Variation of the first eigenvalue with respect topotential in two-dimensional Schr�dinger equation

東吳大學數學系碩士班論文

指導教授:朱啟平 教授

二維薛丁格方程第一特徵值相對位勢的變化

Variation of the first eigenvalue with respect to potential in two-dimensional Schrödinger equation

研究生:蔡富翔 撰

誌謝(Acknowledgment)

這篇論文的完成，首先要先感謝我的指導教授朱啟平 教授，由於教授細心的指導，我才能寫出這篇論文，再來 我需要感謝東吳大學數學系的教授們，林大風教授、林慧 婷教授、簡茂丁教授、符之琪教授、黃文中教授、蔡漢彬 教授、蕭志如教授、張秀瑜教授、葉麗娜教授等，因為在 大學生與研究生期間修了這些教授的課，才建立了數學的 基礎，對往後論文的撰寫有非常大的影響。

父母的支持也是我能完成這篇論文的動力之一，不時 的催促我要好好的完成學業，精神上的推動、物質上的供 給，才使得我可以無後顧之憂的完成這篇論文，所以我必 須好好的感謝我的雙親，能夠讓我有機會能完成在東吳大 學裡的學業。

在大學生與研究生期間與我的一起吃飯、讀書、聊天、

遊玩的，學長姐、同學、學弟妹們，也是我能夠完成學業 的推手之一，他們能夠讓我對大學生生活與研究生生活不 會產生無趣。

最後我要將我的論文獻給所有支持我的人。

摘要

討論二維薛丁格方程的第一方程特徵值隨位勢的水平移 動與旋轉而變化的情況，並且與哈達馬方程式做比較，最後 利用FLEXPDE作模擬。

Abstract

We discuss the variation of the first eigenvalue of two-dimensional Schrödinger equation with respect to potential. And we compare the result with Hadamard formula.

ii

目錄(Table of Contents)

章 節 標 題 頁 次

_____________________________________________________________________

誌 謝 (Acknowledgment)... i

摘 要 (Abstract)... ii

目 錄 (Table of Contents)... iii

第一章 介 紹（Introduction）... 2

第二章 主要結果 (Main results)... 3

第三章 例 子 (Examples)... 14

參考文獻 (References)... 19

Variation of the first eigenvalue with respect to potential in two-dimensional Schr¨ odinger equation

Fu-shiang Tsai

2012.6.12

1 Introduction

Let D be a simply connected domain in R², E0 be properly contained in D and considered as an obstacle in D. For 0 < |ε|  1, let Eε⊆ D be varied from E0 continuously with respect to ε, let V_ε^α= α(ε)χEε, α(ε) > 0 (so Eεis a soft obstacle as in [3]).

We consider Schr¨odinger equation

4uε(x, y) − V_ε^αuε(x, y) + λ(ε)uε(x, y) = 0 ∀ (x, y) ∈ D (1)

with Dirichlet boundary condition. Denote the eigenvalues in (1) in ascending order (with multiplicities included) by {λi(ε)}^∞_i=1. It is known (see [4]) that λ2(ε) > λ1(ε), and λ1(ε) > 0 due to V_ε^α≥ 0.

Fix α(0) > 0, for |ε| ≥ 0, denote α(0) by α, λ1(ε) by λ(ε) for convenience, and let uε(x, y) > 0 in D be the corresponding normalized first eigenfunction (i.e. RR

Du²_ε(x, y)dA = 1). Let

A = {V |V is constant on supp(V ) ⊂ D with Z

D

V dA = α · area(E)},

the potentials in A are interested in quantum mechanices.

In this article, we are interested in the variation of λ(ε) with respect to ε for V_ε^α∈A. Specifically, we compute the differentiation of λ(ε) with respect to ε at ε = 0 when translation or rotation is acted on E0seperately.

The main results are stated in Section 2. In section 3, we give some ex- amples by FLEXPDE.

2 Main results

The following identity (4) will be used frequently. By Green’s identity, for φ, ψ ∈ C^∞₀ (D), from (1) and the bundary condition, we have

 RR

D(5u0· 5ψ + V₀^α· u0· ψ)dA =RR

D(λ(0) · u0· ψ)dA RR

D(5uε· 5φ + V_ε^α· uε· φ)dA =RR

D(λ(ε) · uε· φ)dA Let ψ = uεand φ = u0, it becomes



 RR

D(5u0· 5uε+ V₀^αu0uε)dA = λ(0)RR

Du0uεdA (2)

RR

D(5uε· 5u0+ V_ε^αuεu0)dA = λ(ε)RR

Duεu0dA (3)

and

[λ(ε) − λ(0)]

Z Z

D

u0uεdA = Z Z

D

(V_ε^α− V₀^α)u0uεdA

= Z Z

Eε\E0

V_ε^αu0uεdA+

Z Z

Eε∩E0

(V_ε^α−V₀^α)u0uεdA−

Z Z

E0\Eε

V₀^αu0uεdA, (4)

the last equality is due to V_ε^α= α(ε)χEε.

We consider the actions mentioned before seperately.

2.1 Translation

In this subsection, we consider the action Tt(x, y) = (x, y) + t−→

V for (x, y)

∈ E₀, with−→

V = (v₁, v₂), |~V | = 1. Let E_ε= T_ε(E₀) with E₀= [ξ, η] × [a, b] ⊆ D then Eε = E0+ ε−→

V = [ξ + εv1, η + εv1] × [a + εv2, b + εv2] ⊆ D when |ε|

is small. Since E0 and Eε have the same area, and Vε ∈ A, we see that V_ε^α=

α (x, y) ∈ Eε

0 (x, y) ∈ D\Eε, so V_ε^α= V₀^α in E0∩ Eε. Now (4) becomes, as shown in Figure 1,

[λ(ε) − λ(0)]

Z Z

D

u0uεdA

= α ZZ

E_ε\E₀

u₀u_εdA − α Z Z

E₀\E_ε

u₀u_εdA

= α

ZZ

I

− ZZ

II

 +

ZZ

III

− Z Z

IV

 +

ZZ

V

− Z Z

V I



u0uεdxdy

= αh Z η ξ+εv₁

Z b+εv2

b

u₀u_εdydx − Z η

ξ+εv₁

Z a+εv2

a

u₀u_εdydx

!

+ Z b

a+εv₂

Z η+εv1

η

u₀u_εdxdy − Z b

a+εv₂

Z ξ+εv1

ξ

u₀u_εdxdy

!

+

Z b+εv₂ b

Z η+εv₁ η

u0uεdxdy −

Z a+εv₂ a

Z ξ+εv₁ ξ

u0uεdxdy

!i

Figure 1: Translation

By Mean value theorem for integration, we see that for each x ∈ (ξ + εv1, η) there exist y^x_b(ε) ∈ (b, b + εv2), y^x_a(ε) ∈ (a, a + εv2), such that

( ZZ

I

− Z Z

II

)u0uεdA

= εv2

Z η ξ+εv₁

(u0uε)(x, y^x_b(ε))dx − Z η

ξ+εv₁

(u0uε)(x, y_a^x(ε))dx



. (5)

Similarly, we have for each y ∈ (a + εv₂, b) there exists x^y_η(ε) ∈ (η, η + εv₁), x^y_ξ(ε) ∈ (ξ, ξ + εv₁), such that

( Z Z

III

− ZZ

IV

)u0uεdA

= εv1

Z b a+εv₂

(u0uε)(x^y_η(ε), y)dy − Z b

a+εv₂

(u0uε)(x^y_ξ(ε), y)dy

!

. (6)

From Mean value theorem again, we have (

Z Z

V

− Z Z

V I

)u0uεdA

= ε²v1v2[u0uε(x^∗(ε), y^∗(ε))] − ε²v1v2[u0uε(x^∗∗(ε), y^∗∗(ε))] , (7)

for some (x^∗(ε), y^∗(ε)) ∈ V , (x^∗∗(ε), y^∗∗(ε)) ∈ V I.

Since ε → 0,RR

Du0uεdA → 1 (uεconverges to u0uniformly in ¯D (see[3, Lemma1.2.(b)]), and x^y_η(ε) → η, x^y_ξ(ε) → ξ, y^x_b(ε) → b, y_a^x(ε) → a. From (5),(6),(7) we have

λ⁰(0) = lim

ε→0

λ(ε) − λ(0) ε

= α

"

v2

Z η ξ

u²₀(x, b)dx − Z η

ξ

u²₀(x, a)dx

 + v1

Z b a

u²₀(η, y)dy − Z b

a

u²₀(ξ, y)dy

!#

(8)

Theorem 1. Under the translation action Tε over E0, for V = (v1, v2),

|V | = 1, let−→

n denote the outer unit normal vector on ∂E0, we have

λ⁰(0) = α Z

∂E₀

u²₀−→ n ·−→

V ds, (9)

where ds denotes the integral element with respect to arc length.

Proof. Since on {(x, b)|x ∈ [ξ, η]}, −→n = (0, 1), −→n ·−→

V = v₂, ds = dx on {(x, a)|x ∈ [ξ, η]}, −→n = (0, −1), −→n ·−→

V = −v2, ds = dx on {(η, y)|y ∈ [a, b]}, −→n = (1, 0), −→n ·−→

V = v1, ds = dy on {(ξ, y)|y ∈ [a, b]}, −→n = (−1, 0), −→n ·−→

V = −v1, ds = dy from (8), we have our result._

Remark:(Hadamard formula). (9) is coincide with the result in [3, Proposition 1.4]: Let E0 be an interior soft obstacle (that is: α > 0) with

∂E0 being piecewise smooth, which can be moved rigidly a positive distance in the direction of a unit vecter−→

V , then

∂λ

∂−→ V

:= lim

ε→0

λ(ε) − λ(0)

ε = α

Z

∂E₀

u²₀−→n ·−→ V ds,

with−→n being the outward unit normal vector field on ∂E0.

2.2 Rotation

In this subsection, we consider the action of rotation O_t(x, y) = e^it(x, y) for (x, y) ∈ E₀= [−R, R] × [−R, R] ⊆ D. Let E_θ = O_θ(E₀) be rotated coun- terclockwise by an angle θ from E₀, then V_θ^α ∈ A is the potential defined by V_θ^α=

α (x, y) ∈ Eθ⊆ D 0 (x, y) ∈ D \ E_θ

Let Eθ\ E0 = (I ∪ III ∪ V ∪ V II) , E0\ Eθ = (II ∪ IV ∪ V I ∪ V III) as shown in Figure 2, since V_θ^α= V₀^α in E₀∩ E_θ, from (4) with ε = θ, we have

[λ(θ) − λ(0)]

Z Z

D

u₀u_θdA

= αh ZZ

I

− Z Z

II

 +

ZZ

III

− Z Z

IV



+

ZZ

V

− Z Z

V I

 +

ZZ

V II

− Z Z

V III



u0uθdxdyi

It is sufficiently to consider α RR

I−RR

IIu0uθdxdy

. We see that ∂I \ ∂E0

is rotated from ∂II \ ∂Eθ counterclockwise by an angle 0 < θ  1.

Figure 2: Rotation by an angle θ, ∂Eθ: red solid line; ∂E0: black solid line

Lamme 1.(a) Let ∂II \ ∂Eθ = L^θ₁∪ L^θ₂, where L^θ₁ = AC, L^θ₂ = AP as shown in Figure 3 (a) (b), then L^θ₁, L^θ₂ can be parametrized by

−

→γ1(ϕ) = (r1(ϕ) cos ϕ, r1(ϕ) sin ϕ) with r1(ϕ) = _{cos ϕ}^R , ϕ ∈ (^θ₂,^π₄] and

−

→γ2(ϕ) = (r2(ϕ) cos ϕ, r2(ϕ) sin ϕ) with r2(ϕ) = _{sin ϕ}^R , ϕ ∈ (^π₄,^π₄+^θ₂].

(b) We may represent ∂I \ ∂E0= l^θ₁∪ l₂^θand l^θ₁, l₂^θcan parametrize by

−→

χ1(τ ) = (ρ1(τ ) cos τ, ρ1(τ ) sin τ ) with ρ1(τ ) = _{cos τ}^R , τ ∈ (^π₄ −^θ₂,^π₄] and

−→

χ2(τ ) = (ρ2(τ ) cos τ, ρ2(τ ) sin τ ) with ρ2(τ ) = _{sin τ}^R , τ ∈ (^π₄,^π₂ −^θ₂].

Proof. (a) First we claim that ∠AOC = ∠BOC in Figure 3(b). In fact, ∠AEO = ∠OBD = ^π₄, ∠AOE = ∠BOD, and length(AO) = length(BO) reveals that 4AOE ∼= 4BOD, it follows that length(OD) = length(OE) so length(AD) = length(BE ), and 4ACD ∼= 4BCE. Therefore length(CD) = length(CE), 4COD ∼= 4COE, and the claim is true. Therefore ∠AOC =

∠BOQ−∠AOQ

2 = ^π2−θ₂ = ^π₄ −^θ₂, ∠COF = ∠AOF − ∠AOC = ^π₄ − (^π₄ −^θ₂) =

θ

2, hence L^θ₁ can be parametrized by −→

γ1(ϕ) = (r1(ϕ) cos ϕ, r1(ϕ) sin ϕ) with r1(ϕ) = _{cos ϕ}^R , ϕ ∈ (^θ₂,^π₄]. Since OA is rotated from OQ by the angle - θ, we see that ∠AOP = ∠COF = ^θ₂, and L^θ₂ can be parametrized by −→γ2(ϕ) = (r2(ϕ) cos ϕ, r2(ϕ) sin ϕ) with r2(ϕ) = _{sin ϕ}^R , ϕ ∈ (^π₄,^π₄ +^θ₂].

Figure 3: (a) (b) rotation by an angle θ

(b) As shown in Figure 4, we rotate the original x, y axes both by an angle θ to the new X⁰, Y⁰ axes with the new polar coordinates being (ρ, τ ) so τ = ϕ − θ.

Since in xy coordinates, the polar angle of OP is ^π₄ +^θ₂, we see that in X⁰Y⁰ coordinates, the polar angle of OP is (^π₄ + ^θ₂) − θ = ^π₄ −^θ₂. It follows that l₁ can be parametrized by −χ→₁(τ ) = (ρ₁(τ ) cos τ, ρ₁(τ ) sin τ ) with ρ₁(τ ) = _{cos τ}^R , τ ∈ (^π₄ −^θ₂,^π₄]. Notice that in Figure 3(b), the angle between OC and OP is

π

4, it follows that the angle between OP and OP⁰ is ^π₄, too. That is, the new polar angle of OP⁰ is τ = ^π₄ −^θ₂

+^π₄ = ^π₂ −^θ₂, so l2 can be paramrized by

−→

χ2(τ ) = (ρ2(τ ) cos τ, ρ2(τ ) sin τ ) with ρ2(τ ) = _{sin τ}^R , τ ∈ (^π₄,^π₂ −^θ₂]._

Figure 4: rotation by an angle θ

With notations as in Figure 5, α

ZZ

I

− Z Z

II



u0uθdA

= α

ZZ

Ia

+ Z Z

Ib



−

ZZ

IIa

+ Z Z

IIb



u0uθdA (10)

with I = Ia∪ Ib (disjoint union), II = IIa∪ IIb (disjoint union).

Figure 5: rotated

We use the new polar coordinates ρ, τ = ϕ − θ on Ia, from Lemma1(a), and the mean value theorem of integral with respect to τ , we have

Z Z

Ia

u0uθdA = Z ^π₄

π 4−^θ₂

Z _{cos τ}^R

R sin(τ +θ)

u0uθρdρdτ = θ 2

Z _{cos τ ∗}^R

R sin(τ ∗ +θ)

u0uθ(ρ, τ^∗)ρdρ

where τ^∗ lies between (^π₄ −^θ₂,^π₄]. As θ → 0, τ^∗→ ^π₄, so

lim

θ→0

RR

I_au0uθdA

θ =1

2 Z _{cos π}^R

4 R sin π4

u²₀(ρ,π

4)ρdρ = 0 (11)

Similarly, using the original polar coordinates on IIa, and the mean value theorem of integral with respect to ϕ, we have

ZZ

IIa

u0uθdA = Z ^π₄+^θ₂

π 4

Z _{sin ϕ}^R

R cos(ϕ−θ)

u0uθrdrdϕ = θ 2

Z _{sin ϕ∗}^R

R cos(ϕ∗ −θ)

u0uθ(r, ϕ^∗)rdr

where ϕ^∗ lies between (^π₄,^π₄ +^θ₂]. As θ → 0, ϕ → ^π₄, so

lim

θ→0

RR

II_au0uθdA

θ =1

2 Z _{sin π}^R

4 R cos π4

u²₀(r,π

4)rdr = 0 (12)

Using the new polar coordinates on Ib, and the mean value theorem of inte- gral with respect to ρ, we have

Z Z

Ib

u0uθdA = Z ^π₂^−θ₂

π 4

Z _{sin τ}^R

R sin(τ +θ)

u0uθρdρdτ

= Z ^π₂^−θ₂

π 4

 ( 1

sin τ − 1 sin(τ + θ))R



u0uθ(ρ^∗, τ )ρ^∗dτ,

where ρ^∗ lies between _{sin τ}^R and _{sin(τ +θ)}^R . L⁰hˆopital⁰s rule shows that

lim

θ→0

1 θ

 1

sin τ − 1 sin(τ + θ)



R = cos τ sin²τR

so lim

θ→0

1 θ

Z Z

I_b

u0uθdA = Z ^π₂

π 4

cos τ

sin²τR · u²₀( R

sin τ, τ ) R sin τdτ

= Z ^π₂

π 4

cos τ

sin³τR²· u²₀( R

sin τ, τ )dτ. (13)

Similarly, using the original polar coordinates on IIb, and the mean value the- orem of integral with respect to r, we have

Z Z

II_b

u0uθdA = Z ^π₄

θ 2

Z _{cos ϕ}^R

R cos(ϕ−θ)

u0uθrdrdϕ

= Z ^π₄

θ 2

 ( 1

cos ϕ− 1 cos(ϕ − θ))R



u0uθ(r^∗, ϕ)r^∗dϕ,

where r^∗ lies between _{cos ϕ}^R and _cos(ϕ−θ)^R . L⁰hˆopital⁰s rule shows that

θ→0lim 1 θ

 1

cos ϕ− 1 cos(ϕ − θ)



R = sin ϕ cos²ϕR

so lim

θ→0

1 θ

Z Z

IIb

u0uθdA = Z ^π₄

0

sin ϕ

cos³ϕR²· u²₀( R

cos ϕ, ϕ)dϕ, (14)

Therefore, from (10)-(14), we have Lemma 2.

lim

θ→0

α θ

ZZ

I

− Z Z

II



u0uθdA

= α Z ^π₂

π 4

cos τ

sin³τR²· u²₀( R

sin τ, τ )dτ − Z ^π₄

0

sin ϕ

cos³ϕR²· u²₀( R

cos ϕ, ϕ)dϕ

!

= α Z

l⁰₂

u²₀(x, R)R cot τ dx − Z

L⁰₁

u²₀(R, y)R tan ϕdy

!

with L⁰₁= {(R, y)|y ∈ [0, R]}, l⁰₂= {(x, R)|x ∈ [0, R]}.

Proof. From Lemma 1(a), on L^θ, (x, y) = ( ^R cos ϕ, ^R sin ϕ) = (R, R tan ϕ),

Theorem 2. Let −→

V (ξ) = iξ for ξ ∈ E0 ⊆C ∼= R², in case of rotation, we have

λ⁰(0) = lim

θ→0

λ(θ) − λ(0) θ

= α Z

∂E0

u²₀ −→n ·−→ V ds.

Proof. Represented L⁰₁ with the original polar coordinates, we have L⁰₁= {_{cos ϕ}^R (cos ϕ, sin ϕ)|ϕ ∈ [0,^π₄)}, so

V |~ L1 = {_{cos ϕ}^R (cos(ϕ + ^π₂), sin(ϕ + ^π₂))|ϕ ∈ [0,^π₄)} = {R(− tan ϕ, 1)|ϕ ∈ [0,^π₄)}.

Since−→n = (1, 0) on L⁰₁, we have−→n ·−→

V = R tan ϕ and ds = dy on L⁰₁, that is, Z

L⁰₁

u²₀(R, y)R tan ϕ dy = Z

L⁰₁

u²₀−→n ·−→ V ds.

Secondly, l⁰₂= {_{sin τ}^R (cos τ, sin τ )|τ ∈ [^π₄ −^θ₂,^π₂ −^θ₂)}, we have V |~ _l0

2= {_{sin τ}^R (cos(τ +^π₂), sin(τ +^π₂))|τ ∈ [^π₄ −^θ₂,^π₂ −^θ₂)}

= {R(−1, cot τ )|τ ∈ [^π₄−^θ₂,^π₂ −^θ₂)}.

Since−→

n = (0, 1) on l⁰₂,−→ n ·−→

V = R cot τ , ds=dx on l₂⁰, that is Z

l⁰₂

u²₀(x, R)R cot τ dx = Z

l⁰₂

u²₀−→n ·−→ V ds.

Similar arguments on the others parts on ∂E0reveal the result to be true._ Remark : The result in Theorem 2 is similar to Hadamard formula in [5, Proposition 2]. It tells that Eθ = Φθ(E0) with Φθ(ξ) = e^iθξ for ξ ∈ E0 ⊆ C ∼= R², and−→

V (ξ) = _dθ¹Φθ(ξ)|θ=0= iξ, then

λ⁰(0) = α Z

∂E₀

u²₀ −→n ·−→ V ds.

3 Examples

In [3],there are some results:

Theorem A [3,Theorem 2.1] Assume that Ω has the interior reflection property with respect to a hyperplane P about which the set B is reflection- symmetric. Suppose that B is translated in the direction of a unit vector v perpendicular to P and pointing from the small side to big side. Then, in the case of a hard or soft obstacle, ^dλ_dv > 0.

Corollary B [3,Corollary 2.4] Suppose Ω is convex and that B is a ball of radius ρ, and in addition that Ω is symmetric with respect to reflection through a hyperplane H. Then at the minimizing position the obstacle is in contact with the boundary, while at the maximizing position its center is on H.

Where Ω is not convex, the result sence unpredictable.

We give some examples simulated by FLEXPDE.

Example 1.

This is a translation sample : E₀= [−1, 1] × [−1, 1] is a square inside a disk D = {re^iθ; θ ∈ [0, 2π), 0 ≤ r < 4}, E_ε= [−0.9, 1.1] × [−0.9, 1.1], E_εis translated from E₀ by a unit vector ~V = (v₁, v₂) = (

√ 2 2 ,

√ 2 2 ) Let V₀=

1 (x, y) ∈ E₀ 0 (x, y) ∈ D, V_ε=

1 (x, y) ∈ E_ε 0 (x, y) ∈ D, where ε = 0.1, then λ(0) = 0.560865 > λ(ε) = 0.559479, λ(ε) − λ(0) = −0.001386.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

Example 2.a to 2.e we have the same E0, Eεas in Example 1. but D is not convex and different in each case.

Let V0=

1 (x, y) ∈ E0

0 (x, y) ∈ D, Vε=

1 (x, y) ∈ Eε

0 (x, y) ∈ D. Example 2.a

λ(0) = 0.050584 < λ(ε) = 0.050761, λ(ε) − λ(0) = 0.000177.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

E0 inside D Eεinside D

Example 2.b

λ(0) = 0.050857 < λ(ε) = 0.051019, λ(ε) − λ(0) = 0.000162.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

E0 inside D Eεinside D

Example 2.c

λ(0) = 0.051648 < λ(ε) = 0.051771, λ(ε) − λ(0) = 0.000123.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

E₀ inside D E_εinside D

Example 2.d

λ(0) = 0.052921 < λ(ε) = 0.052977, λ(ε) − λ(0) = 0.000056.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

E0 inside D Eεinside D

Example 2.e

λ(0) = 0.054617 > λ(ε) = 0.054570, λ(ε) − λ(0) = −0.00047.

Y

-3.

-2.

-1.

0.

1.

2.

3.

4.

Y

-3.

-2.

-1.

0.

1.

2.

3.

4.

We rotate E0 in examples 3a to 3e. (D is the same as examples 2a to 2e respectively ) Let V0=

1 (x, y) ∈ E0

0 (x, y) ∈ D, Vθ=

1 (x, y) ∈ Eθ

0 (x, y) ∈ D, where E0= [−1, 1] × [−1, 1] and Eθ=

cos θ − sin θ sin θ cos θ

 E0, θ = ^π₄, D = {re^iϕ; ϕ ∈ [0, 2π), 0 ≤ r < 4}.

Example 3.a

λ(0) = 0.050584 > λ(θ) = 0.050554, λ(θ) − λ(0) = −0.000030

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

E0 inside D Eθinside D

Example 3.b

λ(0) = 0.050857 > λ(θ) = 0.050825, λ(θ) − λ(0) = −0.000032

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

E₀ inside D E_θinside D

Example 3.c

λ(0) = 0.051648 > λ(θ) = 0.051615, λ(θ) − λ(0) = −0.000033.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

E₀ inside D E_θinside D

Example 3.d

λ(0) = 0.052921 > λ(θ) = 0.052886, λ(θ) − λ(0) = −0.000035.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

X

-4. -3. -2. -1. 0. 1. 2. 3. 4.

Y

-4.

-3.

-2.

-1.

0.

1.

2.

3.

4.

E0 inside D Eθinside D

Example 3.e

λ(0) = 0.054617 > λ(θ) = 0.054583, λ(θ) − λ(0) = −0.000034.

Y

-3.

-2.

-1.

0.

1.

2.

3.

4.

Y

-3.

-2.

-1.

0.

1.

2.

3.

4.

References

[1] Mark S. Ashbaugh and Evans M. Harrell II, And Roman Svirsky, On minimal and maximal eigenvalue gaps and their causes, PACIFIC JOURNAL OF MATHEMATICS Vol. 147, No. 1, (1991) 1-24

[2] C.P Chu, On eigenpairs of Schr¨odinger equation with multiwell potential, J. Math. Anal. Appl. 332 (2007) 219-235.

[3] Evans M. Harrell II and Pawel Kr¨oger And Kazuhiro Kurata, On the placement of an obstacle or a well so as to optimize the fundamental eigenvalue, SIAM J. MATH. ANAL. Vol, 33, No. 1, (2001) 240-259 [4] Q. Kong and Zettl, Eigenvalue of Regular Sturm-Liouville Problems,

J.D.E 131, (1996) 1-9

[5] Ahmad El Soufi and Rola Kiwan, Extremal First Dirichlet eigenvalue of doubly connected plane domains and dihedeal symmetry, SIAM J.

MATH. ANAL. Vol. 39, No. 4, (2007) 1112-1119