = 0 in Ω ⊂ Rn, n ≥ 2 is less than C(1 + kV−k1/2_L∞_(Ω)+ (oscΩV )2), where L is a general uniform second order elliptic

operator, V−= max{−V, 0}, and oscΩV = supΩV − infΩV . However, for the equation (1.2)

with W 6≡ 0, Kukavica’s method can not produce an order which is algebraic in kW kL∞ (see

[Ku98, Remark 5.4]). For (1.2) without the presence of W , the method in [BK05] proves that the maximal vanishing order of u is less than CM2/3 (see [Ke07]). In [Da12], Davey used the Carleman method to study the quantitative uniqueness estimate of u to

∆u + W · ∇u + V u = 0 in Rn,

where W and V also satisfy some decaying properties. Her results lead to an order R2 _in

(1.5) under the assumption (1.4) (see also [LW13]). Moreover, Meshkov-type examples are constructed in [Da12] showing that in the presence of W (complex-valued) with V ≡ 0, the exponent 2 is optimal.

The Carleman method has been a powerful technique in studying questions related to Landis’ conjecture and is able to produce optimal bounds in the complex-valued case. Since the Carleman estimate does not seem to distinguish real- or complex-valued functions, a direct use of such estimates to resolve Landis’ conjecture seems likely to fail. In this paper we will take a different approach. The main idea lies in the nice relation between second order elliptic equations in the plane and the Beltrami system. The assumption of V ≥ 0 allows us to construct a global positive multiplier which enables us to convert (1.2) into an elliptic equation in divergence form (see (2.10)). By the Beltrami system, we can derive three-ball inequalities (with precise exponent, constant, and radii) for solutions u of (1.2) without using the Carleman estimate or the frequency function.

For the problem in an exterior domain, the local result like Theorem 1.1 remains true. Unfortunately, the usual scaling argument does not lead us to Landis’ conjecture. One of the obstacles is the lack of simply-connectedness in B2 in the rescaled problem. To overcome

the difficulty, we introduce an appropriate cutoff function and construct an ”approximate” stream function. We reduce the original equation to an inhomogeneous d-bar equation and then apply a Carleman estimate for ¯∂.

The paper is organized as follows. In Section 2, we consider a simple ¯∂ equation and study (1.2) with W ≡ 0 to explain the main ideas of the proof. Based on the ideas in Section 2, we prove Theorem 1.1 and 1.2 in Section 3. We prove Theorem 1.3, 1.4 in Section 4. We prove Theorem 1.5 in Section 5. Finally, we discuss some related questions in Section 6. Throughout the paper, C denotes an absolute positive constant whose dependence will be

specified whenever necessary. The value of C may vary from line to line. Also, in the paper, we use z ∈ R2 _{or (x, y) ∈ R}2_.

2

Main ideas and proofs of Theorem 1.1, 1.2 for W ≡ 0

To motivate the main ideas of our method. We begin with a simple ¯∂ equation. ¯

∂u = V (z)u in B2, (2.1)

with kV kL∞_(B

2) ≤ M . Assume that kukL∞(B1) ≥ 1 and kukL∞(B2) ≤ e

C0M_{. It is clear that}

any solution of (2.1) can written as

u = exp(w)f or f = exp(−w)u, where w(z) = −1 π Z B2 V (ζ) ζ − zdζ. (2.2) Recall that |w(z)| ≤ CkV kL∞_(B 2) ≤ CM for z ∈ B2.

We can see that

kukL∞_(B 1) ≥ e −CM and kukL∞_(B 2)≤ e CM_,

where the second constant C depends on C0.

Since f is holomorphic in B2, by Hadamard’s three-circle theorem, we have

kf kL∞_(B r1)≤ kf k θ L∞_(B r)kf k 1−θ L∞_(B r2), where r < r1 < r2 and θ = log( r2 r1) log(r2 r) .

Taking r1 = 1 and r2 = 3₂ yields

e−CM ≤ kukθ L∞_(B r) and hence kukL∞_(B r)≥ r CM _(2.3) and C depends on C0.

Having derived the estimate of vanishing order, we can prove the following quantitative uniqueness estimate.

Theorem 2.1 Let u be any solution of ¯

∂u = V u in R2.

Assume that kV kL∞_(R2₎ ≤ 1, |u(z)| ≤ eC0|z|, and u(0) = 1. Then

inf

|z0|=R

sup

|z−z0|<1

|u(z)| ≥ exp(−CR log R) for R  1 (2.4)

where C depends on C0.

Proof. This theorem is an easy consequence of (2.3) by the scaling argument used in [BK05]. We include the proof for the sake of completeness. Denote |z0| = R. Let uR(z) = u(R(z +

z0/R)), then uR satisfies (2.1) with a rescaled potential VR(z) = RV (R(z + z0/R)), where

kqRkL∞_(B

2) ≤ R.

It is easy to see that |uR(z)| ≤ eC0R and

uR(− z0 R) = u(0) = 1 with |z0 R| = 1. Hence, we have kuRkL∞_(B 1) ≥ 1.

The quantitative estimate (2.4) follows easily from (2.3) by taking r = R−1.

2

We now would like to prove Theorems 1.1 and 1.2 without the presence of W based on the ideas described above. We first consider the local problem. Let u be a real solution to

∆u − V (z)u = 0 in B2 ⊂ R2, (2.5)

where V (z) ≥ 0 a.e. Likewise, we assume that u satisfies kukL∞_(B

2) ≤ exp(C0

√

M ) (2.6)

for some C0 > 0 and

kV kL∞_(B

2)≤ M. (2.7)

We let M ≥ 1 for simplicity. We will first construct a positive multiplier. Let us consider φ1(z) = φ1(x, y) = exp(λx), then ∆φ1 = λ2φ. Hence, if we choose λ =

√

M , then we have ∆φ1− V φ1 = (λ2− V )φ1 ≥ 0,

that is, φ1 is a subsolution. On the other hand, define φ2 = exp(2

√

M ), then ∆φ2− V φ2 ≤ 0,

i.e., φ2 is a supersolution. Note that φ2 ≥ φ1. Thus, there exists a positive solution φ

satisfying (2.5) and

One way to verify that such solution φ exists is to define

φ(z) = sup{ϕ(z) : ∆ϕ − V ϕ ≥ 0 in B2, ϕ ≤ exp(2

√

M ) on ∂B2}.

Then φ solves (2.5) and satisfies (2.8). We can also see that φ is Lipschitz. Moreover, from the gradient estimate for Poisson’s equation (see, for example, [GT83]), we have that for 0 < a1 < a2 with a2r < 2

k∇φkL∞_(B a1r)≤

CM

r kφkL∞(Ba2r), (2.9)

where C is an absolute constant. If we set u = φv, then v satisfies

∇ · (φ2_{∇v) = 0, in B}

2. (2.10)

Let ˜v with ˜v(0) = 0 be the stream function related to v, i.e., (

∂yv = φ˜ 2∂xv,

−∂xv = φ˜ 2∂yv.

(2.11)

Let g = φ2v + i˜v, then g satisfies ¯

∂g = ¯∂φ2v = ∂φ¯

2

2φ2(g + ¯g) in B2. (2.12)

As usual, we define ¯∂ = 1₂(∂x+ i∂y) and ∂ = 1₂(∂x− i∂y). Let

α = ∂φ¯ 2 2φ2 = ¯ ∂φ φ = ¯∂ log(φ), (2.13) then (2.12) is equivalent to ¯ ∂g = αg + α¯g in B2. (2.14)

We perform one more reduction. Defining

˜ α = ( α + α¯g/g, if g 6= 0, 0, otherwise, (2.15) (2.14) now is reduced to ¯ ∂g = ˜αg in B2. (2.16)

We can solve for (2.16) directly. Before doing so, we need to obtain a precise estimate of α = ¯∂ log(φ). In view of (2.8), if we denote ψ = log φ, then ψ satisfies

|ψ(z)| ≤ 2√M in B2, (2.17)

and solves the following equation

∆ψ + |∇ψ|2 = V in B2. (2.18)

Lemma 2.2

k∇ψkL∞_(B

7/5) ≤ C

√

M , (2.19)

where C > 0 is an absolute constant.

Proof. We begin with an L2 _{estimate. Let θ ∈ C}∞

0 (B2) satisfies 0 ≤ θ ≤ 1 and θ = 1 for

(x, y) ∈ B9/5. Multiplying both sides of (2.18) by θ, using (2.17), and the integration by

parts, we have that Z θ|∇ψ|2 = Z θV − Z (∆θ)ψ ≤ C(M +√M ) i.e., Z B9/5 |∇ψ|2 _{≤ CM,}

where C > 0 is an absolute constant.

To proceed further, we rescale the equation (2.18). Define ϕ = ψ/C√M for some C > 0. Then (2.18) becomes

ε∆ϕ + |∇ϕ|2 = ˜V , (2.20) where ε = 1/C√M and ˜V = V /C2_{M . We can choose C > 10 (so ε < 1/10) sufficiently large}

so that k ˜V kL∞_(B 2) ≤ 1, kϕkL∞(B2)≤ 1 and Z B9/5 |∇ϕ|2 ≤ 1. (2.21)

Claim 2.3 For any z ∈ B7/5 and ε < r < 1/5, we have

Z

Br(z)

|∇ϕ|2 _{≤ Cr}2_.

Proof of Claim 2.3. It suffices to take z = 0. Choose a cutoff function η ∈ C₀∞(B2r) and

η = 1 on Br. Denote m = R B2rϕ/|B2r|. Clearly, we have 0 = ε Z ∆((ϕ − m)η2) = ε Z ∆ϕη2+ 4ε Z ∇ϕ · ∇ηη + ε Z ∆η2(ϕ − m) := I + II + II. (2.22)

Now substituting ε∆ϕ from (2.20) in I yields

I = − Z |∇ϕ|2_η2₊ Z ˜ V η2 ≤ − Z |∇ϕ|2_η2_{+ Cr}2_{. (2.23)}

Next we can estimate

|II| ≤ 4ε Z |∇ϕ|2_η2 1/2Z |∇η|2 1/2 ≤ 1 2 Z |∇ϕ|2_η2_{+ Cε}2_{. (2.24)}

Finally, for III, we obtain that |III| ≤ Cεr−2 Z B2r |ϕ − m| ≤ C Z B2r |ϕ − m|2 1/2Z B2r ε2r−4 1/2 ≤ Cr Z B2r |∇ϕ|2 1/2 (ε2r−2)1/2 ≤ Cε2₊ 1 400 Z B2r |∇ϕ|2_, (2.25)

where we used the Poincar´e inequality in the third inequality above. Putting (2.22)-(2.25) together gives

Z Br |∇ϕ|2 ≤ Cε2+ Cr2+ 1 200 Z B2r |∇ϕ|2 ≤ Cr2+ 1 200 Z B2r |∇ϕ|2. (2.26)

Now if r2 _{≥ 1/200, then (2.26) implies}

Z

Br

|∇ϕ|2 ≤ Cr2

by the last estimate of (2.21). On the other hand, if r2 _{< 1/200, we can pick a k ∈ N such} that 1 5 ≤ 2 k r ≤ 2 5, i.e., r2 _{≥ (1/100)}k_{. It is observed that B}

2k_r(z) ⊂ B_9/5 for z ∈ B_7/5. Iterating (2.26) in k steps yields Z Br |∇ϕ|2 _{≤ Cr}2 ₊ 1 200k Z B_{2k r} |∇ϕ|2 _{≤ Cr}2_.

This ends of the proof of claim.

2

We will use Claim 2.3 to give a pointwise bound of ∇ϕ(z) for z ∈ B7/5. For this end, we

use another scaling. Let ϕε(z) = 1_εϕ(εz), then we have that

∇ϕε(z) = ∇ϕ(εz), ∆ϕε(z) = ε∆ϕ(εz)

and therefore

∆ϕε+ |∇ϕε|2 = ˜V (εz) := ˜Vε(z) for z ∈ B2, (2.27)

k ˜VεkL∞_(B 2)≤ 1.

Moreover, it follows from Claim 2.3 that Z B2 |∇ϕε|2 = Z B2 |∇ϕ(εz)|2 ₌ 1 ε2 Z B2ε |∇ϕ|2 _≤ 1 ε2 · Cε 2 _{= C.}

Now applying the elliptic regularity theorem to (2.27) (see [Gi83, Chapter V, Theorem 2.3 and Proposition 2.1]), we obtain that there exists p > 2 such that

k∇ϕεkLp_(B

1) ≤ C.

Define a new function

˜ ϕε(z) = ϕε(z) − 1 |B1| Z B1 ϕε.

Then ∇ϕε= ∇ ˜ϕε and ˜ϕε satisfies

∆ ˜ϕε= −|∇ ˜ϕε|2+ ˜Vε:= ζ in B1.

It is clear that kζk_Lp/2_(B

1) ≤ C. Moreover, from Poincar´e’s inequality, we have

k ˜ϕεkLp/2_(B

1) ≤ k ˜ϕεkLp(B1)≤ Ck∇ϕεkLp(B1) ≤ C.

Elliptic regularity theorem implies

k ˜ϕεkW2,p/2_(B r) ≤ C

for r < 1. By bootstrapping arguments, we obtain that k∇ ˜ϕεkL∞_(B

r0) = k∇ϕεkL∞(Br0) = k∇ϕkL∞(Bεr0) ≤ C

with r0 < r. The method clearly works for any x ∈ B7/5. The derivation of (2.19) is then

completed.

2

Remark 2.4 In view of the definition of ψ = log φ, if we normalize φ at a point ˆz ∈ B7/5

to be 1, i.e., φ(ˆz) = 1, then the Lipschitz estimate of ψ implies

|ψ(z)| = |ψ(z) − ψ(ˆz)| ≤ C√M |z − ˆz| for z ∈ B7/5.

Therefore, we can see that if φ(ˆz) = 1, then 1

C ≤ φ(z) ≤ C, ∀ z ∈ B√CM

(ˆz).

Using (2.19), we have that kαkL∞_(B

7/5)≤ C

√

M , which immediately implies

k ˜αkL∞_(B

7/5) ≤ C

√

M . (2.28)

Let w(z), z = x + iy, be defined by

w(z) = −1 π Z B7/5 ˜ α(ξ) ξ − zdξ,

then ¯∂w = ˜α in B7/5 and

kwkL∞_(B

7/5) ≤ Ck ˜αkL∞(B7/5) ≤ C

√

M . (2.29)

It is clear that any solution g of (2.16) in B7/5 is given by

g(z) = exp(w(z))h(z), where h is holomorphic in B7/5.

Applying Hadamard’s three-circle theorem to analytic function h, we have that khkL∞_(B r1) ≤ (khkL∞(Br/2)) θ_(khk L∞_(B r2)) 1−θ_{, (2.30)} where r/2 < r1 < r2 < 7/5 and θ = log( r2 r1) log(2r2 r ) . (2.31)

Substituting h = exp(−w)g into (2.30) and using (2.29) implies kgkL∞_(B r1) ≤ exp(C √ M )(kgkL∞_(B r/2)) θ (kgkL∞_(B r2)) 1−θ . (2.32) Recall that g = φ2v + i˜v = φu + i˜v and hence,

|φu| ≤ |g| ≤ (|φu| + |˜v|). (2.33) Following from ˜v(0) = 0 and (2.11) , for z ∈ B2, we have that

|˜v(z)| ≤ Cr exp(2√M )k∇vkL∞_(B

r) ∀ z ∈ Br, r < 2. (2.34)

Assume that u satisfies

kukL∞_(B

2) ≤ exp(C0

√

M ) (2.35)

for some C0 > 0. Now choosing r/2 < 1 < 6/5, i.e., r1 = 1, r2 = 6/5 in (2.30), using (2.33),

v = u/φ, and the interior estimate (2.9) for φ and u, we conclude that kukL∞_(B 1)≤ exp(C √ M )(kukL∞_(B r)) θ_(kuk L∞_(B 7/5)) 1−θ _{≤ exp(C}√_{M )(kuk} L∞_(B r)) θ_,

where C depends on C0. In summary, we have proved that

Theorem 2.5 Let u be a real solution to (2.5). Assume that (2.7), (2.35) hold and further-more

kukL∞_(B 1) ≥ 1.

Then we have that

kukL∞_(B r) ≥ r

C√M_,

For any bounded solution u solving

∆u − V u = 0 in R2, (2.36) with V ≥ 0 and kV k_L∞_(R2₎ ≤ 1, Landis’ conjecture (qualitative) is trivial. Nonetheless, using

the scaling argument of [BK05] as in the proof of Theorem 2.1, Theorem 2.5 immediately implies a quantitative version of Landis’s conjecture.

Theorem 2.6 Let u be a real solution to (2.36) with V ≥ 0. Assume that |u(z)| ≤ exp(C0|z|), |V (z)| ≤ 1, and u(0) = 1. Then u satisfies

inf

|z0|=R

sup

|z−z0|<1

|u(z)| ≥ exp(−CR log R) for R  1, (2.37)

where C depends on C0.

3

Proofs of Theorem 1.1, 1.2

Now we consider

Lu := ∆u − ∇(W (x, y)u) − V (x, y)u = 0 in B2, (3.1)

where W = (W1, W2) and V are real-valued, measurable, and V (x, y) ≥ 0. As above, assume

that

kV kL∞_(B

2) ≤ M, kW kL∞(B2) ≤ K

with M ≥ 1, K ≥ 1. To construct a positive multiplier for (3.1), we consider the adjoint operator of L, i.e.,

L∗u = ∆u + W · ∇u − V u. (3.2) Let φ1 = exp(λx), then ∇φ1 = λ exp(λx)(1, 0), and

L∗φ1 =∆φ1+ W · ∇φ1− V φ1

=(λ2 − V ) exp(λx) + λ exp(λx)W1

≥(λ2 − kV kL∞_(B

2)− λkW kL∞(B2))φ1.

Hence, if λ = (√M + K), then L∗φ1 ≥ 0. On the other hand, let φ2 = exp(2(

√

M + K)), then L∗φ2 = −V φ2 ≤ 0. Therefore, as before, there exists a positive Lipschitz solution φ

satisfying L∗φ = 0 in B2 and estimates

exp(−2(√M + K)) ≤ φ ≤ exp(2(√M + K)), ∀ (x, y) ∈ B2.

As above, if we let v = u/φ, then v satisfies

∇ · (φ2_{(∇v − W v)) = 0 in B}

Denote ˜v with ˜v(0) = 0 the stream function corresponding to v, i.e. (

∂yv = φ˜ 2∂xv − φ2W1v,

−∂xv = φ˜ 2∂yv − φ2W2v.

(3.4)

As before, let g = φ2v + i˜v, then g solves ¯ ∂g = γ(g + ¯g) in B2, (3.5) where γ = ¯ ∂φ2 2φ2 + 1 2(W1+ iW2) = ¯∂ log φ + 1 2(W1+ iW2). Likewise, let ˜ γ = ( γ + γ ¯g/g, if g 6= 0, 0, otherwise, then (3.5) becomes ¯ ∂g = ˜γg in B2. (3.6)

Let φ = exp(ψ), then ψ satisfies

|ψ(z)| ≤ 2(√M + K), for all z ∈ B2 (3.7)

and

∆ψ + |∇ψ|2+ W · ∇ψ = V in B2. (3.8)

Similarly, we prove the following estimate of ∇ψ. Lemma 3.1

k∇ψkL∞_(B

7/5) ≤ C(

√

M + K). (3.9)

Proof. This lemma can be proved in the same way as in Lemma 2.2. We first derive an L2

bound. Let θ ∈ C₀∞(B2) satisfies 0 ≤ θ ≤ 1 and θ = 1 for (x, y) ∈ B9/5. Multiplying both

sides of (3.8) by θ, using (3.7) and the integration by parts, we obtain that Z θ|∇ψ|2 = Z V θ − Z θW · ∇ψ − Z ∆θψ ≤ 1 2 Z θ|∇ψ|2+ C(M + K2 +√M + K) i.e., Z B9/5 |∇ψ|2 _{≤ C(M + K}2_).

To bound ∇ψ(x) for all x ∈ B7/5 by C(

√

M + K), choosing ε = 1/C(√M + K) and ϕ = ψ/C(√M + K), we proceed as in the proof of Lemma 2.2. We avoid repeating the

With the help of estimate (3.9), we have that k˜γkL∞_(B

7/5)≤ C(

√

M + K). Therefore, any solution g of (3.6) in B7/5 is represented by

g = exp( ˜w)h, where h is holomorphic in B7/5 and

k ˜wkL∞_(B

7/5) ≤ C(

√

M + K).

The remaining arguments in proving Theorem 1.1 and 1.2 are exactly similar to those of Theorem 2.5 and 2.6.

4

Proofs of Theorem 1.3, 1.4

Recall that u is a real solution of

∆u + W · ∇u − V u = 0 in B2, (4.1)

where W = (W1, W2) and V ≥ 0 are real-valued and

kW kL∞_(B

2) ≤ K and kV kL∞(B2) ≤ M.

Let φ be the positive solution of (4.1) constructed in Section 3 for L∗u = 0 (see (3.2)). Defining v = u/φ implies that v satisfies

∆v + (2∇ψ + W ) · ∇v = 0 in B2. (4.2)

In view of estimate (3.9), we have

k2∇ψ + W kL∞_(B

7/5)≤ C(

√

M + K).

Note that 4∆ = ¯∂∂. Hence, (4.2) can be written as ¯ ∂(∂v) = fW (∂v) in B2 (4.3) with kfW kL∞_(B 7/5) ≤ C( √ M + K).

Argued as above, using Hadamard’s three-circle theorem, we have that k∇vkL∞_(B r1)≤ exp(C( √ M + K))(k∇vkL∞_(B r/2)) θ_(k∇vk L∞_(B r2)) 1−θ_{, (4.4)}

where r/2 < r1, r1 = 6/5, r2 = 7/5, and θ = log( r2 r1) log(2r2 r ) .

Using the interior estimate again, (4.4) becomes k∇vkL∞_(B 6/5) ≤ exp(C( √ M + K))(r−1kukL∞_(B r)) θ_{. (4.5)}

Now we would like to bound the left hand side of (4.5) from below using the a priori condition kukL∞_(B

1) ≥ 1. Since kukL∞(B1) ≥ 1, there exists z0 ∈ B1 such that |u(z0)| ≥ 1.

It suffices to assume u(z0) ≥ 1. For a real-valued u, it is clear that given any a > 0, either

u(z) ≥ a for all z ∈ B6/5 or there exists z1 ∈ B6/5 such that u(z1) < a. Here we would like

to choose an appropriate a. For the latter case, recalling that

exp(−2(√M + K)) ≤ φ(z) ≤ exp(2(√M + K)), we can see that

u(z1) φ(z1) ≤ a φ(z1) ≤ a exp(2(√M + K)) and u(z0) φ(z0) ≥ exp(−2(√M + K)). Therefore, if we set a exp(2( √ M + K)) = 1 2exp(−2( √ M + K)), i.e., a = 1 2exp(−4( √ M + K)),

then we have that

u(z1) φ(z1) ≤ 1 2exp(−2( √ M + K)) and thus k∇vkL∞_(B 6/5)≥ |v(z0) − v(z1)| ≥ u(z0) φ(z0) − u(z1) φ(z1) ≥ 1 2exp(−2( √ M + K)),

which implies (1.7) with the help of (4.5). Now for the other case, i.e., u(z) ≥ a for all z ∈ B6/5, (1.7) is obviously satisfied. This completes the proof of Theorem 1.3. By

5

Landis’ conjecture in an exterior domain

In this section we prove Landis’ conjecture in an exterior domain, Theorem 1.5. Recall that we consider

∆u − V (x, y)u = 0 in B₁c. (5.1) Assume that the potential V is defined everywhere in R2 _{and satisfies V ≥ 0 and}

kV kL∞_(R2₎ ≤ 1. (5.2)

As mentioned in the Introduction, for (5.1), the local vanishing order result, Theorem 2.5, remains true. However, the scaling argument fails to imply Landis’ conjecture. We have to work harder to prove the conjecture in this case. Here our main tool is a Carleman estimate. But we need to set up everything carefully before applying the Carleman estimate

Let z₀0 ∈ R2 _{with |z}0

0|  1. Since (5.1) is invariant under rotation, we can assume that

z₀0 = |z₀0|e1, where e1 = (1, 0). Translating the origin to −5e1/2, (5.1) becomes

∆u − V (x, y)u = 0 in B₁c(−5e1/2). (5.3)

By abuse of notation, we continue to write u and V in the equation in the new coordinates. Now we denote z0 = (|z00| − 5/2)e1 and set R = |z0|. As before, we define the scaled solution

uR(z) = u(ARz + z0), where A > 0 will be determined later. Therefore, uR solves

∆uR− VRuR= 0 in Bc1 AR (z1), (5.4) where z1 = −( 1 A + 5 2AR)e1 and VR(z) = (AR)2V (ARz + z0), thus,

kVRkL∞_(R2₎ ≤ (AR)2.

Under the above change of coordinates, the origin moves to

ˆ

z = − z0 AR = −

1 Ae1. We choose a large A so that

B 1

AR(z1) ⊂ B7/5.

Let φ be the positive solution of

as constructed in Section 2. Here we normalize φ at ˆz, i.e., φ(ˆz) = 1. Likewise, we define ψ = log φ. Then ψ satisfies

kψkL∞_(B

7/5) ≤ C(AR) and k∇ψkL∞(B7/5) ≤ C(AR). (5.6)

To simplify the notation, we suppress the subscript R of uR and denote uR by u. As before,

let v = u/φ, then v solves

∇ · (φ2_{∇v) = 0 in B} 2\ B 1

AR(z1). (5.7)

Note that here the domain of (5.7) is not simply-connected. The stream function of v, solu-tion to (5.7), may not exist. However, we can construct an ”approximate” stream funcsolu-tion of v. To do this, we choose a cutoff function χ ≡ 1 on |z −z1| ≥ _8AR9 and ≡ 0 for |z −z1| ≤ _16AR17 .

Note that ∇χ is supported on _16AR17 ≤ |z − z1| ≤ _8AR9 . Next we denote

ae1 =  9 8AR − ( 1 A + 5 2AR)  e1 =  −1 A − 11 8AR  e1, i.e., a = −1 A − 11 8AR. Now we define ˜ v(x, y) = Z x a −[χφ2_∂ yv](s, y)ds + Z y 0 [χφ2∂xv](a, s)ds.

It is easy to see that

∂y˜v(x, y) = (χφ2∂xv)(a, y) + Z x a −[∂yχφ2∂yv](s, y)ds + Z x a [χ∂x(φ2∂xv)](s, y)ds = χφ2∂xv(x, y) + Z x a −[∂yχφ2∂yv + ∂xχφ2∂xv](s, y)ds (5.8) and ∂xv(x, y) = −χφ˜ 2∂yv(x, y). (5.9)

As before, we set g = χφ2v + i˜v and hence

¯ ∂g = ( ¯∂φ2)χv + ( ¯∂χ)φ2v + 1 2 Z x a [∂yχφ2∂yv + ∂xχφ2∂xv](s, y)ds, = ¯ ∂φ2 2φ2(g + ¯g) + ( ¯∂χ)φu + 1 2 Z x a [∂yχφ2∂yv + ∂xχφ2∂xv](s, y)ds = αg + α¯g + ( ¯∂χ)φu + 1 2 Z x a [∂yχφ2∂yv + ∂xχφ2∂xv](s, y)ds, (5.10)

where α is given in (2.13). Defining ˜α as in (2.15), (5.10) now is equivalent to ¯ ∂g = ˜αg + ( ¯∂χ)φu +1 2 Z x a [∂yχφ2∂yv + ∂xχφ2∂xv](s, y)ds in B2. (5.11)

We now write ˆz as a point in the complex plane, i.e., ˆz = −_A1 + i0. Let w(z) be defined by w(z) = 1 π Z B7/5 ˜ α ξ − zdξ − 1 π Z B7/5 ˜ α ξ − ˆzdξ, then ¯∂w = − ˜α in B7/5. Recall that

k ˜αkL∞_(B

7/5)≤ C(AR).

In view of [Ve62, (6.9a)], we have the following estimate of w(z).

|w(z)| ≤ C(AR)|z − ˆz| log  C |z − ˆz|  , ∀ z ∈ B7/5. (5.12)

Let h = ew_{g, then it follows from (5.11) that}

¯ ∂h = ew( ¯∂χ)φu +e w 2 Z x a [∂yχφ2∂yv + ∂xχφ2∂xv](s, y)ds = H1+ H2 in B7/5, (5.13) where H1 = ew( ¯∂χ)φu and H2 = ew 2 Z x a [∂yχφ2∂yv + ∂xχφ2∂xv](s, y)ds.

We now come to the Carleman estimate. Here we will use the following estimate for ¯

∂ from [DF90, Proposition 2.1]. Let ϕτ(z) = ϕτ(|z|) = −τ log |z| + |z|2, then for any

h ∈ C₀∞(B7/5\ {0}), we have that Z | ¯∂h|2eϕτ _≥ 1 4 Z (∆ϕτ)|h|2eϕτ = Z |h|2_eϕτ_{. (5.14)}

Note that ϕτ is decreasing in |z| for τ > 8. We introduce another cutoff function 0 ≤ ζ ≤ 1

satisfying ζ(z) =            0, when |z| < 1 4AR, 1, when 1 2AR < |z| < 1, 0, when |z| > 6/5. Hence the following estimates holds

where X = { 1 4AR < |z| < 1 2AR} and Y = {1 < |z| < 6/5}. We also denote Z = { 1 2AR < |z| < 1}.

The relative positions of different domains in the rescaled problem are shown in the following figure, Figure 1. − 1 A ˆ z − 1 A−2AR5 z1 a −1 −6 5 0 z0 size of ball = _ARC

Figure 1: The figure represents the domain in the rescaled problem. ∇χ is supported in the red region and ∇ζ is supported in the green region. All three balls centered at z0, ˆz, and z1

have radius _AR1 , corresponding to balls of radius 1 in the unscaled problem. φ is bounded from above and below (away from zero) with bounds independent of R in the ball of size _ARC centered at ˆz.

Applying the Carleman estimate (5.14) to ζh gives Z Z |h|2eϕτ _{≤ 2} Z (| ¯∂ζh|2 + |ζ ¯∂h|2)eϕτ ≤ C(AR)2 Z X |h|2_eϕτ _{+ C} Z Y |h|2_eϕτ ₊ Z e Z (|H1|2+ |H2|2)eϕτ, (5.16)

where e Z = { 1 4AR < |z| < 6 5}. The termsR_Z|h|2_eϕτ _and R

e Z(|H1|

2_{+ |H}

2|2)eϕτ are the most crucial ones in (5.16). We would

like to study them in more detail. We begin with the first one. It is clear that for A large, R large Z Z |h|2eϕτ _≥ Z B 1 AR (ˆz) |h|2eϕτ_.

From Remark 2.4, we have that 0 < C1 ≤ φ(z) ≤ C2 in the ball B 1

AR(ˆz) with absolute

constants C1, C2. Since χ ≡ 1 on B1/AR(ˆz), we thus obtain that

|g| ≥ |φ2_{v| = |φu| ≥ C|u| for z ∈ B}

1

AR(ˆz). (5.17)

On the other hand, (5.12) implies

|w(z)| ≤ C(AR) 1

ARlog(CAR) = C log(CAR) for z ∈ BAR1 (ˆz),

i.e.,

ew(z) ≥ 1

C(AR)C for z ∈ BAR1 (ˆz). (5.18)

Combining (5.17), (5.18), and using that for z ∈ B1/AR(ˆz), |z| ≤ _AR1 +_A1, we have

Z Z |h|2eϕτ _≥ e ϕτ(_A1+_AR1 ) C(AR)C Z B 1 AR (ˆz) |u|2. (5.19) Next we look at R e Z|H1| 2_eϕτ_{. Recall that H}

1 is supported in _16AR17 ≤ |z − z1| ≤ _8AR9 since

∇χ is. It is clear that

G := { 17 16AR ≤ |z − z1| ≤ 9 8AR} ⊂ {|z − ˆz| ≤ 29 8AR}. As in (5.18), we can see that

ew(z) ≤ C(AR)C _for 17

16AR ≤ |z − z1| ≤ 9

8AR. (5.20) Using (5.20) and the same argument based on Remark 2.4 as above, we have that

Z e Z |H1|2eϕτ ≤ C(AR)C Z 17 16AR≤|z−z1|≤ 9 8AR

Finally, we studyR e Z|H2| 2_eϕτ_{. Observe that} supp H2 ⊂ {− 6 5 < x < − 1 A − 11 8AR, |y| < 9 8AR}.

Argued as above, φ is uniformly bounded from above and below in G. In view of the a priori boundedness assumption on u, Lemma 2.2, and the interior estimate, we obtain that

|H2(z)| ≤ C(AR)Ce|w(z)| for z ∈ supp H2. (5.22)

For z ∈ supp H2, we have that

|z − ˆz| ≤ |y| + |x + 1 A| ≤ 9 8AR + |x + 1 A|, 11 8AR ≤ |x + 1 A| ≤ |z − ˆz|, and so (5.12) implies |H2| ≤ exp(C(AR)|x + 1 A| log(AR)). (5.23) We will multiply by exp(−ϕτ(_A1 +_AR1 )) on both sides of (5.16). Thus, we want to take a

closer look at log 1 A + 1 AR |z|  for z ∈ supp H2. Since 1 A+ 1 AR |z| < 1, because |x| ≥ _A1 + _8AR11 , it suffices to estimate

1 −1 + 1 R |Az| = A|z| − 1 − _R1 A|z| ≥ A|x| − 1 −_R1 A|z| ≥ A(|x| − _A1) A|z| ≥ C(|x| − 1 A)

for z ∈ supp H2. Here C depends on A, but A has been chosen. Note that when z = (x, y) ∈

supp H2, |x + _A1| = |x| − _A1. Thus, we have

log  1 A+ 1 AR |z|  ≤ −C|x + 1 A| and exp(ϕτ(z)) exp(−ϕτ( 1 A + 1 AR))|H2| ≤ exp(C(AR)|x + 1 A| log(AR) − Cτ |x + 1 A|). (5.24)

Now we can put everything together. Multiplying C(AR)Cexp(−ϕτ(_A1 + _AR1 )) on both

sides of (5.16), using (5.19), (5.21), (5.24), the second inequality of (2.33), interior estimates, and the a priori bound of u, we obtain that

Z

B 1 AR(ˆz)

|u|2 _{≤ C(AR)}C_{exp(C(AR) log(AR))} exp(ϕτ( 1 4AR)) exp(ϕτ(_A1 +_AR1 )) Z B 1 AR(0) |u|2

+ C(AR)Cexp(C(AR) log(AR)) exp(ϕτ(1)) exp(ϕτ(_A1 + _AR1 )) + C(AR)Cexp(ϕτ( 1 A + 29 8AR)) exp(ϕτ(_A1 +_AR1 )) + exp  (C(AR) log(AR) − Cτ ) 11 8AR  , (5.25)

where the last term of (5.25) was derived by using |x + _A1| ≥ 11

8AR and taking τ large enough

such that

C(AR) log(AR) − Cτ < 0. (5.26) To fulfill (5.26), it suffices to choose

τ = ˜C(AR) log(AR) (5.27) for an appropriate large fixed constant ˜C. Consequently, the last term of (5.25) satisfies

exp 

(C(AR) log(AR) − Cτ ) 11 8AR



≤ exp(− ˜C log(AR)) = (AR)− ˜C. (5.28)

Rescaling back to the original variables, we observe that Z B 1 AR(ˆz) |uR|2 = 1 (AR)2 Z B1(0) |u|2 _≥ C0

(AR)2 (from (1.9)) and

Z B 1 AR(0) |uR|2 = 1 (AR)2 Z B1(x0) |u|2

Finally, choosing τ as in (5.27) (choosing ˜C larger if necessary) and taking R sufficiently large, it is not hard to see that

                          

C(AR)C+2exp(C(AR) log(AR)) exp(ϕτ(

1 4AR))

exp(ϕτ(_A1 +_AR1 ))

≤ exp(C(AR)(log(AR))2_),

C(AR)C+2exp(C(AR) log(AR)) exp(ϕτ(1)) exp(ϕτ(_A1 +_AR1 )) → 0, C(AR)C+2exp(ϕτ( 1 A+ 29 8AR)) exp(ϕτ(_A1 +_AR1 )) → 0, (AR)2exp  (C(AR) log(AR) − Cτ ) 11 8AR  → 0 (from (5.28)).

Therefore, if R is large enough, then the last three terms on the right hand side of (5.25) can be absorbed by the term on the left. The proof now is completed.

The same method can be used to generalize Theorems 1.2, 1.3, 1.4, to the case of an exterior domain. Before ending this section, we would like to remark that we can express (5.7) as

∆v + 2∇ψ · ∇v = 0 in B2\ B 1

AR(z1), (5.29)

which is exactly in the same form as in (4.2). Thus, using g = ∂v, (5.29) can be transformed into a ¯∂ equation for g in B2\ B 1

AR(z1) without introducing an approximate stream function

as before. However, this reduction does not work for the equation (1.2) (see (3.3)). Our point here is to introduce a unified approach using an approximate stream function to prove Landis’ conjecture (Theorems 1.2, 1.3, 1.4) in an exterior domain.

6

Conclusions and discussions

We proved positive answers to Landis’ conjecture for ∆u−∇(W u)−V u = 0 and ∆u+W ∇u− V u = 0 in the plane or in an exterior domain with or without W under the assumption of V ≥ 0. One key ingredient of our method is the existence of a global positive multiplier which convert the original equation into an equation in divergence form. Another important tool is a global three-ball inequality with an optimal exponent (θ in (2.31)), namely, Hadamard’s three-circle theorem. In the case of an exterior domain, we use a global Carleman estimate for ¯∂. For the general potential V , one is also able to construct a positive multiplier that can be used to convert the original equation into a divergence-form equation (see, for ex-ample, [Al98], [Al10], [Sc98]). However, this positive multiplier exists only locally. In other words, the divergence-form equation is valid only locally, which leads to a local quantitative uniqueness estimate (three-ball inequality). If we try to prove bounds like (1.3) or (1.5), we need to iterate that local estimate, which will give rise to double exponential bounds and are, apparently, worse than the results derived by the Carleman method.

One trick to remove the assumption V ≥ 0 is to consider the new function uξ(x, y) =

cosh(λξ)u(x, y). Then if u satisfies ∆x,yu − V u = 0, uξ satisfies

∂_ξ2uξ+ ∆x,yuξ− (λ2 − V )uξ = 0 in R3. (6.1)

The new potential function λ2 _{− V is non-negative if λ ≥ pkV k}

L∞. Consequently, we can

construct a global positive function to (6.1), still denoted by φ, satisfying similar estimates described in Section 2. Using φ, (6.1) becomes

∇ξ,x,y· (φ2∇ξ,x,yvξ) = 0 in R3 (6.2)

with vξ = uξ/φ. However, at this stage, a global optimal three-ball inequality or a global

Carleman estimate with a suitable weight function for (6.2) is not available. Therefore, new ideas are needed to resolve Landis’ conjecture in the general case.

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