一維薛丁格方程之第二特徵函數其節點位置的變化

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東吳大學數學系碩士班碩士論文

指導教授︰朱啟平 教授

一維薛丁格方程之第二特徵函數其節點位置的變化

Variation of the nodal points of the 2nd eigenfunction of one-dimensional Schrödinger equation.

研究生︰許皓強 撰

中華民國一○三年一月

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誌謝 (acknowledgment)

首先誠摯的感謝我的指導教授朱啟平教授,老師細心的教導使我得以一窺數 學領域的深奧,不時的討論並指點我正確的方向,使我在這些年中獲益良多。老 師對學問的嚴謹更是我學習的典範。也感謝我的口試指導委員林惠婷教授以及謝 忠村教授,使我的論文更加完整。

學生生涯即將劃下句點的同時,也要感謝研究所同學,承軒、與祥、國祥、

孟霖、思瑩以及學長姐、學弟妹,除了知識的學習,令我收穫最豐的便是這段友 誼。感謝你們的一路相伴,讓我這段充斥著酸甜苦辣的碩士生活,增添了更多歡 笑。並感謝助教及秘書的關心,讓我能夠得到需要的資源和幫助。

最後,謹以此論文給我摯愛的家人,因為有你們的關心與支持,才能讓我的 求學生涯能毫無後顧之憂,向前邁進。

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摘要

使用阿達瑪微分公式的手法來研究薛丁格方程第二特徵函數的節點位置與 其勢函數支撐區間位置的相對關係。

ii

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Abstract

With the same idea as Hadamard differentiation formula to investigate the variation of the position of nodal points of the 2nd eigenfunction of Schrödinger equation with respect to the location of the support of its potential.

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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) ……… 1

第二章 主要結果 (Main results) ………… 3

第三章 例 子 (Examples) ……… 10

參考文獻 (References) ………… 18

iii

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1 Introduction

We consider one-dimensional Schr¨odinger equation with Dirichlet boundary condition :

φ00(x) − V (x)φ(x) + λφ(x) = 0, (1)

where x ∈ (−π 2,π

2), φ(−π

2) = φ(π

2) = 0, and V (x) ≥ 0 is the potential function.

Denote λ(V ) to be the 2nd Dirichlet eigenvalue of (1), φV the corresponding 2nd Dirichlet eigenfunction.

We are interested in rectangle-potentials: suppV (that is, support of V) is a closed interval and V ≡ constant on suppV . These potentials are interested in the study of the tunneling effects of the valence band problems.

Specifically, we consider the following potential in this paper: let

A = {V |suppV = [a, b] ⊆ (−π 2,π

2), V ≡ constant on [a, b], Z π2

π2

V(x)dx = 2π}.

It is well-known that φV(x) has exactly one nodal point in (−π 2,π

2) and φ0V is absolutely continuous (see [E, Theorem 0]) for V ∈ A. Moreover, the smooth- ness of the position of nodal point with respect to the translation of support V (see [T]).

We shall use the same idea as Hadamard differentiation formula (Theorem 1) to investigate the variation of the position of nodal points with respect to the location of suppV .

In fact, we demonstrate that for V ∈ A, when the nodal point lies inside suppV, it tends to escape suppV from the left as suppV moved toward right (Corollary 1). For the nodal point lying outside suppV , as suppV moved to- ward right, the variation of the position of nodal points depends on which side are they located with respect to suppV (Theorem 2).

First we fix notations. For α ≥ 0, 0 ≤ ||  1, let Vα ∈ A be the potential defined by

Vα(z) =

( α z∈ [a + , b + ] ⊆ (−π 2,π

2) 0 otherwise

with α(b − a) = 2π.

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Fix α, for 0 ≤ ||  1, denote φVα by u, and z() ∈ (−π 2,π

2) be the nodal point of u. Denote

λ0(0) = lim

→0

λ(Vα) − λ(V0α)

 , z0(0) = lim

→0

z() − z(0)

 .

Mainly we study the relation between λ0(0) and z0(0) in this article.

One purpose is, through these simple-looked V , we can see how the position of nodal points, and the eigenpairs influence each other.

The main results are stated in Section 2. The examples are given in Section 3 simulated by Matlab.

2

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2 Main Results

The main tool we shall use is an improved Hadamard formula which is de- rived with the same idea as Hadamard differentiation formula with respect to the translation of suppV . We state and prove the original Hadamard formula in our case for reader’s convenience.

Theorem 1. (Hadamard formula) (ref. [H, Proposition 1.4]) Suppose (λ(), u) is the 2nd eigenpair of (1) with V ≡ Vα, Rπ2

π2u2dx= 1 and (u0(−π2) · u00(−π2)) > 0, then λ0(0) = α[(u0)2(b) − (u0)2(a)].

Proof.

Denote V= Vα, λ() = λ(Vα) for 0 ≤ ||  1.

Let φ, ψ ∈ C0[−π2,π2], then from (1) Rπ2

π

2[u000ψ(x) − u0V0(x)ψ(x) + λ(0)u0ψ(x)]dx = 0 Integration by parts, we get

Z π2

π2

[u00ψ0+ αχ[a,b]u0ψ]dx = λ(0) Z π2

π2

u0ψdx (2)

Similarly, Z π2

π

2

[u0φ0+ αχ[a+,b+]uφ]dx = λ() Z π2

π

2

uφdx (3)

Substituting φ = u0 and ψ = uinto (2),(3) respectively, Z π2

π

2

[u00u0+ αχ[a,b]u0u]dx = λ(0) Z π2

π

2

u0udx (4)

Z π2

π

2

[u0u00+ αχ[a+,b+]uu0]dx = λ() Z π2

π

2

uu0dx (5) (5) − (4)

[λ() − λ(0)]

Z π2

π2

uu0dx= α[

Z b+

b

uu0dx− Z a+

a

uu0dx]

= α[uu0(x1) − uu0(x2)] (6)

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where x1 ∈ (b, b + ), x2∈ (a, a + ) Knowing that lim→0Rπ2

π2 uu0dx=Rπ2

π2 u20(x)dx = 1 ([H, proof of Proposi- tion 1.4]), we have from (6)

λ0(0) = lim

→0

λ() − λ(0)

 = α[(u0)2(b) − (u0)2(a)]. 

In Theorem 2 we investigate the differentiation of the position of the nodal point when suppV is under a translation action. In some sense, the formula in Theorem 2 can present the ”detail” of the formulas in Theorem 1 (i.e.

(7)+(8) is the Hadamard formula in Theorem 1).

Theorem 2.

Suppose (λ(), u) is the 2nd eigenpair of (1) with V ≡ Vα, Rπ2

π2u2dx= 1 and (u0(−π2) · u00(−π2)) > 0.

(i) Suppose z(0) ∈ [a, b], it holds

λ0(0) Z z(0)

π2

u20dx+ αu20(a) = −z0(0)u002(z(0)) (7)

λ0(0) Z π2

z(0)

u20dx− αu20(b) = z0(0)u002(z(0)) (8) (ii) If z(0) ∈ (−π2, a), then

λ0(0) = −2π2z0(0)

[z(0) +π2]3 (9)

If z(0) ∈ (b,π2), then

λ0(0) = 2π2z0(0)

[π2− z(0)]3 (10)

From (7), (8), we see that the nodal points inside suppV tends to escape from the left of suppV to the outside as suppV moved toward right:

Corollary 1. If z(0) ∈ (a, b), then z0(0) < 0

4

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Proof.

If λ0(0) ≥ 0, then (7) implies z0(0) < 0 If λ0(0) < 0, then (8) implies z0(0) < 0

We prove Theorem 2 through the following Lemmas.

Lemma 1.

If z(0) ∈ (a, b), z() ∈ (a + , b + ), then

λ0(0) Z z(0)

π2

u20dx+ αu20(a) = −z0(0)u002(z(0))

λ0(0) Z

π 2

z(0)

u20dx− αu20(b) = z0(0)u002(z(0)) Proof.

We start by proving the first equality. Viewing u, u0 as the 1st Dirichlet eigenfunction of (1) with V ≡ 0 over [−π2, z()] and [−π2, z(0)] respectively, sim- ilar to the computation as in (4), (5). We have

Z z()

π2

u0u00dx+ αχ[a+,z()]uu0dx= λ() Z z()

π2

uu0dx (11) Z z(0)

π2

u00u0dx+ αχ[a,z(0)]u0udx= λ(0) Z z(0)

π2

u0udx (12) The right hand side of (11) − (12) is:

λ() Z z()

π2

uu0dx− λ(0) Z z(0)

π2

u0udx

= λ() Z z()

π2

uu0dx− λ(0)[

Z z()

π2

+ Z z(0)

z()

]uu0dx

= [λ() − λ(0)]

Z z()

π

2

uu0dx− λ(0) Z z(0)

z()

uu0dx

=  · {[λ() − λ(0)

 ]

Z z()

π

2

uu0dx−λ(0)



Z z(0) z()

uu0dx}

=  · {[λ() − λ(0)

 ]

Z z()

π2

uu0dx− λ(0)z(0) − z()

 uu0(x1)}, where x1 lies between z(0), z() and lim→0uu0(x1) = u20(z(0)) = 0

(11)

The left hand side of (11) − (12) is:

Z z()

π2

u0u00dx+ α Z z()

a+

uu0dx− ( Z z(0)

π2

u00u0dx+ α Z z(0)

a

u0udx)

= −α[

Z a+

a

uu0dx+ Z z(0)

z()

uu0dx] − Z z(0)

z()

u0u00dx

= ·(−α)[(a +  − a)(uu0)(ˆa)

 +(z(0) − z())(uu0)(ˆz)

 ]−(z(0) − z())(u0u00)(˜z)

 },

where ˆalies between a, a + , ˆz, ˜z lie between z(0) and z() and lim→0[z(0)−z()]

 uu0(ˆz) = −z0(0) · u20(z(0)) = 0, Consider lim→0(11)−(12)

 , we have λ0(0)Rz(0)

π

2

u20dx= −αu20(a) − z0(0)u002(z(0))

Similarly, we prove the 2nd equality. Viewing u, u0 as the 1st Dirichlet eigenfunction over of (1) [z(),π2] and [z(0),π2] respectively, we have

Z π2

z()

u0u00+ αχ[z(),b+]uu0dx= λ() Z π2

z()

uu0dx (13) Z π2

z(0)

u00u0+ αχ[z(0),b]u0udx= λ(0) Z π2

z(0)

u0udx (14) The right hand side of (13) − (14) is:

λ() Z π2

z()

uu0dx− λ(0) Z π2

z(0)

u0udx

= λ()[

Z z(0) z()

+ Z π2

z(0)

]uu0dx− λ(0) Z π2

z(0)

uu0dx

= [λ() − λ(0)]

Z π2

z(0)

uu0dx+ λ() Z z(0)

z()

uu0dx

=  · {[λ() − λ(0)]



Z π2

z(0)

uu0dx+λ()

 Z z(0)

z()

uu0dx}

=  · {λ() − λ(0)



Z π2

z(0)

uu0dx+ λ()z(0) − z()

 uu0(x2)},

where x2 lies between z(0), z() and lim→0uu0(x2) = u20(z(0)) = 0

6

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The left hand side of (13) − (14) is:

Z π2

z()

u0u00dx+ α Z b+

z()

uu0dx− ( Z π2

z(0)

u00u0dx+ α Z b

z(0)

u0udx)

= α[

Z z(0) z()

uu0dx+ Z b+

b

uu0dx] + Z z(0)

z()

u0u00dx

= ·α[(z(0) − z())(uu0)(ˆz)

 +(b +  − b)(uu0)(ˆb)

 ]+(z(0) − z())(u0u00)(˜z)

 },

where ˆb lies between b, b + , ˆz, ˜z lie between z(0) and z() and lim→0[z(0)−z()]

 uu0(ˆz) = z0(0) · u20(z(0)) = 0, Consider lim→0(13)−(14)

 , we have λ0(0)R

π 2

z(0)u20dx= αu20(b) + z0(0)u002(z(0)). 

Lemma 2.

(i) If z(0) ∈ [a, b], z() < a + , ∀, then

λ0(0) Z z(0)

π2

u20dx= −z0(0)u002(z(0)) (ii) If z(0) ∈ [a, b], z() > b + , ∀, then

λ0(0) Z π2

z(0)

u20dx= z0(0)u002(z(0)) Proof.

(i) In this case, z(0) = a since z() → z(0) as  → 0. With the same com- putation as in Lemma 1 with the 2nd term vanishes in both (11) and (12), or equivalently, plug u(a) = 0 into (7), we get the result.

(ii) In this case, z(0) = b since z() → z(0) as  → 0. With the same com- putation as in Lemma 1 with the 2nd term vanishes in both (13) and (14), or equivalently, plug u(b) = 0 into (8), we get the result. 

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Lemma 3.

(i) If z(0) < a, z() < a + , 0 < ||  1, then λ0(0) = −2π2z0(0) [z(0) +π2]3 (ii) If z(0) > b, z() > b + , 0 < ||  1, then λ0(0) = 2π2z0(0)

[π2− z(0)]3 Proof.

(i) If z(0) ∈ (−π

2, a), z() < a + , then λ() = ( π

z() +π2)2, so λ0(0) = π2[−2 · (z(0) +π

2)3· z0(0)] = −2π2z0(0) [z(0) +π2]3 (ii) If z(0) ∈ (b,π

2), z() > b + , then λ() = ( π

π

2 − z())2, so λ0(0) = π2[−2 · (π

2 − z(0))3· (−z0(0))] = 2π2z0(0)

[π2− z(0)]3.  Through Lemma 1 to Lemma 3, we prove Theorem 2 to be true.

Remark.

When z(0) = a or z(0) = b, (7) = (9), (8) = (10) in Theorem 2, that is, (i) when z(0) = a, λ0(0) = −2π2z0(0)

(a +π2)3 = −z0(0)u002(a) Z a

π

2

u20dx (ii) when z(0) = b, λ0(0) = 2π2z0(0)

(π2 − b)3 = z0(0)u002(b) Z

π 2

b

u20dx Proof.

(i) Since z(0) = a and suppV = [a, b], we have

u0(x) = sin[

π(x +π 2) a+π

2

] on [−π 2, a],

u00(a) = {[cos

π(x +π 2) a+π

2

] · ( π a+π

2

)}|x=a= − π (a +π

2) and

8

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Z a

π

2

u20dx = Z a

π

2

sin2[π(x +π2) a+π2 ]dx =

Z a

π

2

[

1 − cos2π(x +π2) a+π2

2 ]dx

= x 2 −

sin2π(x + π2) a+π2

4 π

|aπ

2 = a+π2 2

So −u002(a) Z a

π

2

u20dx

=

− π2 (a +π

2)2 a+π

2 2

= −2π2 (a +π

2)3, and the conclusion is true.

(ii) Since z(0) = b and suppV = [a, b], we have

u0(x) = sin[

π(π 2 − x) π

2 − b ] on [b,π 2],

u00(b) = {[cos π(π

2 − x)

π

2 − b ] · ( −π π

2 − b)}|x=b= π (π

2 − b) and Z π2

b

u20dx = Z π2

b

sin2[π(π2 − x)

π

2− b ]dx = Z π2

b

[

1 − cos2π(π2 − x)

π 2 − b

2 ]dx

= x 2 +

sin2π(π2 − x)

π 2 − b

4 π

|

π 2

b =

π 2 − b

2

So u002(b) Z π2

b

u20dx

= π2

2 − b)2 π 2 − b

2

= 2π2

2 − b)3, and the conclusion is true. 

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3 Examples

We give some examples simulated by Matlab.

Example 1.

(i) V0 = 3χ[−π

4,12], V= 3χ[−π

4+,12+] with  = 240π , Rπ2

π2 V0dx=Rπ2

π2 Vdx= 2π.

(ii) In (7), (8), we have α= 3, a = -π

4 ∼ −0.7854; b = 5π

12 ∼ 1.3090, u20(a) = u20(−π

4) = 0.5216; u20(b) = u20(5π

12) = 0.1526 (iii) Estimate z(0), z().

Observe that u0(−14π

240) = -0.0033, u0(−13π

240) = 0.0158 Let P = (−14π

240,−0.0033), Q = (−13π

240,0.0158) We shall use the slope of P Q to estimate u0(z0) and its x-intercept to estimate z0:

the slope of P Q is 0.0158 − (−0.0033) π

240

= 1.4591 and

P Qintersect x-axis at x = −14π

240 + 0.0033

1.4591 = -0.1810 Therefore z(0) ∼ -0.1810 ∈ (a, b) = (−0.7854, 1.3090), u0(z(0)) ∼ 1.4591, u002(z(0)) ∼ 2.1290

(iv) Similarly observe that u(−15π

240) = -0.0126, u(−14π

240) = 0.0066 u0(z()) ∼ 0.0066 − (−0.0126)

π 240

= 1.4668 and

z() ∼ −15π

240 + 0.0126

1.4668 = -0.1878 ∈ (a + , b + ) = (−0.7723, 1.3221) (v) λ() = 6.1388, λ(0) = 6.1538

λ0(0) ∼ λ() − λ(0)

 = 6.1388 − 6.1538 π 240

= -1.1459

z0(0) ∼ z() − z(0)

 = −0.1878 − (−0.1810) π

240

= -0.5195

10

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(vi) Plug there approximate values into

λ0(0) Z z(0)

π2

u20dx+ αu20(a) = −z0(0)u002(z(0)) (7) We estimate the first term:

Z z(0)

π2

u20dx= Z 14π240

π2

u20+ (1.45912) ·[z(0) − (−14π240)]3 3

= 0.4031000777 so the left hand side is

-1.1459 · 0.4031000777 + 3 · (0.5216) = 1.1029 the right hand side is

0.5195 · 2.1290 = 1.1061

(vii) Plug there approximate values into

λ0(0) Z π2

z(0)

u20dx− αu20(b) = z0(0)u002(z(0)) (8) Z π2

z(0)

u20dx= Z π2

14π

240

u20+ (1.45912) · [z(0) − (−14π240)]3 3

= 0.5969000777 the left hand side is

-1.1459 · 0.5969000777 - 3 · (0.1526) = -1.1418 the right hand side is

-0.5195 · 2.1290 = -1.1061

V3= 3χ

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Example 2.

(i) V0 = 4χ[−π6,π3], V= 4χ[−π6+,π3+] with  = 240π , Rπ2

π2 V0dx=Rπ2

π2 Vdx= 2π.

(ii) In (7), (8), we have α= 4, a = -π

6 ∼ −0.5236; b = π

3 ∼ 1.0472, u20(−π

6) = 0.0870; u20

3) = 0.6509 (iii) Estimate z(0), z().

Observe that u0(−23π

240) = -0.0044, u0(−22π

240) = 0.0131 Let P = (−23π

240,−0.0044), Q = (−22π

240,0.0131) We shall use the slope of P Q to estimate u0(z0) and its x-intercept to estimate z0:

the slope of P Q is 0.0131 − (−0.0044) π

240

= 1.3369 and

P Qintersect x-axis at x = −23π

240 + 0.0044

1.3369 = -0.2978 Therefore z(0) ∼ -0.2978 ∈ (a, b) = (−0.5236, 1.0472), u0(z(0)) ∼ 1.3369, u002(z(0)) ∼ 1.7873

(iv) Similarly observe that u(−24π

240) = -0.0140, u(−23π

240) = 0.0036 u0(z()) ∼ 0.0036 − (−0.0140)

π 240

= 1.3445 and

z() ∼ −24π

240 + 0.0140

1.3445 = -0.3037 ∈ (a + , b + ) = (−0.5105, 1.0603) (v) λ() = 6.2599, λ(0) = 6.2312

λ0(0) ∼ λ() − λ(0)

 = 6.2599 − 6.2312 π 240

= 2.1925

z0(0) ∼ z() − z(0)

 = −0.3037 − (−0.2978) π

240

= -0.4507

12

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(vi) Plug there approximate values into

λ0(0) Z z(0)

π2

u20dx+ αu20(a) = −z0(0)u002(z(0)) (7) Z z(0)

π2

u20dx= Z 23π240

π2

u20+ (1.33692) ·[z(0) − (−23π240)]3 3

= 0.2163000214 the left hand side is

2.1925 · 0.2163000214 + 4 · (0.0870) = 0.8222 the right hand side is

0.4507 · 1.7873 = 0.8055

(vii) Plug there approximate values into

λ0(0) Z π2

z(0)

u20dx− αu20(b) = z0(0)u002(z(0)) (8) Z π2

z(0)

u20dx= Z π2

23π

240

u20+ (1.33692) · [z(0) − (−23π240)]3 3

= 0.7837000214 the left hand side is

2.1925 · 0.7837000214 - 4 · (0.6509) = -0.8853 the right hand side is

-0.4507 · 1.7873 = -0.8055

V04= 4χ[−π6,π3]

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Example 3.

(i) V0 = 5χ[−π4,20], V= 5χ[−π4+,20+] with  = 240π , Rπ2

π2 V0dx=Rπ2

π2 Vdx= 2π.

(ii) In (7), (8), we have α= 5, a = −π

4 ∼ −0.7854; b = 3π

20 ∼ 0.4712, u20(−π

4) = 0.9436; u20(3π

20) = 0.0634 (iii) Estimate z(0), z().

Observe that u0(18π

240) = -0.0009, u0(19π

240) = 0.0132 Let P = (18π

240,−0.0009), Q = (19π

240,0.0132)

We shall use the slope of P Q to estimate u0(z0) and its x-intercept to estimate z0:

the slope of P Q is 0.0132 − (−0.0009) π

240

= 1.0772 and

P Qintersect x-axis at x = 18π

240 + 0.0009

1.0772 = 0.2365 Therefore z(0) ∼ 0.2365 ∈ (a, b) = (−0.7854, 0.4712), u0(z(0)) ∼ 1.0772, u002(z(0)) ∼ 1.1604

(iv) Similarly observe that u(17π

240) = -0.0011, u(18π

240) = 0.0130 u0(z()) ∼ 0.0130 − (−0.0011)

π 240

= 1.0772 and

z() ∼ 17π

240 + 0.0011

1.0772 = 0.2236 ∈ (a + , b + ) = (−0.7723, 0.4843) (v) λ() = 5.6550, λ(0) = 5.7119

λ0(0) ∼ λ() − λ(0)

 = 5.6550 − 5.7119 π 240

= -4.3468

z0(0) ∼ z() − z(0)

 = 0.2236 − (−0.2365) π

240

= -0.9855

14

(20)

(vi) Plug there approximate values into

λ0(0) Z z(0)

π2

u20dx+ αu20(a) = −z0(0)u002(z(0)) (7) Rz(0)

π

2

u20dx= Z 18π240

π2

u20+ (1.07722) · [z(0) −18π240]3 3

= 0.8271000003 the left hand side is

-4.3468 · 0.8271000003 + 5 · (0.9436) = 1.1228 the right hand side is

0.9855 · 1.1604 = 1.1436

(vii) Plug there approximate values into

λ0(0) Z π2

z(0)

u20dx− αu20(b) = z0(0)u002(z(0)) (8) Rπ2

z(0)u20dx= Z π2

18π 240

u20+ (1.07722) ·[z(0) −18π240]3 3

=0.1729000003 the left hand side is

-4.3468 · 0.1729000003 - 5 · (0.0634) = -1.0686 the right hand side is

-0.9855 · 1.1604 = -1.1436

V5 = 5χ

(21)

Example 4.

(i) V0 = 8χ[0,π4], V= 8χ[0+,π4+] with  = 240π , Rπ2

π2 V0dx=Rπ2

π2 Vdx= 2π.

(ii) In (9), we have α= 8, a = 0; b = π

4 ∼ 0.7854, u20(0) = 0.0954; u20

4) = 0.9791 (iii) Estimate z(0), z().

Observe that u0(−28π

240) = -0.0082, u0(−27π

240) = 0.0048 Let P = (−28π

240,−0.0082), Q = (−27π

240,0.0048) We shall use the slope of P Q to estimate u0(z0) and its x-intercept to estimate z0:

the slope of P Q is 0.0048 − (−0.0082) π

240

= 0.9931 and

Therefore P Q intersect x-axis at x = −28π

240 + 0.0082

0.9931 = -0.3583 z(0) ∼ −0.3583 < a = 0 , u0(z(0)) ∼ 0.9931

(iv) Similarly observe that u(−29π

240) = -0.0136, u(−28π

240) = 0.0001 u0(z()) ∼ 0.0001 − (−0.0136)

π 240

= 1.0446 and

z() ∼ −29π

240 + 0.0136

1.0446= −0.3666 < a +  = 0.0131 (v) λ() = 6.8060, λ(0) = 6.7124

λ0(0) ∼ λ() − λ(0)

 = 6.8060 − 6.7124 π 240

= 7.1505

z0(0) ∼ z() − z(0)

 = −0.3666 − (−0.3583) π

240

= -0.6341

16

(22)

Plug there approximate values into λ0(0) = −2π2z0(0)

[z(0) +π2]3 (9)

the left hand side is 7.1505

the right hand side is

−2π2(−0.6341) [−0.3583 +π

2]3 = 12.5166

1.7826 = 7.0215

V08 = 8χ[0,π4]

(23)

References

[E] Everitt, W. N. , Kwong, M. K. and Zettl, A. , Oscillation of eigenfunc- tions of weighted regular Sturm-Liouville problem. J. London Math. Soc.

27(1983), 106-120.

[H] Harrell E.M. II, Kr¨oger, P, Kurata, K. , On the placement of an obsta- cle or a well so as to optimize the fundamental eigenvalue, SIAM J. Math.

Anal. Vol.33, No.1, p.240-259.

[T] Trynin, A. Yu, Differential properties of zeros of eigenfunctions of the Sturm-Liouville Problem, Ufa Mathematical Journal, vol.3, no.4, (2011), p.130-140.

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