麥克斯威爾電磁系統的強唯一連續延拓性及其定量分析

國立臺灣大學理學院數學學系 碩士論文

Department of Mathematics College of Science

National Taiwan University Master Thesis

麥克斯威爾電磁系統的強唯一連續延拓性及其定量分析

Quantitative uniqueness estimate of strong unique continuation property for the Maxwell system with anisotropic media

孫裕沛 Yu-Pei Sun

指導教授：王振男 博士 Advisor: Jenn-Nan Wang, Ph.D.

中華民國 101 年 6 月

June 2012

誌謝 誌謝 誌謝 誌謝

感謝王振男教授兩年以來的用心指導，以及同儕們平時的幫忙，彼此花時間討論出一些有趣的數學論述以及想 法，讓我可以完成這篇文章，也同時感謝我的家人及麻糬的陪伴跟鼓勵，以後我會更努力的！！！

ii

中文摘要 中文摘要 中文摘要 中文摘要

在這篇文章中，我們考慮非零解在時諧性麥克斯威爾系統的局部行為，其系統為非等向 性的媒體。而我們主要得到的結果是此系統的強連續延拓性在某些條件之下將會成立，並且 導出強連續延拓性的定量分析，也可以得到非零解趨近到零的速度。

我們主要運用到的工具為 Carleman 估計導出 Three-balls 不等式，再運用另一個 Carleman 估計以及 Three-balls 不等式推導出 Doubling 不等式，因此可得出強連續延拓性的定量分析。

中文關鍵字：卡勒門估計，麥克斯威爾系統，非等向性，強唯一連續延拓性

Abstract

In this article, we consider the local behavior of a non-trivial solution for the time-harmonic Maxwell system with anisotropic media. The main result of this article is the bound on the

vanishing order of the solution of the Maxwell system, which is a quantitative estimate of the strong unique continuation property(SUCP). And the most important tool is Carleman estimate. Our strategy in the proof is to derive doubling inequality through three-balls inequality.

Key words: Carleman estimate, Maxwell system, anisotropic media, strong unique continuation

目 錄

誌謝……….. i

中文摘要………. ii

英文摘要………. iii

第一章 介紹……….. 1

第二章 主要定理……….. 4

第三章 定理一證明……….. 5

定理二以及定理三證明……….. 10

參考文獻……….…… 18

Quantitative uniqueness estimate of strong unique continuation property for the Maxwell system with

anisotropic media

1. Introduction

The Maxwell system is firstly mentioned by J ames C lerk M axwell in the pa- per ”On Physical Lines of Force” which is published in 1861. He derived it from G auss⁰s l aw, F araday⁰s l aw and Ampre⁰s circuital l aw. Furthermore, he derived electromagnetic wave equations in 1865 and claim that light is an electromagnetic wave. In fact, he established the fundamental electrodynamics and had a significant impact on modern physics.

1. G auss⁰s l aw: The total electric flux coming out of a closed surface is equal to the total charge enclosed by that closed surface. It means that

ZZ

₀E · dA = Q

2. G auss l aw for the Magnetic Fields: The total magnetic flux coming out of a closed surface is always zero. It means that

ZZ

µ₀H · dA = 0

3. F araday⁰s l aw: The line integral of electric field over a closed contour is equal to the time rate of change of the total magnetic flux that goes through any arbitrary surface that is bounded by the closed contour. It means that

I

E · ds = ∂

∂t Z Z

µ₀H · dA

4. Ampre⁰s circuital l aw: The line integral of magnetic field over a closed contour is equal to the total current plus the time rate of change of the total electric

flux that goes through any arbitrary surface that is bounded by the closed contour. It means that

I

H · ds = Z Z

J · dA + ∂

∂t Z Z

₀E · dA

where E is the electric field, H is the magnetic field, J is the current density, Q is the total charge, ₀ is the permittivity of vacuum, and µ₀ is the permeability of vacuum.

Actually, Maxwell system can describe more complicated physical phenomenon in real life, so it has a general form, which is depending on the medium. Now we assume J = 0 to simplify the problem.

So we can define that E = (E₁, E₂, E₃) is the electric field, H = (H₁, H₂, H₃) is the magnetic field and ω is the frequency in a domain Ω. Denote the time-harmonic Maxwell system with anisotropic media







curlE = −iωµH curlH = iωεE

in Ω (1.1)

where Ω is an open subset of R³ containing 0, ω ∈ C\ {0},and ε(x),µ(x) are two real symmetric matrix-valued and positive-definite functions in Ω satisfying the following property :

(a) ε(0) = hµ(0) where h is a constant.

(b) ε,µ ∈ C² (Ω)

We can reduce the Maxwell system to a weakly coupled second order elliptic system.

Denote that

γ_jl^k =











1 , if (k,j,l) is an even permutation of (1,2,3)

−1, if (k,j,l) is an odd permutation of (1,2,3) 0 , otherwise

2

From the Maxwell system we can obtain that







∂_kE = ∇E_k− iωγ^kµH

∂_kH = ∇H_k+ iωγ^kεE By simple calculation and (1.1), we know that







div(curlH) = 0 div(curlE) = 0

⇒







div(εE) = 0 div(µH) = 0 So we have that for k = 1, 2, 3, the following formulas is called (1.2)







0 = ∂_kdiv(εE) = div(ε∇E_k) + div(∂_kε · E − iωεγ^kµH) 0 = div(µ∇H_k) + div(∂_kµ · H + iωµγ^kεE)

Now let P (x, D) = X

j,k

a_jk(x)D_jD_k be an elliptic operator in Ω such that a_jk(0) is a symmetric and positive-definite matrix and a_jk(x) ∈ C² (Ω) , so we can rewrite (1.2)















P₁(x, D)E + 2∇E · divε + E · ˜ε −

3

X

k=1

div(iωεγ^kµH) = 0

P₂(x, D)H + 2∇H · divµ + E · ˜µ +

3

X

k=1

div(iωµγ^kεE) = 0

where

P₁(x, D) =

3

X

i,j=1

ε_ij(x)D_iD_j , P₂(x, D) =

3

X

k,l=1

µ_kl(x)D_kD_l,

˜ ε =

3

X

m,n=1

D_mD_nε(x) , ˜µ =

3

X

m,n=1

D_mD_nµ(x)

So it implies that







|P₁(x, D)E| ≤ α₁|E| + α₂|∇E| + α₃|∇H| ≤ α₄|U | + α₅|∇U |

|P₂(x, D)H| ≤ β₁|H| + β₂|∇H| + β₃|∇E| ≤ β₄|U | + β₅|∇U |

(1.3)

where U = (E, H) is the non-trivial solution for the (1.1), α_i, β_i are constants for i = 1, 2, 3, 4, 5, and by (1.3) we can assume that M₁ = max{α₄, α₅} and M₂ = max{β₄, β₅} ⇒







|P₁(x, D)E| ≤ M₁|U | + M₂|∇U |

|P2(x, D)H| ≤ M1|U | + M2|∇U |

(1.4)

2. The main theorems

Theorem 1 There exists a positive number R₁ < 1 such that if 0 < r₁ < r₂ < r₃ ≤ R₀ and r₁/r₃ < r₂/r₃ < R₁ then

Z

|x|<r2

|U |²dx ≤ C

Z

|x|<r1

|U |²dx

τZ

|x|<r3

|U |²dx

1−τ

for U = (E, H) ∈ (L²(B_R0))⁶ where B_R0 ⊂ Ω and U is the non-trivial solution for the (1.1), where C depend on r1/r3, r2/r3 ,P1(x, D) and P2(x, D) and 0 < τ < 1 is only depending on r₁/r₃, r₂/r₃.

And then we want to show the quantitative estimate of strong unique continuation property for the Maxwell system. The strong unique continuation means that

For all U = (E, H) ∈ H_loc¹ (Ω) vanishes of infinite order at 0, then U=0 in Ω

Theorem 2 gives the upper bound on the vanishing order of the solution of the Maxwell system, and theorem 3 is the quantitative estimate of strong unique con- tinuation property.

Theorem 2 If U = (E, H) ∈ (L²_loc(Ω))⁶ is a non-trivial solution of Maxwell system,

then we can find a constant R₂ depending on P₁(x, D) , P₂(x, D) and constant m₁ depending on P₁(x, D), P₂(x, D) and ||U ||_L2(|x|<R²₂)/||U ||_L2(|x|<R⁴₂) satisfying

Z

|x|<R

|U |²dx ≥ KR^m1

where R is sufficient small and the constant K depending on R₂ , U .

Theorem 3 Let U = (E, H) ∈ (L²_loc(Ω))⁶ be a non-trivial solution to the Maxwell system. Then there exists positive constant R3 and C3 depending on P1(x, D) , P₂(x, D) and m₁ such that if 0 < r ≤ R₃,

Z

|x|<2r

|U |²dx ≤ C3

Z

|x|<r

|U |²dx where m₁ is the constant obtained in theorem 2.

4

3. Proofs

Proof of the theorem 1 First we denote that ϕ_β = ϕ_β(|x|) = exp((β

2)(log|x|)²) and recall a Carleman estimate [1]:

Lemma For any β > 0 large enough. Let S be a small neighborhood of 0, and u : S\{0} ⊂ Ω → R and that u ∈ H²(S\{0}) with compact support. Then we have

β³ Z

ϕ²_β|x|⁻ⁿ|u|²dx + β Z

ϕ²_β|x|⁻ⁿ⁺²|∇u|²dx ≤ ˜C₀ Z

ϕ²_β|x|⁻ⁿ⁺⁴|P (x, D)u|²dx (1.5) for some positive constant ˜C₀ depending only on P (x, D). Now ε, µ are C² func- tions, and U = (E, H) ∈ (L²_loc(Ω))⁶ then U = (E, H) ∈ (H_loc¹ (Ω))⁶ [2] and us- ing regularization, Friedrich⁰s Lemma and ellipticity of P (x, D). We can see that U = (E, H) ∈ (H_loc² (Ω\{0}))⁶.

Consider that 0 < r₁ < r₂ < R < 1, B_R ⊂ Ω where R is a constant. Define a cut-off function φ(x) ∈ C₀^∞(Rⁿ) satisfying 0 ≤ φ(x) ≤ 1 and

φ(x) =











0, if |x| ≤ r₁ e 1, if r₁

2 ≤ |x| ≤ er2

0, if |x| ≥ 3r2

where exp(1) = e. And it is easy to know that for all multiindex α and C1,C2 are constants.







|D^αφ(x)| ≤ C₁r₁^−|α|, ∀ r1

e ≤ |x| ≤ r1

2

|D^αφ(x)| ≤ C₂r₂^−|α|, ∀ er₂ ≤ |x| ≤ 3r₂

(1.6)

We assume n = 3 in the lemma because of the domain Ω ∈ R³ and then apply (1.5) to φE and φH. Firstly, we consider φE and use (1.4),(1.5),(1.6) and Cauchy- Schwarz inequality. We obtain

β³ Z

r1/2<|x|<er2

ϕ²_β|x|⁻³|E|²dx + β Z

r1/2<|x|<er2

ϕ²_β|x|⁻¹|∇E|²dx

≤ β³ Z

ϕ²_β|x|⁻³|φE|²dx + β Z

ϕ²_β|x|⁻¹|∇φE|²dx

≤ ˜C0

Z

ϕ²_β|x||P1(x, D)(φE)|²dx

≤ ˜C₀

 Z

r1/2<|x|<er2

ϕ²_β|x|(2M₁²|φU |²+ 2M₂²|φ∇U |²)dx +

Z

r1/e<|x|<r1/2

ϕ²_β|x|⁻³(C1|U |²+ C2|x|²|∇U |²)dx +

Z

er2<|x|<3r2

ϕ²_β|x|⁻³(C₁|U |²+ C₂|x|²|∇U |²)dx



≤ ˜C₁

 Z

r1/2<|x|<er2

ϕ²_β(|x|⁻³|U |²+ |x|⁻¹|∇U |²)dx +

Z

r1/e<|x|<r1/2

ϕ²_β|x|⁻³(|U |²+ |x|²|∇U |²)dx +

Z

er2<|x|<3r2

ϕ²_β|x|⁻³(|U |²+ |x|²|∇U |²)dx



≤ ˜C₂

 Z

r1/2<|x|<er2

ϕ²_β(|x|⁻³|U |²+ |x|⁻¹|∇U |²)dx +r⁻³₁ ϕ²_β(r₁/e)

Z

r1/e<|x|<r1/2

(|U |²+ |x|²|∇U |²)dx +r⁻³₂ ϕ²_β(er₂)

Z

er2<|x|<3r2

(|U |²+ |x|²|∇U |²)dx



(1.7) where ˜C1 = max{2 ˜C0M₁², 2 ˜C0M₂², ˜C0C1, ˜C0C2} and ˜C1e³ = ˜C2

We introduce a corollary in [3]

Corollary For 0 < a3 < a1 < a2 < a4 such that Ba4r ⊂ Ω, we can show the follow- ing inequality

Z

a1r<|x|<a2r

||x|^|α|D^αu|²dx ≤ C⁰ Z

a3r<|x|<a4r

|u|²dx where C⁰ is a constant independent of r and |α| ≤ 2.

So by the corollary, it implies (1.8) β³

Z

r1/2<|x|<er2

ϕ²_β|x|⁻³|E|²dx + β Z

r1/2<|x|<er2

ϕ²_β|x|⁻¹|∇E|²dx

≤ ˜C₃

 Z

r1/2<|x|<er2

ϕ²_β(|x|⁻³|U |²+ |x|⁻¹|∇U |²)dx +r⁻³₁ ϕ²_β(r₁/e)

Z

r1/4<|x|<r1

|U |²dx + r₂⁻³ϕ²_β(er₂) Z

2r2<|x|<4r2

|U |²dx



where ˜C₁, ˜C₂, ˜C₃ are independent of r₁,r₂.

And we have the same conclusion for the φH, so we obtained β³

Z

r1/2<|x|<er2

ϕ²_β|x|⁻³|H|²dx + β Z

r1/2<|x|<er2

ϕ²_β|x|⁻¹|∇H|²dx

6

≤ ˜C₃

 Z

r1/2<|x|<er2

ϕ²_β(|x|⁻³|U |²+ |x|⁻¹|∇U |²)dx +r⁻³₁ ϕ²_β(r₁/e)

Z

r1/4<|x|<r1

|U |²dx + r₂⁻³ϕ²_β(er₂) Z

2r2<|x|<4r2

|U |²dx

 (1.9) Therefore, we can combine the inequality (1.8) and (1.9) such that

β³ Z

r1/2<|x|<er2

ϕ²_β|x|⁻³(|E|² + |H|²)dx + β Z

r1/2<|x|<er2

ϕ²_β|x|⁻¹(|∇E|²+ |∇H|²)dx

= β³ Z

r1/2<|x|<er2

ϕ²_β|x|⁻³|U |²dx + β Z

r1/2<|x|<er2

ϕ²_β|x|⁻¹|∇U |²dx

≤ 2 ˜C3

 Z

r1/2<|x|<er2

ϕ²_β(|x|⁻³|U |²+ |x|⁻¹|∇U |²)dx +r⁻³₁ ϕ²_β(r₁/e)

Z

r1/4<|x|<r1

|U |²dx + r₂⁻³ϕ²_β(er₂) Z

2r2<|x|<4r2

|U |²dx



Now let β₀ ≥ 1 and β³ ≥ β ≥ β₀ ≥ 3 ˜C₃, then we can get another inequality (1.10)

Z

r1/2<|x|<er2

ϕ²_β|x|⁻³|U |²dx + Z

r1/2<|x|<er2

ϕ²_β|x|⁻¹|∇U |²dx

≤ ˜C₄



r⁻³₁ ϕ²_β(r₁/e) Z

r1/4<|x|<r1

|U |²dx + r⁻³₂ ϕ²_β(er₂) Z

2r2<|x|<4r2

|U |²dx

 where ˜C₄ = 1/ ˜C₃, and it is easy to get that

r₂⁻³ϕ²_β(r₂) Z

r1/2<|x|<r2

|U |²dx ≤ Z

r1/2<|x|<r2

ϕ²_β|x|⁻³|U |²dx ≤ Z

r1/2<|x|<er2

ϕ²_β|x|⁻³|U |²dx

≤ ˜C₄



r⁻³₁ ϕ²_β(r₁/e) Z

r1/4<|x|<r1

|U |²dx + r⁻³₂ ϕ²_β(er₂) Z

2r2<|x|<4r2

|U |²dx



Dividing the term r₂⁻³ϕ²_β(r₂), we obtain Z

r1/2<|x|<r2

|U |²dx ≤ ˜C₄



(r₂/r₁)³[ϕ²_β(r₁/e)/ϕ²_β(r₂)]

Z

r1/4<|x|<r1

|U |²dx +[ϕ²_β(er₂)/ϕ²_β(r₂)]

Z

2r2<|x|<4r2

|U |²dx



≤ ˜C₅



(r₂/r₁)³[ϕ²_β(r₁/e)/ϕ²_β(r₂)]

Z

|x|<r1

|U |²dx +(r₂/r₁)³[ϕ²_β(er₂)/ϕ²_β(r₂)]

Z

|x|<4r2

|U |²dx



(1.11) where ˜C₅ = max{ ˜C₄, 1}

By choosing such ˜C5, we know that

C˜₅(r₂/r₁)³[ϕ²_β(r₁/e)/ϕ²_β(r₂)] > 1

for 0 < r₁ < r₂ ≤ 1.

Adding Z

|x|<r1/2

|U |²dx to the both sides of (1.11) and r₂ < 1/4, and then we have Z

|x|<r₂

|U |²dx ≤ ˜C₅



(r₂/r₁)³[ϕ²_β(r₁/e)/ϕ²_β(r₂)]

Z

|x|<r₁

|U |²dx +(r₂/r₁)³[ϕ²_β(er₂)/ϕ²_β(r₂)]

Z

|x|<1

|U |²dx



+ ˜C₅(r₂/r₁)³[ϕ²_β(r₁/e)/ϕ²_β(r₂)]

Z

|x|<r₁/2

|U |²dx

≤ 2 ˜C₅



(r₂/r₁)³[ϕ²_β(r₁/e)/ϕ²_β(r₂)]

Z

|x|<r1

|U |²dx +(r₂/r₁)³[ϕ²_β(er₂)/ϕ²_β(r₂)]

Z

|x|<1

|U |²dx



Assume A = (log r₁ − 1)² − (log r₂)², B = −1 − 2 log r₂, and A > 0, B > 0 by simple computation. Therefore, the above inequality becomes

Z

|x|<r2

|U |²dx ≤ 2 ˜C₅(r₂/r₁)³



exp(Aβ) Z

|x|<r1

|U |²dx + exp(−βB) Z

|x|<1

|U |²dx

 (1.12) By standard argument, we consider two cases

Case1 : If exp(Aβ0) Z

|x|<r₁

|U |²dx < exp(−β0B) Z

|x|<1

|U |²dx and pick β > β0 such that

exp(Aβ) Z

|x|<r₁

|U |²dx = exp(−βB) Z

|x|<1

|U |²dx so we have the following important inequality

Z

|x|<r₂

|U |²dx ≤ 4 ˜C₅(r₂/r₁)³exp(Aβ) Z

|x|<r₁

|U |²dx

= 4 ˜C₅(r₂/r₁)³



exp(Aβ) Z

|x|<r₁

|U |²dx

_A+B^B 

exp(−βB) Z

|x|<1

|U |²dx

_A+B^A

= 4 ˜C₅(r₂/r₁)³

 Z

|x|<r₁

|U |²dx

_A+B^B  Z

|x|<1

|U |²dx

_A+B^A

Case2 : If exp(Aβ₀) Z

|x|<r₁

|U |²dx ≥ exp(−β₀B) Z

|x|<1

|U |²dx, then we have Z

|x|<r₂

|U |²dx ≤

 Z

|x|<1

|U |²dx

_A+B^B  Z

|x|<1

|U |²dx

_A+B^A

≤ exp(β₀B)

 Z

|x|<r₁

|U |²dx

_A+B^B  Z

|x|<1

|U |²dx

_A+B^A

By the arguments, we can take ˜C₆ = maxexp(β₀B), 4 ˜C₅(r₂/r₁)³ and get that

8

Z

|x|<r2

|U |²dx ≤ ˜C₆

 Z

|x|<r1

|U |²dx

_A+B^B  Z

|x|<1

|U |²dx

_A+B^A

(1.13) For the general case, we can assume that R₁ ≤ 1/4 and 0 < r₁ < r₂ < r₃ ≤ R₀ with r₁/r₃ < r₂/r₃ ≤ 1/4.

By scaling, ˜U (y) = U (r₃y) , ˜ε_ij(y) = ε_ij(r₃y) , ˜µ_ij(y) = µ_ij(r₃y) We can have the same conclusion by above argument and obtain

Z

|y|<r2/r3

| ˜U |²dx ≤ C

 Z

|y|<r1/r3

| ˜U |²dx

τ Z

|y|<1

| ˜U |²dx

1−τ

(1.14) where τ = B/(A + B) and C = maxexp(β0B), 4 ˜C₅(r₂/r₁)³







A = log (r₁/r₃) − 1
2

− log (r₂/r₃)
2

B = −1 − 2 log (r₂/r₃)

Providing r₃ < 1 and ˜C₅ can be chosen independent of r₃. So undoing the change of variable of (1.14), we have

Z

|x|<r₂

|U |²dx ≤ C

 Z

|x|<r₁

|U |²dx

τ Z

|x|<r₃

|U |²dx

1−τ

(1.15) The proof is now complete.

And then we are going to prove that the Maxwell system have the strong unique continuation property, so we have to prove the two theorems by using theorem 1.

Proof of the theorem 2 and theorem 3

Without loss of generality, we can use the change of coordinates and property (a) to obtain that

P₁(0, D) =

3

X

i,j=1

ε(0)D_iD_j = ∆

P₂(0, D) =

3

X

i,j=1

µ(0)D_iD_j = P₁(0, D)

h = ∆

h

So we recall another Carleman estimate[1] : For any u ∈ H_loc² (Rⁿ\{0}) with compact support and for any m ∈ {j + 1

2|j ∈ N} we have that X

|α|≤2

Z

m^2−2|α||x|^{−2m+2|α|−n}|D^αu|²dx ≤ C Z

|x|^−2m+4−n|∆u|²dx (2.1)

where C only depends on the dimension n.

And from the previous description, we know that U = (E, H) ∈ (H_loc² (Ω\{0}))⁶, so we can use the Carleman estimate for U .

Define a cut-off function χ(x) ∈ C₀^∞(Rⁿ) satisfying 0 ≤ χ(x) ≤ 1 and

χ(x) =















0, if |x| ≤ δ 3 1, if δ

2 ≤ |x| ≤ (R₀+ 1)R₀R

4 = r₄R 0, if |x| ≥ 2r₄R

where δ ≤ R²₀R/4 , R₀ > 0 is a small number and it will be determined later, and R is sufficiently small satisfying 0 < R ≤ R₀. Using the (2.1) for χE and χH. Now for χE, we can derive that

X

|α|≤2

m^2−2|α|

Z

δ/2≤|x|≤r4R

|x|^{−2m+2|α|−3}|D^αE|²dx

≤ X

|α|≤2

m^2−2|α|

Z

|x|^{−2m+2|α|−3}|D^α(χE)|²dx

≤ C Z

|x|^−2m+1|∆(χE)|²dx

≤ C Z

δ/2≤|x|≤r4R

|x|^−2m+1|∆E|²dx + C Z

|x|>r4R

|x|^−2m+1|∆(χE)|²dx +C

Z

δ/3≤|x|≤δ/2

|x|^−2m+1|∆(χE)|²dx (2.2)

10

On the right hand side of (2.2), the first term we use the triangle inequality (2.3)

C Z

δ/2≤|x|≤r4R

|x|^−2m+1|∆E − P₁(x, D)E + P₁(x, D)E|²dx

≤ C Z

δ/2≤|x|≤r4R

|x|^−2m+1|∆E − P₁(x, D)E|²dx +C

Z

δ/2≤|x|≤r4R

|x|^−2m+1|P₁(x, D)E|²dx

and the first term of the right hand side of (2.3), we can find out that C

Z

δ/2≤|x|≤r4R

|x|^−2m+1|∆E − P₁(x, D)E|²dx

= C Z

δ/2≤|x|≤r4R

|x|^−2m+1|(P₁(0, D) − P₁(x, D))E|²dx

= C Z

δ/2≤|x|≤r4R

|x|^−2m+1|

3

X

i,j=1

(ε_ij(0) − ε_ij(x))D_iD_jE|²dx

≤ C Z

δ/2≤|x|≤r4R

(|x|sup|ε⁰_ij(x)|)²|x|^−2m+1

3

X

i,j=1

|D_iD_jE|²dx

≤ C⁰X

α=2

r²₄R² Z

δ/2≤|x|≤r4R

|x|^−2m+1|D^αE|²dx (2.4)

since ε_ij(x) is C²-function and C,C⁰ are constants.

So by (2.2),(2.3),(2.4) and (1.4) we obtain X

|α|≤2

m^2−2|α|

Z

δ/2≤|x|≤r4R

|x|^{−2m+2|α|−3}|D^αE|²dx

≤ C⁰X

α=2

r₄²R² Z

δ/2≤|x|≤r4R

|x|^−2m+1|D^αE|²dx +2CM₁²

Z

δ/2≤|x|≤r4R

|x|^−2m+1|U |²dx+2CM₂² Z

δ/2≤|x|≤r4R

|x|^−2m+1|∇U |²dx +C

Z

|x|>r4R

|x|^−2m+1|∆(χE)|²dx+C Z

δ/3≤|x|≤δ/2

|x|^−2m+1|∆(χE)|²dx (2.5) And we can have the same argument for χH to get that

X

|α|≤2

m^2−2|α|

Z

δ/2≤|x|≤r4R

|x|^{−2m+2|α|−3}|D^αH|²dx

≤ ˜C⁰X

α=2

r₄²R² Z

δ/2≤|x|≤r4R

|x|^−2m+1|D^αH|²dx +2 ˜CM₁²

Z

δ/2≤|x|≤r4R

|x|^−2m+1|U |²dx+2 ˜CM₂² Z

δ/2≤|x|≤r4R

|x|^−2m+1|∇U |²dx + ˜C

Z

|x|>r₄R

|x|^−2m+1|∆(χH)|²dx+ ˜C Z

δ/3≤|x|≤δ/2

|x|^−2m+1|∆(χH)|²dx(2.6) And then we can derive (2.7) from (2.5),(2.6)

X

|α|≤2

m^2−2|α|

Z

δ/2≤|x|≤r4R

|x|^{−2m+2|α|−3}|D^αU |²dx

≤ C⁰⁰X

α=2

r²₄R² Z

δ/2≤|x|≤r4R

|x|^−2m+1|D^αU |²dx +C_M

Z

δ/2≤|x|≤r4R

|x|^−2m+1|U |²dx + C_M Z

δ/2≤|x|≤r4R

|x|^−2m+1|∇U |²dx + ˆC

Z

|x|>r4R

|x|^−2m+1|∆(χU )|²dx+ ˆC Z

δ/3≤|x|≤δ/2

|x|^−2m+1|∆(χU )|²dx (2.7) where C⁰⁰ = C⁰+ ˜C⁰, C_M = 2((C + ˜C)M₁²+ (C + ˜C)M₂²) and ˆC = C + ˜C

By choosing

R = 1

m√

C⁰⁰ and r²₄R² = R²₀(R0+ 1)² 16m²C⁰⁰ Choosing R₀ < 1 (such that R²₀(R₀+ 1)²

16 < 1

2) and m = m(R₀) large enough such that

X

|α|≤2

m^2−2|α|

Z

δ/2≤|x|≤r4R

|x|^{−2m+2|α|−3}|D^αU |²dx

≤ 2C Z

δ/3≤|x|≤δ/2

|x|^−2m+1|∆(χU )|²dx+2C Z

r4R<|x|<2r4R

|x|^−2m+1|∆(χU )|²dx(2.8)

The first three terms on right hand side of (2.7) is absorbed by the left hand side when the |x| is small enough.

And then by the definition of χ, it is easy to obtain that for all multiindex α







|D^αχ(x)| ≤ C₃δ^−|α|, ∀ δ

3 ≤ |x| ≤ δ 2

|D^αχ(x)| ≤ C₄(r₄R)^−|α|, ∀ r₄R ≤ |x| ≤ 2r₄R

(2.9)

where C₃,C₄ are constants. Now we provide R₀ ≤ 1/16 such that R²₀ ≤ r₄, so by the corollary in theorem 1 and (2.9), we can derive (2.10) from (2.8)

m²(2δ)^−2m−3 Z

δ/2≤|x|≤2δ

|U |²dx + m²(R₀²R)^−2m−3 Z

2δ<|x|≤R²₀R

|U |²dx

≤ X

|α|≤2

m^2−2|α|

Z

δ/2≤|x|≤r4R

|x|^{−2m+2|α|−3}|D^αU |²dx

≤ 2CC₃ X

|α|≤2

δ^−4+2|α|

Z

δ/3≤|x|≤δ/2

|x|^−2m+1|D^αU |²dx

+2CC₄ X

|α|≤2

(r₄R)^−4+2|α|

Z

r4R<|x|<2r4R

|x|^−2m+1|D^αU |²dx

≤ C₃⁰(δ/3)^−2m−3 Z

|x|≤δ

|U |²dx+C₄⁰(r₄R)^−2m−3 Z

|x|≤R0R

|U |²dx (2.10)

where C₃⁰,C₄⁰ are independent of R₀,R and m.

12

Adding m²(2δ)^2m−3 Z

|x|<δ/2

|U |²dx to both sides of (2.10), we can get the left hand side

m²(2δ)^−2m−3 Z

|x|≤2δ

|U |²dx + m²(R²₀R)^−2m−3 Z

2δ<|x|≤R²₀R

|U |²dx

= 1

2m²(2δ)^−2m−3 Z

|x|≤2δ

|U |²dx + 1

2m²(2δ)^−2m−3 Z

|x|≤2δ

|U |²dx +m²(R²₀R)^−2m−3

Z

2δ<|x|≤R²₀R

|U |²dx

≥ 1

2m²(2δ)^−2m−3 Z

|x|≤2δ

|U |²dx + m²(R²₀R)^−2m−3 Z

|x|≤2δ

|U |²dx +m²(R²₀R)^−2m−3

Z

2δ<|x|≤R²₀R

|U |²dx

= 1

2m²(2δ)^−2m−3 Z

|x|≤2δ

|U |²dx+m²(R²₀R)^−2m−3 Z

|x|≤R²₀R

|U |²dx(2.11) Combine (2.10) and (2.11), it implies that

1

2m²(2δ)^−2m−3 Z

|x|≤2δ

|U |²dx + m²(R²₀R)^−2m−3 Z

|x|≤R²₀R

|U |²dx

≤ (C₃⁰+ m²)(δ/3)^−2m−3 Z

|x|≤δ

|U |²dx + C₄⁰(r₄R)^−2m−3 Z

|x|≤R0R

|U |²dx We can rewrite C₄⁰(r₄R)^−2m−3 and get m²(R²₀R)^−2m−3C₄⁰m⁻²(R²₀

r4

)^2m+3. Therefore,

C₄⁰m⁻²(R²₀ r4

)^2m+3 = C₄⁰m⁻²( 4R₀

R0+ 1)^2m+3 ≤ C₄⁰m⁻²(4R₀)^2m+3≤ exp(−2m) For all R0 ≤ 1/16 and m² ≥ C₄⁰. Thus, we can derive

1

2m²(2δ)^−2m−3 Z

|x|≤2δ

|U |²dx + m²(R²₀R)^−2m−3 Z

|x|≤R²₀R

|U |²dx

≤ (C₃⁰+m²)(δ/3)^−2m−3 Z

|x|≤δ

|U |²dx+m²(R²₀R)^−2m−3e^−2m Z

|x|≤R0R

|U |²dx The above argument is valid for all m = m(R₀) = j + 1/2 with j ∈ N, and m large enough. So we can assume a j₀ which depends on R₀, and R_j = (√

C⁰⁰(j + 1/2))⁻¹. For all j ≥ j₀, we have the following inequality (2.12)

1

2m²(2δ)^−2m−3 Z

|x|≤2δ

|U |²dx + m²(R²₀R_j)^−2m−3 Z

|x|≤R²₀Rj

|U |²dx

≤ (C₃⁰+m²)(δ/3)^−2m−3 Z

|x|≤δ

|U |²dx+m²(R²₀Rj)^−2m−3e

−2c Rj

Z

|x|≤R₀Rj

|U |²dx where c = 1

√C⁰⁰

We can observe that R_j+1 < R_j < 2R_j+1 for all j ∈ N by simple calculation. If we can find a R such that Rj+1 < R ≤ Rj, it implies R0Rj ≤ 2Rj+1/16 ≤ Rj+1 since R₀ < 1/16 and then we have such relation R₀R_j ≤ R_j+1 < R < R_j < 2R_j+1. So we

can get a conclusion









 Z

|x|≤R²₀R

|U |²dx ≤ Z

|x|≤R²₀Rj

|U |²dx e

−2c Rj

Z

|x|≤R0Rj

|U |²dx ≤ e^−cR Z

|x|≤R

|U |²dx

(2.13)

If there exists a s ∈ N such that

R_j+1 < R^2s₀ ≤ R_j for some j ≥ j₀ (2.14) It is just replacing R by R^2s₀ on the above description. By (2.12) and (2.13), we can obtain (2.15)

1

2m²(2δ)^−2m−3 Z

|x|≤2δ

|U |²dx + m²(R²₀R_j)^−2m−3 Z

|x|≤R^2s+2₀

|U |²dx

≤ (C₃⁰+m²)(δ/3)^−2m−3 Z

|x|≤δ

|U |²dx+m²(R²₀R_j)^−2m−3exp(−cR^−2s₀ ) Z

|x|≤R^2s₀

|U |²dx

So now our goal is to find a appropriate s and R₀ to claim an inequality which is

exp(−cR^−2s₀ ) Z

|x|≤R^2s₀

|U |²dx ≤ 1 2

Z

|x|≤R^2s+2₀

|U |²dx (2.16)

By theorem 1, we assume r₁ = R₀^2s+2,r₂ = R^2s₀ and r₃ = R^2s−2₀ where s ≥ 1.

And r₁/r₃ < r₂/r₃ ≤ R²₀ ≤ 1/4, then we divide

 Z

|x|<R^2s₀

|U |²dx

1−τ

to both sides of (1.15)

 Z

|x|<R^2s₀

|U |²dx

τ

≤ C

 Z

|x|<R^2s+2₀

|U |²dx

τ Z

|x|<R^2s−2₀

|U |²dx

 Z

|x|<R^2s₀

|U |²dx

1−τ

and it implies (2.17) Z

|x|<R^2s₀

|U |²dx

 Z

|x|<R^2s+2₀

|U |²dx ≤ C^1/τ

 Z

|x|<R^2s−2₀

|U |²dx

 Z

|x|<R^2s₀

|U |²dx

a

where C = maxexp(β₀(−1 − 4logR₀)), 4 ˜C₅(R₀)⁻⁶

a = 1 − τ

τ = (4logR0− 1)² − (2logR0)²

−1 − 4logR₀ by definition of τ in the proof of theorem 1.

We can see that







1 < C ≤ ˜C₅R^−β₀ ¹ 2 < a ≤ −5logR0

(2.18)

14

where β₁ = max{6, 4β₀}.

Because exp(β0(−1 − 4logR0) = e^−β0R^−4β₀ ⁰ and ˜C5 ≥ 1, we have first inequality.

For the second inequality, we consider that

a = −4logR₀+ 2 − 1/4logR₀+ logR₀

1/4logR₀+ 1 > −3logR₀+ 2 > 2 a < −3logR₀− logR₀

1 − 1/5 < −5logR₀

And we use (2.17) recursively Z

|x|<R^2s₀

|U |²dx

 Z

|x|<R^2s+2₀

|U |²dx ≤ C^1/τ

 Z

|x|<R^2s−2₀

|U |²dx

 Z

|x|<R^2s₀

|U |²dx

a

≤ C

as−1−1 τ (a−1)

 Z

|x|<R₀²

|U |²dx

 Z

|x|<R⁴₀

|U |²dx

a^s−1

(2.19) For all s ∈ N. And by definition of a we know that τ = 1/(a + 1), from (2.18) we have

a^s−1− 1

τ (a − 1) = (a + 1)(a^s−1− 1)

a − 1 ≤ 3a^s−1 Then we derive the following inequality from (2.19)

Z

|x|<R^2s₀

|U |²dx

 Z

|x|<R^2s+2₀

|U |²dx ≤ C^3(−5logR0)

s−1 Z

|x|<R²₀

|U |²dx

 Z

|x|<R⁴₀

|U |²dx

a^s−1

≤ ( ˜C₅³R^−3β₀ ¹)^{(−5logR0)s−1}

 Z

|x|<R²₀

|U |²dx

 Z

|x|<R⁴₀

|U |²dx

a^s−1

Multiply exp(−cR₀^−2s) on both sides, we obtain (2.20) exp(−cR^−2s₀ )

Z

|x|<R^2s₀

|U |²dx

 Z

|x|<R^2s+2₀

|U |²dx

≤ exp(−cR^−2s₀ )( ˜C₅³R^−3β₀ ¹)^{(−5logR0)s−1}

 Z

|x|<R²₀

|U |²dx

 Z

|x|<R⁴₀

|U |²dx

a^s−1

Let κ = −logR₀, and compute log( ˜C₅³R^−3β₀ ¹)^(5κ)s−1 = (5κ)^s−1(log ˜C₅³+ 3β₁κ).

So we can find out that if R0 small enough, it means that κ sufficient large, then we have (2.21)

cR^−2s₀

4 > (5κ)^s−1(log ˜C₅³+ 3β₁κ) ⇒ ( ˜C₅³R^−3β₀ ¹)^(5κ)s−1 < exp(cR^−2s₀ 4 ) < 1

2exp(cR^−2s₀ 2 ) The (2.21) holds for all s ∈ N, and now we should fix R0 such that (2.21) holds and the m(R₀) and j(R₀) are fixed as well. Fixing R₀, we can derive from (2.20)

exp(−cR^−2s₀ ) Z

|x|<R^2s₀

|U |²dx ≤ 1

2exp(−cR^−2s₀

2 )

 Z

|x|<R²₀

|U |²dx

 Z

|x|<R⁴₀

|U |²dx

a^s−1

× Z

|x|<R^2s+2₀

|U |²dx

Our goal is (2.16), so coping with the term

 Z

|x|<R²₀

|U |²dx

 Z

|x|<R₀⁴

|U |²dx

a^s−1

If we have an estimate that

 Z

|x|<R²₀

|U |²dx

 Z

|x|<R⁴₀

|U |²dx

a^s−1

≤ exp(cR^−2s₀ /2) (2.22) Then (2.16) is proved for appropriate s. So now we have to find s such that (2.22) holds.

Assume N =

 Z

|x|<R²₀

|U |²dx

 Z

|x|<R⁴₀

|U |²dx



, take loglog for (2.22), we have log2 − log(ac) + loglogN ≤ s(−2logR₀− loga)

By (2.18), we know −2logR0−loga > 0 for all R0 ≤ 1/16, so we can define a number s₀ as

s0 = min{s ∈ N|s ≥ (log2 − log(ac) + loglogN)(−2logR⁰− loga)⁻¹} So the claim (2.16) holds for all s ≥ s₀.

But now s should also be chosen to assure (2.14) holds.

Let s₁ be the smallest positive integer such that R^2s₀ ¹ ≤ R_j0, then we can find a j₁ ∈ N with j1 ≥ j₀ such that R_j1+1 < R^2s₀ ¹ ≤ R_j1. We now can define s_p depending on P₁(x, D) , P₂(x, D) and N as

s_p = max{s₀, s₁}

For this s_p, (2.14),(2.21),(2.22) hold. Thus, we set m = j₁+ 1

2 and m₁ = 3 + 2m plus into (2.15).

(m₁− 3

8 )²(2δ)^{−2(m1−32 )−3} Z

|x|≤2δ

|U |²dx+(m₁− 3

8 )²(R²₀R_j1)^{−2(m1−32 )−3} Z

|x|≤R^2sp+2₀

|U |²dx

≤ (C₃⁰ + (m₁− 3

2 )²)(δ/3)^{−2(m1−32 )−3} Z

|x|≤δ

|U |²dx (2.23) So consider the second term of the left hand side of (2.23)

1

8(m₁− 3)²(R²₀R_j1)^−m1 Z

|x|≤R^2sp+2₀

|U |²dx ≤ (C₃⁰ + (m₁− 3

2 )²)(δ/3)^−m1 Z

|x|≤δ

|U |²dx

16

then it implies (m₁− 3)²

8C₃⁰ + 2(m₁− 3)²(3R²₀R_j1)^−m1 Z

|x|≤R^2sp+2₀

|U |²dx ≤ (δ)^−m1 Z

|x|≤δ

|U |²dx (2.24)

(2.24) is valid for all δ ≤ R^2s₀ ^p+2/4 because of the definition of δ. So the proof of theorem 2 is complete with R₂ = R₀. And the first term of the left hand side of (2.23)

1

8(m₁− 3)²(2δ)^−m1 Z

|x|≤2δ

|U |²dx ≤ (C₃⁰ + (m₁− 3

2 )²)(δ/3)^−m1 Z

|x|≤δ

|U |²dx then it implies

Z

|x|≤2δ

|U |²dx ≤ 8C₃⁰ + 2(m₁− 3)² (m₁− 3)² 6^m1

Z

|x|≤δ

|U |²dx (2.25)

(2.25) is valid for all δ ≤ R^2s₀ ^p+2/4 because of the definition of δ. So the proof of theorem 3 is complete with R₃ = R^2s₀ ^p+2/4 and C₃ = 8C₃⁰ + 2(m₁− 3)²

(m₁− 3)² 6^m1.

References

[1] REGBAOUI, R.: Strong uniqueness for second order differential operators.

J.Differential Equations 141(1997), no. 2,201-217.

[2] C. Weber, Regularity theorems for Maxwell equation, Math. Methods Appl.

Sci., 3(1981),523-536.

[3] HORMANDER, L.: The analysis of linear partial differential opera- tors.III. Grundlehren der Mathematischen Wissenschaften 274. Springer- Verlag, Berlin, 1985.

[4] CHING-LUNG LIN, GEN NAKAMURA and JENN-NAN WANG: Quanti- tative uniqueness for second order elliptic operators with strongly singular coefficients, Rev. MAT. IBEROAMERICANA 27(2011), no 2, 475-491 [5] TU NGUYEN and JENN-NAN WANG: Quantitative uniqueness estimate

for the Maxwell system with Lipschitz anisotropic media, Proc. Amer. Math.

Soc. 140 (2012), 595-605

18